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From the line, x = 3y − 1. Substitute into 3x2 − 7xy = 5 to get 6y2 − 11y − 2 = 0, so y = 2 or y = −1/6. The matching x-values are 5 and −3/2, and their sum is 3.5.
F is the larger x-value on its own, and C is the smaller one. A is the negative of the larger x-value, and B is the negative of the sum. E is 5 − 1/2, from taking the second x-value as −1/2. The two x-values add to 3.5.
The inequality is (x − 1)(x + 1)(x − 2) > 0. The product changes sign at −1, 1 and 2, and it is positive for −1 < x < 1 and for x > 2.
A includes 1 < x < 2, where the product is negative, and drops x > 2. B keeps only the two outer rays. C is the single interval −1 < x < 2. D places the sign change at 1 on the wrong side. The positive set is −1 < x < 1 together with x > 2.
Since x + 2 is a factor, the polynomial is 0 at x = −2. That gives c2 − 6c + 8 = 0, so c = 2 or c = 4. The sum of those values is 6.
A is −10 and E is 10. B is the negative of the sum, and C is 0. The roots of the quadratic in c add to 6.
For 2x2 − 11x + c = 0 the difference of the roots is √(121 − 8c) / 2. Set that equal to 2: 121 − 8c = 16, so c = 105/8.
B, C and D are 113/8, 117/8 and 119/8, from taking 121 − 8c equal to 8, 4 or 2. The squared gap of the roots is 16, so c = 105/8.
The equation is (x2 − 2x)2 = 10. Then x2 − 2x − √10 = 0 has discriminant 4 + 4√10 > 0, while x2 − 2x + √10 = 0 has discriminant 4 − 4√10 < 0. There are two real roots.
E counts four real roots, as if both quadratics had real solutions. D counts three and B counts one. A is the count from the quadratic whose discriminant is negative. Only one of the two quadratics contributes real roots, so there are two.
The equation is a x2 + (a − 2)x − 2 = 0. Its discriminant is (a − 2)2 + 8a = (a + 2)2, which is positive for every a except −2. Since a is already non-zero, the roots are real and distinct for every a ≠ −2.
B and C keep only one side of −2, but (a + 2)2 is positive on both sides. A includes a = −2, where the root is repeated. E says no such a exists. Every non-zero a except −2 works.
p(1) = 2 means that p(x) − 2 is 0 at x = 1, so x − 1 divides p(x) − 2. Therefore p(x) = (x − 1)q(x) + 2 for some polynomial q.
C has constant term −2, which would make p(1) = −2. B and D use the factor x + 1, so they constrain p(−1). E through H constrain p(2) or p(−2). The condition p(1) = 2 is the remainder 2 on division by x − 1.
If a ≥ b, multiplying by −1 gives −b ≥ −a, so statement 1 holds. Also (a − b)2 ≥ 0 rearranges to a2 + b2 ≥ 2ab, so statement 2 holds. Statement 3 fails for a = 2, b = 1 and c = −1, because ac = −2 and bc = −1.
F, G and H include statement 3, which depends on the sign of c. B keeps only statement 1, and C keeps only statement 2. The inequalities that follow from a ≥ b are statements 1 and 2.
If r satisfies x3 + a x2 − b x − c = 0, then −r satisfies x3 − a x2 − b x + c = 0. The new roots are the negatives of the old ones. One positive root and two negative roots become two positive roots and one negative root, and all three remain real.
A keeps the original sign pattern. D, E and F claim a single real root, but negating the three real roots leaves three real roots. C and G allow the number or the signs to depend on a, b and c. The new roots are always two positive and one negative.
Expand (ax + b)3 = a3 x3 + 3a2 b x2 + 3a b2 x + b3. The real cube root of the x3 coefficient 8 is a = 2, and b3 = −3√3 gives b = −√3. The x2 coefficient is then 3(4)(−√3) = −12√3. The printed expansion writes that coefficient as −p, so p = 12√3. The x coefficient 3(2)(3) = 18 matches the given expansion.
A is −12√3, the x2 coefficient itself, before the minus sign in −p x2 is removed. B is 3ab = −6√3 and G is −3ab = 6√3, both using a once instead of a2. C is a2 b = −4√3 and F is −a2 b = 4√3, both missing the factor 3 in 3a2 b. D is b = −√3 and E is −b = √3.
Since x + 2 is a factor, the value at x = −2 is 0, so 3(−8) + 13(4) + 8(−2) + a = 0 and a = −12. Synthetic division by x + 2 leaves the quotient 3x2 + 7x − 6. That factors as (3x − 2)(x + 3), so the complete factorisation is (x + 2)(x + 3)(3x − 2).
A uses (x − 1) and B uses (x + 1) in place of (x + 3). C uses both (x + 1) and (3x + 2). D is (x + 2)(x − 3)(3x + 2), which expands to a quadratic factor 3x2 − 7x − 6, the correct quotient with both non-leading signs reversed. F keeps (x + 3) but replaces (3x − 2) by (3x + 2).
In (1 + 2x + 3x2)6, the power x3 arises in two ways. Three factors of 2x and three factors of 1 contribute C(6; 3, 3) · 23 = 20 · 8 = 160. One factor of 2x, one factor of 3x2 and four factors of 1 contribute C(6; 4, 1, 1) · 2 · 3 = 30 · 6 = 180. The coefficient is 340. In (1 − a x2)5 the coefficient of x4 is C(5, 2) a2 = 10a2. Twice that equals 340 when 20a2 = 340, so a2 = 17 and a = ±√17.
A is ±2√2, from setting 20a2 equal to 160, the contribution of the three factors of 2x alone. C is ±√34, from setting 10a2 = 340 and omitting the factor of two. D is ±2√17, from reversing the comparison and setting 10a2 = 2 · 340. E says there are no possible values, but a2 = 17 has two real solutions.
The map t ↦ t3/5 is strictly increasing on the non-zero reals, because the fifth root is strictly increasing and cubing preserves order. So x3/5 < y3/5 if and only if x < y, and E is sufficient.
A and B compare fourth powers, so they compare absolute values: x = 1 and y = −2 satisfies A but not x < y, and x = 3 and y = 1 satisfies B but not x < y. C and D compare reciprocals, which reverse the order when x and y have the same sign; x = −1 and y = −2 satisfies C but not x < y, and x = 1 and y = −1 satisfies D but not x < y. F is the reverse of E, so it gives y < x.
The roots are p ± √(p2 − q), so their difference is 2√(p2 − q). The condition 2 < 2√(p2 − q) < 4 is equivalent to 1 < p2 − q < 4, hence p2 − 4 < q < p2 − 1. That is the same as q < p2 − 1 < q + 3.
A is p2 − 4 < q < p2, so the difference lies between 0 and 4. C uses ≤ and ≥, so it includes differences equal to 2 or 4. E is p2 − 5 < q < p2 − 1, so the difference lies between 2 and 2√5. B forces p > 0 and q ≥ 0 and replaces the bound 1 by the shifted square (p − 1)2.
The remainder on division by x + 1 is the product at x = −1. That value is (3 − 8 − 3)(−p − 1) = (−8)(−p − 1) = 8(p + 1). Set this equal to 24, so p + 1 = 3 and p = 2.
A solves 8(p + 1) = −24, which changes the sign of the remainder. C evaluates the product at x = 1: (3 + 8 − 3)(p − 1) = 24 gives p = 4, which is not the remainder for the factor x + 1. D and E do not satisfy 8(p + 1) = 24. The remainder condition gives p = 2.
x2 − 8x + 12 < 0 factors as (x − 6)(x − 2) < 0, so 2 < x < 6. The linear inequality is x > 4. The overlap is 4 < x < 6, which is (x − 6)(x − 4) < 0, or x2 − 10x + 24 < 0.
A is x < −1/2 or 2 < x < 6, where the product of the two original left-hand sides is negative, and B is the complementary product inequality. D is x < 4 or x > 6, the opposite of the correct quadratic. E is 2 < x < 4 and F is its complement. G and H are the single rays x < 2 and x > 6. The intersection of the two given inequalities is 4 < x < 6.
The first inequality says x2 + bx + c = (x − p)(x − q), so b = −(p + q) and c = pq. Since c < 0 and p < q, it follows that p < 0 < q. The new quadratic is x2 + b c x + c3 = (x − p c)(x − q c). Multiplying p < q by c reverses the inequality, so q c < p c, and the quadratic is negative for q c < x < p c.
C multiplies the roots by c but keeps the order p c < x < q c, which is empty because c < 0. A and B divide the roots by c; B is q/c < x < p/c. E and F multiply the roots by c2, and F also reverses those ends. The factorisation (x − pc)(x − qc) is negative between q c and p c.
Substitute y = a − x into 3x2 + 2xy = 4 to get x2 + 2ax − 4 = 0. The discriminant is 4a2 + 16 = 4(a2 + 4), which is positive for every real a. Each such x gives one y, so there are two distinct real solutions for every real a.
B is −2 < a < 2, from completing the square as (x + a)2 = 4 − a2, and F is the outside of that interval. C is −1 < a < 1 and E is a < −1 or a > 1, the same split with 1 in place of 2. D is a = 0, and A says there are no such values of a. The discriminant 4(a2 + 4) is positive for every real a.
The remainders are f(−2) and f(−3). Their difference is 19 − 5a + b, and the constant term cancels. With {a, b, c} = {1, 2, 3}, this is largest when a = 1 and b = 3, and then R − S = 17.
B is 5, the value of 19 − 5a + b when a = 3 and b = 1, which is the smallest of the six legal assignments. E is 29, the largest possible f(−2) minus the smallest possible f(−3) taken from different assignments. A is −26, the sum of the two remainders when (a, b, c) = (1, 2, 3). C is 7, which is not a value of R − S for any allowed assignment.
|2 − x| ≤ 6 gives −4 ≤ x ≤ 8, and |y + 2| ≤ 4 gives −6 ≤ y ≤ 2. Both ranges are closed, so |xy| attains its maximum at a corner. The corner products have absolute values 24, 8, 48 and 16, and the greatest is 48.
A is 16, the product of the positive ends x = 8 and y = 2. B is 24, from the corner (−4, −6). C is 32, from taking |y| as large as 4. D is 40. F says there is no greatest value, which would be the case if a range were unbounded.
The general term of (x6 − x−2)12 that takes the second summand r times is C(12, r) (−1)r x72 − 8r. The power is 0 when r = 9, so the constant term is C(12, 9)(−1)9 = −C(12, 3) = −220.
E is C(12, 3) with the sign of (−1)9 dropped. A and F are ±C(12, 4) = ±495, and C and D are ±C(12, 2) = ±66, from choosing the wrong number of factors of x−2.
The left side is non-negative, so p ≥ 0. For p = 0, x = 0 works. For p > 0, isolate √(x + p) = p − √x, which requires √x ≤ p, and square to get 2√x = p − 1. A non-negative square root exists exactly when p ≥ 1, and then x = ((p − 1)/2)2 satisfies the original equation. For 0 < p < 1 the right side p − 1 is negative, so there is no solution. The possible values are p = 0 and p ≥ 1.
A omits every p > 1. F omits p = 0, although x = 0 works in that case. E includes (0, 1), where 2√x = p − 1 cannot hold. C and D still contain x, so they do not describe a set of values of p.
A turning point at (−1, 3) gives f(x) = a(x + 1)2 + 3. The graph passes through (1, −1), so 4a + 3 = −1 and a = −1. Then f(x) = −(x + 1)2 + 3 = −x2 − 2x + 2.
C has its turning point at (1, −1) rather than at (−1, 3). D and F have the correct axis x = −1 and pass through (1, −1), but f(−1) is not 3. Expanding the vertex form gives −x2 − 2x + 2.
The expression is x2 + (k + 2)x + (1 − 2k). Its leading coefficient is positive, so it is positive for every real x exactly when the discriminant is negative: (k + 2)2 − 4(1 − 2k) = k(k + 12) < 0. That is −12 < k < 0.
B is the complement of that interval, where the quadratic takes negative values. E requires k + 2 and 1 − 2k both to be positive, which is −2 < k < 1/2, and F is the complement of E. C and D use the bounds −√6 − 3 and √6 − 3. The discriminant condition is −12 < k < 0.
The general term of (2x + 1/x)6 is C(6, k) 26-k x6-2k. Multiplying by the outer x2 raises the power of x to 8 - 2k. That power is 4 when k = 2, and the coefficient is C(6, 2) * 24 = 15 * 16 = 240.
A is the binomial coefficient 15 with the power of 2 left out. B, C and D are 15 * 2, 15 * 22 and 15 * 23, using too small a power of 2. The x4 term uses 24, so the coefficient is 240.
There is no x term, so the roots are -1/2, 2 and a third root r. The sum of the products of the roots in pairs is 0, which gives r = 2/3. The sum of the roots is 13/6 = -p/2, so p = -13/3, and the constant term is q = 4/3. Then 2p + q = -22/3.
B is the value obtained by taking the root of 2x + 1 to be +1/2 instead of -1/2. A is the value when the sign of q is reversed. D, E and F are the negations of C, B and A. With the root -1/2, 2p + q = -22/3.
Since a < 0, multiplying a < b < c by a reverses both comparisons and gives a2 > ab > ac, so I is true. Both a and c are negative, so a + c < 0, and multiplying by b < 0 gives II. Dividing III by b < 0 would reverse it to c < a, which contradicts a < c, so III is false. For a = -3, b = -2 and c = -1, c/b = 1/2 and a/b = 3/2.
D, F, G and H include III, which fails for a = -3, b = -2, c = -1. A, B and C each omit I or II, both of which follow from multiplying by a negative number. The true statements are I and II.
The quadratic p x2 + q x + p has two distinct real roots when p > 0 and q2 - 4p2 > 0. That is |q| > 2p. On the printed axes the boundary lines have gradient 2 and pass through (2, 4) and (2, -4). The shaded region in F is exactly q > 2p together with q < -2p.
B keeps only q > 2p and misses q < -2p. E shades the strip |q| < 2p, where the discriminant is negative. H shades |q| > p, the same shape with gradient 1 instead of 2, and D and G are the one-sided and interior versions of that gradient. A and C shade below a line rather than outside the pair. The region is the one in F.
At x = 2 the inequality says 2a + 1 ≤ 0, so a ≤ -1/2. This is also sufficient. For x ≥ 2 it becomes x(1 - a) ≥ 3, whose smallest value on that ray is 2(1 - a), and a ≤ -1/2 makes that at least 3. For 0 ≤ x ≤ 2 it becomes 1 ≥ x(1 + a), and the largest right-hand side on the interval is at most 1. For x < 0 the left side of the rearranged inequality is negative while the right side is 1. So the inequality holds for every real x exactly when a ≤ -1/2.
A, B, C and D all allow values greater than -1/2, and at x = 2 those values give 2a + 1 > 0. F and G stop at -1 and -3/2, but a = -1/2 gives equality at x = 2 and still works for every other x. H says no such a exists. The complete set is a ≤ -1/2.
The expression is √(9 - 4√2) + √(17 - 12√2). The first square root is 2√2 - 1, because (2√2 - 1)2 = 8 - 4√2 + 1. The second is 3 - 2√2, because (3 - 2√2)2 = 9 - 12√2 + 8. Their sum is 2√2 - 1 + 3 - 2√2 = 2.
C is -2, from taking both square roots in the opposite order; a principal square root is nonnegative. B is 4√2 - 4, from using 2√2 - 3 for the second root. D is 4 - 4√2, from using 1 - 2√2 for the first root. A is the square root of the sum of the two radicands, √(26 - 16√2), and F replaces 16√2 by 4√2. The value of the sum is 2.
The cubic has no x term, so it is (2x + 1)(x − 2)(x − r) = (2x2 − 3x − 2)(x − r). The coefficient of x is 3r − 2, and that must be 0, so r = 2/3. Then p = −2r − 3 = −13/3 and q = 2r = 4/3, and 2p + q = −22/3.
A is −10, which is 2p − q for these same p and q. D is 22/3, the correct value with the sign changed. The missing linear term forces the third root to be 2/3, and that gives 2p + q = −22/3.
(x + 2)(x − 2) < 0 is −2 < x < 2. The cubic factor (x + 4)(x + 3)(1 − x) is positive for x < −4 and for −3 < x < 1. The overlap of the two conditions is −2 < x < 1.
C is −2 < x < 2, the quadratic inequality on its own. F is x < −4 or −3 < x < 1, the cubic inequality on its own. A is 1 < x < 2, the part of −2 < x < 2 in which the cubic product is negative. G keeps −4 < x < −2, which fails (x + 2)(x − 2) < 0. The values that satisfy both inequalities at once are −2 < x < 1.
The axis of px2 + 6x − q is x = −3/p. Set that equal to −1/4, so p = 12. Touching the x-axis means the discriminant is zero: 36 + 4pq = 0, so q = −3/4. Then p + 8q = 12 − 6 = 6.
B is 18, from taking q = +3/4, which treats the constant term as +q inside the discriminant. C is 21, from using the axis −b/a instead of −b/(2a): that gives p = 24 and q = −3/8, and 24 − 3 = 21. The axis formula with the factor 2 gives p = 12 and p + 8q = 6.
f(x) = (p − x)(x + 2) is a downward parabola with roots p and −2. Its maximum is (p + 2)2/4. The inequality (p + 2)2/4 < 4 is |p + 2| < 4, so −6 < p < 2.
B is |p + 2| < 2√2, from taking the maximum to be (p + 2)2/2 and setting that less than 4. A widens the same interval to 4√2. E is −4 < p < 0, which is |p + 2| < 2. The factor 4 in the completed-square maximum gives −6 < p < 2.
The roots satisfy α + β = 14 and αβ = 9. Then √α + √β has square α + β + 2√(αβ) = 20, and the product of the square roots is 3. The quadratic is x2 − √20 x + 3.
B is x2 − √14 x + 3, using α + β as the square of the new sum and dropping the cross term 2√(αβ). A uses √10, half of that cross term short of the correct 20. D is x2 − 178x + 81, which is (x − α2)(x − β2): the roots have been squared instead of square-rooted. The sum of the square roots is √20 and the product is 3.
The term x2 y4 comes from choosing the summand x twice, the summand y2 twice, and the summand 1 three times. The number of ways is 7!/(2! 2! 3!) = 210.
E is 105, half of 210, from dividing by an extra 2. D is 35, which is C(7, 3), the ways of placing only the three constant factors. C is 21, which is C(7, 2). Counting all three kinds of factor gives 210.
Let f(x) = x2 − 52x − 52. Then f(52) = −52 < 0. And f(53) = 532 − 52×54 = 532 − (532 − 1) = 1 > 0, because 52×54 = (53 − 1)(53 + 1). The graph opens upwards, so 53 is the smallest positive integer at which f is positive.
D is 52, where the expression equals −52 and is still negative. A is 26, the axis of symmetry, where the value is more negative still. F is 54, the next integer after the expression has already become positive. The first positive value is at 53.
The solutions are x = a and the roots of x2 − x + a = 0, whose discriminant is 1 − 4a. When a = 1/4 the quadratic has the double root x = 1/2, distinct from a, so there are exactly two distinct solutions. When a itself is a root of the quadratic, substituting x = a gives a2 = 0, so a = 0, and the distinct solutions are 0 and 1. If 1 − 4a > 0 and a is not one of the quadratic roots, there are three distinct real solutions. If 1 − 4a < 0, the only real solution is x = a. Exactly two values of a work.
B counts only one of a = 0 and a = 1/4. D is 3, from counting a further parameter value that still produces three distinct solutions for x. A says no such a exists. The two values a = 0 and a = 1/4 are the only ones that leave exactly two distinct solutions for x.
Set x - 2 = x2 + kx + 2. This rearranges to x2 + (k - 1)x + 4 = 0. The line meets the curve when the discriminant is non-negative: (k - 1)2 - 16 >= 0, so |k - 1| >= 4, and k <= -3 or k >= 5.
B is those same bounds with the inequality reversed, which is where the line misses the curve. A and D come from (k - 1)2 >= 4 instead of 16. C and F use |k| >= 4, leaving out the shift from k to k - 1.
Adding 0 < a + b < c + d and 0 < a + c < b + d gives 2a + b + c < b + c + 2d, so a < d. Also a + b > 0 and c + d > a + b, so both parts are positive and a + b + c + d > 0. Thus I and III must hold. II need not: a = 1, b = 2, c = 1, d = 100 satisfies both given inequalities with b > c.
II fails for a = 1, b = 2, c = 1, d = 100, which removes C, E, G and H. I follows by adding the inequalities, and III follows because a + b and c + d are both positive, so A, B and D each drop a statement that must hold.
The difference is (a - p)x3 + (b - q)x2 + (c - r)x + (d - s). If a ≠ p it is a cubic and tends to -∞ in one direction, so a = p and I is false. If b = q but c ≠ r, the difference is linear and takes negative values, so II is true. The value at x = 0 is d - s, which must be positive, so III is true.
B, E, F and H include I. For f(x) = x3 + 1 and g(x) = x3 the difference is the positive constant 1, so a = p, while II and III hold. C drops III, but d - s is the value at 0. D drops II; if b = q and c ≠ r the difference changes sign. A drops both II and III.
Since 9 is a multiple of 3, f(9) = f(3)f(3). Since 16 is not a multiple of 3, f(16) = 16. Since 8 is not a multiple of 3, f(8) = 8, and 24 is a multiple of 3, so f(24) = f(3)f(8) = 8f(3). The given equation is f(3)2 − 8f(3) + 16 = 0, hence f(3) = 4.
B is 2√2, from treating f(24) as 24, which produces f(3)2 + 16 − 24 = 0. D is 16/5, from writing f(9) = 3f(3) instead of f(3)2, which produces 3f(3) + 16 − 8f(3) = 0.
In descending powers the fourth term has power xn−3. That power is 3, so n = 6, and the term in x3 is then both the fourth ascending term and the fourth descending term. The x2 coefficient gives C(6, 2) a4 b2 = 105, and the x3 coefficient gives C(6, 3) a3 b3 = 210. Dividing these equations yields a/b = 2/3, so (a/b)2 = 4/9.
A is 1/4, from taking (a/b)2 = (105/210)2 and leaving out the binomial factor C(6, 3)/C(6, 2) = 4/3. C is 25/36 and D is 5/6, from reading the fourth descending power as xn−4, so n = 7; the term ratio then gives a/b = (n − 2)/6 = 5/6, and the square of that ratio is 25/36.
A cubic with nonzero leading coefficient tends to opposite infinities, so it has a real root: I is sufficient. A stationary point need not be a root: x2 + 1 has derivative 0 at x = 0 and is never 0, so II is not sufficient. If f(u)f(v) < 0, the intermediate-value theorem gives a root between u and v, so III is sufficient. The row is Yes, No, Yes.
B and E mark II as sufficient, but x2 + 1 has a stationary point and no real root. D and H mark III as not sufficient, but a sign change forces a root. The matching row is Yes for I, No for II and Yes for III.
