54 questions. Try each one, then open the answer to check yourself and read the explanation.
Which of the following sets is equivalent to (P ∪ Q) ∩ (P ∪ Q)?
Answer: C (P ∪ Q)
Any set intersected with itself is the same set, so (P ∪ Q) ∩ (P ∪ Q) = P ∪ Q.
Simplify (cos 2θ − 1)/sin 2θ
Answer: A (−tanθ)
cos 2θ − 1 = −2sin2θ and sin 2θ = 2sinθcosθ. Dividing gives −sinθ/cosθ = −tanθ.
Solve the inequality x2 − 2x ≥ 3
Answer: B (x ≥ 3 or ≤ −1)
x2 − 2x − 3 = (x − 3)(x + 1) ≥ 0. Both factors have the same sign (or one is zero) when x ≥ 3 or x ≤ −1.
Give that √6, 3√2, 3√6, 9√2, ...are the first four terms of an exponential sequence (G.P). Find, in its simplest form, the 8th term.
Answer: C (81√2)
r = 3√2 ÷ √6 = √3. T8 = ar7 = √6 × (√3)7 = √6 × 27√3 = 27√18 = 81√2.
Given that sin x = −√3/2 and cos x > 0, find the value of x.
Answer: A (300°)
sin negative and cos positive puts x in the 4th quadrant. The reference angle is 60°, so x = 360° − 60° = 300°.
Evaluate log10(⅓ + ¼) + 2log102 + log10(3/7).
Answer: B (0)
⅓ + ¼ = 7/12. The sum is log10(7/12 × 4 × 3/7) = log101 = 0.
Simplify x3n+1/(x2n+5/2√(x2n−3))
Answer: C (1)
√(x2n−3) = xn−3/2, so the denominator is x2n+5/2+n−3/2 = x3n+1. The fraction equals 1.
From the diagram, which of the following represents the vector V in component form?
Answer: B (3i + 3√3j)
V has magnitude 6 at 60° to the i-axis: 6cos60° i + 6sin60° j = 3i + 3√3j.
QRS is a triangle such that QR = (3i + 2j) and SR = (−5i + j). find SQ.
Answer: D (−8i − j)
SQ = SR + RQ = SR − QR = (−5i + j) − (3i + 2j) = −8i − j.
If (x + 1) is a factor of the polynominal x3 + px2 + x + 6 = 0, find the value of p.
Answer: B (−4)
By the factor theorem f(−1) = 0: −1 + p − 1 + 6 = 0, so p = −4.
The equation of a circle is 3x2 + 3y2 + 24x − 12y = 15. Find it radius.
Answer: D (5)
Divide by 3: x2 + y2 + 8x − 4y = 5, so (x + 4)2 + (y − 2)2 = 5 + 16 + 4 = 25 and r = 5.
A polynominal is defined by f(x + 1) = x3 + 3x2 − 4x + 2, find f(2).
Answer: C (2)
x + 1 = 2 when x = 1, so f(2) = 1 + 3 − 4 + 2 = 2.
If the mid-point of the line joining (1 − k, −4) and (2, k + 1) is (−k, k), find the value of k.
Answer: B (−3)
x-coordinate: (1 − k + 2)/2 = −k gives 3 − k = −2k, so k = −3. Check y: (−4 + k + 1)/2 = (−3 − 3)/2 = −3 = k.
Evaluate ∫−23 (3x2 − 2x − 12) dx.
Answer: A (−30)
∫ = [x3 − x2 − 12x] from −2 to 3 = (27 − 9 − 36) − (−8 − 4 + 24) = −18 − 12 = −30.
If y = x3 − x2 − x + 6, find the values of x at the turning points.
Answer: C (1, −⅓)
dy/dx = 3x2 − 2x − 1 = (3x + 1)(x − 1) = 0 at turning points, so x = 1 or x = −⅓.
Given that P = (2 1 / 5 −3) and Q = (4 −8 / 1 −2), find (2P − Q). (Matrices are written row by row.)
Answer: D ((0 10 / 9 −4))
2P = (4 2 / 10 −6). Subtract Q entry by entry: (4 − 4, 2 + 8 / 10 − 1, −6 + 2) = (0 10 / 9 −4).
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 & 10. Find the mean mark
Answer: B (5.50)
The marks add up to 55, so the mean is 55 ÷ 10 = 5.50.
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 & 10. Find the variance
Answer: A (8.25)
With mean 5.5 the squared deviations add up to 82.5, so the variance is 82.5 ÷ 10 = 8.25.
