52 questions. Try each one, then open the answer to check yourself and read the explanation.
If P = {x : −2 < x < 5} and Q = {x : −5 < x < 2}, are subsets of μ = {x : −5 ≤ x ≤ 5}, where x is a real number, find P ∪ Q.
Answer: A ({x : −5 < x < 5})
Q covers −5 < x < 2 and P covers −2 < x < 5; together they cover −5 < x < 5 (neither endpoint is included).
Express (8 − 3√6)/(2√3 + 3√2) in the form p√3 + q√2
Answer: B (−(17/3)√3 + 7√2)
Multiply by (2√3 − 3√2)/(2√3 − 3√2): numerator 16√3 − 24√2 − 18√2 + 18√3 = 34√3 − 42√2; denominator 12 − 18 = −6. Result −(17/3)√3 + 7√2.
An operation * is defined on the set, R, of real numbers by p * q = p + q + 2pq. If the identity element of the operation is zero, find the value of p for which the operation has no inverse.
Answer: A (−½)
The inverse q of p satisfies p + q + 2pq = 0, so q = −p/(1 + 2p). This fails only when 1 + 2p = 0, i.e. p = −½.
Consider the statements: p: Musa is short. q: Musa is brilliant. Which of the following represents the statement "Musa is short but not brilliant"?
Answer: C (p ∧ ~q)
'But' means 'and', and 'not brilliant' is ~q: p ∧ ~q.
If f(x) = 4/x − 1, x ≠ 0, find f−1(7)
Answer: C (½)
Solve 4/x − 1 = 7: 4/x = 8, so x = ½.
If y = 4x − 1, list the range of the domain {−2 ≤ x ≤ 2}, where x is an integer.
Answer: D ({−9, −5, −1, 3, 7})
x = −2, −1, 0, 1, 2 give y = −9, −5, −1, 3, 7.
Factorize completely: x2 + x2y + 3x − 10y + 3xy − 10.
Answer: C ((x − 2)(x + 5)(y + 1))
Group by (1 + y): x²(1 + y) + 3x(1 + y) − 10(1 + y) = (y + 1)(x² + 3x − 10) = (x − 2)(x + 5)(y + 1).
If the solution set of x2 + kx − 5 = 0 is {−1, 5}, find the value of k.
Answer: B (−4)
Sum of roots = −k = 4, so k = −4.
The remainder when x3 − 2x + m is divided by x − 1 is equal to the remainder when 2x3 + x − m is divided by 2x + 1. Find the value of m.
Answer: C (1/8)
1 − 2 + m = 2(−⅛) − ½ − m, so m − 1 = −¾ − m, 2m = ¼ and m = ⅛.
If (2t + 3s)(t − s) = 0, find t/s
Answer: D (−3/2 or 1)
2t = −3s gives t/s = −3/2; t = s gives t/s = 1.
Solve the equation 5x × 5x+1 = 25
Answer: C (½)
5^(2x + 1) = 5², so 2x + 1 = 2 and x = ½.
If log10 y + 3 log10 x ≥ log10 x, express y in terms of x.
Answer: D (y ≥ 1/x2)
log(x³y) ≥ log x, so x³y ≥ x and (x > 0) y ≥ 1/x².
Simplify: nP5/nC5
Answer: D (120)
ⁿP₅ = 5! × ⁿC₅, so the ratio is 5! = 120.
Given that n = 3, evaluate 1/(n − 1)! − 1/(n + 1)!
Answer: D (11/24)
1/2! − 1/4! = 1/2 − 1/24 = 11/24.
Find the coefficient of x3 in the expansion of [⅓(2 + x)]6.
Answer: D (160/729)
(1/729)(2 + x)⁶; the x³ term of (2 + x)⁶ is ⁶C₃ × 2³ = 160. Coefficient 160/729.
Find the fourth term in the expansion of (3x − y)6.
Answer: A (−540x3y3)
The fourth term is ⁶C₃(3x)³(−y)³ = 20 × 27 × (−1)x³y³ = −540x³y³.
The third and sixth terms of a geometric progression (G.P) are 8/3 and 64/81, respectively. Find the common ratio.
Answer: B (2/3)
r³ = (64/81) ÷ (8/3) = 8/27, so r = 2/3.
Given that −6, −2½, …, 71, is a linear sequence, calculate the number of terms in the sequence.
Answer: D (23)
d = 3½. 71 = −6 + (n − 1)(3.5) gives n − 1 = 22, so n = 23.
If P = (1 2 / 5 1) and Q = (0 1 / 1 3), find PQ. (Matrices written row by row; / separates the rows.)
