50 questions. Try each one, then open the answer to check yourself and read the explanation.
Find the domain of g(x) = (4x2 − 1)/√(9x2 + 1).
Answer: D ({x : x ∈ R})
9x² + 1 is always positive, so the square root is defined and never zero. g is defined for every real x.
Given that f(x) = 3x2 − 12x + 12 and f(x) = 3, find the values of x.
Answer: A (1, 3)
3x² − 12x + 9 = 0 gives x² − 4x + 3 = 0, (x − 1)(x − 3) = 0: x = 1 or 3.
If the polynomial f(x) = 3x3 − 2x2 + 7x + 5 is divided by (x − 1), find the remainder.
Answer: D (13)
Remainder = f(1) = 3 − 2 + 7 + 5 = 13.
P = {1, 3, 5, 7, 9, 11}, Q = {2, 4, 6, 8, 10, 12} and R = {2, 3, 5, 7, 11} are subsets of U = {1, 2, 3, …, 12}. Which of the following statements is true?
Answer: C ((R ∩ P) ⊂ (R ∩ U))
R ∩ P = {3, 5, 7, 11} and R ∩ U = R = {2, 3, 5, 7, 11}, so C is true. Q ∩ R = {2}, 2 ∉ P, and P′ ∩ R = {2} has 1 member.
If log3 a + 2 = 3 log3 b, express a in terms of b.
Answer: D (a = b3/9)
log₃ a − log₃ b³ = −2, so a/b³ = 3⁻² = 1/9 and a = b³/9.
If α and β are the roots of 2x2 − 5x + 6 = 0, find the equation whose roots are (α + 1) and (β + 1).
Answer: C (2x2 − 9x + 13)
α + β = 5/2, αβ = 3. New sum = 5/2 + 2 = 9/2; new product = αβ + (α + β) + 1 = 13/2. Equation: x² − (9/2)x + 13/2 = 0, i.e. 2x² − 9x + 13 = 0.
Resolve (3x − 1)/(x − 2)2, x ≠ 2 into partial fractions
Answer: B (3/(x − 2) + 5/(x − 2)2)
3x − 1 = A(x − 2) + B: A = 3, and x = 2 gives B = 5. So 3/(x − 2) + 5/(x − 2)².
If α and β are the roots of the equation 2x2 − 5x + n = 0 such that αβ = 2, find the value of n.
Answer: D (4)
αβ = n/2 = 2, so n = 4.
Solve log2(12x − 10) = 1 + log2(4x + 3).
Answer: B (4.00)
log₂[(12x − 10)/(4x + 3)] = 1, so 12x − 10 = 2(4x + 3) = 8x + 6 and x = 4.
The general term of an infinite sequence 9, 4, −1, −6, … is ur = ar + b. Find the values of a and b.
Answer: B (a = −5, b = 14)
u₁ = a + b = 9 and u₂ = 2a + b = 4 give a = −5, b = 14.
If |k k / 4 k| + |2 3 / −1 k| = 6, find the value of the constant k, where k > 0. (Determinants written row by row; / separates the rows.)
Answer: C (3)
(k² − 4k) + (2k + 3) = 6 gives k² − 2k − 3 = 0, (k − 3)(k + 1) = 0, so k = 3 (k > 0).
How many numbers greater than 150 can be formed from the digits 1, 2, 3, 4, 5 if no repetition is allowed?
Answer: C (291)
3-digit: first digit 2–5 gives 4 × 4 × 3 = 48, and 152, 153, 154 add 3 (total 51). 4-digit: 5 × 4 × 3 × 2 = 120. 5-digit: 120. Total 291.
The first term of a Geometric Progression (G.P.) is ¾. If the product of the second and third terms of the sequence is 972, find its common ratio.
Answer: D (12)
ar × ar² = a²r³ = (9/16)r³ = 972, so r³ = 1728 and r = 12.
If sin θ = 3/5, where 0° < θ < 90°, evaluate cos(180 − θ).
Answer: D (−4/5)
cos θ = 4/5, and cos(180° − θ) = −cos θ = −4/5.
Find the radius of the circle x2 + y2 − 8x − 2y + 1 = 0.
Answer: C (4)
Centre (4, 1); r² = 16 + 1 − 1 = 16, so r = 4.
In how many ways can the letters of the word 'ELECTIVE' be arranged?
Answer: C (6720)
8 letters with E three times: 8!/3! = 6720.
If the determinant of the matrix (2 x / 3 5) = 13, find the value of x. (Matrix written row by row; / separates the rows.)
Answer: B (−1)
10 − 3x = 13, so x = −1.