If r denotes the correlation co-efficient between two variables, which of the following is always true?
Answer: D (−1 ≤ r ≤ 1)
A correlation coefficient can never be below −1 or above 1, and it can take either end value, so −1 ≤ r ≤ 1.
A stone is dropped from a height of 45m. Find the time it takes to hit ground. [Take g = 10ms−2]
Answer: A (3.0seconds)
Dropped means u = 0, so 45 = ½ × 10 × t2, t2 = 9 and t = 3.0 s.
two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x-component of their resultant.
Answer: A (5√3 − 3)
x-components from bearings: 10 sin 60° = 5√3 and 6 sin 330° = −3. Their sum is 5√3 − 3.
In the diagram, g : x → y and h : y → z. Use the diagram to answer questions 25 and 26. Find h[g(3)].
Answer: B (β)
From the diagram g(3) = s, and h(s) = β.
In the diagram, g : x → y and h : y → z. Use the diagram to answer questions 25 and 26. g o h is
Answer: B (one onto)
Following g then h, every element of Z is reached (onto), but 2 and 3 both end at β, so it is not one-to-one.
Given that nPr = 90 and nCr = 15, find the value of r.
Answer: B (3)
nPr = r! × nCr, so r! = 90 ÷ 15 = 6 and r = 3.
Which one of the following options is not measure of Central Tendency
Answer: B (Variance)
Mode, mean and median are averages; variance measures spread (dispersion).
A fair die tossed twice. Find the probability of obtaining a 3 and a5
Answer: B (1/18)
There are 36 equally likely outcomes, and two of them are (3, 5) and (5, 3). 2/36 = 1/18.
Find the values of x at the point of intersection of the curve y = x2 + 2x − 3 and the line y + x = 1
Answer: D ((1, −4))
Substitute y = 1 − x: x2 + 2x − 3 = 1 − x, so x2 + 3x − 4 = (x − 1)(x + 4) = 0 and x = 1 or −4.
if P(x − 3) + Q(x + 1) = 2x + 3, find the value of (P + Q)
Answer: C (2)
Compare the coefficients of x: P + Q = 2.
Find the constant term in the binomial expansion of (2x − 3/x)8.
Answer: A (90720)
The general term is 8Cr(2x)8−r(−3/x)r; the constant term has r = 4: 70 × 16 × 81 = 90720.
The diagram above is a velocity time graph of a moving object. Calculate the distance travelled when the acceleration is zero.
Answer: C (150m)
Acceleration is zero on the flat part, from t = 3 to t = 8 s at 30 ms−1. Distance = 30 × 5 = 150 m.
A straight line makes intercepts of −3 and 2 on the x and y axes respectively. Find the equation of the line.
Answer: D (−2x + 3y − 6 = 0)
Intercept form: x/(−3) + y/2 = 1. Multiply by 6: −2x + 3y = 6, i.e. −2x + 3y − 6 = 0.
Find the number of different arrangements of the letters of the word IKOTITINA.
Answer: A (30240)
9 letters with I three times and T twice: 9!/(3! × 2!) = 362880/12 = 30240.
Find the acute angle between the line 2x + y = 4 and −3x + y + 7 = 0
Answer: C (45°)
Gradients are −2 and 3. tan θ = |(3 − (−2))/(1 + (−2)(3))| = |5/(−5)| = 1, so θ = 45°.
A box contains 4 red and 3 blue identical balls. If two balls are picked at random, one after the other without replacement, find the probability that one is red and the other is blue.
Answer: A (4/7)
P(red then blue) + P(blue then red) = (4/7)(3/6) + (3/7)(4/6) = 2/7 + 2/7 = 4/7.
The distance s in metre covered by a particle in t seconds is s = 3/2 t2 − 3t. Find its acceleration.
Answer: C (3ms−2)
v = ds/dt = 3t − 3 and a = dv/dt = 3 ms−2.
The angle of a sector of circle is 0.9 radians. If the radius of the circle is 4cm, find the length of the arc of the sector.
Answer: A (3.6cm)
Arc length = rθ = 4 × 0.9 = 3.6 cm.
Two functions g and h are defined on the set R of real numbers by g : x → x2 − 2 and h : x → 1/(x + 2), x ≠ −2. Find: (a) h−1, the inverse of h; (b) g∘h when x = −½
(a) Let y = 1/(x + 2); then x + 2 = 1/y, x = 1/y − 2
h−1(x) = 1/x − 2 = (1 − 2x)/x, x ≠ 0
(b) h(−½) = 1/(3/2) = 2/3
g(2/3) = 4/9 − 2 = −14/9
For g∘h apply h first, then g.