Answer: C ((2 7 / 1 8))
PQ = (0 + 2, 1 + 6 / 0 + 1, 5 + 3) = (2 7 / 1 8).
Evaluate cos 75°, leaving the answer in a surd form.
Answer: B ((√2/4)(√3 − 1))
cos(45° + 30°) = (√2/2)(√3/2) − (√2/2)(½) = (√2/4)(√3 − 1).
Given that tan x = 5/12 and tan y = 3/4, find tan(x + y).
Answer: D (56/33)
(5/12 + 3/4)/(1 − 15/48) = (14/12)/(33/48) = 56/33.
Find the equation of the line which passes through (−4, 3) and parallel to line y = 2x + 5.
Answer: A (y = 2x + 11)
Gradient 2: y − 3 = 2(x + 4), so y = 2x + 11.
Points E(−2, −1) and F(3, 2) are the ends of the diameter of a circle. Find the equation of the circle.
Answer: D (x2 + y2 − x − y − 8 = 0)
(x + 2)(x − 3) + (y + 1)(y − 2) = 0 gives x² + y² − x − y − 8 = 0.
The lines 2y + 3x − 16 = 0 and 7y − 2x − 6 = 0 intersect at point P. Find the coordinates of P.
Answer: A ((4, 2))
3x + 2y = 16 and −2x + 7y = 6 give y = 2 and x = 4.
Find limx→−3 ((2x2 + x − 21)/(x − 3)).
Answer: B (1)
Substituting x = −3: (18 − 3 − 21)/(−6) = −6/−6 = 1. (Factorising, 2x² + x − 21 = (2x + 7)(x − 3), so the expression is 2x + 7 = 1 at x = −3.)
Find the gradient of the normal to the curve y = x3 − x2 at the point where x = 2.
Answer: A (−1/8)
dy/dx = 3x² − 2x = 8 at x = 2, so the normal has gradient −1/8.
Find the minimum value of y = 3x2 − x − 6.
Answer: B (−6 1/12)
dy/dx = 6x − 1 = 0 at x = 1/6; y = 3/36 − 1/6 − 6 = −6 1/12.
The radius of a circle increases at a rate of 0.5 cms−1. Find the rate of change in the area of the circle with radius 7 cm. [Take π = 22/7]
Answer: B (22 cm2s−1)
dA/dt = 2πr dr/dt = 2 × 22/7 × 7 × 0.5 = 22 cm²s⁻¹.
Find an expression for y given that dy/dx = x2√x
Answer: C ((2/7)x7/2 + c)
x²√x = x^(5/2), so y = x^(7/2)/(7/2) + c = (2/7)x^(7/2) + c.
Given that n = 10 and Σd2 = 20, calculate the spearman's rank correlation coefficient.
Answer: C (0.879)
ρ = 1 − 6(20)/(10 × 99) = 1 − 120/990 ≈ 0.879.
Find the vaiance of 11, 12, 13, 14 and 15.
Answer: A (2)
Mean 13; squared deviations 4, 1, 0, 1, 4 add to 10. Variance = 10/5 = 2.
A fair coin is tossed three times. Find the probability of obtaining two heads.
Answer: B (3/8)
HHT, HTH, THH: 3 of 8 outcomes.
A box contains 14 white balls and 6 black balls. Find the probability of first drawing a black ball and then a white ball without replacement.
Answer: B (0.22)
6/20 × 14/19 = 84/380 ≈ 0.22.
Given that r = 3i + 4j and t = −5i + 12j, find the acute angle between them.
Answer: C (59.5°)
r·t = −15 + 48 = 33; |r| = 5, |t| = 13. cos θ = 33/65, θ ≈ 59.5°.
Find the unit vector in the direction of −2i + 5j.
Answer: B ((1/√29)(−2i + 5j))
|−2i + 5j| = √29, so the unit vector is (−2i + 5j)/√29.
A force 10N acts in the direction 060° and another 6N acts in the direction 330°. Find the y-component of their resultant force.
Answer: C ((5 + 3√3) N)
The y (north) components are 10 cos 60° = 5 and 6 cos 330° = 3√3, so the resultant's y-component is (5 + 3√3) N.
A man of mass 80 kg stands in a lift. If the lift moves upwards with acceleration 0.5 ms−2, find the reaction from the floor of the lift on the man. [Take g = 10 ms−2]
Answer: D (840 N)
R − mg = ma, so R = 80(10 + 0.5) = 840 N.
A ball falls from a height of 18 m above the ground. Find the speed with which the ball hits the ground. [Take g = 10 ms−2]
Answer: D (18.97 ms−1)
v = √(2gh) = √(2 × 10 × 18) = √360 ≈ 18.97 ms⁻¹.