Express (13/4)π radians in degrees.
Answer: C (585°)
(13/4) × 180° = 585°.
Find the equation of the tangent to the circle x2 + y2 − 4x − 2y = 0 at the point (1, 3)
Answer: A (2y − x − 5 = 0)
Centre (2, 1). The radius to (1, 3) has gradient −2, so the tangent has gradient ½: y − 3 = ½(x − 1), i.e. 2y − x − 5 = 0.
Given that y = x(x + 1)2, calculate the maximum value of y.
Answer: B (0)
dy/dx = 3x² + 4x + 1 = (3x + 1)(x + 1) = 0. d²y/dx² = 6x + 4 is negative at x = −1, so the maximum value is y(−1) = 0.
The midpoint of M(4, −1) and N(x, y) is P(3, −4). Find the coordinates of N.
Answer: B ((2, −7))
(4 + x)/2 = 3 gives x = 2; (−1 + y)/2 = −4 gives y = −7.
Find the stationary point of the curve y = 3x2 − 2x3.
Answer: B ((1, 1))
dy/dx = 6x − 6x² = 6x(1 − x) = 0 at x = 0 or 1. The points are (0, 0) and (1, 1); only (1, 1) is offered.
Evaluate ∫½1 (x3 − 4)/x3 dx.
Answer: A (−5.5)
∫(1 − 4x⁻³) dx = x + 2/x². From ½ to 1: (1 + 2) − (½ + 8) = −5.5.
Calculate the standard deviation of 30, 29, 25, 28, 32 and 24.
Answer: B (2.8)
Mean = 168/6 = 28. Squared deviations 4, 1, 9, 0, 16, 16 add to 46. SD = √(46/6) ≈ 2.8.
Evaluate ∫−11 (x + 1)2 dx.
Answer: A (8/3)
[(x + 1)³/3] from −1 to 1 = 8/3 − 0 = 8/3.
Out of 70 schools, 42 of them can be attended by boys and 35 can be attended by girls. If a pupil is selected at random from these schools, find the probability that he/she is from a mixed school.
Answer: B (1/10)
Mixed schools: 42 + 35 − 70 = 7. P = 7/70 = 1/10.
The marks scored by 4 students in Mathematics and Physics are ranked as shown in the table below.
Mathematics: 3, 4, 2, 1
Physics: 4, 3, 1, 2
Calculate the Spearman's rank correlation coefficient.
Answer: C (0.6)
Differences −1, 1, 1, −1, so Σd² = 4. ρ = 1 − 6(4)/(4 × 15) = 1 − 0.4 = 0.6.
Given that a = i − 3j, b = −2i + 5j and c = 3i − j, calculate |a − b + c|.
Answer: B (3√13)
a − b + c = (1 + 2 + 3)i + (−3 − 5 − 1)j = 6i − 9j. |6i − 9j| = √117 = 3√13.
What is the probability of obtaining a head and a six when a fair coin and a die are tossed together?
Answer: A (1/12)
½ × 1/6 = 1/12.
If OX = (−7, 6) and OY = (16, −11) (column vectors), find YX.
Answer: D ((−23, 17))
YX = OX − OY = (−7 − 16, 6 + 11) = (−23, 17).
A body of mass 28 g initially at rest is acted upon by a force, F Newtons. If it attains a velocity of 5.4 ms−1 in 18 seconds, find the value of F.
Answer: B (0.0084 N)
a = 5.4/18 = 0.3 ms⁻²; F = 0.028 × 0.3 = 0.0084 N.
Find the angle between forces of magnitudes 7 N and 4 N if their resultant has a magnitude of 9 N.
Answer: B (73.40°)
9² = 7² + 4² + 2(7)(4) cos θ, so 81 = 65 + 56 cos θ, cos θ = 16/56 and θ ≈ 73.40°.
Find the constant term in the binomial expansion of (2x2 + 1/x)9.
Answer: D (672)
The term ⁹Cᵣ(2x²)^(9 − r)(1/x)ʳ has power 18 − 3r, which is 0 when r = 6: ⁹C₆ × 2³ = 84 × 8 = 672.
A particle starts from rest and moves through a distance S = 12t2 − 2t3 metres in time t seconds. Find its acceleration in 1 second.
Answer: C (12 ms−2)
v = 24t − 6t² and a = 24 − 12t; at t = 1, a = 12 ms⁻².
A car is moving at a speed of 120 kmh−1. Find its speed in ms−1
Answer: A (33.3 ms−1)
120 km/h = 120 000 m ÷ 3600 s ≈ 33.3 ms⁻¹.
