53 questions. Try each one, then open the answer to check yourself and read the explanation.
Simplify: (1/(2 − √3) + 2/(2 + √3))−1
Answer: C (1/33 (6 + √3))
1/(2 − √3) = 2 + √3 and 2/(2 + √3) = 2(2 − √3) = 4 − 2√3, so the sum is 6 − √3. Its reciprocal is (6 + √3)/(36 − 3) = (6 + √3)/33.
For what range of values of x is x2 − 2x − 3 ≤ 0?
Answer: A (−1 ≤ x ≤ 3)
x² − 2x − 3 = (x − 3)(x + 1) ≤ 0 between the roots: −1 ≤ x ≤ 3.
Given that M = (3 2 / −1 4) and N = (5 6 / −2 −3), calculate (3M − 2N). (Matrices written row by row; / separates the rows.)
Answer: B ((−1 −6 / 1 18))
3M = (9 6 / −3 12) and 2N = (10 12 / −4 −6). 3M − 2N = (−1 −6 / 1 18).
Simplify: ⅓log 8 + ⅓log 64 − 2log 6.
Answer: C (log 2/9)
⅓log 8 = log 2, ⅓log 64 = log 4, 2log 6 = log 36. log(2 × 4/36) = log(2/9).
Solve (1/9)x+2 = 243x−2.
Answer: B (6/7)
3^(−2x − 4) = 3^(5x − 10), so −2x − 4 = 5x − 10, 7x = 6 and x = 6/7.
If g(x) = 2x + 3 and f(x) = 3x2 − 2x + 4, find f[g(−3)].
Answer: A (37)
g(−3) = −3, then f(−3) = 27 + 6 + 4 = 37.
Using binomial expansion of (1 + x)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6, find, correct to three decimal places, the value of (1.998)6.
Answer: B (63.617)
Put x = 0.998 and add the terms (or note 1.998⁶ = 64 × 0.999⁶): 63.61696…, which is 63.617 to three decimal places.
In how many ways can 8 persons be seated on a bench if only three seats are available?
Answer: C (336)
Seating order matters and a person can't sit twice: ⁸P₃ = 8 × 7 × 6 = 336.
If α and β are the roots of 3x2 − 7x + 6 = 0, find the value of 1/α + 1/β.
Answer: A (7/6)
α + β = 7/3, αβ = 2. 1/α + 1/β = (α + β)/αβ = 7/6.
If f(x) = 4x3 + px2 + 7x − 23 is divided by (2x − 5), the remainder is 7. Find value of p.
Answer: B (−8.0)
f(5/2) = 62.5 + 6.25p + 17.5 − 23 = 57 + 6.25p = 7, so p = −8.
For what value of k is 4x2 − 12x + k, a perfect square?
Answer: D (9)
A perfect square needs b² = 4ac: 144 = 16k, so k = 9 (4x² − 12x + 9 = (2x − 3)²).
A binary operation * is defined on the set of real numbers, R, by p * q = (q2 − p2)/(2pq). Find 3 * 2
Answer: C (−5/12)
p = 3, q = 2: (4 − 9)/12 = −5/12.
Evaluate ∫−23 (4x3 + 6x2 − 7) dx.
Answer: B (100)
[x⁴ + 2x³ − 7x] from −2 to 3 = (81 + 54 − 21) − (16 − 16 + 14) = 114 − 14 = 100.
Given that P = {x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3}, and Q = {x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P ∩ Q.
Answer: D ({36})
P = {3, 6, 9, 12, 18, 36} and Q = {0, 4, 16, 36}. The only common member is 36.
A box of mass 15 kg is placed on a rough plane inclined at 60° to the horizontal. If the box is at rest, calculate the normal reaction to the plane. [Take g = 10 ms−2]
Answer: B (75.0 N)
R = mg cos 60° = 150 × ½ = 75 N.
A fair die is tossed 60 times and the results are recorded in the table.