The point (3, 0) satisfies both defining inequalities and has x outside (-1, 2), so I does not hold throughout R. The point (0, -10) is in R and gives (y - x)(y - x2) = 100, so II does not hold throughout R. The point (10, 12) is in R and has y = 12, so III does not hold throughout R. None of the three statements is true for every point of R.
B, E, F and H include I, but (3, 0) lies in R. C, E, G and H include II, but at (0, -10) the product of the two differences is 100. D, F, G and H include III, but (10, 12) lies in R. No listed statement holds for every point.
f and g are roots of t2 − (f + g)t + cos2 x = 0, and (f + g)2 = (f − g)2 + 4fg = 4. So f + g is 2 or −2. Then f is 1 + sin x or −1 + sin x. The second of these reaches −2.
C is the minimum of 1 + sin x, the other choice of sign. D is −1, the value of that choice at sin x = 0. The smaller function reaches −2.
Let t = a3 b3. Expanding the product gives −t + 4/t = √2, so t2 + √2 t − 4 = 0. The solutions are t = √2 and t = −2√2. Then ab is a cube root of t, and the smaller one is the cube root of −2√2, which is −√2.
H is the positive cube root of √2. C is −2√2, the value of t rather than of ab. The smaller value of ab is −√2.
The expression is x + y(1 − x). If x ≠ 1, the coefficient of y is not zero, so y can be chosen to make the expression negative. If x = 1, the expression equals 1 for every y. The statement is therefore true for every real x except x = 1.
F includes x = 1, where the expression is 1. D excludes two values, but only one value fails. Every x other than 1 works.
|x + 5| < |x + 11| means x is closer to −5 than to −11, so x > −8. |x + 11| < |x + 1| means x is closer to −11 than to −1, so x < −6. Both hold for −8 < x < −6, an interval of length 2.
C is half of that length. E is the distance from −8 to −5. The two midpoint conditions leave an interval of length 2.
Let u = √x, with u > 6. The difference of the two fractions is 12/(u2 − 36) = 3/11, so u2 − 36 = 44 and x = 80. At this value both denominators are positive, so it is in the domain.
B is 4√5, which is √x rather than x. E, F and G are the integers next to 36 + 44 that come from an arithmetic slip in that last step. The value that satisfies the equation is 80.
Set u = x2. Four distinct real values of x need two distinct positive values of u. The quadratic u2 + bu + c = 0 must therefore have positive discriminant, positive product and positive sum, so c > 0, b < 0 and b2 > 4c. Together these are c > 0 and b < −2√c. Equality would repeat a value of u and would not give four distinct roots.
C has the same bound with the inequality reversed, which makes both values of u negative. A is only the discriminant condition, so the roots in u need not be positive. E and F do not force the two values of u to be distinct.
The terms 4, 4r and 4r3 are in arithmetic progression, so r3 − 2r + 1 = 0. One factor is r − 1, and the root in (0, 1) is r = (−1 + √5)/2. The sum to infinity is 4/(1 − r) = 2(3 + √5).
A is (−1 + √5)/2, the common ratio. B is 2(3 − √5), the conjugate left after rationalising 8/(3 − √5). C is 2(1 + √5). The sum is 2(3 + √5).
Up to degree 2, (1 + 4x3)5 equals 1. With u = 2x + 3x2, (1 + u)6 = 1 + 6u + 15u2 + terms beyond x2, which is 1 + 12x + 78x2. The difference of the two brackets is 12x + 78x2. Multiplying by 4 − x2, the coefficient of x2 is 4 × 78 = 312.
C is 78, the coefficient inside the bracket before multiplication by 4. A, B, D, E and F are all smaller than 4 × 78. The x2 coefficient is 312.
a1 = 2, and each odd index adds −1 while the next even index adds 1, so the terms alternate 2, 1, 2, 1, …. From n = 1 to 100 there are 50 terms equal to 2 and 50 equal to 1. The sum is 150.
B is 250. C and D are large totals with the wrong sign or the wrong pairing. E and F are closed forms of geometric series, and this sequence is not geometric. The sum of the fifty 2s and fifty 1s is 150.
Since (−1)n+2 = (−1)n and (−1)n−1 = −(−1)n, each term simplifies to an = (−1)n + (−1)n + (−1)n = 3(−1)n. From n = 1 to n = 39 there are 20 odd values and 19 even values, so the sum of (−1)n is −1. The required sum is 3(−1) = −3.
A is −39 and G is 39, from adding −1 or +1 once for every term. C is −1, the sum of (−1)n with the factor 3 left out, and E is the opposite of that sum. D is 0, from pairing terms and dropping the unpaired final term. F is 3, the size of a single term.
Let the common ratios be r and s. The first term of S is 4 + 4 = 8. The next two terms give r + s = 3/4 and r2 + s2 = 5/16. Then (r + s)2 − (r2 + s2) = 2rs, so rs = 1/8. Thus r and s are 1/2 and 1/4. Both series converge, with sums 4/(1 − 1/2) = 8 and 4/(1 − 1/4) = 16/3. The sum of S is 8 + 16/3 = 40/3.
E is 16, the result of giving both series the ratio 1/2. B is 20/3, the average of the separate sums 8 and 16/3. A is 32/5, the geometric sum with first term 4 and ratio 3/8. C is 64/5, the geometric sum with first term 8 and ratio 3/8, although the third term of S is 5/4, so S is not geometric. F is 32 = 8/(1 − 3/4), using r + s as if it were the common ratio of S.
f1(x) = x10 and fn+1(x) = x fn'(x). Differentiating x10 and multiplying by x gives 10x10, then 102 x10, and in general fn(x) = 10n−1 x10. The sum from n = 1 to 20 is x10(1 + 10 + … + 1019) = x10(1020 − 1)/9.
A and B sum powers of x instead of powers of 10; B also runs one power too far. D uses (1021 − 1)/9, which is 21 powers of 10. E and F treat the common ratio as 10x. G drops the coefficients 10n−1. H sums the falling products from repeated differentiation and never multiplies back by x.
S8 = 4(2a + 7d) and S6 = 3(2a + 5d). The inequality S8 > 3S6 simplifies to 10a + 17d < 0. This holds for a = 17 and d = −11, for a = −18 and d = 10, and for a = d = −1. Each of a and d can be positive and each can be negative, so neither sign is determined.
A, B and C each name one sign pair that can occur, but the other pairs above also satisfy 10a + 17d < 0. D says a must be negative, but a = 17 and d = −11 works. E says d must be negative, but a = −18 and d = 10 works.
The arithmetic condition is 2q = p + p2. The geometric condition is p4 = p q, so q = p3 because p ≠ 0. Then 2p3 − p2 − p = 0, and p(2p + 1)(p − 1) = 0. The condition p < 0 gives p = −1/2 and q = −1/8. The arithmetic progression has first term −1/2 and common difference 3/8, so the sum of 10 terms is 5(−1 + 27/8) = 95/8.
A is the tenth term, −1/2 + 9 × 3/8 = 23/8, rather than the sum. C is 5 × 23/8 = 115/8, which is (n/2) times that tenth term and leaves out the first term in the sum formula. D is the same sum with common difference 5/8 instead of 3/8. The sum with difference 3/8 is 95/8.
The rule sends 7 to 3, 3 to 1, 1 to −5, and −5 back to 7. The sequence has period 4. Since 100 is a multiple of 4, x100 = x4 = −5.
E is the first term 7, from treating a multiple of the period as a return to the start, or from using the three given terms as a cycle of length 3. C and D are the third and second terms of that shorter list. B is not in the cycle. The fourth term repeats at every multiple of 4, so x100 = −5.
In (a + bx)5 the coefficient of x4 is 5 a b4 and the coefficient of x2 is 10 a3 b2. The given ratio is 5 a b4 = 80 a3 b2. Since a and b are non-zero, b2 = 16 a2, so b = 4a. Then a + b = 5a, and the smallest value for positive integers is 5, when a = 1 and b = 4.
B is 4, the value of b when a = 1, rather than a + b. A is the sum for b = 2a. D is the sum for b = 8a, reading the factor 8 as the ratio b/a. F is the sum for b = 16a, taking b = 16a from b2 = 16a2. E is 5 + 8. The smallest a + b is 1 + 4 = 5.
∫ from 0 to n of (n − x) dx = n2/2, so F(n) = n/2. Then G(n) = Σ from r = 1 to n of r/2 = n(n + 1)/4. The inequality n(n + 1) > 600 first holds at n = 25, since 24 × 25 = 600 gives G(24) = 150, and 25 × 26 = 650 gives G(25) = 325/2 > 150.
C is n = 24, where G(n) = 150, which does not satisfy the strict inequality. A and B give G(22) = 126.5 and G(23) = 138, both below 150. E is n = 26, which does satisfy G(n) > 150 but is larger than 25. The smallest such integer is 25.
The fourth term gives 2√3 · r3 = 9/4, so r = √3/2. Then |r| < 1 and the sum to infinity is 2√3/(1 − √3/2) = 4√3/(2 − √3). Rationalising the denominator gives 4(2√3 + 3).
B is the sum for common ratio −√3/2, but the fourth term is positive, so the ratio is positive. D and E divide one of those sums by 7. A and F are smaller expressions that are not equal to 2√3/(1 − √3/2). The sum is 4(2√3 + 3).
Each term un−1 is a constant in the integral, and ∫ from 0 to 1 of 4x dx equals 2. So un = 2 un−1. Starting from u0 = 1 gives u1000 = 21000.
B replaces the factor 2 by 4, as if the integral of 4x from 0 to 1 were 4. C to H divide by a factorial, which this recurrence does not produce. The term is 21000.
The sequence for f is 5, 16, 8, 4, 2, 1 and then the cycle 4, 2, 1. For n ≥ 4, f(n) = 4 when n ≡ 1 (mod 3), so f(1000) = 4. The sequence for g is the cycle 3, 8, 4, 2, 1, 6 of length 6. Since 1000 ≡ 4 (mod 6) and g(4) = 2, the difference is 4 − 2 = 2.
A and B are negative, but f(1000) is larger than g(1000). C is 1, which appears inside the cycles but is not this difference. E and F are 4 and 8, terms of the cycles rather than f(1000) − g(1000).
The sum of the first n terms is n/2 times (2a + (n − 1)d). Setting the 5-term sum equal to the 8-term sum gives 5(a + 2d) = 4(2a + 7d), so 5a + 10d = 8a + 28d and a = −6d. The same relation is the statement that terms 6, 7 and 8 add to zero.
B is a = −7d, from setting the 8th term alone to zero. D and E are those two relations with the sign of d reversed. A is a = −(38/3)d, from adding the d coefficients 10 and 28 after moving the a terms, and F is that relation with the sign reversed.
In (1 + kx2)7 the x6 term comes from (kx2)3, with coefficient C(7, 3) k3 = 35k3. In (k + x)10 the x6 term has coefficient C(10, 6) k4 = 210k4. Then 35k3 = 210k4. Since k is non-zero, k = 35/210 = 1/6.
B is 6, the reciprocal, from dividing the binomial coefficients the other way. C is √6/6 and D is √6, from treating (kx2)3 as a k6 term, so that k2 = 1/6 or k2 = 6. E is √30/30 and F is √30, the same square-root pattern with 30 in place of 6.
Let the first term be a and the common ratio be r. Then a/(1 − r) = 6 and a2/(1 − r2) = 12. Dividing gives 6(1 − r)/(1 + r) = 2, so r = 1/2 and a = 3. The cubes form a geometric series with sum 27/(1 − 1/8) = 216/7.
A is 8, which is 23, from taking the first term to be 2. C is 24, the cube sum in the model where the squares and the cubes keep the original ratio r, which forces a = 2 and r = 2/3. E is 72, the product of the two given sums. F is 216, which is a3 × 8, using 1/r3 in place of 1/(1 − r3). B is 18.
The sequences are 11 + 3k and 2 + 5m. Setting them equal gives 3(k + 3) = 5m, so the common terms are 15t + 2 for positive integers t. The 20th is 15*20 + 2 = 302. Since 301 = 43*7, the remainder is 1.
A is the remainder of 15*20 = 300. F is the remainder of the 20th term of the first sequence, 11 + 19*3 = 68. G is the remainder of the 20th term of the second sequence, 2 + 19*5 = 97. E is the remainder of the first term 11. The common term 302 leaves remainder 1.
An integer arithmetic progression has integer first term a and integer common difference d. If n is odd, the median of the first n terms is a term. If n is even, that median is an integer only when d is even. The same dichotomy shows that the median of the first n + 2 terms is an integer, so I holds. The terms in even positions form a progression with difference 2d, so the median of the first n of them is an integer whether n is odd or even, and III holds. II need not: for 1, 2, 3, … and n = 1, the median of the first term is 1, but the median of the first two terms is 3/2.
II fails for the progression of positive integers with n = 1. I uses the even common difference forced by an even value of n, and III uses the even difference 2d. The statements that must hold are I and III.
The coefficient of x in (1 + x)n is n for n ≥ 1, and the n = 0 term contributes nothing. The required coefficient is 1 + 2 + … + 80 = 80 × 81 / 2 = 3240.
A is the single coefficient from (1 + x)80. B is 81, the number of powers in the sum. G is 80 × 81, the triangular sum before dividing by 2. F is 81 × 82 / 2 and H is 81 × 82, both of which run the sum one term too far. The sum of the first 80 positive integers is 3240.
Each step takes a square root, so xn = 101/2n − 1 = 1021 − n. With the printed convention that abc means abc, the hundredth term is 102−99.
A is 10299, using 99 square roots in the wrong direction. D takes one extra root and gives 102−100. E and F put the minus sign on a large power of 2, so the power of 10 is a large negative integer. The exponent after 99 square roots is 2−99.
For a common ratio r ≠ 1, the sum condition is (r6 − 1)/(r3 − 1) = 9, so r3 + 1 = 9 and r = 2. The ratio r = 1 does not satisfy the sum condition. The seventh term is a × 26 = 360, so a = 360/64 = 45/8.
D is 360/27 and F is 360/25, from taking the seventh term to be a r7 or a r5. B is 360/81, from replacing r3 + 1 = 9 by r3 = 9 and then using a(r3)2 = 360. With r = 2 the first term is 45/8.
The series 1, 19 has integer terms, first term odd, and sum 20, so I need not be true. The series 2, 3, 4, 5, 6 has sum 20, five terms and common difference 1, so neither II nor III must be true. None of the three statements must hold.
B, E, F and H include I, but 1 + 19 = 20. C, E, G and H include II, and D, F, G and H include III; the five-term series 2 + 3 + 4 + 5 + 6 defeats both. None of the statements is forced.
The terms a, ar and ar2 are the 1st, 4th and 6th terms of an arithmetic progression, so a + 3d = ar and a + 5d = ar2. Eliminating d gives 3r2 − 5r + 2 = 0, hence r = 1 or r = 2/3. The ratio r = 1 does not give a finite sum to infinity. For r = 2/3, a/(1/3) = 12, so a = 4.
F is 6, from putting the common ratio equal to 1/2 in the sum formula. C is 3, which is 1/(1 − r) for r = 2/3 rather than the first term. The arithmetic spacing forces r = 2/3, and the infinite sum then forces the first term to be 4.
The first m and first 2m sums are equal exactly when 2a + (3m - 1)d = 0, so -2a/d equals 3m - 1 for some positive integer m. That value must be one of 2, 5, 8, .... Opposite signs make -2a/d positive, but a = 1 and d = -2 give -2a/d = 1, and 1, -1, -3, ... never has the property. Even d is not required: a = 1 and d = -1 give m = 1, because the first term 1 equals 1 + 0. Neither statement is true.
B and D include I, but a = 1, d = -2 has ad < 0 and no such m. C and D include II, but a = 1, d = -1 has the property with d odd.
Both progressions start at 1/2. The second-term condition gives d + r/2 = 0. The third-term condition then becomes r2 − 2r + 3/4 = 0, so r = 1/2 or r = 3/2. Only |r| = 1/2 is less than 1. The sum to infinity is (1/2)/(1 − 1/2) = 1.
B is −1, the value of (1/2)/(1 − r) at the other root r = 3/2, whose modulus is greater than 1, so that geometric progression does not converge. F is 1/2, the first term. E is 1/3, from putting r = −1/2 into (1/2)/(1 − r).
The first two steps give (3 + p)/(3 + q) = 5 and (5 + p)/(5 + q) = 7. Solving, q = −9 and p = −33. Then x4 = (7 − 33)/(7 − 9) = 13.
A is the value of q. F is −q. The next term, using both constants, is 13.
The geometric sum gives S30 − S20 = a r20 (r10 − 1)/(r − 1) and S10 = a(r10 − 1)/(r − 1). Their ratio is r20. The smallest integer r > 1 is 2, so the smallest k is 220.
A is 210, using the block of ten terms instead of twenty. C is 230. The exponent on the common ratio is 20.
Up to the x5 term, multiplying by 1 + x + … + x5 is the same as multiplying by 1/(1 − x). So (1 + x)5 times that series contributes every binomial coefficient of (1 + x)5 to the coefficient of x5. Those coefficients add to 32.
B is the single coefficient C(5, 1). C is 24. The sum of all six binomial coefficients of (1 + x)5 is 32.
The terms are 3k − 2 for k = 1 to 24. Two distinct terms add to 74 exactly when their indices add to 26, giving the 11 pairs (2, 24) through (12, 14). The terms 1 and 37 are in no such pair. Choosing one term from each pair, plus those two, gives 13 terms with no pair summing to 74. A 14th term must complete a pair.
B is 13, the largest selection that can avoid the pair. A is one smaller still. The pigeonhole forces the pair at 14.
sin(nπ + π/3) = (−1)n sin(π/3) = (−1)n √3/2. The series is therefore (√3/2) times a geometric series with first term 1 and common ratio −1/2. The sum is (√3/2) / (1 + 1/2) = √3/3.
B is 1/3, from using sin(π/3) = 1/2 instead of √3/2. D is √3, from treating the common ratio as +1/2, so the denominator is 1/2 rather than 3/2. A is 0, from cancelling the alternating terms before summing the geometric series. The closed sum is √3/3.
In (2 + 3x)12 the coefficient of xk is C(12, k) × 212 − k × 3k. For k = 0 the coefficient is 212, which has no factor of 3. For k = 12 it is 312, which has no factor of 2. For every k from 1 to 11 the coefficient is divisible by 12. That is 11 coefficients.
G counts all 13 terms, including the two that are not divisible by 12. F drops only one of those two. D drops one of the interior coefficients as well, even though each of them is divisible by 12.
The series is geometric with first term 4 and common ratio 2k/7. It has a finite sum when |2k/7| < 1. Among the 11 integers from −5 to 5, that is k = −3, −2, −1, 0, 1, 2, 3. The sum is 28/(7 − 2k), and this is greater than 3 when k > −7/6. The integers that do both are −1, 0, 1, 2, and 3, so the probability is 5/11.
G is 7/11, the probability that the series converges, without asking for a sum greater than 3. C is 3/11, counting only the positive values of k that work. H is 7/10, the seven convergent values out of a sample that has dropped k = 0.
The midpoint is (7/2, −1) and the segment has gradient 10/3, so the perpendicular has gradient −3/10. Its equation is y + 1 = (−3/10)(x − 7/2). Setting y = 0 gives x = 1/6.
D is 19/5, the x-intercept of the line through the midpoint with the segment's own gradient. E is 7/2 + 10/3. A and C are 1/20 and 1/3. The perpendicular meets the x-axis at 1/6.
Divide the equation by 2 and complete the square: (x − 2)2 + (y + 3)2 = 4 + 9 − 15/2 = 11/2. The radius is √(11/2).
A is √(5/2). C is √(41/2), D is √37 and E is √67, all larger than the radius left after the constants 4 and 9 are reduced by 15/2. The radius is √(11/2).
The centre is the midpoint (5, 4). Half of the diameter from (3, 3) to (7, 5) has squared length 22 + 12 = 5. Translation by −3 in x moves the centre to (2, 4), and reflection in the x-axis moves it to (2, −4). Enlargement by scale factor 4 about that centre leaves the centre fixed and multiplies the squared radius by 16, giving 80. The final equation is (x − 2)2 + (y + 4)2 = 80.
C has the correct squared radius 80 but keeps (y − 4)2, the centre before reflection. E and F use squared radius 20, which is 5 multiplied by 4 rather than by 42; E also keeps the unreflected centre. A and B use squared radius 320, from taking the full diameter squared, 20, and then multiplying by 16. A again keeps (y − 4)2.
The line y = 6 − 2x has gradient −2 and meets the x-axis at x = 3. The perpendicular through (−6, 0) has gradient 1/2, so its equation is y = x/2 + 3. The lines meet at (6/5, 18/5). The enclosed region is a triangle of base 9 and height 18/5, so the area is (1/2) × 9 × 18/5 = 81/5.
D is (1/2) × 9 × 6 = 27, using the y-intercept 6 as the height. C is (1/2) × 12 × 18/5 = 108/5, taking the positive intercept as 6 rather than 3. E is 81/2, the area when the second gradient is 2 rather than 1/2. B is 18, the area of a triangle on legs of length 6. The base is 9 and the height is 18/5, so the area is 81/5.
The circle has centre the origin and radius 12. The line through (20, 0) and (0, c) is c x + 20 y = 20c. Its distance from the origin is 12, so 20c = 12 √(c2 + 400). Then 25c2 = 9(c2 + 400), so 16c2 = 3600 and c2 = 225. The positive intercept is c = 15.
A is the radius 12, the intercept of the horizontal tangent, which does not pass through (20, 0). D is the x-coordinate of the given point. E is (202 − 122)/12 = 64/3, the square of the tangent length divided by the radius. F is 80/3, the intercept of the line through (20, 0) with gradient 4/3 rather than 3/4. C does not satisfy 16c2 = 3600. The positive tangency condition gives c = 15.
PQ = (1, 2) has length √5. The line through Q perpendicular to PQ has gradient −1/2 and meets the x-axis at R(17, 0), so QR has length 8√5. The area of the rectangle is √5 × 8√5 = 40.
A is much smaller than either side of this rectangle. B and D are surds on the scale of a single length, not the product of the two sides. C is half of 40. The area is 40.
The centres are (−2, 3) and (7, −6), so they are 9√2 apart. The radii are √18 = 3√2 and √2. The circles are separate, and the gap between them is 9√2 − 3√2 − √2 = 5√2.
C is 18 − 2, the difference of the squared radii. D is 3√2 − √2, the difference of the radii, with the gap between the centres left out. B is 4, from taking the centre separation as 9 and the radii as 3 and 2. A is 0, the distance when the circles meet.
The centres are (−4, −1) and (8, 4), so the distance between them is √(122 + 52) = 13. The circles touch when r + 8 = 13 or r − 8 = 13, giving r = 5 and r = 21. The difference is 16.
A is 12 − 8, the horizontal gap with the first radius removed. D is 5 + 21, the sum of the two possible radii. The two positive solutions are 5 and 21, and they differ by 16.
The distance from (t, t2 + 4) to the line 2x − y − 2 = 0 is (t2 − 2t + 6)/√5. The quadratic t2 − 2t + 6 has minimum 5 at t = 1, so the shortest distance is 5/√5 = √5. The curve stays above the line, since t2 − 2t + 6 ≥ 5.