Two functions f and g are defined on the set of real numbers by f: x → x2 + 1 and g: x → x − 2. Find f o g.
Answer: B (x2 − 4x + 5)
f(g(x)) = (x − 2)² + 1 = x² − 4x + 5.
Given that log3 x − 3logx 3 + 2 = 0, find the values of x.
Let t = log₃ x; then log_x 3 = 1/t.
t − 3/t + 2 = 0, so t² + 2t − 3 = 0, (t + 3)(t − 1) = 0.
t = 1: x = 3. t = −3: x = 3⁻³ = 1/27.
log_x 3 is the reciprocal of log₃ x.
(a) Using the substitution u = x − 2, write (x3 + 5)/(x − 2)4 as an expression in terms of u.
(b) Using the answer in 3(a), express (x3 + 5)/(x − 2)4 in partial fractions.
(a) x = u + 2: ((u + 2)³ + 5)/u⁴ = (u³ + 6u² + 12u + 13)/u⁴ = 1/u + 6/u² + 12/u³ + 13/u⁴.
(b) 1/(x − 2) + 6/(x − 2)² + 12/(x − 2)³ + 13/(x − 2)⁴.
Expand (u + 2)³ and divide each term by u⁴.
The sum of the first twelve terms of an Arithmetic Progression (A.P) is 168. If the third term is 7, find the values of the common difference and the first term.
S₁₂ = 6(2a + 11d) = 168, so 2a + 11d = 28.
a + 2d = 7, so 2a + 4d = 14.
Subtracting: 7d = 14, d = 2; a = 3.
Set up two equations in a and d.
Two panel of judges, X and Y, rank 8 brands of cooking oil as follows:
Cooking oil type: A, B, C, D, E, F, G, H
X: 8, 5, 1, 7, 2, 6, 3, 4
Y: 6, 3, 4, 8, 5, 7, 1, 2
Calculate the Spearman's rank correlation coefficient.
d = X − Y: 2, 2, −3, −1, −3, −1, 2, 2; Σd² = 36.
ρ = 1 − 6(36)/(8 × 63) = 1 − 216/504 ≈ 0.57.
The values are already ranks, so subtract directly.
(a) The probability that Kunle solves a particular questions is 1/3 while that of Tayo is 1/5. If both them attempt the question, find the probability that only one of them will solve the question. (b) A committee of 8 is to be chosen from 10 persons. In how many ways can this be done if there is no restriction?
(a) Kunle only: (1/3)(4/5) = 4/15; Tayo only: (2/3)(1/5) = 2/15. Total 6/15 = 2/5.
(b) ¹⁰C₈ = 45 ways.
'Only one' means one succeeds and the other fails.
Given that m = 3i − 2j, n = 2i − 3j and p = −i − 6j, find 4m − 2n − 3p.
4m = 12i − 8j; −2n = −4i + 6j; −3p = 3i + 18j.
4m − 2n − 3p = 11i + 16j.
Multiply each vector, then add the i and j parts separately.
A body of mass 20 kg moving with a velocity of 80 ms−1 collides with another body of mass 30 kg moving with a velocity of 50 ms−1. If the two bodies moved together in the same direction after collision, find their common velocity if they moved in the: (a) same direction before collision; (b) opposite direction before collision.
(a) 20(80) + 30(50) = 50v, so 3100 = 50v and v = 62 ms⁻¹.
(b) 20(80) − 30(50) = 50v, so 100 = 50v and v = 2 ms⁻¹ (in the 20 kg body's direction).
Momentum is conserved; give opposite velocities opposite signs.
A circle is drawn through the points (3, 2), (−1, −2) and (5, −4). Find the: (a) coordinates of the centre of the circle; (b) radius of the circle; (c) equation of the circle.
(a) Centre (h, k) is equidistant from the three points:
from (3, 2) and (−1, −2): h + k = 1; from (3, 2) and (5, −4): h − 3k = 7.
k = −3/2, h = 5/2: centre (5/2, −3/2).
(b) r² = (3 − 5/2)² + (2 + 3/2)² = 1/4 + 49/4 = 25/2, so r = 5/√2 ≈ 3.54.
(c) (x − 5/2)² + (y + 3/2)² = 25/2, i.e. x² + y² − 5x + 3y − 4 = 0.
Equal distances from the centre give two linear equations.
(a) Solve 23y+2 − 7(22y+2) − 31(2y) − 8 = 0, y ∈ R.