Number on Die: 1, 2, 3, 4, 5, 6
Frequency: 15, 10, 14, 2, 8, 11
Find the probability of obtaining a prime number
Answer: D (8/15)
Primes on a die are 2, 3 and 5: 10 + 14 + 8 = 32 of 60 throws. P = 32/60 = 8/15.
If 2y2 + 7 = 3y − xy, find dy/dx.
Answer: C (−y/(4y + x − 3))
4y dy/dx = 3 dy/dx − x dy/dx − y, so dy/dx(4y + x − 3) = −y and dy/dx = −y/(4y + x − 3).
In ΔPQR, PQ = 5i − 2j and QR = 4i + 3j. Find RP.
Answer: B (−9i − j)
Going round the triangle, PQ + QR + RP = 0, so RP = −(5i − 2j) − (4i + 3j) = −9i − j.
A stone is thrown vertically upwards and its distance, S metres, after t seconds is given by S = 12t + (5/2)t2 − t3. Calculate the maximum height reached.
Answer: C (31.5 m)
dS/dt = 12 + 5t − 3t² = 0 gives (3t + 4)(t − 3) = 0, t = 3. S = 36 + 22.5 − 27 = 31.5 m.
A stone is thrown vertically upwards and its distance, S metres, after t seconds is given by S = 12t + (5/2)t2 − t3. Calculate the distance travelled in the third second
Answer: A (5.5 m)
The third second is from t = 2 to t = 3: S(3) − S(2) = 31.5 − (24 + 10 − 8) = 5.5 m (the stone is still rising).
Given that f′(x) = x3√x, find f(x).
Answer: A (2x9/2/9 + c)
x³√x = x^(7/2), so f(x) = x^(9/2)/(9/2) + c = 2x^(9/2)/9 + c.
If (1 − 2x)4 = 1 + px + qx2 − 32x3 + 16x4, find the value of (q − p).
Answer: D (32)
p = 4(−2) = −8 and q = 6(4) = 24, so q − p = 32.
If sin x = 12/13 and sin y = 4/5, where x and y are both acute angles, find cos(x + y).
Answer: C (−33/65)
cos x = 5/13 and cos y = 3/5. cos(x + y) = (5/13)(3/5) − (12/13)(4/5) = 15/65 − 48/65 = −33/65.
The first term of an Arithmetic Progression (A.P.) is 4 and the sum of the first three terms is 18. Find the product of the first three terms.
Answer: C (192)
(3/2)(8 + 2d) = 18 gives d = 2, so the terms are 4, 6, 8 and their product is 192.
A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?
Answer: D (30)
Choose the other 2 boys from 5 and the other girl from 3: ⁵C₂ × ³C₁ = 10 × 3 = 30.
The common ratio of a Geometric Progression (G.P.) is 2. If the 5th term is greater than the first term by 45, find the sum of the first five terms.
Answer: C (93)
16a = a + 45 gives a = 3. S₅ = 3(2⁵ − 1) = 93.
If √5 cos x + √15 sin x = 0, for 0° < x < 360°, find the values of x.
Answer: C (150° and 330°)
tan x = −√5/√15 = −1/√3. tan is negative in the 2nd and 4th quadrants: x = 150° or 330°.
If 2i + pj and 4i − 2j are perpendicular, find the value of p.
Answer: C (4)
Perpendicular vectors have zero dot product: 8 − 2p = 0, so p = 4.
Consider the following statements
x: Benita is polite
y: Benita is neat
z: Benita is intelligent
Which of the following symbolizes the statement: 'Benita is neat if and only if she is neither polite nor intelligent'?
Answer: C (y ⇔ ~x ∧ ~z)
'Neither polite nor intelligent' means not polite AND not intelligent: ~x ∧ ~z. So the statement is y ⇔ (~x ∧ ~z).
A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.
Answer: D (32/91)
Red then blue or blue then red: (8/14)(4/13) + (4/14)(8/13) = 32/182 + 32/182 = 32/91.
The gradient of y = 3x2 + 11x + 7 at P(x, y) is −1. Find the coordinates of P.