C is 6√5/5, the distance from the vertex (0, 4) to the line. F is 5, the vertical gap at t = 1, and G is 6, the vertical gap at t = 0. The perpendicular distance is smallest at t = 1 and equals √5.
The circles have centres (−2, 1) and (4, 1) and the same radius √3, so the external tangents are horizontal. The common tangent of positive gradient crosses between them and passes through the midpoint (1, 1). Its distance from (−2, 1) equals √3, which gives 3|m|/√(m2 + 1) = √3. The positive solution is m = √2/2, so tan θ = √2/2.
G is √3/3, which is sin θ in the right triangle with opposite side √3 and hypotenuse 3. F is √6/3, which is cos θ in that triangle. D is √2, which is cot θ. A is 1/2, which is the square of the gradient. The acute angle with the x-axis has tangent √2/2.
The spiral turns left in steps of length 2, 2, 4, 4, 6, 6, ..., so the same corner pattern repeats at every larger even size. The pair 99 and 100 matches the pair 3 and 4 on the drawing. The lines y = 4, y = -4 and x = 4 contain (3, 4), (-3, 4), (3, -4), (-3, -4), (4, 3) and (4, -3), and x = -4 from y = 2 down to y = -4 contains (-4, -3). The point (-4, 3) sits in the gap above the corner (-4, 2). The corresponding missing point is (-100, 99).
A and C correspond to (3, 4) and (-3, 4) on the top side. B and D correspond to (3, -4) and (-3, -4) on y = -4. E and F correspond to (4, 3) and (4, -3) on x = 4. H corresponds to (-4, -3), which lies on the downward side at x = -4.
The circles have equal radii, so their common chord is the perpendicular bisector of the segment joining (−2, 1) and (3, −2). The midpoint is (1/2, −1/2) and the line of centres has slope −3/5, so the chord has slope 5/3. Its equation is y + 1/2 = (5/3)(x − 1/2), which rearranges to 5x − 3y = 4.
B is 3x + 5y = −1, the line of centres itself: slope −3/5 through the same midpoint. A is 3x − 5y = 4, from using slope 3/5 and so swapping the coefficients 5 and 3. E is 5x − 3y = 1, the correct left-hand side with the constant taken from the midpoint coordinates before they are cleared.
Each line meets the parabola once, so the discriminants vanish: (b − 2)2 = 4(c − 3) and (b − 4)2 = 4(c + 2). Subtracting gives (b − 4)2 − (b − 2)2 = 20, so −4(b − 3) = 20 and b = −2. Then 16 = 4c − 12, so c = 7 and b − c = −9.
D is 5, the value of b + c for the same pair b = −2 and c = 7. F is 14, which is 2c. E is 6, which is c − 1. The required combination is b − c.
Completing the square gives x = (y − 3)2 + 2, a rightward parabola with vertex (2, 3). Relative to P(−2, 3), a 90° clockwise rotation sends (u, v) to (v, −u). The image of the vertex is (−2, −1), and substituting the rotated coordinates into the original equation produces y = −x2 − 4x − 5, whose vertex is that image point.
A is y = −x2 − 4x − 3, the same downward parabola shifted so that its vertex is only 2 units below P. That 2 is the constant in x = (y − 3)2 + 2, while the horizontal distance from P to (2, 3) is 4. G is y = x2 − 6x + 11, the original equation with x and y interchanged, which reflects the curve in y = x instead of rotating it about P.
The logarithms require x − y > 0, x < 1 and y > −5. Squaring the left side after combining the right side gives (x − y)2 = 2(1 − x)(y + 5), which simplifies to (x + 5)2 + (y − 1)2 = 36. This is the circle of centre (−5, 1) and radius 6. It meets the line y = x at (1, 1) and (−5, −5), the two points excluded by the logarithms, and the radii to those points are perpendicular. The condition y < x selects the quarter-circle between them, whose length is (1/4) · 2π · 6 = 3π.
F is 12π, the circumference of the whole circle of radius 6. E is 9π, the complementary three-quarter arc, on which y > x, so log10(x − y) is not defined. The domain keeps only the quarter-circle.
The midpoint of A(0, 2) and C(4, 0) is (2, 1), and AC has direction (4, -2), so slope -1/2. The other diagonal is perpendicular to AC, so its slope is 2, and it passes through the midpoint: y - 1 = 2(x - 2), which is y = 2x - 3.
C is the line AC itself, y = -(1/2)x + 2. A uses slope -2, the slope of the vector (4, -2), for the diagonal that should be perpendicular to it. F has slope 2 but passes through A rather than through (2, 1). The perpendicular through the midpoint is y = 2x - 3.
A line bisects the circle only if it passes through the centre (9, -2), and it bisects the square only if it passes through the centre (0, 1). The unique such line has slope (1 - (-2))/(0 - 9) = -1/3, so y - 1 = -x/3. It meets the x-axis at x = 3.
D is 4.5, the x-coordinate of the midpoint of the two centres, which is not the x-intercept of the line through them. G would be right if some other line bisected both areas, but each of these regions is bisected only by lines through its centre. The line through both centres meets the x-axis at 3.
The centre is (-a/2, -b/2) and the radius squared is (a2 + b2)/4 - c. The distance from the centre to the y-axis is |a|/2. Setting that equal to the radius gives a2/4 = (a2 + b2)/4 - c, so b2 = 4c. Since a ≠ 0, the radius is then |a|/2 > 0. This equality is necessary and sufficient.
A is a2 = 4c, the same calculation for the x-axis. C equates a/2 with the radius, so it covers only a > 0; the circle with a = -2, b = 2 and c = 1 has centre (1, -1) and radius 1, hence is tangent to the y-axis, while a/2 = -1 is not the radius. D and F compare a coordinate of the centre with the radius in the y-direction, which is tangency to the x-axis.
Completing the square gives (x − p)2 + (y − 3)2 = 2p(p − 4). This is a circle when the right side is positive, so p < 0 or p > 4.
B is the interval where the right side is negative, which is not a circle. E shifts the boundary from 4 to 9. The radius squared is 2p(p − 4), so the circle exists for p < 0 or p > 4.
The centre is (−f, −g). The distance from (p, q) to that centre is determined by f, g, p and q. The constant h changes the radius, not the centre, so it is not needed.
A includes h and omits the coordinates of P. C and D each omit one coordinate of the centre. The four values f, g, p and q are enough on their own.
Place the intersection of the diagonals at the origin, with P at (−x, 0), R at (x, 0), Q at (0, −y) and S at (0, z). The vectors from P to S and from P to Q are (x, z) and (x, −y). Their dot product is x2 − yz, so the angle at P is a right angle exactly when x2 = yz.
A forces the kite to be a rhombus, which is stronger than a right angle at P. E is the condition for a right angle at Q or S, not at P. The right angle at P is x2 = yz.
The circles have centres (2, 1) and (4, −5), and both have radius 4. The distance between the centres is √(22 + 62) = 2√10. The greatest distance between a point on each circle is that distance plus both radii, 8 + 2√10.
B is 14, which adds the vertical gap of 6 to both radii and ignores the horizontal gap. C is 16, the two diameters, with no gap between the centres. The centre distance plus both radii is 8 + 2√10.
The right angle can sit at any of the three vertices (2, 3), (9, −1), and (5, k). The dot products give k = 33/4, k = −8, and k = 5 or k = −3. None of these four positions is collinear. The sum is 33/4 − 8 + 5 − 3 = 9/4.
F is 8.25, only the right angle at (2, 3). A is −8, only the right angle at (9, −1). D is 2, only the two positions of the right angle at (5, k). C is 0.25, the first two values and not the third vertex. G is 10.25, which drops the −8. B is −6, which drops the 8.25. All four values add to 2.25.
The equation x2 + y2 = 2n(x + y) is the circle (x − n)2 + (y − n)2 = 2n2, so its area is 2π n2. The shaded pieces are the regions between C1 and C2, C3 and C4, and so on up to C99 and C100. The piece between C2k − 1 and C2k has area 2π(4k − 1). Summing k from 1 to 50 gives 2π × 5050 = 10100π.
D is 5050π, which is the same sum with area π n2 instead of 2π n2. A is 100π, one π for each circle. C is 2500π, which is 502 π. F is four times the correct area.
The gradients are −a/b, −b/c and −c/a. The product of the first two is a/c, so those lines are perpendicular only when a = −c, and then the third gradient is −c/a = 1. The other two pairings work the same way: whenever two lines are perpendicular, the third has gradient 1, so it is parallel to y = x.
A says the third line would also be parallel to the first two, but equal gradients are a different condition, b2 = ac, and they do not force the third gradient to match. The perpendicular case forces the third gradient to be 1.
Replace sin2 θ by 1 − cos2 θ. The equation becomes (cos θ − 1)(cos θ − 2) = 0. cos θ = 2 has no solution. cos θ = 1 at θ = 0, 2π and 4π, and all three lie in 0 ≤ θ ≤ 4π.
A, B and C leave out at least one of those three angles. E, F and G count extra solutions, including the impossible value cos θ = 2. The closed interval contains three solutions.
On [0, π], −1 ≤ tan x ≤ 1 holds on [0, π/4] and on [3π/4, π]. Also sin 2x ≥ 1/2 holds on [π/12, 5π/12]. The overlap is [π/12, π/4], which has length π/6.
A is π/12, the left endpoint of the overlap. C is π/4, the right endpoint. D, E, F and G are longer than the overlap and include angles where tan x lies outside [−1, 1] or sin 2x is below 1/2. The total length is π/6.
Write cos2(2x) = 1 − sin2(2x). With u = sin(2x) the equation becomes 1 − u2 + √3 u − 7/4 = 0, so u2 − √3 u + 3/4 = 0. The discriminant is 3 − 3 = 0, hence sin(2x) = √3 / 2. For 0° ≤ 2x ≤ 720° the solutions are 2x = 60°, 120°, 420° and 480°, so x = 30°, 60°, 210° and 240°. The greatest is 240°.
A and B are the smaller solutions 30° and 60°. E is 210°, another solution of the same equation. C is 120°, a value of 2x. D is 150°, from solving sin x = 1/2. G is 300°, from solving sin x = −√3/2. H is 330°, from solving sin x = −1/2.
The equation x tan x = 1 is equivalent to tan x = 1/x where cos x ≠ 0. The left side is an even function, so the roots are symmetric about 0, and x = 0 is not a root. On (0, π/2), tan x climbs from 0 to +∞ while 1/x falls, so there is one root. On (π, 3π/2), tan x climbs from 0 to +∞ and crosses 1/x once. On (π/2, π) and on (3π/2, 2π), tan x is negative while 1/x is positive. That gives two positive roots and two negative roots, so four roots in [−2π, 2π].
A counts no intersections. B counts only one positive root, and C counts only the positive pair. D drops one of the four symmetric roots. F counts one root in each of the five tan branches inside (−2π, 2π), including the two outer branches, where tan x and 1/x have opposite signs. G counts two roots in the central branch and one in each of the other four, again including those two outer branches.
The product (1 + 2 cos x) cos 2x is zero at x = π/4, x = 2π/3 and x = 3π/4 in (0, π). Now 1 + 2 cos x is positive on (0, 2π/3) and negative on (2π/3, π). Also cos 2x is positive on (0, π/4), negative on (π/4, 3π/4) and positive on (3π/4, π). The product is therefore negative on π/4 < x < 2π/3 and on 3π/4 < x < π.
A is the set where the product is positive, namely 0 < x < π/4 and 2π/3 < x < 3π/4. E keeps only the first negative interval, and F joins the two negative intervals across 2π/3 < x < 3π/4, where both factors are negative and the product is positive. B and C each mix one positive end interval with part of the negative set.
Replace cos2 x by 1 − sin2 x. The equation becomes 4 sin2 x + 4 = 7, so sin2 x = 3/4. The solutions of sin x = −√3/2 in [0, 2π] are 4π/3 and 5π/3, and 5π/3 is the larger.
A is the larger solution of sin x = √3/2, so it stops in [0, π]. C is the other solution of sin x = −√3/2. B is the larger solution of sin x = 1/2 in [0, π]. E is the larger solution of sin x = −√2/2, from sin2 x = 1/2. F is the larger solution of sin x = −1/2.
Solutions are the levels c of f(x) = 3 sin x + 2 − x. On [0, π], f rises from 2 to a maximum 2 + 2√2 − arccos(1/3) and then falls to 2 − π, so any c in [2 − π, 2) meets the graph once there. On (−π, 0), f has a minimum 2 − 2√2 + arccos(1/3) ≈ 0.40. Thus c = 1 meets (−π, 0) as well, while c = −1 does not. For every x > π, f(x) < 2 − π, because later peaks are 2π lower and the first of them is about −2.69. So c = −1 also has no solution beyond π. All three statements are true.
B, E and F omit statement 2, but c = −1 has one intersection on [0, π] and none on (−π, 0). C, E and G omit statement 1, but c = 1 has one intersection on [0, π] and another on (−π, 0). D, F and G omit statement 3, but that same c = −1 stays above every value of f(x) for x > π. A omits all three.
On 0 ≤ x ≤ π the factor 1 − 2 sin x is non-negative on 0 ≤ x ≤ π/6 and on 5π/6 ≤ x ≤ π, and cos x is non-negative on 0 ≤ x ≤ π/2. The product is non-negative when the factors have the same sign, including zeros: 0 ≤ x ≤ π/6 or π/2 ≤ x ≤ 5π/6.
B replaces the second interval by 5π/6 ≤ x ≤ π, where cos x is negative and 1 − 2 sin x is positive. C is the set on which the product is negative, apart from the zeros. D is π/6 ≤ x ≤ 5π/6, where 1 − 2 sin x is negative, with the factor cos x ignored. The product is non-negative on 0 ≤ x ≤ π/6 and π/2 ≤ x ≤ 5π/6.
Angle PRQ is the angle at R, opposite the side a + 2d. The cosine rule gives cos R = (a − 3d)/(2a). The condition 3d > 2a makes this cosine less than −1/2. The triangle inequality a > d makes it greater than −1. So −1 < cos R < −1/2, and 120° < angle PRQ < 180°.
A is 0° to 60°, the range obtained if the side a + 2d is added in the cosine rule instead of subtracted, which makes the cosine greater than 1/2. B stops at the boundary 120°, and C is the interval on the other side of that boundary. D includes angles whose cosine is greater than −1/2, which need 3d ≤ 2a. The strict conditions give 120° < angle PRQ < 180°.
With t = tan x, sin 2x = 2t/(1 + t2) and cos 2x = (1 − t2)/(1 + t2). Then sin 2x < cos 2x only for t between −1 − √2 and √2 − 1. On that range cos 2x is at least sin 2x, and sin 2x is already greater than t, so cos 2x < tan x never holds there. The ordering in C does not occur.
A holds at tan x = 2, B at tan x = 3/4, D at tan x = −1/2, E at tan x = −2, and F at tan x = 1/2. The ordering in C does not occur anywhere in the interval.
Where cos 2x = 0 the right side is 0 and the left side is ±x, so there is no solution. Elsewhere tan 2x = 1/x. For x from 0 to 2π, the angle 2x runs from 0 to 4π, and tan is positive on four branches. On each of those branches tan rises from 0 to infinity while 1/x stays positive, so each branch contributes one root. The endpoints x = 0 and x = 2π fail. The total is 4.
D is 3, one branch short of the four positive branches in (0, 4π). C is 2 and B is 1, from stopping 2x at 2π or counting a single intersection. A is 0. The negative branches of tan do not meet the positive graph of 1/x.
sin θ is symmetric about θ = π/2 + kπ. Set 2x − 4π/3 = π/2 + kπ, which gives x = 11π/12 + kπ/2. The positive values start at 5π/12, then 11π/12. The smallest positive one is 5π/12.
D is 11π/12, the axis for k = 0, one step of π/2 later than the smallest positive axis. A is π/12, the absolute value of the next axis to the left, which is negative. C is 7π/12 and E is 19π/12, neighbouring twelfths that are not in the arithmetic progression 11π/12 + kπ/2.
Let AC = b. The cosine rule at A gives b2 − 20b cos θ + 51 = 0. The two possible triangles share sin θ, so their areas are in the ratio of their values of b. That ratio is 3, so the roots are 3t and t. Their sum and product give 4t = 20 cos θ and 3t2 = 51, hence 75 cos2 θ = 51 and cos θ = √17/5.
C is 2√2/5, which is sin θ, since 1 − 17/25 = 8/25. B is 151/200, from (100 + 51)/200, putting the constant term 51 into the cosine rule with both known sides taken as 10. A is 5/7. E is √51/8 and F is √34/8.
sin2 θ + sin2(90° − θ) = 1. Pair 0° with 90°, 1° with 89°, and so on up to 44° with 46°. That is 45 pairs, each summing to 1. The unpaired middle term is sin2 45° = 1/2. The total is 45.5.
D is 45, the 45 pairs with sin2 45° left out. F is 46, from counting that middle term as 1. C is 1.5, one pair plus sin2 45°. B is 1, a single pair, and A is 0.5, the middle term alone.
On (0, 3π/2), cos x = c has two solutions exactly when −1 < c < 0. Then cos2(2x) = c2 means cos(2x) = |c| or cos(2x) = −|c|, with 2x running through (0, 3π). Each value in (−1, 0) ∪ (0, 1) is attained three times there: twice in (0, 2π) and once in (2π, 3π). The two values are distinct because c ≠ 0, so there are 6 solutions.
A is the number of solutions of the original equation. B counts only one of the two values of cos(2x). C counts two solutions for each sign and misses the third in (2π, 3π). F counts four for each sign, which is the count for 2x in an interval of length 4π. E is one more than 6, from including an endpoint of (0, 3π).
The sine rule gives sin Q = 1/p, so a triangle exists only for p ≥ 1. Let α = arcsin(1/p). Then Q = α always gives a triangle. The second possibility Q = 180° − α gives a positive third angle exactly when 1 < p < 2, and the two choices give different lengths PQ. At p = 1 the two choices coincide, and for p ≥ 2 the second choice is impossible. The length is unique for p = 1 and for p ≥ 2.
C and D are ranges in which the ambiguous case produces two triangles. A keeps only the right-angled case p = 1, and H keeps only p ≥ 2. B and F use √3, the side that appears in a single 30°–60°–90° triangle, in place of the boundary p = 1. G includes the whole ambiguous interval.
Let θ = 2x. Adding the first equation to √3 times the second gives 4 sin θ = 2, so sin θ = 1/2. The second equation then gives cos θ = −√3/2. For x from 0° to 360°, θ runs from 0° to 720°, so θ = 150° or 510°. The values of x are 75° and 255°, and their sum is 330°.
D is 150° + 510°, the sum of the values of 2x rather than of x. The cosine condition excludes every other angle with sine 1/2. The two solutions in range are 75° and 255°, and they add to 330°.
On [0, π], sin 2θ = 1/2 at θ = π/12 and θ = 5π/12, and sin θ − cos θ = 0 at θ = π/4. The product is non-negative on [0, π/12] and on [π/4, 5π/12]. Those lengths are π/12 and π/6, so the total length is π/4. The fraction of the interval is 1/4.
A is the first piece alone and B is the second piece alone. F is 3/4, the fraction of the interval where the inequality fails. Together the two pieces occupy 1/4 of [0, π].
The terms repeat every four steps: sin 10°, sin 100°, sin 190° = −sin 10°, and sin 280° = −sin 80°. Each block sums to 0. From k = 0 to k = 90 there are 91 terms, which is 22 full blocks plus k = 88, 89 and 90. Those three terms are sin 10°, sin 100° and −sin 10°, so the sum is sin 100°.
A is the sum of the complete blocks, with the three leftover terms omitted. B is sin 10°, the first leftover term before it cancels the last one. D is sin 190°, the final term on its own. After that cancellation the sum is sin 100°.
Let u = |cos x|, so u ≥ 0 and cos2 x = u2. The equation becomes 2u2 + u − 1 = 0, or (2u − 1)(u + 1) = 0. The only admissible root is u = 1/2. In the interval from 0° to 180°, |cos x| = 1/2 at x = 60° and x = 120°. There are two solutions and their sum is 180°.
B still says two solutions, but the sum 240° is 60° + 180°. That pair is what you get by solving 1 − 2 cos2 x = cos x and dropping the absolute value, which loses 120° and keeps the endpoint 180°, where the two sides are −1 and 1. C and D claim three solutions. E and F claim four. Only 60° and 120° satisfy the equation in the given interval.
tan θ = 2 > 0 and 180° < θ < 360°, so θ is in the third quadrant and cos θ is negative. Opposite 2 and adjacent 1 give hypotenuse √5, so |cos θ| = 1/√5 = √5/5. Thus cos θ = -√5/5.
E is the positive value from the first quadrant. G and H put the opposite side 2 in the numerator, giving 2√5/5. A to D are built from √3, which is neither side of the 1-2-√5 triangle.
f(4) = 4 sin(4) with the 4 in radians. A degree-mode sine returns that value when its input is 4 × 180/π. Multiplying the result by 4 gives 4 × sin(180 × 4 ÷ π).
A and B convert 4 as though it were a number of degrees, using π × 4 ÷ 180. C keeps the factor 4 but still feeds the sine that degree-to-radian conversion. D and E multiply by the degree measure 180 × 4 ÷ π instead of by 4; D also takes the sine of 4 degrees.
sin(kπ/3) repeats every six terms: √3/2, √3/2, 0, -√3/2, -√3/2, 0, and each block sums to 0. The partial sums inside a block are √3/2, √3, √3, √3/2, 0, 0. The sum equals √3/2 exactly when the remainder on division by 6 is 1 or 4, which is the same condition as n being 1 more than a multiple of 3.
A keeps only n = 1, but n = 4 also works. B and C are multiples of 3 or of 6, and those sums are √3 or 0. E keeps only remainder 1 on division by 6, so it misses n = 4. F adds remainder 2, whose sum is √3, and still misses n = 4.
With c = cos x, the denominator is 7 + 5c − (1 − c2) = (c + 2)(c + 3). The numerator is c + 3, and c is never −3, so f(x) = 1/(cos x + 2). As cos x runs from −1 to 1, f runs from 1 to 1/3. The positive difference is 1 − 1/3 = 2/3.
B is 1/3, the minimum value of 1/(cos x + 2). E is 1, the maximum value. F is 2, the difference between the largest and smallest values of the denominator cos x + 2, which are 3 and 1.
The right side is 4 sin2(πx/2), hence lies in [0, 4]. The left side is 4 − (x − 1)2, which is negative outside [−1, 3], so every solution lies in that interval. Setting t = x − 1 gives cos2(πt/2) = 1 − t2/4, or with u = t/2 in [−1, 1], cos2(πu) = 1 − u2. The semicircle √(1 − u2) meets |cos(πu)| at u = 0 and at exactly one point in (1/2, 1); the evenness of both graphs supplies the partner in (−1, −1/2). Those are three values of x.
A is 2, the symmetric pair in (−1, −1/2) and (1/2, 1), leaving out the solution x = 1 where both sides equal 4. C is 4, from also counting x = −1 and x = 3, where the left side is 0 and the right side is 4. The graphs meet at three points.