Express 3x2 − 6x + 10 in the form a(x − b)2 + c where a, b and c are integers. Hence state the minimum value of 3x2 − 6x + 10 and the value of x for which it occurs.
3x2 − 6x + 10 = 3(x2 − 2x) + 10 = 3(x − 1)2 − 3 + 10
= 3(x − 1)2 + 7, so a = 3, b = 1, c = 7
Since 3(x − 1)2 ≥ 0, the minimum value is 7, when x = 1
Complete the square after taking out the factor 3.
The twenty-first term of an Arithmetic progression (AP) is 5½ and the sum of the first twenty-one terms is 94½. Find the: (a) first term; (b) common difference; (c) sum of the first thirty terms.
a + 20d = 5.5
S21 = (21/2)(2a + 20d) = 94.5, so a + 10d = 4.5
Subtract: 10d = 1
(a) a = 3.5
(b) d = 0.1
(c) S30 = 15(2 × 3.5 + 29 × 0.1) = 15 × 9.9 = 148.5
S₂₁ = (21/2)(2a + 20d) simplifies to 21(a + 10d).
Write down the first three terms of the binomial expansion (1 + ax)n in ascending powers of x. If the coefficients of x and x2 are 2 and 3/2 respectively, find the various values of a and n.
(1 + ax)n = 1 + nax + [n(n − 1)/2]a2x2 + ...
na = 2 and n(n − 1)a2/2 = 3/2, i.e. n(n − 1)a2 = 3
a = 2/n: n(n − 1)(4/n2) = 3, so 4(n − 1) = 3n
n = 4, a = ½
Substitute a = 2/n into the second equation.
The gradient function of y = ax2 + bx + c is 8x + 4. If the function has a minimum value of 1, find the values of a, b and c.
dy/dx = 2ax + b = 8x + 4, so a = 4, b = 4
Minimum where 8x + 4 = 0: x = −½
y(−½) = 4(¼) + 4(−½) + c = c − 1 = 1
c = 2
Compare coefficients, then use the minimum point where dy/dx = 0.
Three forces −63j, (32.14i + 38.3j) and (14i − 24.25j) act on a body of mass 5kg. Find, correct to one decimal place, the:
(a) magnitude of the resultant force;
(b) acceleration of the body.
(a) R = (32.14 + 14)i + (38.3 − 63 − 24.25)j = 46.14i − 48.95j
|R| = √(46.142 + 48.952) = √4525.0 ≈ 67.3 N
(b) a = |R|/m = 67.27/5 ≈ 13.5 ms−2
Add the i parts and the j parts separately.
Simplify n+1C4 − n−1C4
n+1C4 = (n + 1)n(n − 1)(n − 2)/24, n−1C4 = (n − 1)(n − 2)(n − 3)(n − 4)/24
Difference = (n − 1)(n − 2)[n(n + 1) − (n − 3)(n − 4)]/24
= (n − 1)(n − 2)(8n − 12)/24
= (n − 1)(n − 2)(2n − 3)/6 = (2n3 − 9n2 + 13n − 6)/6
Take out the common factor (n − 1)(n − 2) before expanding.
the marks scored by 35 students in a test are given in the table below:
Marks: 1−5, 6−10, 11−20, 21−30, 31−35, 36−40
No. of students: 2, 7, 12, 8, 5, 1
Draw a histogram for the distribution
Class boundaries 0.5, 5.5, 10.5, 20.5, 30.5, 35.5, 40.5; the class widths are unequal (5, 5, 10, 10, 5, 5)
Bar heights must be frequency densities, e.g. per 5 marks: 2, 7, 6, 4, 5, 1 (or frequency ÷ width: 0.4, 1.4, 1.2, 0.8, 1.0, 0.2)
Draw touching bars over the boundaries with these heights, with a stated scale
With unequal class widths the area, not the height, of each bar shows the frequency.
(a) The polynomial f(x) = x3 + px2 − 10x + q is exactly divisible by (x2 + x − 6). Find the: (i) values of p and q; (ii) third factor.
(b) The volume of a cube is increasing at the rate of 2½cm3s−1. Find the rate of change of the side of the base when its length is 2cm.