Answer: B ((−2, −3))
6x + 11 = −1 gives x = −2; y = 12 − 22 + 7 = −3. P = (−2, −3).
Find the equation of the normal to the curve y = 2x2 − 5x + 10 at P(1, 7).
Answer: C (y − x − 6 = 0)
dy/dx = 4x − 5 = −1 at x = 1, so the normal has gradient 1. y − 7 = x − 1 gives y − x − 6 = 0.
Find the value of the derivative of y = 3x2(2x + 1) with respect to x at the point x = 2.
Answer: B (84)
y = 6x³ + 3x², so dy/dx = 18x² + 6x = 72 + 12 = 84 at x = 2.
Find the radius of the circle 2x2 − 4x + 2y2 − 6y − 2 = 0.
Answer: C (√17/2)
Divide by 2: x² − 2x + y² − 3y − 1 = 0, i.e. (x − 1)² + (y − 3/2)² = 17/4. r = √17/2.
Given that f: x → x2 − x + 1 is defined on the set Q = {x: 0 ≤ x < 20, x is a multiple of 5}, find the range of f.
Answer: C ({1, 21, 91, 211})
Q = {0, 5, 10, 15} (20 is excluded). f(0) = 1, f(5) = 21, f(10) = 91, f(15) = 211.
If (15 − 2x)/((x + 4)(x − 3)) ≡ R/(x + 4) + 9/(7(x − 3)), find the value of R.
Answer: B (−23/7)
15 − 2x = R(x − 3) + (9/7)(x + 4). Put x = −4: 23 = −7R, so R = −23/7.
The table shows the distribution of marks obtained by some students in a test.
Marks: 0−9, 10−19, 20−29, 30−39, 40−49
Frequency: 4, 12, 16, 6, 2
What is the upper class boundary of the upper quartile class?
Answer: C (29.5)
There are 40 students, so the upper quartile is the 30th mark. Cumulative frequencies 4, 16, 32 put it in 20−29, whose upper boundary is 29.5.
The table shows the distribution of marks obtained by some students in a test.
Marks: 0−9, 10−19, 20−29, 30−39, 40−49
Frequency: 4, 12, 16, 6, 2
Find the modal class mark.
Answer: C (24.5)
The modal class is 20−29 (frequency 16); its class mark is (20 + 29)/2 = 24.5.
The polynomial f(x) = 2x3 + px2 + qx − 5 has (x − 1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.
f(1) = 0: 2 + p + q − 5 = 0, so p + q = 3.
f(−2) = 27: −16 + 4p − 2q − 5 = 27, so 2p − q = 24.
Adding: 3p = 27, p = 9 and q = −6.
Use the factor theorem and remainder theorem to get two equations.
Evaluate: ∫19 x(2x − 3)/√x dx
x(2x − 3)/√x = 2x^(3/2) − 3x^(1/2).
∫ = [4x^(5/2)/5 − 2x^(3/2)] from 1 to 9.
At 9: 4(243)/5 − 54 = 140.4. At 1: 4/5 − 2 = −1.2.
Answer = 141.6 (= 708/5).
Split the fraction into powers of x first.
Given that (p + ½√3)(1 − √3)2 = 3 − √3, find the value of p.
(1 − √3)² = 4 − 2√3.
p + ½√3 = (3 − √3)/(4 − 2√3) = (3 − √3)(4 + 2√3)/4 = (6 + 2√3)/4 = 3/2 + ½√3.
So p = 3/2.
Rationalise the denominator, then compare.
If 5yC2 = 190, find the value of y.
5y(5y − 1)/2 = 190, so 25y² − 5y − 380 = 0, i.e. 5y² − y − 76 = 0.
(5y + 19)(y − 4) = 0.
y must make 5y a positive integer, so y = 4.
ⁿC₂ = n(n − 1)/2.
The table shows the distribution of monthly income (in thousands of naira) of workers in a factory.
Monthly Income (₦'000): 135−139, 140−149, 150−154, 155−164, 165−169
Number of workers: 20, 42, 28, 38, 22
(a) Draw a histogram for the distribution. (b) Use your graph to estimate the mode of the distribution.