Set r2 = x2 + y2. Then sin(r2) = 1/2 with 0 ≤ r2 ≤ 8π, so r2 is one of π/6, 5π/6, 13π/6, 17π/6, 25π/6, 29π/6, 37π/6, 41π/6. That is eight circles centred at the origin. The gaps in r2 alternate between 2π/3 and 4π/3, so the circles form four close pairs. sin 0 = 0, so there is no point at the origin. Sketch A is the one with those eight paired radii.
C draws the circles with equal steps in the radius, whereas equal steps belong to r2. E puts a dot at the origin, but sin 0 = 0, so the origin is not a solution. B includes a ring much smaller than √(π/6) beside the next ring, which breaks the ratio √5 between the first two radii.
Let φ = 60° · 4cos θ. Then sin2 φ = 3/4. Since cos θ lies in [−1, 1], φ lies in [15°, 240°], and the only available solutions are φ = 60°, 120° and 240°. They occur at θ = 0° (where φ = 240°), θ = 60° (where φ = 120°) and θ = 90° (where φ = 60°). The next solution is θ = 270°. Exactly three solutions lie in 0° ≤ θ ≤ x° when 90 ≤ x < 270.
D is 270 ≤ x < 300, from discarding θ = 0. There sin(240°) = −√3/2, but the square is still 3/4, so θ = 0 is a solution and the fourth solution is 270°. A is 90 ≤ x < 120, which stops at 120° even though no new solution appears until 270°.
The identity √(1 + sin θ) √(1 - sin θ) = |cos θ|, and likewise with sine and cosine swapped, turns the expression into sin θ |cos θ| + cos θ |sin θ|. This sum is 0 whenever sin θ and cos θ have opposite signs or either one is 0, which is exactly the range 90° through 180° and the range 270° through 360°. Each block contains 91 of the listed degree values, so the total is 182. From 1° to 89° the sum is 2 sin θ cos θ > 0, and from 181° to 269° it is -2 |sin θ cos θ| < 0.
D counts only the four axis angles 90°, 180°, 270° and 360°. E is 93, which is 90° through 180° together with 270° and 360°, leaving out 271° through 359°. G is 271, which also counts 181° through 269°, where the expression is negative. The two blocks together contain 182 values.
Let u = cos2 θ. Then 2u2 − 5u + 3 = 0, so (2u − 3)(u − 1) = 0. u = 3/2 is impossible because cos2 θ is at most 1. u = 1 gives cos θ = ±1, so θ = 0, π, or 2π. There are 3 solutions.
D counts a fourth solution from u = 3/2, which is outside the range of cos2 θ. A and B miss one of the endpoints 0 and 2π. The valid solutions are exactly those three.
The squared cosines are the roots of 7u2 − 6u + 1 = 0. The squared sines are 1 minus those roots, so their sum is 8/7 and their product is 2/7. The quadratic is 7u2 − 8u + 2 = 0, and the equation in x is 7x4 − 8x2 + 2 = 0.
D is the original equation. E changes the signs as if sine and cosine swapped without replacing the roots by 1 minus those roots. The transformed equation is 7x4 − 8x2 + 2 = 0.
The side −x2 + 6x − 5 is positive for 1 < x < 5, and the side x − 1 is then positive as well. The 30° angle is acute and the opposite side is x − 1, so there are two triangles when (1/2)(−x2 + 6x − 5) < x − 1 < −x2 + 6x − 5. Those inequalities simplify to 3 < x < 4.
A is the whole interval on which both lengths are positive. E keeps x > 3 but not the upper bound x < 4. The ambiguous case is exactly 3 < x < 4.
The maximum of cos x is 1. sin(cos x) has maximum sin 1, since sine is increasing on [−1, 1]. cos(sin(cos x)) reaches 1 when sin(cos x) = 0. sin of that cosine reaches sin 1 again, when the cosine inside it reaches 1. The fifth function is cos of a value that stays at least sin(cos(sin 1)) > 0, so its maximum is less than 1. Thus m1 = m3 = 1, m2 = m4 = sin 1, and 0 < m5 < 1.
F says m4 < m2, but both equal sin 1. C says m5 = 1, but the argument of the outer cosine never reaches 0. The fifth maximum is strictly less than 1, and the second and fourth are equal.
The angle at X is 30°, the adjacent side XY is √3 a, and the opposite side YZ is a. The sine rule gives sin Z = √3 / 2, so Z is 60° or 120°. The larger triangle has XZ = 2a and area (√3 / 2) a2. The smaller has XZ = a and area (√3 / 4) a2. The ratio of the areas is 2 : 1.
A treats the two triangles as the same size. D is the side ratio XY : YZ. E is what you get by reading the adjacent side as 3a, which cannot occur: the sine of Z would then be 3/2.
The equation (1 + 3 cos 3θ)2 = 4 splits into cos 3θ = 1/3 or cos 3θ = −1, with θ from 0° to 180°, so 3θ runs from 0° to 540°. cos 3θ = −1 at 180° and 540°, two values. cos 3θ = 1/3 at α, 360° − α, and 360° + α, three values. The total is 5.
B counts only the two solutions of cos 3θ = −1. C counts only the three solutions of cos 3θ = 1/3. D stops 3θ at 360° and misses 360° + α. F counts a further solution of cos 3θ = 1/3 beyond 540°.
Write 4sin2 x as 22 sin2 x. The equation becomes tan2 x − 2 sin2 x = 0, and tan x is undefined at π/2 and 3π/2, so those values are not solutions. If sin x = 0, then x = 0, π, or 2π, and each satisfies the original equation. If sin x ≠ 0, then cos2 x = 1/2, giving x = π/4, 3π/4, 5π/4, and 7π/4. That is 7 solutions in total.
C counts only the four solutions of cos2 x = 1/2. B counts only the three solutions of sin x = 0. E drops one of the endpoints 0 and 2π. G counts an extra value where tan x is undefined. All seven listed values satisfy the equation.
The equation factors as sin x (cos2 x − p2) = 0. In [0, 2π], sin x = 0 gives three solutions. If |p| > 1 there are no further solutions, and if |p| = 1 the cosine condition repeats those same three. If 0 < |p| < 1 there are four further solutions, seven in all. If p = 0 the cosine condition adds only two solutions, five in all. Thus n = 3 does not force p > 1, because p ≤ −1 also gives n = 3. And n = 7 does force −1 < p < 1.
B accepts the first statement, but n = 3 is also possible for p < −1. A rejects the second statement, although every case with seven solutions does lie strictly between −1 and 1. Only II is true.
y = −log10(1 − x) means 10−y = 1 − x, so x = 1 − 10−y.
F is 101−y, with the minus absorbed into the exponent as 1 − y. E is 10−y − 1. B and C replace the power of 10 by log10 y. A inverts a logarithm of 1 − y. The rearrangement is 1 − 10−y.
ax b2x c3x = (a b2 c3)x = 2. Taking log10 of both sides gives x = log10 2 / log10(a b2 c3).
B puts the sum a + 2b + 3c inside the logarithm. E is the logarithm of the single fraction 2/(a b2 c3). G and H drop one or both logarithms. The exponent is the quotient of the two logarithms.
Let u = 2x. Then u2 − 8u + 15 = 0, so u = 3 or u = 5. The roots are log2 3 and log2 5, and their sum is log2 15 = log10 15 / log10 2.
A is 3 + 5, the sum of the values of 2x. B is the coefficient 8. C is log10 4 and D is log10(15/4). The sum of the x-roots is log10 15 / log10 2.
If y = a xb with a and b constant, then log y = log a + b log x. That is a straight line when log y is plotted against log x.
A gives b log y = x log a, linear in x. B gives log y = log a + x log b, again linear in x. C does not rearrange into a linear relation between log y and log x. E makes x log y constant. The log-log graph is a straight line for y = a xb.
From 25 ≈ 33, take logarithms base 3: 5 log3 2 ≈ 3, so log3 2 ≈ 3/5.
D is 5/3, which approximates log2 3. C is 3/2 and B is 2/3. E is 1/2 and F is 2. The comparison of the powers gives 3/5.
tan(3π/4) = −1. log10 100 = 2. sin(π/2) = 1, so its tenth power is 1. Since √2 − 1 < 1, (√2 − 1)10 < 1. log2 10 > log2 8 = 3, so it is larger than each of the others.
A is −1, B is 2, C is 1, and E is less than 1. The value above 3 is log2 10.
Let u = 22x > 0. Then 42x = u2 and 22x+3 = 8u, so u2 − 8u + 12 = 0, and (u − 6)(u − 2) = 0. Thus 22x = 6 or 22x = 2. The roots are x = log4(6) and x = 1/2, with log4(6) larger because 6 > 2. Their difference is log4(6) − log4(2) = log4(3) = log10(3) / log10(4).
C is 6 − 2, the difference of the two values of 22x. B is 1, from taking the values of u to be 8 and 2, so the roots are x = 3/2 and x = 1/2. D is −1/2 + log10(3/2), from replacing log4(6) by log10(6) − log10(4). F is log10(3)/log10(2) = log2(3), which is twice log4(3). A is 3/4, which is not log4(3).
The second equation says y + 3 − 3x = 100 = 1, so y = 3x − 2. The first equation says 2(y − 1) = x2, with x > 0 and y > 1. Substituting gives 2(3x − 3) = x2, so x2 − 6x + 6 = 0 and x = 3 ± √3. Both roots are positive. Then y = 3(3 ± √3) − 2 = 7 ± 3√3, and both values are greater than 1, so both are valid.
B is 3 ± √3, the two values of x. D is 3 and 9, from setting y + 3 = 3x, so that the argument of the second logarithm is 0, and then solving x2 − 6x + 8 = 0. E is 1 and 13; y = 1 makes log10(y − 1) undefined. A is (5 ± 3√5)/2, the roots of t2 − 5t − 5 = 0, whereas the quadratic for y is y2 − 14y + 22 = 0.
Let a = 2x and b = 2y. Then a + 3b = 3 and a2 − 9b2 = 6. The second equation is (a − 3b)(a + 3b) = 6, so a − 3b = 2. Solving the linear pair gives a = 5/2 and b = 1/6. Hence p − q = log2(5/2) − log2(1/6) = log2 15.
A is 2p × 2q = 5/12, and C is its base-2 logarithm, which is p + q. B is 2p − 2q = 7/3, and D is the logarithm of that difference. E is log2 9, which would be p − q if 2p − q were 9. The ratio (5/2)/(1/6) is 15, so p − q = log2 15.
The solid graph is increasing, so a > 1. For x > 0 the dashed graph lies above it, and for x < 0 it lies below it, which is y = bx with b > a. Also akx = (ak)x, and k > 1 gives the larger base ak. A smaller base, or a power k < 1, would lie below the solid graph for x > 0.
Any option that includes statement 2 or 4 describes a curve that grows more slowly than y = ax. An option that keeps only one of statements 1 and 3 drops a second description of the same curve. Both 1 and 3 fit the dashed graph.
log2 7 lies between 2 and 3. Option B is (1/8 + 1/4)−1 = 8/3. Since π/3 > 1, option C is greater than 2. Option D equals (5√2 + 7)/4, which is greater than 3. Option E is 4 sin2(π/4) = 4 × 1/2 = 2, and that is the smallest.
A is greater than 2, B is 8/3, and D is greater than 3. C is a power of 2 with exponent greater than 1, so it is greater than 2. The smallest value is 2.
Let t = log2 p. The points (3, t) and (t, 4) lie on y = mx + 4, so t = 3m + 4 and 4 = mt + 4. The second equation says mt = 0, so m = 0 or t = 0. If t = 0 then p = 1 and m = −4/3, and both points fit. If m = 0 then the line is y = 4, so t = 4 and p = 16. The possible values are p = 1 and p = 16.
A keeps p = 1 and replaces 16 by 4. C and D include p = 1/4, and E and F include p = 1/64. Those are other powers of 2. The two lines that work have log2 p equal to 0 and equal to 4.
(√3)x+4 = 3(x+4)/2 = 9 × 3x/2. With u = 3x/2 the equation is u2 − 9u + 20 = 0, so u = 4 or u = 5. Then x = 2 log3 4 or x = 2 log3 5, and the sum is 2 log3 20.
D is log3 20, the sum of log3 4 and log3 5 with the factor 2 omitted. F is 4 log3 20, from taking x = 4 log3 u. B is 4 and C is 9, the root u = 4 and the factor (√3)4 = 9. A is 1.
The graph of y = 2x meets the horizontal line y = 1/2 at x = −1, so a negative solution does not force c > 1 and I is false. It meets y = 2 only at x = 1, so c > 1 does not force two solutions and II is false. It meets y = 1 only at x = 0, so c ≤ 1 does not force two distinct positive solutions and III is false. None of the three statements is true.
I fails for m = 0 and c = 1/2. II fails for m = 0 and c = 2. III fails for m = 0 and c = 1. Every option from B to H includes at least one of these three claims.
Since the base 5 is greater than 1 and x > y > z, the inequality in C rearranges to 2*5y > 5z + 52z − x. Here 2z − x < z, so 52z − x < 5z, and the right-hand side is less than 2*5z. The left-hand side is greater than 2*5z because y > z. The inequality is therefore strict.
A fails for x = 4, y = 3, z = 2: the left side is 1/2 and the right side is 5/2. B fails for the same numbers, because (34 + 32)/33 = 10/3 > 2. D fails for x = 2.1, y = 2, z = 1.1, where 70.1 + 7−0.9 is less than 2.
Let a = log3 x and b = log3 y. Then a + 2b = 1 and ab = −3. Substituting a = 1 − 2b gives 2b2 − b − 3 = 0, so b = 3/2 or b = −1. The matching values are a = −2 and a = 3. Thus x = 1/9 or x = 27, and the sum is 27 + 1/9.
G keeps only the value 27. B is 1, the sum of the two logarithms −2 and 3. E is 9 + 1/27, the pair obtained by reversing both signs of those logarithms. The x-values are 1/9 and 27, and their sum is 27 1/9.
Let u = 2sin x. Then 4sin x = u2, and sin x runs from −1 to 1, so u runs from 1/2 to 2. The expression is u2 − 4u + 17/4 = (u − 2)2 + 1/4. On this interval the largest value is at u = 1/2: (1/2 − 2)2 + 1/4 = 5/2.
A is 1/4, the minimum, reached at sin x = 1. C is the value of (u − 2)2 + 17/4 at u = 1/2, which keeps 17/4 outside the completed square. F says there is no maximum, but u lies in a closed interval and the quadratic in u attains 5/2.
The equation is 29x / 83x = 1/4. Since 83x = 23x + 1, the exponents of 2 satisfy 9x − 3 × 3x = −2. With u = 3x this is (u − 1)(u − 2) = 0. Then u = 1 gives x = 0, which is excluded, and u = 2 gives x = log3 2.
B is 2 log3 2, twice the solution. E is log2 3, the reciprocal obtained by swapping the base and the argument. C and D are the integers 1 and 2, neither of which satisfies 3x = 2. The non-zero solution is log3 2.
Let A = log a, B = log b and C = log c. The three equations are A + 2B + C = 7, 2A + B + 2C = 11 and 2A + 2B + 3C = 15. Solving gives A = 2, B = 1 and C = 3. Thus log b = 1, so the common base is b. Then a = b2 and c = b3, and both sides of each given equation agree.
A would be the base if log a were 1, but log a = 2. C would be the base if log c were 1, but log c = 3. D and E say the base is something else or cannot be found, but log b = 1 fixes the base as b.
Let u = 5x, so u > 0. The denominator is u2 − 4u + 7 = (u − 2)2 + 3, whose least value is 3, at u = 2. The denominator is always positive, so the greatest value of the reciprocal is 1/3.
D is 3, the least value of the denominator rather than the greatest value of the function. A is 1/7, the value obtained by keeping only the constant 7 in the denominator. F is 7, that constant itself. E is 4, the coefficient of 5x. The denominator cannot fall below 3, so the maximum is 1/3.
Write every power with base 2. The first equation is 23x = 23y+9, so x = y + 3. The second is 22x+2 = 2y−5, so 2x + 2 = y − 5. Substituting x = y + 3 gives y = −13 and x = −10, and x + y = −23.
F is −10, the value of x on its own. The second equation forces y = −13, and the sum of the two is −23.
log2(1/x) = −log2 x, so the squared middle term is (log2 x)2. Let u = (log2 x)2. The equation becomes u2 + 12u − 64 = 0, and the non-negative root is u = 4. Then log2 x = ±2, so x = 4 or x = 1/4. The positive difference is 15/4.
A is 4, the larger solution, which is also the difference of the logarithms 2 and −2. D is 17/4, the sum of the two solutions rather than the difference. E is 255/16, the difference of 16 and 1/16, from taking (log2 x)2 = 16 instead of 4. The two solutions are 4 and 1/4, and they differ by 15/4.
Set t = 2x > 0. The expression becomes t2 − 8t + 4 = (t − 4)2 − 12. The square is zero at t = 4, which is attained when x = 2, so the minimum value is −12.
A is −16, the minimum of t2 − 8t with the constant +4 omitted. E is 4, the constant term, which would require t = 0, and 2x is never 0. F is 20, from evaluating 42 + 4 at the vertex and dropping the negative term −8t.
Set u = 2x > 0 and v = log2(y). Subtracting the equations gives (p - 1)u = 1. For p = 1 the left-hand sides agree and the right-hand sides are 2 and 1, so there is no solution. For p ≠ 1, u = 1/(p - 1), and this is positive exactly when p > 1. Then v = 1 - u is a real number and y = 2v is positive, so every p > 1 works.
A takes the opposite inequality, which makes 1/(p - 1) negative and so not a power of 2. B keeps those negative values and only excludes p = 1. F and G impose a bound of 2, but p = 3 gives 2x = 1/2 and log2(y) = 1/2. The equations have a real solution exactly when p > 1.
For x > 1, let t = log2(x) > 0. Then f(x) = log2(t/2) = log2(t) - 1 and g(x) = log2(√t) = (1/2) log2(t). The difference is f - g = g - 1, so f ≥ g exactly when g ≥ 1. Thus either 1 ≤ g(x) ≤ f(x), or f(x) ≤ g(x) ≤ 1. That is option F, and both functions are defined for every x > 1.
B fails at x = 4, where f = 0 and g = 1/2, so g is neither at most 0 nor at most f. A and E fail at x = 216, where f = 3 and g = 2: the value 2 lies above 1, and 3 is larger than 2, so neither claimed order about 0 or about 1 holds. The bound that works for every x > 1 is 1, with g between 1 and f.
The integral of x from log2 5 to log2 20 is (1/2)((log2 20)2 − (log2 5)2). That factors as (1/2)(log2 4)(log2 100) = log2 100. So M = 100.
D is 20 and E is 25, the limits of the integral rather than M. G is 1002. The factored difference of squares is log2 100.
log10(31 − n) = (1 − n) log10 3. The sum of (1 − n) from n = 1 to 100 is −(0 + 1 + … + 99) = −4950. The sum is −4950 log10 3.
B drops the minus sign. C and D use the triangular number 5050, which is 1 + … + 100 rather than 0 + … + 99. The sum of those coefficients is −4950.
The chain rule for logarithms gives logx y · logy z · logz x = 1. The first two factors are z and x, so logz x = 1/(xz).
A is y, as if the three logarithms were equal. D is 1/(xy), swapping the roles of the three variables. The product of the three logarithms is 1, so the missing one is 1/(xz).
The ratio 4 : 1 means 2x + 4 = 4Q(x). With u = 2x, Q(x) = u2/4 − 4u + 16, so u + 4 = u2 − 16u + 64. Then u2 − 17u + 60 = 0, and (u − 12)(u − 5) = 0. Both u = 12 and u = 5 make Q positive, so both are valid. The larger x is log2 12.
E is log2 5, the smaller valid solution. B is 12 and A is 5, the values of 2x rather than of x. The larger solution of the quadratic is log2 12.
For f(x) = acos x with a > 0, the exponent runs through [−1, 1]. If a > 1 the greatest value is a and the least is 1/a, so a − 1/a = 3 and a = (3 + √13)/2. If 0 < a < 1 the greatest and least swap, so 1/a − a = 3 and a = (−3 + √13)/2. The two possible bases add to √13.
D is 3, the sum of both roots of a2 − 3a − 1 = 0, including the negative root, which cannot be a base. A is 0, but a = 1 makes the function constant and the difference is then 0, not 3. The two valid bases add to √13.
For x = 1 and y = 16 both sides equal 1. For x = 3 and y = 4 both sides equal 9. For x = 2 and y = 8 the left side is 16 and the right side is 2√8, which is not 16. Only the second pair is a counterexample.
B and F include the first pair, and D and G include the third pair, but both of those pairs satisfy the equation. The only counterexample is the second pair.
For a > 1 the equation ax = x has 0, 1 or 2 positive solutions, depending on a. Statement I is the same equation written with a logarithm, so it has the same solutions. Statement III, with u = 2x, is au = u, so it has the same number of solutions. Statement II is ax = |x|, which always has one extra negative solution, so its number is different.
Any option that includes II counts that extra negative solution. Any option that drops I or III drops an equation with the same solution set. The matching equations are I and III.
Write y = (9x2 − 12x + 4) x−3/2 = 9x1/2 − 12x−1/2 + 4x−3/2. The derivative is (9/2)x−1/2 + 6x−3/2 − 6x−5/2. At x = 2 this is 9/(2√2) + 6/(2√2) − 6/(4√2) = 3√2.
A is (3/2)√2, the first term at x = 2. D is (9/2)√2. C is 4√2 and E is 6√2. The three terms together equal 3√2.
The derivative of a quintic with non-zero leading coefficient is a degree-4 polynomial, so it has the same sign at both ends. Turning points therefore come in an even number. One local minimum and two local maxima would be three turning points, which cannot occur. Four turning points, two turning points, and no turning points, as for y = x5, can all occur.
A is four turning points, C is two, and D is none. Each of those is possible for a quintic. B asks for three turning points, and the derivative cannot change sign an odd number of times.
For y = 2x−2, dy/dx = −4x−3. At x = 1 the point is (1, 2) and the tangent gradient is −4, so the normal gradient is 1/4. The normal is y − 2 = (1/4)(x − 1), or y = x/4 + 7/4. It meets the axes at P(−7, 0) and Q(0, 7/4). Then PQ = √(49 + 49/16) = 7√17 / 4.
A is 3√5 / 2, the same intercept length for the curve y = 2/x, whose derivative at x = 1 is −2. F is 3√17 / 2, the intercept length of the tangent of gradient −4, and B is half of that length. D is 7 + 7/4 = 35/4, the sum of the normal’s intercept sizes. E is larger than the distance between (−7, 0) and (0, 7/4).
In the central cross-section the cylinder is a rectangle inscribed in a circle of radius 5. If its height is H, then r2 = 25 − (H/2)2 and the volume is V = π(25H − H3/4). Then dV/dH = π(25 − 3H2/4) = 0 gives H2 = 100/3, so H = 10/√3 and r2 = 50/3. Hence V = π(50/3)(10/√3) = 500π/(3√3) = 500√3 π / 9.