(b) Find ∫(√(x2 + 1))x dx
(a) Let p = 2ʸ: 4p³ − 28p² − 31p − 8 = 0.
p = 8 is a root: (p − 8)(4p² + 4p + 1) = (p − 8)(2p + 1)² = 0.
2ʸ > 0, so p = 8 and y = 3.
(b) Let u = x² + 1, du = 2x dx: ∫ ½√u du = ⅓u^(3/2) + c = ⅓(x² + 1)^(3/2) + c.
A substitution turns (a) into a cubic in p = 2ʸ.
(a)(i) Write down the binominal expansion of (2 − ½x)5 in ascending powers of x. (ii) Using the expansion in 11(a)(i), find, correct to two decimal places, the value of (1.99)5.
(b) The polynomial x3 + qx2 + rx + 9, where q and r are constants, has (x + 1) as a factor and has a remainder −17 when divided by (x + 2). Find the values of q and r.
(a)(i) 32 − 40x + 20x² − 5x³ + (5/8)x⁴ − (1/32)x⁵.
(ii) 2 − ½x = 1.99 gives x = 0.02: 32 − 0.8 + 0.008 − 0.00004 + … ≈ 31.21.
(b) f(−1) = 0: −1 + q − r + 9 = 0, so q − r = −8.
f(−2) = −17: −8 + 4q − 2r + 9 = −17, so 2q − r = −9.
q = −1, r = 7.
Choose x so that the bracket equals 1.99.
Ten coins were tossed together a number of times. The distribution of the number of heads obtained is given in the following table:
Number of Heads: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Frequency: 2, 7, 23, 36, 11, 61, 100, 12, 8, 5, 3
Calculate, correct to three decimal places, the: (a) mean number of heads. (b) probability of getting an even number of heads. (c) probability of getting an odd number of heads.
Σf = 268; Σfx = 0 + 7 + 46 + 108 + 44 + 305 + 600 + 84 + 64 + 45 + 30 = 1333.
(a) Mean = 1333/268 ≈ 4.974.
(b) Even (0, 2, 4, 6, 8, 10): 2 + 23 + 11 + 100 + 8 + 3 = 147; P = 147/268 ≈ 0.549.
(c) Odd: 121/268 ≈ 0.451.
Use the relative frequencies from the table.
The probabilities that Ali, Baba and Katty will gain admission to College are 2/3, 3/4 and 4/5 respectively. Find the probability that: (a) only Katty and Baba will gain admission. (b) none of them will gain admission. (c) at most two of them will gain admission.
(a) (1/3)(3/4)(4/5) = 1/5.
(b) (1/3)(1/4)(1/5) = 1/60.
(c) 1 − P(all three) = 1 − (2/3)(3/4)(4/5) = 1 − 2/5 = 3/5.
'At most two' is everything except 'all three'.
The position vectors of points A, B and C with respect to the origin are (8i − 10j), (2i + 6j) and (−10i + 4j) respectively. If ABCN is a parallelogram, find (a) the position vector of N. (b) |AN| and |AB|. (c) correct to two decimal places, the acute angle between AN and AB.
(a) In ABCN, AN = BC, so N = A + C − B = (8 − 10 − 2)i + (−10 + 4 − 6)j = −4i − 12j.
(b) AN = −12i − 2j, |AN| = √148 ≈ 12.17; AB = −6i + 16j, |AB| = √292 ≈ 17.09.
(c) AN·AB = 72 − 32 = 40; cos θ = 40/(√148 × √292) ≈ 0.1924, θ ≈ 78.91°.
Opposite sides of a parallelogram are equal vectors.
A uniform beam, XY, 4 m long and weighing 350 N rests on two pivots P and Q. It is kept in equilibrium by weights of 80 N attached at X and 1000 N attached at a point between P and Q such that it is 0.6 m from Q. If |XP| = 0.8 m and |PQ| = 2.2 m: (a) calculate the reactions at P and Q. (b) if the 1000 N weight is replaced with a 1200 N weight, at what point from Q should it be placed in order to maintain the equilibrium.
Distances from X: P 0.8 m, centre 2 m, 1000 N at 2.4 m, Q 3.0 m.
(a) Moments about P: 2.2R_Q + 80(0.8) = 350(1.2) + 1000(1.6), so 2.2R_Q = 1956 and R_Q ≈ 889.09 N.
R_P = 80 + 350 + 1000 − 889.09 ≈ 540.91 N.
(b) Keeping the reaction at P the same (moments about Q): 1200d = 1000(0.6), so d = 0.5 m from Q (between P and Q). Then R_Q = 1630 − 540.91 ≈ 1089.09 N.
Take moments about one support to find the other reaction.