(a) Boundaries (widths): 134.5–139.5 (5), 139.5–149.5 (10), 149.5–154.5 (5), 154.5–164.5 (10), 164.5–169.5 (5).
Unequal widths: plot frequency per 5 (₦'000) – heights 20, 21, 28, 19, 22 – as touching bars on the boundaries.
(b) Tallest bar 149.5–154.5. Cross lines from its top corners to the neighbouring bars (21 and 19): mode ≈ 149.5 + 5 × 7/(7 + 9) ≈ 151.7, i.e. about ₦151 700.
With unequal class widths, bar heights must be frequency densities.
A bag contains 24 mangoes out of which six are bad. If 6 mangoes are selected randomly from the bag with replacement, find the probability that not more than 3 are bad.
P(bad) = 6/24 = ¼, P(good) = ¾, n = 6.
P(X ≤ 3) = 1 − [P(4) + P(5) + P(6)].
P(4) = 15(¼)⁴(¾)² = 135/4096; P(5) = 6(¼)⁵(¾) = 18/4096; P(6) = 1/4096.
P(X ≤ 3) = 1 − 154/4096 = 3942/4096 ≈ 0.962.
It's quicker to subtract the 4, 5 and 6 cases from 1.
(a) The speed of a moving bus reduced from 45 ms−1 to 5 ms−1 with a uniform retardation of 10 ms−2. Calculate the distance covered. (b) A bucket full of water with mass 16 kg is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is 240 N. [Take g = 10 ms−2]
(a) v² = u² + 2as: 25 = 2025 − 20s, so s = 100 m.
(b) T − mg = ma: 240 − 160 = 16a, a = 5 ms⁻² (upwards).
A retardation is a negative acceleration.
Given that x = (−4, 3) and y = (−9, −15) (column vectors), calculate, correct to the nearest degree, the angle between the vectors.
x·y = 36 − 45 = −9.
|x| = 5, |y| = √306 ≈ 17.49.
cos θ = −9/(5 × 17.49) ≈ −0.1029, so θ ≈ 96°.
cos θ = x·y/(|x||y|); a negative cosine means an obtuse angle.
(a) A jogger is training for a 15 km Charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres. (i) How many days will it take the jogger to reach a distance of 15 km in training? (ii) Calculate the total distance he would have run in the training. (b) The second term of a Geometric Progrssion (G.P.) is −3. If its sum to infinity is 25/2, find its common ratios.
(a)(i) 500 + 250(n − 1) = 15 000 gives n − 1 = 58, n = 59 days.
(ii) S = 59/2 × (500 + 15 000) = 457 250 m = 457.25 km.
(b) ar = −3 and a/(1 − r) = 25/2. With a = −3/r: −6 = 25r(1 − r), so 25r² − 25r − 6 = 0, (5r + 1)(5r − 6) = 0.
r = −1/5 or 6/5; a sum to infinity needs |r| < 1, so r = −1/5 (a = 15).
A sum to infinity exists only when −1 < r < 1.
P and Q are two linear transformations in the x − y plane defined by P: (x, y) → (−3x + 6y, 4x + y) and Q: (x, y) → (2x − 3y, −4x − 6y). (a) Write down the matrices of P and Q. (b) What is the image of (−2, −3) under the transformation Q? (c) Obtain a single transformation representing the transformation Q followed by P. (d) Find the image of (1, 4) when transformed by Q followed by P. (e) Find the image P1 of the point (−√2, 2√2) under an anticlockwise rotation of 225° about the origin.
Matrices written row by row:
(a) P = (−3 6 / 4 1), Q = (2 −3 / −4 −6).
(b) Q(−2, −3) = (−4 + 9, 8 + 18) = (5, 26).
(c) Q followed by P is PQ = (−30 −27 / 4 −18).
(d) PQ(1, 4) = (−30 − 108, 4 − 72) = (−138, −68).