F is 1000√3 π / 9, from using height 20/√3 in place of 10/√3. D is 250√3 π / 3, from keeping the correct height but using the sphere’s radius 5 as the cylinder radius. A is 250π, B is 500π and C is 1000π, the volumes π(52)(10), π(52)(20) and π(102)(10), using the sphere’s radius or diameter as a cylinder dimension that does not sit on the sphere.
Let f(x) = 3x5 − 10x3 − 120x + 30. Then f'(x) = 15x4 − 30x2 − 120 = 15(x2 − 4)(x2 + 2). The only stationary points are x = −2 and x = 2. Since f'(x) > 0 for |x| > 2 and f'(x) < 0 for |x| < 2, x = −2 is a local maximum and x = 2 is a local minimum. f(−2) = 254 > 0 and f(2) = −194 < 0. The graph therefore crosses the axis once to the left of −2, once between the stationary points, and once to the right of 2, so there are 3 real roots.
A counts a single real root. B counts the two stationary points as roots. D counts four real roots, and E counts all five roots of the quintic as real. The two stationary values have opposite signs, so exactly three of the five roots are real.
Differentiate y = (2x + a)(x − 2a)2 by the product rule. At x = 1 the gradient is 2(1 − 2a)2 + 2(2 + a)(1 − 2a) = 2(1 − 2a)(3 − a) = 4a2 − 14a + 6. This quadratic in a opens upwards, so its least value is at a = 14/8 = 7/4. Completing the square gives 4(a − 7/4)2 − 25/4, and the least value is −25/4.
D is 7/4, the value of a at which the gradient is least. A is −49/4, the vertex value of 4a2 − 14a with the constant +6 omitted. E is 47/16 = 6 − (7/4)2, from subtracting (7/4)2 once instead of 4(7/4)2 when completing the square. B is −8, the value of −14 + 6 with the a2 term dropped.
The denominator (x2)1/3 equals |x|2/3. For x > 0, f(x) = (1 − x) x−2/3 and f'(x) = −x−5/3(x + 2)/3, which is negative. For x < 0, f'(x) = (−x)−5/3(x + 2)/3, so f'(x) ≤ 0 when x ≤ −2 and f'(x) > 0 when −2 < x < 0. The function is decreasing exactly for x ≤ −2 and for x > 0.
B is −2 ≤ x < 0, where the derivative is positive except at the stationary point, so the function is increasing there. D is x ≥ 1 and F is x ≤ −2 together with x ≥ 1, both starting the positive part at the root of the numerator and leaving out 0 < x < 1. C is x ≤ 1 with x ≠ 0, and E runs from the stationary point to that root.
For x > 0 the fourth root of x3 is x3/4, so f(x) = 2(x3 + 5x) x−3/4 = 2x9/4 + 10x1/4. Differentiating gives (9/2)x5/4 + (5/2)x−3/4.
A multiplies the two terms of f by 4 and by 4/3 and leaves the powers unchanged. C does that to the first term and replaces the second power by −1/4. D is an antiderivative of 2x9/4 + 10x1/4, namely (8/13)x13/4 + 8x5/4.
Let u = 2x−1 − (1/2)x−2, so f = u2. Then u' = −2x−2 + x−3 and u'' = 4x−3 − 3x−4. At x = 1, u = 3/2, u' = −1 and u'' = 1, so f'' = 2(u')2 + 2u u'' = 2 + 3 = 5.
A is f'(1) = 2u u' = −3. B is u'(1) = −1. E is 24 + 5, the second derivative of the expanded form 4x−2 − 2x−3 + (1/4)x−4 with the middle contribution −24 left out. D and F are not equal to 24 − 24 + 5. The second derivative at x = 1 is 5.
f'(x) = −6p2 + 6p x − 3x2, so f'(−1) = −3(2p2 + 2p + 1). The quadratic 2p2 + 2p + 1 has minimum 1/2, so the tangent gradient is at most −3/2 and is always negative. The normal gradient M = −1/f'(−1) therefore lies in (0, 2/3], and its greatest value is 2/3.
C is p = −1/2, the value that makes the tangent least steep, not the normal gradient. A is that tangent gradient, −3/2, and F is its absolute value 3/2. B is −2/3, the normal gradient with the wrong sign. D is 1/4, the square of the minimum 1/2 of 2p2 + 2p + 1. The greatest normal gradient is 2/3.
f'(x) = 6(x + 2), so f is decreasing for x ≤ −2 and increasing for x ≥ −2. g'(x) = 3(x + 1)(x + 3), so g is increasing for x ≤ −3 or x ≥ −1, and decreasing for −3 ≤ x ≤ −1. One increasing and the other decreasing means f' ≥ 0 and g' ≤ 0, or f' ≤ 0 and g' ≥ 0. That set is x ≤ −3 or −2 ≤ x ≤ −1.
G keeps only −2 ≤ x ≤ −1 and misses x ≤ −3. A is x ≥ −1, where both derivatives are non-negative. C includes −3 ≤ x ≤ −2, where both are decreasing, together with x ≥ −1, where both are increasing. D and F each add one of those same-sign regions to part of the correct set. B is the single ray x ≤ −1. The opposite-sign set is x ≤ −3 together with −2 ≤ x ≤ −1.
Expanding the numerator gives y = (1 − 6x + 9x2)/(2x3/2) = (1/2)x−3/2 − 3x−1/2 + (9/2)x1/2. Differentiating term by term gives (9/4)x−1/2 + (3/2)x−3/2 − (3/4)x−5/2.
B and C change the sign of the middle term, which is the derivative of −3x−1/2. D, E and F put a minus sign on (9/4)x−1/2, the derivative of (9/2)x1/2. The three signs that match these derivatives are those in A.
The derivative 3x(x − 2) gives a local maximum of a at x = 0 and a local minimum of a − 4 at x = 2. There is exactly one real root when a < 0 or a > 4. At the endpoints, a = 0 gives the roots 0 and 3, and a = 4 gives the roots −1 and 2. The condition |a| > 4 is the part a > 4 or a < −4 of that region, so it is sufficient.
A and D include values between 0 and 4, where there are three real roots. B includes a = 0, and C includes a = 4, each with two distinct real roots. F includes those same values. G is 9/4, between 0 and 4. H includes a = 3/2, which has three real roots. Only |a| > 4 forces exactly one real root.
The derivative is 6x2 − 6x − 12 = 6(x − 2)(x + 1), so there is a local maximum at x = −1 and a local minimum at x = 2. The values there are c + 7 and c − 20. Three distinct real roots need the local maximum above the axis and the local minimum below it, so −7 < c < 20.
A is −20 < c < 7, the two critical heights with the signs reversed. D is only c > −7, and E is only c < 20. C is c > 7 and F is c < −20, which use one critical height as a one-sided bound.
On y = 10 − x2 the tangent gradient at x = p is −2p, so the normal gradient is 1/(2p). Since that gradient is positive, p > 0. The normal is y = mx + 5 and passes through (p, 10 − p2), so 5 − p2 = mp = 1/2. Then p2 = 9/2 and p = 3√2/2.
A is √2/6, which equals the normal gradient m = 1/(3√2). B is the negative of that gradient and D is −p. A positive normal gradient needs p > 0. E is √5 and F is −√5, from 5 − p2 = 0, which drops the normal-gradient term.
A local minimum of f occurs where f' changes from negative to positive. On the sketch, f' crosses the axis downwards at A, stays negative through B, and crosses upwards at C. After C the derivative is positive through D and E, and it meets the axis at F without going negative. The sign change from negative to positive is at C.
A is the crossing from positive to negative, which is a local maximum of f. B is a minimum of the derivative while f' is still negative, so f is still decreasing. D lies on the positive part of the graph, and E is a maximum of f' with f' still positive. F is a touch on the axis after which f' stays non-negative.
The stationary point is where 2x + b = 0, so x = −b/2 and y = 2 − b2/4. With u = b2 the squared distance from the origin is u2/16 − (3/4)u + 4. For u ≥ 0 the minimum is at u = 6, so b = √6.
D is √6/2, the distance of that stationary point from the y-axis. The stationary points lie on y = 2 − x2, and the closest has x2 = 3/2, so b = −2x = √6. E is √2 and B is 1. C is 2, the value where the stationary point is equally far from both axes. A is b = 0, where the stationary point is (0, 2) and the distance is larger.
For x > 0, f(x) = (1/2) x5/2 − 2 x1/2. Differentiating gives f'(x) = (5/4) x3/2 − x−1/2. At x = 4 this is (5/4)*8 − 1/2 = 10 − 1/2 = 9.5.
D is f(4) = 12, the value of the function rather than the derivative. B is 10 − 1, from treating x−1/2 at x = 4 as 1. E is 5*8 − 1/2 = 39.5, from differentiating x5/2 as 5x3/2. F is 2(3*16 − 4) = 88, twice the derivative of the numerator. A is (3/2)*√4 = 3, the derivative of x3/2 evaluated on its own.
The inequality says that the midpoint value is at most the average of the endpoint values. For a polynomial, put a = t − h and b = t + h. The inequality becomes f(t + h) + f(t − h) ≥ 2f(t) for every t and every h > 0. Dividing by h2 and letting h tend to 0 forces f''(t) ≥ 0. If f''(t) < 0 at any point, the inequality reverses on a small interval about that point. So f''(x) ≥ 0 for all x is necessary.
A and D fail for f(x) = x2 − 1 and f(x) = x2 + 1, both of which satisfy the midpoint inequality. B and E fail for f(x) = x2, whose first derivative changes sign. F is the condition for the reversed inequality, as with f(x) = −x2.
Expanding gives y = 4q2 x + 6q − 2q x3 − 3x2, so dy/dx = 4q2 − 6q x2 − 6x. At x = −1 the gradient is 4q2 − 6q + 6. This quadratic in q opens upwards and its vertex is at q = 6/8 = 3/4.
B is −3/4, the vertex of 4q2 + 6q + 6, which is the same expression with the linear sign changed. E is 1/2, from setting the factor 2q − x2 equal to 0 at x = −1. A is the given x-value. The gradient at x = −1 is least when q = 3/4.
If p is increasing on [a, b], then p(a) ≤ p(b), so I is necessary. It is the only one. For p(x) = x3, the derivative 3x2 is nonnegative, but on [-2, -1] one has p'(-2) = 12 > 3 = p'(-1), so II can fail. For a polynomial with derivative (x2 - 1)2, which is nonnegative, p''(-1/2) = 3/2 > -3/2 = p''(1/2), so III can fail.
C, E, G and H include II, but x3 is increasing on [-2, -1] while its derivative decreases there. D, F, G and H include III, which fails for a polynomial whose derivative is (x2 - 1)2. A drops I, which the definition of increasing requires. Only I is necessary.
The denominator is 2x3/2, so the quotient is (1/2)(x3/2 − 5x1/2). Differentiating gives (3/4)√x − 5/(4√x), which is (3x − 5)/(4√x).
A is −√x/2, the derivative of the second term alone with the sign kept. B is √x/4, half of the first term with the second term dropped. D puts the square root on the 3x in the numerator. F differentiates the numerator and divides by the derivative of the denominator. The power form differentiates to (3x − 5)/(4√x).
Write f(x) = x15/7 − x8/7 + x1/7. For x > 0 the factor 7x6/7 is positive, and multiplying f'(x) by it produces 15x2 − 8x + 1 = 15(x − 1/5)(x − 1/3). This quadratic is negative between the roots, so f decreases on (1/5, 1/3). That subinterval has length 1/3 − 1/5 = 2/15, which is the fraction of (0, 1).
G is 13/15, the fraction of (0, 1) on which 15x2 − 8x + 1 is positive, so f is increasing there. B is 1/5 and C is 1/3, the endpoints of the decreasing interval. The fraction is the length of that interval.
Let u = x2 ≥ 0. Then u2 − p2 u has vertex u = p2/2, and the value there is −p4/4. Setting −p4/4 = −9 gives p2 = 6. The quadratic x2 − px + 6 has minimum 6 − p2/4 = 6 − 6/4 = 9/2, whether p is √6 or −√6.
A is −3, from treating the minimum of x4 − p2 x2 as −p2/4. That forces p2 = 36, and then 6 − p2/4 = −3. D is 3, from subtracting p2/2 instead of p2/4 when completing the square for x2 − px + 6, since 6 − 6/2 = 3.
f''(x) = a gives f(x) = (a/2)x2 + bx + 1, using f(0) = 1. Then f(1) = 2 gives a/2 + b = 1. The integral from 0 to 1 equals 1, so a/6 + b/2 = 0 and b = −a/3. Substituting gives a/6 = 1, so a = 6.
D is 2, which is f(1) rather than a. E is 3, from losing the factor 1/2 in the quadratic term. The two conditions together give a = 6.
Each curve is a parabola opening upwards, with its lowest point at x = k/2 and height k2/2 + 4k + 3. That height is least when k = −4, where it equals −5, and the x-coordinate is −2. The sum of the coordinates is −7.
C is the height alone. A and B are nearby values of the quadratic in k. The lowest point is (−2, −5), so the sum is −7.
A rectangle from x = −t to x = t fits between the curves with height 10 − 3t2. Its area is 20t − 6t3. The maximum for t > 0 is at t2 = 10/9, and the area there is 40√10 / 9.
G is half of that area, from using the width t instead of 2t. B is 52/9, from dropping the square root when t = √(10/9) is substituted. The maximum area is 40√10 / 9.
The derivative is 12x3 + 12x2 + 12x = 12x(x2 + x + 1). The quadratic x2 + x + 1 has discriminant −3, so it has no real root. The only stationary point is x = 0.
C and D count the roots of the quadratic as extra stationary points. That quadratic never meets the axis, so there is one stationary point.
f(x) = (2/3)x3 + 2m x2 + n with m > 0. Then f'(x) = 2x(x + 2m), so the stationary points are x = −2m and x = 0. The first is a local maximum and the second is a local minimum. Three distinct real roots need the local maximum above the axis and the local minimum below it: f(−2m) = (8/3)m3 + n > 0 and f(0) = n < 0. So −(8/3)m3 < n < 0.
B replaces the stationary point x = −2m with x = −m and gets the smaller bound −(4/3)m3. Any option with n > 0 puts the local minimum above the axis, so the cubic cannot cross three times. Equality at either end gives a repeated root, so the inequalities have to be strict.
For f(x) = x2 + 1 there are no real roots, but g(x) = 2x2 has one, so M < N is possible. For f(x) = x, both f and g have exactly one real root, so M = N is possible. For f(x) = x2 − 1 there are two real roots, while g(x) = 2x2 has one, so M > N is possible.
Any row that drops one of the three comparisons excludes a polynomial that realises it. All three comparisons occur.
∫ from 0 to 1 of (x − a)2 dx equals a2 − a + 1/3. This quadratic in a is least at a = 1/2, and the least value is 1/4 − 1/2 + 1/3 = 1/12.
B is 1/3, the value of the integral at a = 0. C is 1/2, the value of a that minimises it. D is 7/12 and E is 2. The minimum of a2 − a + 1/3 is 1/12.
The curve y = x2 − 1 meets the x-axis at x = ±1, so the enclosed area is the absolute area on [−2, 2]. Each outer piece, from 1 to 2 and from −2 to −1, has area 4/3. The middle piece, where the curve is below the axis, has area 4/3. The total is 8/3 + 4/3 = 4.
A is 4/3, the signed integral from −2 to 2, in which the middle area cancels part of the outer area. B is 8/3, the two outer pieces alone. D is 16/3, four copies of the outer piece 4/3. E is 12, the area of the 4 by 3 rectangle up to the endpoint height, and F is 16 = 42.
Expand (x2 − 4/x2)2 as x4 − 8 + 16x−4. The antiderivative is x5/5 − 8x − 16/(3x3). From 1 to 2 this is 31/5 − 8 + 14/3 = 43/15.
E drops the cross term −8 and leaves 163/15. D uses a cross term of −4 rather than −8 and gets 103/15. C reverses the sign of the integral of x−4, which produces −97/15, and then takes the positive value. F adds the integrand at the endpoints, 9 + 9. B is 43/15 rounded to the nearest integer.
Rewrite the derivative as 3x2 − 2x−3 + 3x−2. Integrating gives y = x3 + x−2 − 3x−1 + C. The condition y = 5 at x = 1 gives 1 + 1 − 3 + C = 5, so C = 6. Thus y = x3 + x−2 − 3x−1 + 6.
A integrates 3x2 as x3/3, so its derivative begins with x2. B has (1/2)x−2 where the integral of −2x−3 should contribute x−2, and it does not pass through (1, 5). D integrates 3x−2 as if it were x−2, leaving −x−1 and constant 4. E and F also pass through (1, 5), but their derivatives are 3x2 − 4x−3 + x−2 and 9x2 − 2x−3 + x−2. Only C has both the correct antiderivative and C = 6.
The substitution u = x + 2 gives ∫ from 0 to 2 of f(x + 2) dx = ∫ from 2 to 4 of f(u) du = A. The added 1 integrates to the length of the interval, which is 2. So ∫ from 0 to 2 of [f(x + 2) + 1] dx = A + 2, for every such f.
A has the same integral equal to A + 1, which replaces the interval length 2 by 1. C and D shift the limits the other way and compare ∫ from 4 to 6 of f with A. E and F do the same with ∫ from 6 to 8 of f. Positivity of f does not make those integrals equal to A or A − 1. Only the integral from 0 to 2 equal to A + 2 must hold.
f(x) = −2x2 + 10 has second derivative −4, so it is concave down, and so is y = f(x + 1). Each trapezium lies under the curve, and the rule underestimates (1) and (2). Reflection in y = 6 sends y to 12 − y, which reverses the sign of the second derivative. Curve (3) is concave up, so the trapezium rule overestimates it.
The three results are underestimate, underestimate, overestimate. A marks (3) as an underestimate, missing the reversal of concavity. Every option that marks (1) or (2) as an overestimate puts the chord of a concave-down graph above the curve. Only B matches under, under, over.
Since f is increasing and f(0) = 0, f is non-negative on [a, b], so R is the integral of f from a to b. Adding the constant 2f(b) adds a rectangle of width b − a and height 2f(b). The new area is R + 2(b − a)f(b).
A uses height f(b) rather than 2f(b). C and D omit the width b − a. E and F are differences of squares. G multiplies 2f(b) by f(b) − f(a) rather than by b − a.
Write (3 − 2x)/(x√x) as 3x−3/2 − 2x−1/2. An antiderivative is −6x−1/2 − 4x1/2. From 1 to 4 this is (−3 − 8) − (−6 − 4) = −1.
E is −1/4, from multiplying by the new exponent instead of dividing by it, which gives the antiderivative −3/(2√x) − √x. G is 7, from the antiderivative −6/√x + 4√x, the sign error in the second term. A is −13/2, B is −85/16, C is −13/8 and F is 7/4, which are other evaluations of the same limits.
The roots are 0, p, q and r. The graph is positive on (0, p), negative on (p, q) and positive on (q, r). Let those signed areas be A, B and C. Then A + B + C = 0, A + B = −2 and B + C = −3, so A = 3, B = −5 and C = 2. The enclosed area is 3 + 5 + 2 = 10.
A is 0, the signed integral from 0 to r. D is 5, the sum of the absolute values of the two given integrals, which counts the middle region in both. B is 1, the difference of the two positive pieces. C is 4 and E is 6, partial combinations of 2, 3 and 5.
The trapezium rule is exact for a constant function. If T(f) underestimates the integral of f from 0 to 1, then T(1 − f) = 1 − T(f) is larger than 1 − the integral of f. So the same partition overestimates the integral of 1 − f(x).
A adds a constant and B scales by 2, so both keep the same underestimate. C shifts the graph onto [−1, 0] and D reflects it; the sampled values, and therefore the error, are unchanged. Replacing f by 1 − f reverses the error.
The curves are y = p√x and y = x2/p2. They meet at x = 0 and x = p2, and p√x is above the other curve between those points. The area is the integral from 0 to p2 of (p√x − x2/p2) dx = (2/3)p4 − (1/3)p4 = p4/3.
C is half of that area. A and B still contain p5/2, which is the power obtained by stopping the integral at an upper limit of order p rather than p2. The two areas are (2/3)p4 and (1/3)p4, so the enclosed area is p4/3.
On [−1, 0] the integrand is −x(1 − x), and its integral is 5/6. On [0, 3] it is x(1 − x), and its integral is −9/2. The total is 5/6 − 9/2 = −11/3.
A is 17/3, the integral of |x(1 − x)|, which also folds the sign change of 1 − x at x = 1. D is −16/3, the integral of x(1 − x) with no modulus at all. E is the positive value 11/3. The signed integral of |x|(1 − x) is −11/3.
With u = √t, (t − 1)/(1 + √t) dt = 2u(u − 1) du. Integrating from t = 1 to t = 9 gives 24π[(2/3)t3/2 − t] from 1 to 9 = 24π(28/3) = 224π. Since V = 7 at t = 1, the value at t = 9 is 224π + 7.
B is 216π + 7, which is 24π × 9 + 7 and leaves out the constant fixed by V(1) = 7. A is 208π + 7, the same increase with 8π subtracted. The definite integral from 1 to 9 is 224π, so V = 224π + 7.
The substitution u = x + 1 shows that the integral of f(x + 1) from 0 to 1 equals the integral of f from 1 to 2, so that piece is 6. The first statement becomes 2 times the integral from 0 to 1, plus 30, equals 14. That integral is −8, and adding the piece from 1 to 2 gives −2.
A is −8, the integral from 0 to 1 alone. H is 14, the right-hand side of the first given equation. The two pieces are −8 and 6, so the integral from 0 to 2 is −2.
sin2 x is even, and the four-strip sample points on [-b, -a] are the reflections of those on [a, b]. The trapezium weights are symmetric, so the approximation and the integral are unchanged, and I is an overestimate as well. The trapezium rule is exact for the constant 1, and sin2 x + cos2 x = 1, so the error for cos2 x is the negative of the error for sin2 x. An overestimate for sin2 x is therefore an underestimate for cos2 x.
B keeps only the reflected sine integral. C keeps only the cosine integral. A keeps neither. Both the reflection and the identity sin2 x + cos2 x = 1 apply, so I and II are both necessary.
The curve is y = k(x − 2)(x − q) with k > 0. Region R is the area between the curve and the x-axis from 0 to 2, and S is the area between them from 2 to q. Those areas are equal precisely when the signed integral from 0 to q is zero. That integral is −q3/6 + q2. Since q > 2, q = 6.
D is 4, from giving the two regions the same width 2. A is √6 and B is 3; neither makes the signed integral from 0 to q equal zero. The areas match when q = 6.
For m > 0 the graphs meet at x = 0 and x = ±√m. Both functions are odd, so the enclosed area is twice the area on the positive side: 2 ∫ from 0 to √m of (mx − x3) dx = m2/2. Set m2/2 = 6. Then m2 = 12 and the positive value is m = 2√3.