Given that (5, 2), (−4, k) and (2, 1) lie on a straight line, find the value of k.
Gradient from (5, 2) to (2, 1) = 1/3.
(k − 2)/(−4 − 5) = 1/3, so k − 2 = −3 and k = −1.
Points on one line have equal gradients between any pair.
(a) If f(x + 2) = 6x2 + 5x − 8, find f(5).
(b) Express (7√2 + 3√3)/(4√2 − 2√3) in the form p + q√r, where p, q, and r are rational numbers.
(a) x + 2 = 5 when x = 3: f(5) = 54 + 15 − 8 = 61.
(b) Multiply by (4√2 + 2√3): numerator 56 + 14√6 + 12√6 + 18 = 74 + 26√6; denominator 32 − 12 = 20.
Answer 37/10 + (13/10)√6.
In (a) choose x so that x + 2 = 5.
When f(x) = 2x3 + mx2 + nx + 11 is divided by x2 + 5x + 1, the quotient is 2x − 5 and the remainder is 30x + 16. Find the values of m and n.
f(x) = (x² + 5x + 1)(2x − 5) + 30x + 16 = 2x³ + 5x² − 23x − 5 + 30x + 16 = 2x³ + 5x² + 7x + 11.
m = 5, n = 7.
Dividend = divisor × quotient + remainder.
The probabilities that Ago, Sulley and Musa will gain admission to a certain university are 4/5, 3/4 and 2/3, respectively. Find the probability that: (a) none of them will gain admission; (b) only Ago and Sulley will gain admission.
(a) (1/5)(1/4)(1/3) = 1/60.
(b) (4/5)(3/4)(1/3) = 1/5.
Multiply the probabilities of independent events.
The mean of the numbers 1, 4, k, (k + 4) and 11 is (k + 1). Calculate the: (a) Value of k; (b) standard deviation.
(a) (20 + 2k)/5 = k + 1, so 20 + 2k = 5k + 5 and k = 5.
(b) Numbers 1, 4, 5, 9, 11; mean 6. Squared deviations 25, 4, 1, 9, 25 add to 64.
SD = √(64/5) = √12.8 ≈ 3.58.
Find k, then list the actual numbers.
(a) A boy of mass 3 kg moves with a velocity of 8 ms−1. It collides with a second body moving in the same direction with velocity of 5 ms−1. After collision the bodies move together with a velocity of 6 ms−1. Find the mass of the second body.
(b) If the second body in 7(a) moves with a velocity of 5 ms−1 in the opposite direction as that of the 3 kg body with a velocity of 8 ms−1, find, correct to two decimal places, the common velocity of the two bodies if they move together after collision.
(a) 3(8) + m(5) = (3 + m)(6), so 24 + 5m = 18 + 6m and m = 6 kg.
(b) 3(8) + 6(−5) = 9v, so v = −6/9 ≈ −0.67 ms⁻¹: 0.67 ms⁻¹ in the direction the second body was moving.
Momentum is conserved; opposite directions get opposite signs.
Given that p = (5, 3), q = (−1, 2), r = (17, 5) (column vectors) and r = αp + βq, where α and β are scalars, express q in terms of r and p.
5α − β = 17 and 3α + 2β = 5. Doubling the first and adding: 13α = 39, α = 3, β = −2.
r = 3p − 2q, so q = ½(3p − r).
Compare the top and bottom components.
(a) Without using mathematical tables or calculator, evaluate ((3/2)log 27 − 3 log 5√5)/log 0.6.
(b) Two linear transformations A and B in the Oxy plane, are defined by A: (x, y) → (x + 2y, −x + y), B: (x, y) → (2x + 3y, x + 2y). (i) Write down the matrices A and B. (ii) Find the image of the point P(−2, 2) under the linear transformation A followed by B.
(a) (3/2)log 27 = (9/2)log 3 and 3 log 5√5 = (9/2)log 5. Numerator = (9/2)log(3/5) = (9/2)log 0.6, so the value is 9/2.
(b)(i) A = (1 2 / −1 1), B = (2 3 / 1 2) (rows separated by /).
(ii) A(−2, 2) = (2, 4); then B(2, 4) = (16, 10).
Write 27 and 5√5 as powers of 3 and 5.
(a)(i) Write down the expansion of (1 + x)7 in escending powers of x. (ii) If the coefficients of the fifth, sixth and seventh terms in the expansion in 10(a)(i) form a linear sequence (A.P.), find the common difference of the A.P.