(a)(i) x2 + x − 6 = (x + 3)(x − 2), so f(2) = 0 and f(−3) = 0
f(2) = 8 + 4p − 20 + q = 0: 4p + q = 12
f(−3) = −27 + 9p + 30 + q = 0: 9p + q = −3
5p = −15: p = −3, q = 24
(ii) x3 − 3x2 − 10x + 24 = (x2 + x − 6)(x − 4); third factor (x − 4)
(b) V = s3, dV/dt = 3s2(ds/dt)
2.5 = 3(2)2(ds/dt), ds/dt = 2.5/12 = 5/24 ≈ 0.21 cm s−1
Each factor of the divisor gives a zero of f(x).
(a) Write down the matrix A of the linear transformation A(x, y) → (2x − y, −5x + 3y).
(b) If B = (3 1 / 5 2), find (i) A2 − B2; (ii) matrix C = B2A (iii) the point M(x, y) whose image under the linear transformation C is M(10, 18).
(c) What is the relationship between matrix A and matrix C? (Matrices are written row by row.)
(a) A = (2 −1 / −5 3)
(b)(i) A2 = (9 −5 / −25 14), B2 = (14 5 / 25 9)
A2 − B2 = (−5 −10 / −50 5)
(ii) C = B2A = (14 5 / 25 9)(2 −1 / −5 3) = (3 1 / 5 2)
(iii) 3x + y = 10 and 5x + 2y = 18 give x = 2, y = 4; M(2, 4)
(c) AC = (2 −1 / −5 3)(3 1 / 5 2) = (1 0 / 0 1) = I, so C is the inverse of A (C = A−1, which here equals B)
Multiply row by column carefully; check (c) by working out AC.
(a) Evaluate ∫12 x(3x − 2)/(2√x) dx
(b) The equation of a circle is given by 2x2 + 2y2 − 8x + 5y − 10 = 0. Find the: (i) coordinates of the centre; (ii) radius of the circle; (iii) coordinates of P and Q, if the circle cuts the x-axis at the points P and Q.
(a) x(3x − 2)/(2√x) = (3/2)x3/2 − x1/2
∫ = [(3/5)x5/2 − (2/3)x3/2] from 1 to 2
= (12√2/5 − 4√2/3) − (3/5 − 2/3) = 16√2/15 + 1/15 = (16√2 + 1)/15 ≈ 1.575
(b) Divide by 2: x2 + y2 − 4x + (5/2)y − 5 = 0
(i) Centre (2, −5/4)
(ii) r2 = 4 + 25/16 + 5 = 169/16, r = 13/4
(iii) y = 0: x2 − 4x − 5 = 0, (x − 5)(x + 1) = 0; P(−1, 0) and Q(5, 0)
Split the fraction into powers of x before integrating; the limits are 1 and 2.
(a) (i) Find the sum of the series A(1 + r) + A(1 + r)2 + ... + A(1 + r)n.
(ii) Given that r = 8% and A = GH¢40.00, find the sum of the 6th to 10th terms of the series in 12a(i).
(b) Find the equation of the tangent to the curve y = 1/x at the point on the curve, when x = 2.
(a)(i) G.P. with first term A(1 + r) and common ratio (1 + r)
Sn = A(1 + r)[(1 + r)n − 1]/r
(ii) Sum of T6 to T10 = S10 − S5 = 40(1.08)[(1.08)10 − (1.08)5]/0.08
= 540(2.15892 − 1.46933) ≈ GH¢372.38
(b) y = 1/2 at x = 2; dy/dx = −1/x2 = −1/4
y − ½ = −¼(x − 2)
x + 4y − 4 = 0
The common ratio is 1 + r, so the sum formula divides by (1 + r) − 1 = r.
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining (i) exactly three sixes; (ii) at most three sixes.
(b) Eight percent of screw produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that: (i) exactly two will be defective; (ii) not more than two will be defective.
(a) Binomial n = 6, p = 1/6
(i) 6C3(1/6)3(5/6)3 = 2500/46656 ≈ 0.054
(ii) 1 − [P(4) + P(5) + P(6)] = 1 − (375 + 30 + 1)/46656 = 46250/46656 ≈ 0.991
(b) Binomial n = 10, p = 0.08
(i) 10C2(0.08)2(0.92)8 ≈ 0.148
(ii) P(0) + P(1) + P(2) = 0.4344 + 0.3777 + 0.1478 ≈ 0.960
For 'at most three' it is quicker to subtract P(4), P(5) and P(6) from 1.
The table gives the distribution of heights in metres of 100 students.
Height: 1.40 − 1.42, 1.43 − 1.45, 1.46 − 1.48, 1.49 − 1.51, 1.52 − 1.54, 1.55 − 1.57, 1.58 − 1.60, 1.61 − 1.63
Freq.: 2, 4, 19, 30, 24, 14, 6, 1
(a) Calculate the (i) mean height (ii) mean deviation of the distribution
(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?