(e) Rotation matrix (cos 225° −sin 225° / sin 225° cos 225°) = (−√2/2 √2/2 / −√2/2 −√2/2).
Image = (1 + 2, 1 − 2) = (3, −1).
'Q followed by P' means apply Q first, so the matrix is PQ.
(a) Find the equation of the normal to the curve y = (x2 − x + 1)(x − 2) at the point where the curve cuts the x-axis. (b) The coordinates of the points P, Q and R are (−1, 2), (5, 1) and (3, −4) respectively. Find the equation of the line joining Q and the midpoint of PR.
(a) x² − x + 1 has no real roots, so the curve cuts the x-axis only at (2, 0).
dy/dx = (2x − 1)(x − 2) + (x² − x + 1) = 3 at x = 2; normal gradient −1/3.
y = −(x − 2)/3, i.e. x + 3y − 2 = 0.
(b) Midpoint of PR = (1, −1). Gradient to Q(5, 1) = 2/4 = ½.
y + 1 = ½(x − 1), i.e. x − 2y − 3 = 0.
Use the product rule for the gradient.
A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement. Calculate the probability of picking (a) same colour of bulbs; (b) different colour of bulbs; (c) at least one red bulb.
16 bulbs; 16 × 15 = 240 ordered pairs.
(a) (5×4 + 7×6 + 4×3)/240 = 74/240 = 37/120.
(b) 1 − 37/120 = 83/120.
(c) 1 − P(no red) = 1 − (11/16)(10/15) = 1 − 11/24 = 13/24.
'Different colour' is the complement of 'same colour'.
The table shows the frequency distribution of heights (in cm) of pupils in a certain school.
Heights (cm): 100−109, 110−119, 120−129, 130−139, 140−149, 150−159, 160−169
Frequency: 27, 58, 130, 105, 50, 25, 5
(a)(i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve. (b) Using the curve, estimate the: (i) median height; (ii) interquartile range; (iii) percentage of students whose heights are at most 130 cm.
(a)(i) Upper boundaries 109.5, 119.5, …, 169.5; cumulative frequencies 27, 85, 215, 320, 370, 395, 400.
(ii) Plot these (start at (99.5, 0)) and join with a smooth curve.
(b)(i) Median = 200th value ≈ 119.5 + (115/130) × 10 ≈ 128.4 cm.
(ii) Q₁ (100th) ≈ 120.7 cm, Q₃ (300th) ≈ 137.6 cm; IQR ≈ 16.9 cm.
(iii) About 220 pupils are at most 130 cm: 220/400 ≈ 55%.
(Values read from a graph may differ slightly.)
Read the median at half the total frequency on the cumulative curve.
The position vectors of P, Q and R with respect to the origin are (4i − 5j), (i + 3j) and (−5i + 2j) respectively. If PQRM is a parallelogram, find: (a) the coordinates of M; (b) the acute angle between PM and PQ, correct to the nearest degree.
(a) In PQRM, PM = QR, so M = P + R − Q = (4 − 5 − 1, −5 + 2 − 3) = (−2, −6).
(b) PM = −6i − j, PQ = −3i + 8j. Dot product = 18 − 8 = 10.
cos θ = 10/(√37 × √73) ≈ 0.1924, θ ≈ 79°.
Opposite sides of a parallelogram are equal vectors.
(a) A girl threw a stone horizontally with a velocity of 30 ms−1 from the top of a cliff 50 m high. How far from the foot of the cliff does the stone strike the ground? [Take g = 10 ms−2]
(b) A body A, of mass 2 kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal. Find the tensions T1 and T2 in the strings. [Take g = 10 ms−2]
(a) Fall time: 50 = 5t², t = √10 ≈ 3.16 s. Horizontal distance = 30 × 3.16 ≈ 94.9 m.
(b) Vertically: T₁ cos 56° = 20, so T₁ ≈ 35.8 N.
Horizontally: T₂ = T₁ sin 56° = 20 tan 56° ≈ 29.7 N.
Horizontal and vertical motion are independent; resolve the tensions in two directions.