F is 2√6, from setting the positive-side area m2/4 equal to 6 and forgetting the matching region on the negative side. D is √6, from setting m2 = 6 and dropping the factor 1/2 in the area. The full enclosed area is m2/2, so m = 2√3.
The equality asks for equal signed areas on [-5, 0] and [0, 5]. An even function has that property, and a constant is even, so I and II are sufficient. Neither is necessary. f(x) = sin(2πx) completes five periods on each interval, so both integrals are 0, but f(x) is not equal to f(-x). f(x) = x2 has equal integrals and is neither constant nor odd. None of the three conditions is necessary.
B, E, F and H include I, but sin(2πx) meets the integral condition and is not even. C, E, G and H include II, but x2 meets it and is not constant. D, F, G and H include III, but x2 meets it and is not odd.
If f''(x) > 0 on (a, b), the graph bends upwards and each chord lies above the curve, so the trapezium rule overestimates the integral. That happens whenever f'(x) < 0 and f''(x) > 0 throughout the interval. The completion in D is therefore correct. The sign of f' is not required: f(x) = x2 + 1 on (0, 1) has f' > 0 and f'' > 0, and one trapezium still overestimates.
A, B and C require f'' < 0. For f(x) = 1 - x2 on (-1, 0), f' > 0 and f'' < 0, and one trapezium lies under the curve. E and F say the decreasing condition is necessary, but f(x) = x2 + 1 on (0, 1) overestimates while f' stays positive.
Rearrangement gives ∫ from 0 to 3 of (f(x))2 dx + ∫ from 1 to 3 of f(x) dx = 0. The square integral is at least 0, so ∫ from 1 to 3 of f(x) dx <= 0, which is II. If that integral is negative, f is negative somewhere on [1, 3]. If it is zero, the square integral is zero, so f is zero on [0, 3]. Either way f(x) <= 0 for some x in [1, 3], which is I.
B keeps I and drops II, but II is the rearrangement of the given equation. C keeps II and drops I; an integral that is negative or zero on [1, 3] forces f(x) <= 0 somewhere there. A drops both.
The derivative is 3x2 − 6, so the turning points are x = −√2 and x = √2. On that symmetric interval the odd part x3 − 6x integrates to 0, and ∫ 3 dx from −√2 to √2 equals 3 · 2√2 = 6√2.
D is 0, the integral of the odd part x3 − 6x with the constant term left out. A is −8√2, the negative of the difference between the function values at the two turning points. G is 12, from taking the turning points to be x = ±2 and integrating the constant 3 across an interval of length 4.
The condition xf(x) > 0 means f is positive on (0, 2] and negative on [−2, 0). If P and N are the integrals of f over those two halves, then P + N = 4 and P − N = 8, so P = 6. For x in [−2, 0], |x| lies in [0, 2], and the substitution u = −x turns ∫ from −2 to 0 of f(|x|) dx into P. The value is 6.
D is −2, the integral of f(x) from −2 to 0, which uses f(x) on the negative side instead of f(|x|). B is −6, the positive-side integral with the sign reversed because the limits run through negative x. H is 8, the given integral of |f|.
The interval from 1/2 to 2 has length 3/2, so three strips have width h = 1/2. The ordinates of 2 log10 x at 1/2, 1, 3/2 and 2 are −2 log10 2, 0, 2 log10(3/2) and 2 log10 2. The trapezium rule is (h/2) times (−2 log10 2 + 4 log10(3/2) + 2 log10 2). The endpoint terms cancel, leaving log10(3/2).
C is log10(9/4), from using the factor h in place of h/2, which doubles the estimate to 2 log10(3/2). E is log10(81/16), from dropping h/2 altogether and keeping 4 log10(3/2). D is log10 3, from reading log10(1/2) as log10 2, so the endpoint terms add to 4 log10 2 and the result is log10 3.
Split ∫ from 0 to r of f into the unit pieces from n to n + 1 for n = 0, …, r − 1. Those pieces sum to 1 + 2 + … + r = r(r + 1)/2. The required sum is Σ from r = 1 to 8 of r(r + 1)/2 = (1/2)(Σ r2 + Σ r) = (1/2)(204 + 36) = 120.
A is 36, which is ∫ from 0 to 8 of f, equal to 8 · 9/2, rather than the sum of the eight integrals. B is 84, that sum of triangular numbers with the r = 8 term removed. E is 204, Σ r2 from 1 to 8, before Σ r is added and the total is halved.
The antiderivative of 3√x + 4/x2 is 2x3/2 - 4/x. At the upper limit this is 2 × 8 - 1 = 15, and at the lower limit it is 2 - 4 = -2. The definite integral is 15 - (-2) = 17.
C is 11, which is what you get if 4/x2 is integrated to +4/x: (16 + 1) - (2 + 4) = 11. E is 18, which is 16 - (-2) after the -4/x term at x = 4 is dropped. The signed antiderivative evaluates to 17.
If f(x) ≥ g(x) for every x ≥ 0, then f - g is a non-negative polynomial on [0, ∞), so its integral from 0 to any x ≥ 0 is non-negative. Thus I is true. II fails for f(x) = (x - 1)2 and g(x) = 0: the inequality of values holds everywhere, but f'(1/2) = -1 < 0 = g'(1/2). III fails for f(x) = 0 and g(x) = 1: the derivatives are equal, while f(x) < g(x). Only I is true.
E, G and H include II, but (x - 1)2 stays above 0 while its derivative is negative on part of [0, ∞). D, F and H include III, but equal derivatives leave room for a constant gap, as with 0 and 1. The integral comparison is the one that follows.
For odd n, fn(x) = nx when x ≥ 0 and fn(x) = -x when x < 0. This holds for n = 1, and if it holds for an odd n then fn+1 is nx + x = (n + 1)x on the positive side and 0 on the negative side, after which one further step returns (n + 2)x on the positive side and -x on the negative side. Since 99 is odd, the integral is ∫ from -1 to 0 of -x dx plus ∫ from 0 to 1 of 99x dx, which is 1/2 + 99/2 = 50.
B is 0.5, the integral on [-1, 0] alone. D is 49.5, the integral of 99x on [0, 1] alone. F is 99, the value of f99(1) rather than the integral. Adding the two halves gives 50.
The curve meets y = 0 at x = −6 and x = 6. On the negative side the integrand is x2 + 4x − 12, and on the positive side it is x2 − 4x − 12. Each piece lies below the axis and encloses area 72. The total is 144.
D is 108, from using one of the two pieces and then adding an extra 36. F is twice the total, as if both pieces were counted twice. Each side contributes 72, so the area is 144.
On [0, 1], x/2 ≤ x ≤ √x, with strict inequality on part of the interval. The base 2 is greater than 1, so (√2)x = 2x/2 ≤ 2x ≤ 2√x. The integrals keep that order: R < Q < P.
A reverses the comparison of √x and x. E swaps Q and P. The increasing exponential preserves x/2 ≤ x ≤ √x, so the order is R, Q, P.
Both integrals run from 0 to 1 and equal 1. The first gives a/2 + b = 1. The second gives a/3 + b/2 = 1. Substitute b = 1 − a/2 into the second: a/3 + 1/2 − a/4 = 1, so a/12 = 1/2 and a = 6. Then b = −2, and a + b = 4.
The options are the integers from −1 to 5. C is 1, which is the value of each given integral rather than of a + b. D is 2, which is −b. E is 3, which is a/2. The pair that satisfies both equations is a = 6 and b = −2, so the sum is 4.
Each unit interval from n to n + 1 contributes n + 1. The five integrals run from 0 to 3, 1 to 3, 2 to 3, 4 to 3, and 5 to 3. Those values are 6, 5, 3, −4, and −9. The last two are negative because the limits run backwards. The total is 1.
F is 27, from adding 4 and 9 instead of subtracting them. D is 4, the single forward piece from 3 to 4. The signed sum of all five pieces is 1.
The integral of √(4 − x2) from −2 to 2 is the area of a semicircle of radius 2, which is 2π. Four strips have width 1. The heights at x = −2, −1, 0, 1, 2 are 0, √3, 2, √3, 0. The trapezium estimate is 2 + 2√3. The positive difference is 2π − (2 + 2√3) = 2(π − 1 − √3).
A is 2(π − 2 − 2√3), from forgetting to halve the trapezium sum, so the estimate becomes 4 + 4√3. C replaces the semicircle 2π with the full circle 4π. D is twice the correct difference. E and F count three copies of √3 in the estimate instead of two.
dy/dx = |−6x| = 6|x|. For x ≥ 0 an antiderivative is 3x2. For x < 0 it is −3x2, whose derivative is −6x = 6|x|. One expression that covers both sides is 3x|x|.
F is 3x2, which matches only for x ≥ 0. E is −3x2, which matches only for x < 0. C is −3x|x|, and its derivative is −6|x| rather than 6|x|. A and B are ±6x/|x|, which are not antiderivatives of 6|x|.
Expanding the two squares gives 1/x + 2 + x and 1/x − 2 + x. Their difference is 4. Integrating 4 from 9 to 16 gives 4 × 7 = 28.
A treats the two integrals as equal. D is the length of the interval, with the factor 4 left out. B is that factor before it is integrated. The difference of the integrands is 4, so the value is 28.
On each interval [k − 1, k), for k = 1, 2, …, 99, the ceiling of x is k, apart from endpoints that do not affect the integral. The integral is the sum of 2k from k = 1 to 99, which is 2100 − 2.
E stops one power early in the closed form, giving 2100 − 1. A, B and C use 299, which is the last term rather than the sum. The sum of 2 + 4 + … + 299 is 2100 − 2.
For x ≥ 0, |x| = x, so the integrand is 0 on any interval that starts at a non-negative number. Thus a positive value of the integral forces p < 0, and statement 3 is true. Statement 1 fails for an even polynomial, where the integrand is 0 on every interval. Statement 2 fails for f(x) = −x: the derivative is negative everywhere, but the integrand is 0, so the integral on a negative interval is not negative.
B, E, F and H include statement 1, which an even polynomial disproves. C, E and G include statement 2, which f(x) = −x disproves. Only statement 3 must be true.
Reflection in the line y = 1 sends the point (x, cos x) to (x, 2 − cos x). A translation of π/4 in the positive x-direction then replaces x by x − π/4, so the new curve is y = 2 − cos(x − π/4).
A and B keep a plus before cos, which is reflection in the x-axis followed by a shift up of 2. C uses x + π/4, a shift in the negative x-direction. The reflection in y = 1 followed by the rightward shift is y = 2 − cos(x − π/4).
(x + y)(x2 − xy + y2) = x3 + y3, so the curve is x3 + y3 = 1. It passes through (0, 1) with a horizontal tangent and through (1, 0) with a vertical tangent, and y decreases as x increases. That is graph C.
A crosses both axes near the origin, so it misses (0, 1) and (1, 0). B increases and meets the x-axis near −1, with a horizontal tangent on the positive y-axis. D increases, with a vertical tangent at (1, 0), so it lies below the x-axis for x < 1. The decreasing graph through (0, 1) and (1, 0) is C.
logx 2 = ln 2 / ln x. For x > 1 this is positive. It becomes large as x approaches 1 from above, and it decreases toward 0 as x increases. That is graph E.
C increases from the origin, like log2 x. A also increases and is already above 2. B and D are negative and decrease. F is negative and rises toward 0. The graph that stays positive and falls toward 0 is E.
The equation −f(−x) = 0 holds exactly when f(−x) = 0, so −x is p or −p. Those are the only real roots of f, hence y = −f(−x) meets the x-axis only at x = −p and x = p. Statement 3 must be true. Statement 1 need not: f(x) = x4 − 3x2 − 4 has only the real roots ±2, but f'(x) = 2x(2x2 − 3) = 0 at three points of (−2, 2). Statement 2 need not: for f(x) = x2 − p2 the integral from 0 to p is negative, so twice it is not the area between the curve and the axis.
B, E, F and H include statement 1, which fails for x4 − 3x2 − 4. C, E, G and H include statement 2, which fails for x2 − p2. A excludes statement 3, but the roots of −f(−x) are exactly the reflections of the real roots of f.
A quartic has at most three turning points. If the leading coefficient is positive, the local maximum lies above both local minima. As the horizontal level rises, the number of distinct intersections runs through 0, 1, 2, 3, 4, 3, 2, hitting the odd values only at the turning levels. The sequence 1, 3, 2, 4 would need a 2 after the single intersection at the lower minimum, then a return to 4 after dropping to 2, and neither move fits that order. If the leading coefficient is negative, a single intersection occurs only at the highest turning value, after which the count is 0, so it cannot be followed by 3, 2, 4. The triple (p, q, r, s) = (1, 3, 2, 4) is impossible.
A is the lower minimum, then the two-root band, then the four-root band, then the local maximum. C is the lower minimum, the four-root band, the local maximum, then the region above it. D and E run up a quartic with negative leading coefficient: a two-root or four-root band, then the turning values, ending at the single touch on the highest turning point.
A translation by 2 in the positive y-direction gives y = log10(x) + 2 = log10(100x). A stretch of factor k parallel to the x-axis replaces x by x/k, giving y = log10(x/k). These agree when 1/k = 100, so k = 0.01.
F is 100, from writing the stretch as log10(kx) instead of log10(x/k); that factor translates the graph down by 2. D is the translation distance 2 itself. B is log10 2 and E is log2 10, from changing the base of the added 2. C is 1/2. The stretch factor is 1/100.
The vertex is (b, c − b2). In the given graph it lies to the right of the y-axis and above the x-axis, so b > 0 and c > b2. Replacing b by B > b moves the vertex to (B, c − B2), to the right of P and below it, while the y-intercept stays equal to c. If B is only a little larger than b, the vertex stays above the x-axis. That is graph F.
A and B leave the vertex at the height of P. C, D and G leave it on the vertical line through P. E moves it to the right but upwards. H moves it to the left of that line. Only F moves it right and down.
Symmetry in the line x = a means f(a + t) = f(a − t) for every t. Condition I is that statement. Condition II compares x with 2a − x, and those two inputs are the same distance either side of a, so it is the same condition. Condition III says f(a − x) = f(x), which is symmetry in x = a/2. For a = 2, f(x) = (x − 1)2 satisfies III but is symmetric about x = 1, while f(x) = (x − 2)2 is symmetric about x = 2 and fails III. So I and II are necessary and sufficient, and III is not.
A, C, E and G mark III as necessary and sufficient, but III is symmetry in x = a/2. D drops II, and F and H drop I. The two equivalent conditions are I and II.
f'(x) = 3x(x + 2q). Three distinct real roots, with f(0) = 2 > 0, means the other turning value 4q3 + 2 is negative, so q < −(1/2)1/3. Replacing x by x + 1 only shifts the graph, so II keeps three distinct roots. The equation f(−x) − 1 = 0 asks where y = f meets y = 1. That line lies strictly between the negative local minimum and the local maximum 2, so III has three distinct roots. For I, q3 = −3/5 gives turning value −2/5, which is negative but above −1, so y = −1 meets the graph only once.
I needs the local minimum to be below −1, but the given information only puts it below 0. II is a horizontal shift and III compares the graph with the line y = 1, which sits between the turning values. The statements that must hold are II and III.
The curves meet where h(x) = x3 + x2 − 16x + 4 equals k. Then h'(x) = 0 at x = −8/3 and x = 2, with a local maximum 940/27 and a local minimum h(2) = −16. Three distinct roots occur for −16 < k < 940/27: one below −8/3, one between −8/3 and 2, and one above 2. The outer roots have opposite signs. The middle root is positive exactly when k < h(0) = 4. Both conditions hold for −16 < k < 4.
D uses 0 as the upper bound, which is the value of x3 − 12x at x = 0 rather than h(0) = 4. F uses 16 as the upper bound, the local maximum of x3 − 12x. At k = 4 the middle intersection is on the y-axis, so it is not a positive x-coordinate. The strict range is −16 < k < 4.
Reflecting y = f(x) in the y-axis and then shifting the graph 2 units left gives g(x) = f(-x - 2). Shifting 2 units left and then reflecting gives h(x) = f(-x + 2). These agree for every x exactly when f(t) = f(t + 4) for every t, by setting t = -x - 2. That period-4 condition is necessary and sufficient.
A is the stronger condition f(x) = f(x + 2). It implies period 4, but cos(π x / 2) has period 4 and not period 2, and for that function g and h still agree. C is period 8, which follows from period 4 but does not force it. D is evenness, and x2 is even, but then g(x) = (x + 2)2 while h(x) = (x - 2)2. E, F and G are symmetries about 1, 2 and 4, and (x - 1)2, (x - 2)2 and (x - 4)2 satisfy them without making g equal h. The necessary and sufficient condition is f(x) = f(x + 4).
Translate y = 4x2 by (3, −5) to get y = 4(x − 3)2 − 5. Reflection in the x-axis gives y = 5 − 4(x − 3)2. A stretch of scale factor 2 parallel to the x-axis replaces x by x/2, so y = 5 − (x − 6)2 = −x2 + 12x − 31.
B is −x2 + 12x − 41, from leaving the −5 unflipped when reflecting, so the constant stays negative. C is the same curve as A without the reflection. E is −16x2 + 48x − 31, from replacing x by 2x instead of x/2. The three steps in the stated order give y = −x2 + 12x − 31.
√x = 3 cos x needs x ≥ 0 and √x ≤ 3, so 0 ≤ x ≤ 9, and cos x must be non-negative. On [0, π/2] the decreasing curve 3 cos x meets the increasing curve √x once. On [3π/2, 5π/2], which still lies inside x ≤ 9, 3 cos x starts below √x, rises above it, and falls below it again, giving two further roots. The next interval where cosine is non-negative begins at 7π/2 > 9. There are three roots.
C is 2, from missing the root on the way down between 2π and 5π/2. E and F count a further crossing beyond x = 9, where √x is already greater than 3, so 3 cos x cannot catch it. G would need every later arch of the cosine to meet √x. Inside 0 ≤ x ≤ 9 there are three solutions.
Let t = √x with t ≥ 0. The graphs meet when mt2 − t + c = 0. Two distinct positive roots need sum 1/m > 0 and product c/m > 0, so m > 0 and c > 0, and they need discriminant 1 − 4mc > 0. That is 0 < m < 1/(4c). If c < 0, the product of the roots is negative whenever m > 0, so only one root is positive and the graphs meet once. A negative gradient also meets y = √x at most once.
C is m > 1/(4c). For c > 0 that is where the discriminant is negative, so the graphs do not meet. D is m < 1/(4c), which includes negative gradients, and those meet the square-root graph at most once. B and F bound m by 4c2, the x-coordinate of the point of tangency, rather than by the critical gradient 1/(4c). The two-positive-root condition is 0 < m < 1/(4c).
As x → ∞, 2x dominates the denominator, so y → 1. As x → -∞, 2x → 0, so y → 0. At x = 0, y = 1/2. The graph increases from the asymptote y = 0 to the asymptote y = 1, which is graph A.
B and C rise without bound for large positive x. D and E fall towards 0 for large positive x. F decreases towards its asymptote, so it is higher at x = 0 than for large positive x, whereas this function is 1/2 at x = 0 and close to 1 when x is large.
The graph is the square with vertices (1, 0), (−1, 0), (0, 1) and (0, −1). Its diagonals both have length 2, so the area is (2 · 2)/2 = 2. The same value is 4 ∫ from 0 to 1 of (1 − x) dx = 4 · 1/2.
A is 1/2, the area of one of the four right triangles. B is 1, the area of two of them. D is 4, four times the integral of 1 over [0, 1]. F is √2, the side length of the square, and G is 2√2, that side multiplied by an extra √2.
The graph has period 2, and each period is a triangle of base 2 and height 1, so its area is 1. For m = 2r−1, the substitution u = mx gives ∫ from 0 to 1 of f(mx) dx = (1/m) ∫ from 0 to m of f(u) du. For every r from 1 to 10, m is a positive integer and the integral of f up to m is m/2, so each term equals 1/2. The sum of the ten terms is 5.
B is 1023/512, the sum 1 + 1/2 + … + 1/512, which is what appears if each extra factor of 2 in the argument is assumed to halve the integral. D is 10, from taking each of the ten integrals to be 1. F is 55, the tenth triangular number 10 · 11/2.
The graph of y = x|x| is strictly increasing, with derivative 2|x|, from -∞ to ∞. A line of slope p ≤ 0 therefore meets it exactly once. For p > 0 the difference h(x) = x|x| - px rises from -∞ to a local maximum p2/4, falls to a local minimum -p2/4, and rises to ∞. A horizontal line at height q meets this graph once if |q| > p2/4, twice if |q| = p2/4, and three times if |q| < p2/4. The possible numbers of distinct solutions are 1, 2 and 3.
A, B and D include 0, but h runs from -∞ to ∞, so every real q is attained at least once. C and F include 4, but h has only three monotonic pieces. For p = 2 the choices q = 2, q = 1 and q = 0 give one, two and three solutions. Those are all the possible counts.
A translation of y = x3 has the form (x − h)3 + k, so the coefficient of x2 is −3h and the coefficient of x is 3h2. Only the second cubic has that relationship, with h = 3. The third cubic has leading coefficient 27, which a translation cannot change.
B is the first cubic, whose x coefficient is 9 rather than 3h2 = 3. D includes the third cubic. Only the second graph is a translation.
f is non-positive on (−∞, 2] and touches the axis at 0 and at 1. g is non-positive and touches the axis at q and r. With 0 < q < 1 and 1 < r < 2, one touch of g lies in each of those two intervals. The equation f(x) = g(x) is degree 5, and for every p > 0 it then has five distinct real roots. Every other placement of q and r drops below five roots for some p.
A puts both touches on the same side of 1, and the count falls to three for large p. C and E put one touch beyond 2, where f is positive and g is not. Only q < 1 and 1 < r < 2 keeps five roots for every p.
Squaring both sides factors the equation as x(x − a2)(x2 + a2 x − 2a2) = 0. For a = 0 the only intersection is the origin, so one point is possible. For a ≠ 0 the roots x = 0 and x = a2 are distinct, and the quadratic has discriminant a2(a2 + 8) > 0. Its roots are new except when a2 = 1, in which case one of them repeats x = a2 and the other is new, giving three points. Otherwise there are four points. Two points never occur, and two is smaller than five, which also never occurs.
A is possible when a = 0. C is possible when a2 = 1. D is possible for every other a. The missing value is 2.
On (0, π/2), (cos x)cos x falls from 1 and returns to 1, which is graph R. (sin x)sin x does the same with its dip earlier, which is graph S. (cos x)sin x falls from 1 to 0, which is graph P. (sin x)cos x rises from 0 to 1, which is graph Q.
A assigns the falling graph P to (cos x)cos x, but that function returns to 1. The matching row is R, S, P, Q.
The parabola y = x2 + 5x + 6 and the line y = mx − 3 fail to meet when x2 + (5 − m)x + 9 = 0 has a negative discriminant. That is (5 − m)2 < 36, so −6 < 5 − m < 6, and then −1 < m < 11.