Class marks 1.41, 1.44, 1.47, 1.50, 1.53, 1.56, 1.59, 1.62
(a)(i) Σfx = 151.23, mean = 151.23/100 ≈ 1.51 m (1.5123)
(ii) Σf|x − x̄| ≈ 3.333, mean deviation ≈ 3.333/100 ≈ 0.033 m
(b) The classes above the mean are 1.52 − 1.54 upwards: 24 + 14 + 6 + 1 = 45 students
Probability = 45/100 = 0.45
Use the class mark (middle value) of each class for x.
(a) Two items are selected at random from four items labeled (p, q, r, s) (i) list the sample space if sampling is done (α) with replacement; (β) without replacement. (ii) Find the probability that r is at least one of the two objects selected: (α) in a(i)
(b) How many whole numbers from 100 to 999 are divisible by: (i) 4; (ii) both 3 and 4
(a)(i)(α) With replacement, 16 ordered pairs: pp, pq, pr, ps, qp, qq, qr, qs, rp, rq, rr, rs, sp, sq, sr, ss
(β) Without replacement, 12 ordered pairs: pq, pr, ps, qp, qr, qs, rp, rq, rs, sp, sq, sr
(ii) With replacement: pairs containing r are pr, qr, rp, rq, rr, rs, sr = 7, probability 7/16
Without replacement: pr, qr, rp, rq, rs, sr = 6 of 12, probability ½
(b)(i) 100, 104, ..., 996: 100 + 4(n − 1) = 996, n = 225
(ii) Divisible by both 3 and 4 means divisible by 12: 108, 120, ..., 996; 108 + 12(n − 1) = 996, n = 75
Divisible by both 3 and 4 is the same as divisible by 12.
(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° respectively to the downward vertical, find the tension in each of the strings. [Take g = 10ms−1]
(b) A constant force F acts on the toy car of mass 5kg and increases its velocity from 5ms−1 to 9ms−1 in 2 seconds. Calculate the: (i) magnitude of the force; (ii) velocity of the toy car 3 seconds after attaining a velocity of 9ms−1.
(a) W = 50 N; the angle between the strings is 70°
Horizontal: TAP sin 40° = TBP sin 30°; vertical: TAP cos 40° + TBP cos 30° = 50
(Lami) TAP = 50 sin 30°/sin 70° ≈ 26.6 N
TBP = 50 sin 40°/sin 70° ≈ 34.2 N
(b)(i) a = (9 − 5)/2 = 2 ms−2, F = ma = 5 × 2 = 10 N
(ii) v = 9 + 2 × 3 = 15 ms−1
The string nearer the vertical carries the larger tension.
(a) given that P = (4i − 3j) and q = (−i + 5j), find r such that |r| = 15 and is in the direction of (2p + 3q)
(b) Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the diagram. Find the: (i) magnitude; (ii) direction of the resultant force.
(a) 2p + 3q = (8i − 6j) + (−3i + 15j) = 5i + 9j, |5i + 9j| = √106
r = 15(5i + 9j)/√106 ≈ 7.28i + 13.11j
(b) Take i to the right and j along the 6N force; 8N is 30° left of it and 4N is 60° right of it
Σx = −8 sin 30° + 4 sin 60° = −4 + 2√3 ≈ −0.536
Σy = 6 + 8 cos 30° + 4 cos 60° = 8 + 4√3 ≈ 14.928
(i) R = √(0.5362 + 14.9282) ≈ 14.9 N
(ii) tan α = 0.536/14.928, α ≈ 2.1° to the left of the 6N force (about 92.1° to the horizontal line through P, measured from the right)
Resolve each force along and across the 6N direction.
(a) Find the angle between the vectors a = (−3 / 4) and b = (−8 / −15). (Column vectors written top / bottom.)
(b) Given that a = (4N, 060°) and b = (3N, 120°), Find in component form, the unit vector along a − b.
(a) a · b = 24 − 60 = −36; |a| = 5, |b| = 17
cos θ = −36/85 ≈ −0.4235, θ ≈ 115.1°
(b) Using bearings, (F, θ) = F sin θ i + F cos θ j
a = 2√3i + 2j, b = (3√3/2)i − (3/2)j
a − b = (√3/2)i + (7/2)j, |a − b| = √(3/4 + 49/4) = √13
Unit vector = (√3i + 7j)/(2√13) ≈ 0.240i + 0.971j
Find a − b first, then divide it by its own length.