(b) Using the trapezium rule with ordinates at 1, 2, 3, 4 and 5, calculate, correct to two decimal places, ∫15 √(2x + 8x2) dx.
(a)(i) 1 + 7x + 21x² + 35x³ + 35x⁴ + 21x⁵ + 7x⁶ + x⁷.
(ii) Fifth, sixth, seventh coefficients 35, 21, 7: d = −14.
(b) h = 1; y = 3.1623, 6.0000, 8.8318, 11.6619, 14.4914.
Area ≈ ½[3.1623 + 14.4914 + 2(6 + 8.8318 + 11.6619)] ≈ 35.32.
Trapezium rule: (h/2)[first + last + 2(middle ordinates)].
(a) If kP2 = 72, find the value of k.
(b) Solve the equation 2 cos 2θ − 5 cos θ = 3, for 0° ≤ θ ≤ 360°.
(a) k(k − 1) = 72, (k − 9)(k + 8) = 0, k = 9.
(b) 2(2cos²θ − 1) − 5cos θ = 3, so 4cos²θ − 5cos θ − 5 = 0.
cos θ = (5 ± √105)/8 = 1.906 (impossible) or −0.6559.
θ ≈ 131.0° or 229.0°.
Use cos 2θ = 2cos²θ − 1 to get a quadratic in cos θ.
The table shows the distribution of marks scored by some students in a test.
Marks: 1 − 10, 11 − 20, 21 − 30, 31 − 40, 41 − 50, 51 − 60, 61 − 70, 71 − 80, 81 − 90, 91 − 100
Number of students: 3, 17, 41, 85, 97, 115, 101, 64, 21, 6
(a)(i) Construct a cumulative frequency table for the distrbution. (ii) Draw a cumulative frequency curve for the distribution. (b) Use the curve to estimate the: (i) number of students who scored marks between 32 and 74; (ii) pass mark, if 18% of the students failed; (iii) lowest mark for distinction, if 8% of the students passed with distinction.
(a)(i) Upper boundaries 10.5, 20.5, …, 100.5; cumulative frequencies 3, 20, 61, 146, 243, 358, 459, 523, 544, 550.
(ii) Plot these (start at (0.5, 0)) and join with a smooth curve.
(b)(i) About 74 scored up to 32 and about 481 up to 74: about 407 students.
(ii) 18% of 550 = 99: pass mark ≈ 35.
(iii) Top 8% = 44, so cumulative frequency 506: lowest distinction mark ≈ 78.
(Graph readings may differ slightly.)
Read values off the curve at the required cumulative frequency.
(a) Two different Mathematics books, 5 different Physics books and 3 different Chemistry books are to be arranged on a shelf. How many arrangements are possible if: (i) books on the same subject must stand together? (ii) only the physics books must stand together?
(b) In a certain community, 13 out of every 20 persons speak English. If 8 persons are selected at random from the community, find, correct to three significant figures, the probability that at least 3 of them speak English.
(a)(i) 3! (order of subjects) × 2! × 5! × 3! = 6 × 2 × 120 × 6 = 8640.
(ii) Treat the 5 Physics books as one block: 6! × 5! = 86 400.
(b) p = 0.65, n = 8: P(X ≥ 3) = 1 − [P(0) + P(1) + P(2)] ≈ 1 − 0.0253 = 0.975.
Glue books that must stay together into one block, then arrange inside it.
Four vectors r = αi + βj where α and β are positive constants, s = 2i − j, m = 3i + 2j and n = i + j are such that the magnitude of r is three time as s and is parallel to the vector (m − n). (a) Find the values of α and β. (b) Calculate the magnitude and direction of (r − s).
(a) m − n = 2i + j, so r = k(2i + j). |r| = 3|s| = 3√5 = k√5, so k = 3: α = 6, β = 3.
(b) r − s = 4i + 4j; magnitude 4√2 ≈ 5.66; direction 45° to the positive x-axis (bearing 045°).
Parallel vectors are multiples of each other.
A particle is projected vertically upwards from the ground with speeed 30 ms−1. Calculate the: (a) maximum height reached by the particle; (b) time taken by the particle to return to the ground; (c) time(s) taken for the particle to attain a height of 40 m above the ground. [Take g = 10 ms−2]
h = 30t − 5t².
(a) Top when v = 30 − 10t = 0, t = 3 s: h = 90 − 45 = 45 m.
(b) h = 0 again at t = 6 s.
(c) 30t − 5t² = 40 gives t² − 6t + 8 = 0: t = 2 s (going up) and t = 4 s (coming down).
The particle passes 40 m twice, once each way.