B is the outside of the correct interval, where the graphs meet twice, and it also leaves out the two values of m where they touch. C and D use |m| < 11, dropping the shift of 5 inside (5 − m). E and F reverse the inequality when m is isolated, giving −11 < m < 1 and its complement.
f(x) = x2 − 6x has its minimum at x = 3. The minimum of y = f(kx) is at x = 3/k. The minimum of y = f(x − c) is at x = c + 3. These are the same point when 3/k = c + 3, so k = 3/(c + 3).
A divides the translation by 3 instead of dividing 3 by the new x-coordinate. C and D place the original minimum at x = 6. E and F use 9, the depth of the minimum, in place of the x-coordinate 3.
The graph shows that f takes every value from −2 to 2. Let t = f(x) and g(t) = t2 − t. On [−2, 2] the parabola opens upwards, so its least value is g(1/2) = −1/4 and its greatest value is g(−2) = 6. The difference is 6 − (−1/4) = 25/4.
E is 6, the greatest value of g, with the least value left out. C is 4, the gap between g(−2) and g(2), which ignores the dip below zero at t = 1/2. B is 9/4, the gap between g(2) and −1/4, using the value at the maximum of f as if it were the maximum of g.
Write 10 = 2×5, 20 = 22×5, 8 = 23 and 125 = 53. The expression simplifies to 22c / 54d. For non-zero integers c and d, this is an integer precisely when c > 0 and d < 0.
F leaves a positive power of 5 in the denominator. D and G also require d > 0, which does the same. A, B, C and H each leave one exponent able to put 2 or 5 in the denominator. The expression is an integer when c > 0 and d < 0.
An area of 5600 cm2 to 2 significant figures lies in [5550, 5650). A width of 80 cm to the nearest centimetre lies in [79.5, 80.5). The greatest height is the upper area bound divided by the lower width bound, 5650/79.5 cm.
D divides that area bound by the upper width 80.5, which gives the least height. C uses width 85 and E uses width 75. A is 70.5 cm, below 5650/79.5, and B is 75 cm, above every height the bounds allow. The greatest height is 5650/79.5 cm.
The sum of four consecutive integers starting at n is 4n + 6. This is a multiple of 6 exactly when 4n is a multiple of 6, which, because 4 and 6 share a factor 2, holds exactly when n is a multiple of 3. For n = 3 the sum is 18, and for n = 1 it is 10.
A includes n = 1, whose sum is 10. B keeps the odd value n = 1 and excludes the even multiple n = 6, whose sum is 30. D excludes n = 3, which works. E excludes every n, including n = 3. The sum is a multiple of 6 exactly for multiples of 3.
Standard form means 1 ≤ a, b, c < 10. The equation is a + 10b = 100c, so 0.1 ≤ 10c − b < 1. Combined with b < 10 this forces c < 1.1 and b > 10c − 1 ≥ 9, so b > 9. The values a = 5, b = 9.5, c = 1 satisfy the equation and standard form, and they make I, III and IV false.
I fails for a = 5, b = 9.5 and c = 1, and III fails for the same values. IV compares b with c, but b > 9 while c < 1.1, so b < c never holds. Only II must be true.
A square has every exponent even, so every awkward prime already has even exponent, and n2 = n2 + 02. If N and M each give even exponents to awkward primes, so does NM. If N fails, multiply by the awkward primes whose exponents are odd; the product kN then gives those primes even exponents. The converse fails: 9 = 02 + 32 is a squaresum, but 9 = 3*3 and 3 is an awkward prime to the first power, so 3 is not a squaresum. Statement C is the one that is not true.
A holds because 0 is allowed as one of the two integers. B adds even exponents to even exponents. D multiplies by one copy of each awkward prime that occurs to an odd power. C fails for N = M = 3.
Since 16 = 42, the identity 122 + 162 = 202 is 122 + 44 = 202. So a = 4, b = 12 and c = 20 are positive integers satisfying a4 + b2 = c2. That is a counterexample to B.
A does not apply: 12, 16 and 20 have a common divisor, so they miss the hypothesis. C is not an instance of a4 + b4 = c4. D is not broken by this triple, because 16 is the arithmetic mean of 12 and 20. The identity counters B.
The given expansion is the block 0011 repeated in base 2. Its value is (0/2 + 0/4 + 1/8 + 1/16) multiplied by the geometric series 1 + 1/16 + 1/256 + … . That is (3/16) / (1 − 1/16) = 1/5.
F is 3/16, the value of one block with the repeating tail left out. A, C, D and E are nearby unit fractions. The geometric sum turns 3/16 into 1/5.
Since 7 divides both a and b, it divides every later term, including u2023. The second statement fails for a = 14 and b = 7: 14 does not divide 7, but u4 = 28, which 14 does divide. The third fails for a = 21 and b = 7: the highest common factor of u1 and u5 = 63 is 21, not 7.
E, F and H include the second statement, which the pair 14 and 7 shows to be false. D, F and G include the third statement, which the pair 21 and 7 shows to be false. Only the first statement must be true.
Take 12, 5 and 3 units of P, Q and R. With p the red percentage in R, the red paint is 12(30) + 5(20) + 3p = 460 + 3p, and the total volume is 20 units. Setting (460 + 3p)/20 = 25 gives 3p = 40, so p = 40/3 = 13 1/3. This lies between 0 and 100, so the mixture is possible.
A is 25%, the red percentage of the finished mixture. B is 23%, which is 460/20, the red percentage if R contained no red paint. F says the result is impossible, but 40/3 lies between 0 and 100, and P alone is already above 25% while Q is below it. D is 19 1/2% and E is 9 3/4%; the weighted equation gives 40/3.
Write u = x + y > 6 and v = x − y > −4. Then x = (u + v)/2 > (6 − 4)/2 = 1, so x > 1 everywhere in the region. The point (7, 0) is in the region and has y = 0, so y > 5 is not always true. Also u = 13 and v = −3 give (5, 8), which is in the region, but uv = −39, so (x + y)(x − y) > −24 is not always true. Only statement 1 holds at every point.
C, E, G and H include statement 2, but y can be made arbitrarily small by taking v large. D, F, G and H include statement 3, but a large positive u and a v just above −4 make the product largely negative. A says none of them hold, but x is always greater than 1.
An equilateral cross-section of side 2x has area √3 x2, so the volume is √3 x2 d. The surface area is 2√3 x2 + 6xd. Setting them equal gives d(√3 x − 6) = 2√3 x, and dividing numerator and denominator by √3 gives d = 2x/(x − 2√3).
E has denominator x − √3. C has denominator x − 4√3. B is 3x/(3x − 2√3). A has no square-root term. After simplifying 2√3 x / (√3 x − 6), the length is 2x/(x − 2√3).
Every face meeting the apex is an equilateral triangle of side 20. Let M be the midpoint of OR, so OM = 10. Unfold triangles OPQ and OQR along OQ. In that net, angle POR is 120°. The straight-line distance is given by PM2 = 202 + 102 − 2(20)(10)cos 120° = 500 + 200 = 700, so PM = 10√7. The line meets OQ between O and Q, so it lies on the two faces. The symmetric route across OSP and ORS has the same length, and a route across the square base is longer.
B is 10√3, the cosine-rule length for 60°, the angle inside one equilateral face; P and M do not lie on a common face. C is 10√5, from treating the angle in the net as 90°. A is 10√(5 − 2√3), the cosine-rule length for 30°, and E is 10√(5 + 2√3), the cosine-rule length for 150°. The flat angle between the two faces is 120°.
Equal areas and AB = XY, BC = YZ give sin∠ABC = sin∠XYZ, so the included angles may be supplementary. Sides 2 and 3 with included angles 30° and 150° have equal area and unequal third sides, so condition (1) does not force congruence. Condition (2) gives (1/2)AB·BC·sin∠ABC = (1/2)XY·YZ·sin∠XYZ with AB = XY and equal angles, so BC = YZ and SAS applies. Condition (3) makes the third angles equal, so the triangles are similar, and equal areas force scale factor 1. Thus (2) and (3) imply congruence and (1) does not.
E, F, G and H say condition (1) is enough, but supplementary included angles share a sine and need not give congruent triangles. A, B and C say condition (2) is not enough, but the equal area forces the second pair of sides to match. A, C, E and G say condition (3) is not enough, but similarity together with equal area fixes the scale.
Because PQ is parallel to SR, the triangles on the two bases are similar and SR/PQ = 3/12 = 1/4. The intersection X therefore divides each diagonal in the ratio 1:4, with the shorter piece towards SR. The parallel through X is 1/5 of the way from SR to PQ, so its length is 3 + (1/5)(12 − 3) = 4.8.
D uses the ratio 1/4 itself as the fraction of the height, giving 3 + 9/4 = 5.25. The upper piece is one part out of five, not one out of four. E is √(3 × 12), the geometric mean. B is (12 − 3)/2. A is (3 × 3 + 12 × 1)/5, using weights 3 and 1 instead of 4 and 1.
Take the front of the plan as the bottom row of the plan. The plan then occupies the whole front row and the right-hand cell in the middle and back rows. The front elevation gives maximum heights 2, 1 and 3 from left to right, and the side elevation, read from the right with the front on the left, gives maximum heights 2, 3 and 1 from front to back. The only free height is the front-right column: it must be 1 or 2, since the front row is already 2 high on the left and the right-hand column is already 3 high in the middle. The fixed cubes are 2 + 1 + 3 + 1 = 7, so the total is 8 or 9.
E is the total when the front-right column has height 1 only, and G is the total when it has height 2 only. Both fit the elevations. D pairs 8 with 7, but the front row and the stack of three already force at least 8. A, B and C are smaller still.
The interior angles give (n − 2)/n = (3/4)(m − 2)/m, so n = 8m/(m + 6) = 8 − 48/(m + 6). Thus m + 6 is a divisor of 48. Also n ≥ 3 and m ≥ 3, so m + 6 is one of 12, 16, 24 and 48. The pairs (n, m) are (4, 6), (5, 10), (6, 18) and (7, 42). There are four.
B, C and D drop one or more of the four divisors 12, 16, 24 and 48. A allows no polygon, but the square and the regular hexagon work: 90° is three quarters of 120°. F also counts m + 6 = 8, which gives n = 2. G counts that and m + 6 = 6, which gives n = 0. H treats m as continuous.
Completing the square gives (x − 9)2 + (y − 11)2 = 24, so the radius squared is 24. A regular hexagon inscribed in the circle is six equilateral triangles of side √24. Its area is (3√3/2) × 24 = 36√3.
B is the area of one of those triangles, (√3/4) × 24 = 6√3, and A is the same triangle with √3 omitted. E is six copies of 24/4, again omitting √3. D is half of the hexagon, 18√3, and C is 18. H is 2√3 × 24 = 48√3, and G is 2 × 24. The six triangles together have area 36√3.
Perpendicular diagonals do not force either diagonal to bisect the other, so neither need be a line of symmetry. The quadrilateral with vertices (−3, 0), (0, −1), (1, 0) and (0, 2) has perpendicular diagonals and no reflection symmetry. The figure formed by the midpoints of the sides has sides parallel to the diagonals, so it is a rectangle; it is a square only when the diagonals are also equal. In this example the side lengths are 2 and 1.5. Neither statement must be true.
I is the extra property that makes the quadrilateral a kite. II describes the midpoint rectangle in the special case of equal diagonals. B, C and D each claim that one or both of these stronger properties are forced.
In a parallelogram the opposite sides are equal and parallel, so PQ = SR and PS is parallel to QR. That makes A necessary. It is not sufficient: P(0, 0), Q(2, 2), R(3, 2), S(5, 0) has PQ = SR and PS parallel to QR, but only one pair of sides is parallel, so it is an isosceles trapezium rather than a parallelogram.
B is one pair of opposite sides both equal and parallel, which is enough to force a parallelogram, so it is sufficient as well. C forces a rhombus, which is sufficient but not necessary. D, equal diagonals, fails in a parallelogram that is not a rectangle. The necessary but not sufficient condition is A.
Opposite sides of a parallelogram are equal, and consecutive angles sum to 180°. Condition I makes adjacent sides equal, so PQRS is a rhombus, which need not have a right angle. Condition II holds for every rhombus, including one that is not a square. Condition III makes the consecutive angles equal, so each is 90° and PQRS is a rectangle, which need not have equal adjacent sides. None of the three is sufficient on its own.
A to G mark at least one condition as sufficient. I and II both hold for a rhombus that is not a square. III holds for a rectangle that is not a square.
Let α = arcsin x and β = arcsin y, so 0 < α < β < 90°. The pair A = α, B = β gives third angle 180° - α - β > 0. The pair A = α, B = 180° - β gives third angle β - α > 0. Both are determined by ASA once AB = 1 is fixed, and their angles differ, so they are not congruent. Any pair with A = 180° - α has A + B > 180°. Every allowed choice of x and y therefore gives exactly two triangles.
I claims that some choice gives exactly one triangle, but both admissible angle pairs occur for every such x and y. III claims that some choice gives three; making A obtuse pushes the angle sum past 180°. Only II is correct.
The radii differ by 3. If the smaller radius is r, the smaller arc 6 gives the angle 6/r. The area difference is then (6/r)(2r + 3) = 14, so r = 9. The perimeters are 24 and 32, and the positive difference is 8.
B is 7, half of the angle equation before it is solved. D is 9, the smaller radius rather than the difference of the perimeters. The perimeters differ by 8.
The area 18 sin θ = 9√3 gives sin θ = √3/2. The central angle at least π/2 is 2π/3. The chord PQ has length 6√3, and the greatest distance from that chord to a point on the circle is 9. The greatest area is (1/2) × 6√3 × 9 = 27√3.
B is 18√3, using height 6 instead of 9. F is 36√3, using the diameter as the height. The greatest height from the chord is 9, so the area is 27√3.
Equal areas from the centre mean the central angles have equal sines. With the centre inside the polygon, each angle is θ or π − θ, and the angles add to 2π. For a triangle the only possibility is three angles of 2π/3, so the triangle is equilateral. For every n ≥ 4 the angles π/(n − 3), repeated n − 1 times, and one angle π − π/(n − 3), have equal sines and add to 2π, but the side lengths are not all equal.
C includes n = 4, but the angles π/3, π/3, π/3 and 2π/3 give a non-regular example. E claims every n works. Only n = 3 is forced to be regular.
The square has side 10. Two lines of symmetry place the rectangle’s vertices at (0, x), (x, 0), (10 − x, 10) and (10, 10 − x), taking one corner of the square as the origin. Adjacent sides are x√2 and (10 − x)√2, so the area is 2x(10 − x). Set that equal to 20: x2 − 10x + 10 = 0, so x = 5 ± √15. The larger root is 5 + √15.
E is 5 + √5, from taking the area as x(10 − x) and missing the factor 2 that comes from (√2)(√2). C is 2√15, the square root in the quadratic formula before it is combined with 5. The larger root of the correct equation is 5 + √15.
A line splits a triangle into two triangles only when it joins a vertex to the opposite side. In an acute triangle the three altitudes do this, and each creates a right angle at its foot. In an obtuse triangle the two altitudes from the acute vertices fall outside the opposite sides, but the obtuse vertex contributes the altitude to the opposite side and the two lines perpendicular to the adjacent sides, again three lines. No non-right triangle has exactly one or exactly two.
I and II claim that exactly one or exactly two such lines can occur. Every acute or obtuse triangle has exactly three, so only III is true.
Order the five numbers and let the median be m. The range is 20 and the sum is 0. The median is largest when the two smallest entries are equal, the entry just above the median equals m, and the largest entry is exactly 20 above the smallest. Those conditions give the list −12, −12, 8, 8, 8, whose median is 8, and the ordering stops any larger median.
F is the range, 20. D is 6.5 and C is 4.5. B is 4 and A is 0. The list −12, −12, 8, 8, 8 has mean 0, range 20 and median 8.
Let the sets have n and m members. Exchanging a value a from the first set with a value b from the second raises the first sum by b − a and lowers the second sum by b − a. The mean changes then give b − a = n and b − a = 3m, so n = 3m. The combined sum is 15n + 20m = 65m and the combined size is 4m, so the mean is 65/4 = 16 1/4.
C is 16 1/2, the average of the original means 15 and 20, and also the average of the new means 16 and 17. That average is the combined mean only when the sets have equal size, while the changes +1 and −3 force the sizes to be in the ratio 3:1. D is 16 2/3, the combined mean when the size ratio is taken to be 2:1. B is 16 1/3 and E is 16 3/4.
The first triple sums to 30 with median 8, so it is a, 8, 22 - a with a <= 7. The second sums to 36 with median 9, so it is c, 9, 27 - c with c <= 8. All six integers are distinct, so c cannot be 8 and cannot equal a. The second large value is then at least 20, when c = 7. Taking a = 6 gives 6, 8, 16 and 7, 9, 20, with range 20 - 6 = 14. If c <= 6, the partner is at least 21 and the range is at least 15.
A to D are smaller than 14. They need the triples to share 7 or 8, or the second large value to be 19, but 8 and 9 are already the two medians. F is 15, the range of the next arrangement, such as 7, 8, 15 with 6, 9, 21.
If n is odd, the median is the middle value of the ordered list, so it belongs to the list. If n is even, the median is the average of the two middle values: for 1, 2 that average is not in the list, but for 1, 2, 2, 3 it is. An odd length is therefore sufficient for the median to be in the list, and not necessary.
A says the condition is also necessary, but an even list can still have its median in the list. B reverses the two directions. An odd number of entries is sufficient and not necessary.
There are 3n balls. Whatever the first colour, the chance the second matches it is (n − 1)/(3n − 1). The chance the colours differ is 1 − (n − 1)/(3n − 1) = 2n/(3n − 1).
A is the probability that the second ball matches the first. B has numerator 2n − 2. E has numerator 3(n − 1). D and F are built from cubed counts, as for a draw of three balls. The complementary probability for two draws is 2n/(3n − 1).
Fred, Hermione, Lavender and Ron are all ahead of George: Hermione is ahead of Lavender, and Lavender is ahead of George. George is therefore last. The other four can finish in 4! = 24 orders, and half of those have Hermione ahead of Lavender. That leaves 12 orders.
F is 5!, with no restrictions. E is 4!, the orders with George last before Hermione and Lavender are ordered. B is 6 and D is 18. A is a single order. Half of the 24 orders with George last is 12.
Take the club as one member. The men are 2/5 of the club, and 2/5 of them play cricket, so the male cricketers are 4/25. They are 1/3 of all cricketers, so the cricketers are 12/25 and the female cricketers are 8/25. The women are 3/5 = 15/25, so the women who play tennis are 15/25 − 8/25 = 7/25.
D is 11/25, from subtracting the male cricketers 4/25 from all the women. A is 1/5, from subtracting the male proportion 2/5 from the female proportion 3/5. C is 1/3, the fraction of cricketers who are men. E is 3/5, the proportion of members who are women.
All three chosen students study mathematics only if the second does, and the second does only if at least one of them does. Most students study mathematics, so that proportion m is greater than 1/2, and the college still contains other students: otherwise the three probabilities would be equal. For 1/2 < m < 1 the probabilities are m3, m and 1 - (1 - m)3, so T < S < R.
Any option with T above S treats all three studying mathematics as more likely than the second student doing so. Any option with R below S treats at least one as less likely than the second student. The nested events run in the opposite order, so the true comparison is T < S < R.
If the probability is 4/15, the class size must be 15k and the number who wear glasses must be 4k for some positive integer k. Then 15k is always divisible by 3, so II must hold. I forces k = 1, and III forces k = 2, but every k works, so only II is necessary.
E, F and H include I. A class of 30 with 8 wearing glasses has probability 4/15, and 22 students do not wear glasses. D, F, G and H include III, but a class of 15 with 4 wearing glasses also has probability 4/15. Only II is forced.
The centre lies in a 4 by 6 rectangle, of area 24. The circles meet unless the centre is strictly inside the circle of radius 1 about the origin, and that disk lies wholly inside the rectangle. Its area is π. The probability of an intersection is (24 − π)/24.
E uses 16 in place of 24, the area of a square on the shorter side. D uses 6 as the area of the rectangle. The rectangle has area 24, so the probability is (24 − π)/24.
The claim applies only to cards with a vowel on the front, which are A and E. Card E has 8, which is even. Card A has 3, which is odd, so card A is a counterexample.
E fits the claim. B, C and D have consonants on the front, so an odd number on the back does not contradict it. The card that breaks the claim is A.
The negation of “every day, Fred does at least one maths problem” is “on at least one day, Fred does no maths problem”.
A and B say that he does more than one problem, every day or on some day. C says that he never does more than one. D says that he does none on every day, which is stronger than the negation. F says that there is no day on which he does none. The negation is that on some day he does none.
n is an S-number when every factor m > 1 of n is a multiple of some member of S. The negation is that there is some factor m > 1 which is a multiple of no member of S.
A, B and C require the failing condition for every factor m > 1, but one such factor is enough. D says that some member of S fails to divide m, which can happen while another member does divide m. F says that every member of S divides m. The negation is option E.
Q and T cannot both be true, and R is true only if P is true and R's first name is Robert. The consistent cases are: P, Q and S true, with R not named Robert; and P, R and T true, with R named Robert. Each case has exactly three true statements, and every other count contradicts at least one of the five claims.
A, B, C, E, F, G and H allow a count other than 3. Both consistent situations have three true statements, so the only possible number is 3.
Let n be the common positive number of balls. For n = 1 the true labels are P, S and T. For n = 2 they are Q, S and T. For n = 4 they are P, Q and R. For every n ≥ 5 every label is false. Only n = 3 makes exactly one label true, and that label is R: 3 is more than 2 and fewer than 5, while P, Q, S and T all fail.
A is true for n = 1 and n = 4, always together with other true labels. B is true for n = 2 and n = 4, again with others. D and E are true for n = 1 and n = 2, each time alongside two further true labels. None of those four labels is ever the only true one.
The statement says that if n is 1 or 5 less than a multiple of 6, then n is prime. A counterexample is a whole number in that form which is not prime. For 0 < n < 50 the numbers are 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47 and 49. The ones that are not prime are 1, 25, 35 and 49, so there are four.
A counts 2 and 3, the primes that fail the converse. B omits 1. D adds 55 by reading the upper bound as 60. E adds 55 and 65 by reading it as 70.
A counterexample must satisfy f(x)2 ≤ 1 on [−1, 1] and have ∫ f2 > ∫ f. For f(x) = x − x3, the derivative 1 − 3x2 vanishes at ±1/√3, and the extreme value has absolute value 2/(3√3) < 1, while f(±1) = 0. So the bound holds. The integrand x − x3 is odd, so its integral on [−1, 1] is 0, whereas the integral of its square is positive. The claimed comparison fails.
A, B, C and E all exceed 1 in absolute value somewhere on [−1, 1], so the hypothesis does not apply. F stays between 0 and 1/4, but ∫ f2 = 16/315 and ∫ f = 4/15, so the claimed inequality still holds.
For the odd prime p = 3, 10p2 + 1 = 91 = 7 × 13, which is not prime, so statement 1 is false. Statement 2 is about primes, and 91 is not prime; 7 and 13 are both of the form 6n + 1, so this factorisation does not refute it. Statement 3 says that no multiple of 7 greater than 7 is prime, and 91 is a composite example of that, so it is not a counterexample either.
A misses the failure of statement 1 at p = 3. Every option that includes 2 or 3 treats 91 as if it were prime, or treats 7 or 13 as if it were not of the form 6n + 1. Only statement 1 is refuted.
The relation f(x + 2) = −f(2 − x) says f(2 + h) = −f(2 − h). The interval from 1 to 3 is symmetric about 2, so the two sides cancel and the integral is 0.
A and B fix f at one or two points and leave the integral free. C and E make f odd about 0. The function f(x) = x satisfies both, and its integral from 1 to 3 is 4. Only D forces that integral to vanish.
A counterexample must be an integer at every integer, with a derivative that is not. Option C equals x(x + 1)(x2 + 1)/2. The product of two consecutive integers is even, so this is an integer at every integer. Its derivative is (4x3 + 3x2 + 2x + 1)/2, which equals 1/2 at x = 0.
A equals 3/4 at x = 1, and B equals 3/2 at x = 1, so neither is integer-valued. D equals [x(x + 1)/2]2, and its derivative x(x + 1)(2x + 1)/2 is an integer whenever x is. Only C breaks the implication.
Stapled means that every member has a prime factor which divides some other member. The negation is that some member has no prime factor which divides any other member. That is F.
A to D say this of every member, but one exceptional member is enough. E replaces 'at least one other number' by 'every other number'. G and H only say that some prime factor fails to divide another member, which can still leave a different prime factor that does. The negation is F.
The three scores of zero leave four possible passwords: bcead, dceab, beacd and deacb. Entering bcead distinguishes them, because the number of matching positions is 5, 3, 2 or 0 according as the truth is bcead, dceab, beacd or deacb. One further attempt is enough, and none of the four can yet be ruled out.
A leaves all four of those passwords open. C, D and E ask for more attempts than this test needs. One attempt separates the four possibilities.
The claimed average is (50 + 40)/2 = 45. Journey I covers 200 km in 100/50 + 100/40 = 9/2 hours, so its average is 400/9. Journey III covers 180 km in 80/50 + 100/40 = 41/10 hours, so its average is 1800/41. Both differ from 45. Journey II takes one hour at each speed, so its average really is 45 and it is not a counterexample. The counterexamples are I and III only.
II has equal times, so the arithmetic mean of the two speeds is the overall average. Any option that includes II, or that omits I or III, misses the pair of unequal-time journeys.
A counterexample needs f'(x) > 0 for every real x, but f(x) ≤ 0 for some x. For f(x) = x3 + x + 1, the derivative is 3x2 + 1, which is at least 1. Also f(−1) = −1. This function breaks the implication.
A and B have derivative 2x, which is negative for x < 0, so the hypothesis fails. D has derivative −1. E has positive derivative and positive values, so the implication holds for that function.
A valid code starts with U. Reversing it, or swapping U and D, produces a code that starts with D, so the profile immediately goes below sea level. The code UD is a counterexample to both I and II. Adding U at the front and D at the end lifts the original profile by one unit and returns to sea level at the end, so the new code stays non-negative. Only III is true.
I and II both force the new code to start with D. Options that include either of them treat a reflection or a reversal as if it preserved the sea-level condition. III is the only operation that keeps every valid code valid.
The statement is: for every positive integer N there exists a positive integer K such that for every positive integer m, N(Km + 1) − 1 is not prime. Negating flips each quantifier and the final claim, giving: there exists an N such that for every K there exists an m for which N(Km + 1) − 1 is prime.
E has those quantifiers but still says the expression is not prime. G requires it to be prime for every m, rather than for some m. A, B and C keep “for any N there is a K”. D quantifies everything universally. H puts the existence on K as well as N.
ab = ac rearranges to a(b - c) = 0, so a = 0 or b = c. Neither alternative is forced. a = 1, b = 2, c = 2 has a ≠ 0, so I need not hold. a = 0, b = 1, c = 2 has ab = ac, but b and c are nonzero and unequal, so II and III need not hold. None of the three must be true.
B, F and H include I, but a = 1, b = c = 2 is a counterexample. C, G and H include II, and D, F, G and H include III; a = 0, b = 1, c = 2 defeats both. No one of the three statements must follow.
The conjecture applies only when the number of 0 digits between the two 1s is odd. 101 has one 0 and is prime, so it fits the conjecture. 1001 has two 0s, so it is not an instance of the conjecture. 10001 has three 0s and equals 73 × 137, so it meets the hypothesis and is composite. Only III is a counterexample.
Any option that includes I uses a number that is prime, so it does not break the conjecture. Any option that includes II uses a number with an even number of 0s, which the conjecture does not claim is prime. The only counterexample is III.
If a is divisible by n, then ab is divisible by n, and likewise if b is. The condition is therefore sufficient for (*). It is not necessary: a = 2, b = 6 and n = 4 give ab = 12, which is divisible by 4, while neither a nor b is.
A reverses the two directions. C says the condition is also necessary, but 2 × 6 is divisible by 4. D says it is not sufficient, but a factor n in either a or b does force n to divide ab. The condition is sufficient but not necessary.
A new person can avoid sitting next to someone only by taking the middle seat of three consecutive empty chairs. So every gap between occupied chairs must contain at most two empty chairs. Thirteen people give 13 gaps and 27 empty chairs, and 13 × 2 = 26, so one gap still has three empty chairs. Fourteen people give 14 gaps and 26 empty chairs, which fit as twelve gaps of 2 and two gaps of 1. Thus 14 is the smallest such number.
C is 13, which still leaves a block of three empty chairs. B is 10, the repeating pattern of one person and three empty chairs, and A is smaller still. F is 20, from alternating occupied and empty chairs, and E is one less than that; both are larger than necessary. The minimum is 14.
If the statement is true exactly when x > 1, then it is true whenever x > 1 and whenever x > 2, so A would make B and D true as well. If it is true exactly when x > 2, then D is true as well. If it is true whenever x > 1, then it is true whenever x > 2, so B would make D true. The only claim that can stand alone is D: the statement may be true for every x > 2, false at x = 1.5, and true at x = 0. Then A, B and C all fail, and D holds.
A forces B and D. B forces D, because x > 2 is a special case of x > 1. C forces D, because an equivalence includes the one-way implication. Each of those choices makes more than one option correct. The option that can be the only correct one is D.
A counterexample must be an integer N > 6 that cannot be written as a sum of two integers greater than 1 that are not prime. N = 5 is not greater than 6, so I is not a counterexample. Those integers start 4, 6, 8, ..., and the smallest sum of two of them is 8, so 7 cannot. An odd total needs an odd summand, and the smallest odd integer greater than 1 that is not prime is 9, so 9 cannot either. II and III are counterexamples.
A, B, E, F and H treat N = 5 as a counterexample, but the claim is only about integers greater than 6. C keeps only 7, and D keeps only 9. Both 7 and 9 are greater than 6 and cannot be written as such a sum.
A truth-teller’s neighbours are all liars, so no two truth-tellers share a side. A liar’s claim is false, so every liar has a truth-telling neighbour. The four corners and the centre share no sides, and each edge-centre borders a corner and the centre, so five truth-tellers are possible. Two are not: every pair of non-adjacent cells leaves someone with no truth-telling neighbour. Three are possible, for example the top-left corner, the middle of the right edge, and the middle of the bottom edge. The smallest number is 3 and the largest is 5.
A, B and C give a smallest number of 1 or 2, but one cell meets at most four others and no pair of cells covers every liar. D and F give a largest number of 4, but the four corners and the centre work. G and H give a smallest number of 4 or 5, but three truth-tellers are possible.
As claims about f, A is “if x >= 0 then f(x) < 0”, and B is its contrapositive “if f(x) >= 0 then x < 0”, so they are true together or false together. D, “if x < 0 then f(x) < 0”, is the contrapositive of E. F includes the implication “if f(x) >= 0 then x < 0”, which is B. Exactly one statement is true, so A, B, D, E and F are false, and C is the true one. For f(x) = |x|, “if x >= 0 then f(x) >= 0” holds, and the other five claims fail.
A and B are contrapositives, so either one being true makes the other true. D and E are contrapositives in the same way. F contains B, so it is true only when B is also true. For f(x) = |x| the only one of the six claims that holds is C.
A counterexample needs a dividing bc while a divides neither b nor c. For I, 5 divides 10. For III, 6 divides 12. For II, 8 divides 4 × 4 = 16, but 8 divides neither factor 4. Only II is a counterexample.
B and E include I, where 5 already divides b. D and F include III, where 6 already divides c. In both of those triples the claimed implication holds. The only counterexample is II.
For f(x) = (x - 1)2(x + 1), the derivative (x - 1)(3x + 1) is zero at exactly two real values, while f itself is zero at only two real values. So Q can hold when P fails, and P is not necessary for Q. For f(x) = x(x2 - 1)2 there are exactly three distinct real roots, while f'(x) = (x2 - 1)(5x2 - 1) has four real zeros. So P can hold when Q fails, and P is not sufficient for Q.
A and C say that Q requires P, but (x - 1)2(x + 1) has exactly two stationary points and only two distinct roots. B and C say that P forces Q, but x(x2 - 1)2 has three distinct roots and four stationary points. Neither direction holds.
Rolle's theorem says that p(a) = p(b) produces some c strictly between a and b with p'(c) = 0, so the condition is sufficient for (*). It is not necessary: p(x) = x3 on [-1, 1] has p'(0) = 0, while p(-1) = -1 and p(1) = 1.
A and B say the condition is necessary, but x3 has a stationary point in (-1, 1) with unequal endpoint values. D says it is not sufficient, but Rolle's theorem supplies the stationary point whenever the endpoint values agree. The condition is sufficient and not necessary.
A counterexample is a prime n for which un is divisible by neither 3 nor 5. The value 1 is not prime. u2 = 21 and u3 = 30 are both multiples of 3. The next prime is 5, and u5 = 44 is divisible by neither 3 nor 5. So the smallest counterexample is 5.
A uses n = 1, which is not prime, so the implication does not fail there. B and C use primes whose terms are multiples of 3. G uses n = 7, and u7 = 59 is another counterexample, but 5 is smaller.
A counterexample needs n prime and n2 + 2 prime. For n = 2, n2 + 2 = 6, which is not prime. For n = 3, n2 + 2 = 11, which is prime. For n = 4, n is not prime, so the implication does not fail. Only the second value is a counterexample.
B includes n = 2, but 6 is not prime, so the implication holds. D includes n = 4, which is not prime. Only n = 3 breaks the implication.
A line through (1, 2) has y-intercept 2 − m. That is negative only when m > 2, and then the x-intercept is 1 − 2/m, which is positive. The contrapositive says the same thing. The converse fails for the line of gradient −1: its x-intercept is positive and its y-intercept is positive.
E includes the converse, which that line shows to be false. D keeps only the contrapositive and drops the original statement, but the two are equivalent. The true statements are the original and its contrapositive.
For every real k, choose x = −|k| − 1. Then x < k, but x2 = (|k| + 1)2 is greater than k. The implication fails for every k.
B, E and F keep some positive k, but a large negative x is still less than that k and its square is not. There is no value of k for which the statement holds.
The first statement says every real number is less than every positive integer, which fails for x = 2 and n = 1. The second is true: for any real x there is a larger positive integer. The third is true: x = 0 is less than every positive integer.
C drops the third statement, but x = 0 satisfies it. H includes the first statement, which is false. The true statements are the second and the third.
The largest a is some ak, and ak ≤ bk + ck ≤ (maximum of the b's) + (maximum of the c's), so the third statement is true. The first fails for a = (5, 5), b = (0, 10), c = (10, 0): the minimum of a is 5, while the minima of b and c are 0. The second fails for a = (0), b = (5), c = (5).
B and F include the first comparison, which those two lists show to be false. C includes the second. Only the comparison of maxima must hold.
The integral of sin(2x) from 0 to k is (1 − cos(2k))/2. This is 0 exactly when cos(2k) = 1, so 2k = 2nπ and k = nπ for an integer n. That is statement R. R is therefore necessary and sufficient for S.
B would be right if some multiples of π failed to make the integral 0. C would be right if some other values of k also worked. Neither happens: the integral vanishes exactly on the integer multiples of π.
The line is y = −(a/b)x + c/b. A positive gradient and a positive intercept mean a/b < 0 and c/b > 0, so a and c have opposite signs. That sign condition is necessary. It is not sufficient: a = 1, b = 1, c = −1 has opposite signs for a and c, but both the gradient and the intercept are negative.
A is the pair of inequalities itself, so it is necessary and sufficient, which is stronger than the question asks for. B reverses both inequalities. F requires a and c to have the same sign, which is incompatible with the two inequalities.
The interior angles of a pentagon add to 540°. A pentagon with angles 108°, 90°, 90°, 126° and 126°, in that order, has a 108° angle but the angles are not an arithmetic progression, so (*) is false. Its contrapositive is false as well. The converse is true: five angles in arithmetic progression have middle term equal to their average, which is 108°.
E and H include (*), which the pentagon above shows to be false. C includes only the contrapositive, which is equivalent to (*) and so is also false. Only the converse is true.
The theorem is “if 2k + 1 is prime, then k is a power of 2”. Statement I is the converse. Statement II says that 2k + 1 is not prime only if k is not a power of 2, which is also the converse. Statement III says that 2k + 1 being prime is sufficient for k to be a power of 2, which is the theorem again. Only III is equivalent to it.
The rows that mark I or II as equivalent are treating the converse as the same statement. The correct row is No, No, Yes.
Taken as a statement about every real x, “x = 0 or x = 1” implies x ≥ 0, implies the tautology “x ≥ 0 or x ≤ 1”, and implies 0 ≤ x ≤ 1. It does not imply x = 1, because x = 0 is a counterexample. That is exactly three of the other four. x ≥ 0 implies only the tautology. x = 1 implies all four. 0 ≤ x ≤ 1 implies two of them. The tautology implies none of the others.
A implies only one of the others. B implies all four. D is always true and forces none of the rest. E implies two. Only C implies exactly three.
Set x = loga b and y = loga c, so b = ax and c = ay. The given equation is logc d = x2, hence d = cx2. Substituting c = ay gives d = (ay)x2 = ax2 y = (ax)xy = bxy. Therefore logb d = xy = (loga b)(loga c). That is the chain (1), (3), (5), (9), (7), (4).
A and B start from (2). (cx)2 equals c2x, not cx2. D goes from (3) straight to (7): replacing the base c by ay produces (5), and (7) is a later regrouping of the power of a. E, F, G and H begin at (4), which is the equation being proved, so it is not yet available. F and G also pass through (6) and (8), which square an exponent instead of using x2.
The claim says |a + b| < |a| + |b| for every real a and b. For a = 3 and b = 4 both sides equal 7, so the strict inequality fails and the claim is false. One counterexample refutes a “for all” statement, and these values are a genuine counterexample, so the proof is fully correct.
A says the claim is true, but the two sides are equal for a = 3 and b = 4. B says that pair is not a counterexample, but both sides really are 7. C asks for every pair where the inequality fails; one pair is enough. D replaces the claim by the weak triangle inequality, which is a different statement.
The task is to deduce existence of x and y from a2 - 4b3 ≥ 0. The attempt starts from (x - y)2 ≥ 0, rewrites it as (x + y)2 - 4xy ≥ 0, and multiplies by x2 y2 ≥ 0 to reach a2 - 4b3 ≥ 0. Each of those steps is valid, and together they show that every real pair x, y satisfies the inequality. That is the converse of the required implication. Reversing the lines still does not construct x and y from a and b.
A treats the attempt as a proof of the stated direction. B says the reversed order would be a proof, but the reversed lines still assume x and y instead of producing them. D, E and F place an algebraic error on line II, III or IV, and each of those lines follows from the one before it. The algebra proves the converse.
Squaring both sides of √(x + 5) = x + 3 requires x + 3 ≥ 0, so step (1) can add an extra root. The algebra from there is correct and gives x = −4 or x = −1. Only x = −1 satisfies the original equation, because at x = −4 the right side is −1. The extra root is produced at step (1).
A accepts x = −4, where the right side is negative. B rejects x = −1, but √4 = 2 equals −1 + 3. D and E place the fault on the rearrangement or the factorisation, both of which follow from the squared equation. The false root comes from squaring at step (1).
Line 1 is the contrapositive, so it is equivalent to the claim. Lines 2 and 3 are the division algorithm, and line 4 expands bc correctly as a multiple of a plus rs. Line 5 is where the argument fails: rs need not be the remainder, because rs may be at least a. For a = 4, b = c = 6, the remainders are r = s = 2 and rs = 4, which a divides, while bc = 36 is also divisible by a.
A rejects the contrapositive, which really is equivalent to the original claim. B, C and D are valid steps about remainders and the expansion of bc. F is true as a statement about positive r and s; the false inference that a does not divide bc comes from treating rs as the remainder on line 5.
The identity sin 2x = 2 tan x/(1 + tan2 x) gives 2√3/(1 + 3) = √3/2 whenever tan x = √3. Every such x differs from 60° by a multiple of 180°, and sin 2x is √3/2 for all of them. The value is unique, but the student's answer only checks x = 60°.
A treats that single angle as a complete argument. C and E look for other x with sin 2x = √3/2, but those angles need not satisfy tan x = √3. D says √3/2 is not the only value, whereas the identity shows that it is. The right comment is B.
Lines I and II are the difference of cubes. Line III says the factors must be a and a2. A factorisation of a3 can split in other ways, and a ≤ a2 does not force that particular split. That is the first false step.
A says the proof is correct. B and C mark lines that do follow from a3 + b3 = c3. E, F and G are later consequences of the false factorisation, so they are not where the argument first fails.
Step (I) takes logarithms base 1/2. That base is less than 1, so the logarithm is decreasing and the inequality must reverse. The later steps follow from the line they start from: the power rule, division by log1/2(1/4) = 2, which is positive, the evaluation 10 × 5/2 = 25, and the positive integers strictly below 25.
A says the argument is correct, but the first inequality has the wrong direction. C to F place the fault on a later step that does follow from the previous line. The invalid step is (I).
Step (I) rewrites 10x − 5 as 5(2x − 1). On the domain x ≥ 1/2 both sides of that equation are non-negative, so the squaring in step (II) does not remove a root. Both x = 5/8 and x = 5/2 satisfy the original equation. Step (III) divides by 2x − 1 and discards x = 1/2. At x = 1/2 both sides of the original equation are 0, so this is a genuine further solution. The first invalid step is (III).
B, C and D say that only one of 5/8 and 5/2 works, but both check in the original equation. E blames the squaring in step (II), which still holds at x = 1/2. G blames step (IV), but that step only rearranges an equation from which x = 1/2 has already been removed.
If tan θ > 0 and cos θ ≠ 0, then sin θ cos θ = tan θ cos2 θ > 0, so lines (I) and (II) are valid. Adding sin2 θ + cos2 θ = 1 to the left-hand side produces (sin θ + cos θ)2 > 1, so lines (III) and (IV) are valid. A square greater than 1 gives sin θ + cos θ > 1 or sin θ + cos θ < −1. At θ = 5π/4, tan θ = 1 > 0 but sin θ + cos θ = −√2. The first false step is line (V), and the conjecture is false.
A accepts the proof, but θ = 5π/4 is a counterexample. Lines (I) to (IV) correctly reach (sin θ + cos θ)2 > 1. The unjustified step is dropping the possibility that the sum is less than −1.
Line I rewrites the original equation as cos x - 2 sin x + sin x tan x = -1. Expanding line II gives cos x - sin x + sin x tan x - cos x tan x, and cos x tan x = sin x, so line II matches line I wherever tan x is defined. Line III is the first false step: a product equal to -1 does not mean that one of the factors equals -1.
B and C place the first error on a line that is a valid rearrangement. E, F and G describe later faults, including the squaring on line IV and the undefined values of tan x on line VI, but the argument has already failed on line III.
Line I only rewrites 4 as 2 × 2. Line II replaces 2((9n + 1)/2 - (3n - 1)/2) by 9n + 1 - 3n - 1. Cancelling the factor 2 gives (9n + 1) - (3n - 1) = 6n + 2, so the expression equals 2(6n + 2) = 12n + 4. That leaves remainder 1 on division by 3, so the claim is false and the first error is line II.
B places the error on line I, where 4 = 2 × 2 is valid. D, E, F and G place it on lines III to VI, which only continue from the incorrect expansion on line II. A says the argument is correct, but 12n + 4 is not divisible by 3.
Line I splits the integral at 0, and line II is the fundamental theorem with upper limit x. Both are correct. On line III the upper limit is 2x, so the chain rule multiplies (2x)2 by 2 and the derivative is 8x2, not 4x2. The first error is line III. The original derivative is 8x2 - x2 = 7x2.
B and C place the first error on lines I and II, which are valid. E and F place it on lines IV and V, which only subtract x2 from the incorrect 4x2. A says the calculation is correct, but it concludes 3x2.
Line A is the identity sin2 x + cos2 x = 1. Line B concludes that cos x equals √(1 - sin2 x) for every x. That square root equals |cos x|, so at x = π it equals 1 while cos π = -1. Line B is the first false step. Lines C and D only rearrange B, and line E substitutes x = π into that rearranged equation.
C and D name lines that follow from B by algebra. E names the substitution that produces the contradiction 0 = 4; that substitution is arithmetic from the equation on line D. The square root first replaces cos x on line B.
The expansion on line I and the difference on line II are correct. The discriminant of 3x2 + 3x + 1 is −3, so line III is correct as a statement about the polynomial. Line IV does not follow: a value of that polynomial can still be a product of two integers. For x = 5 the value is 91 = 7 × 13. The first false step is line IV, and the claim is false.
E places the first error on line III, but the discriminant really is negative. G places it on the final sentence, which only repeats the false conclusion of line IV. The argument first fails on line IV.
The stationary points and the sign condition in steps I and II are correct. Step III says that three distinct real roots imply opposite signs at the stationary points. The proof needs the other direction: opposite signs imply three distinct real roots. Step IV uses the direction that was not proved.
B says the student proved the converse of the whole statement. What is missing is the converse of step III, which is the direction the argument uses. A treats the gap as if it were not there.
Lines I to V are correct: two consecutive odd primes exist, three consecutive odd integers have the form n − 2, n, n + 2, and exactly one of those three is a multiple of 3. Line VI is the first false step, because that multiple of 3 may be 3 itself, which is prime. The numbers 3, 5 and 7 are three consecutive odd primes, so the conclusion on line VII is also false, but it is not the first error.
A treats the argument as correct. B to F place the first error on a line that is true. H names the false conclusion rather than the step that first fails.
Adding 4 to both sides of x4 − 2x2 − 3 < 0 gives (x2 − 1)2 < 4, so lines I and II are equivalent rearrangements. Line III is |x2 − 1| < 2. The left half, x2 − 1 > −2, is true for every real x because x2 − 1 ≥ −1, so dropping it in line IV does not change the solution. Lines V and VI then solve x2 < 3. Every line is equivalent to the original inequality.
D, E, F and G each name a later line, but those lines stay equivalent once x2 − 1 > −2 is recognised as automatic. There is no first error.
