53 questions. Try each one, then open the answer to check yourself and read the explanation.
A binary operation * is defined on the set of real numbers, R, by x * y = x2 − y2 + xy, where x, y ∈ R. Evaluate (√3) * (√2)
Answer: D (1 + √6)
(√3)² − (√2)² + √3 × √2 = 3 − 2 + √6 = 1 + √6.
If cos x = −0.7133, find the values of x between 0° and 360°.
Answer: D (135.5°, 224.5°)
The reference angle is cos⁻¹ 0.7133 ≈ 44.5°. Cosine is negative in the 2nd and 3rd quadrants: 180° − 44.5° = 135.5° and 180° + 44.5° = 224.5°.
If ∫03 (px2 + 16) dx = 129, find the value of p.
Answer: A (9)
[px³/3 + 16x] from 0 to 3 = 9p + 48 = 129, so p = 9.
Given that x: R → R is defined by x = (y + 1)/(5 − y), y ∈ R, find the domain of x.
Answer: C ({y : y ∈ R, y ≠ 5})
The expression is undefined only when 5 − y = 0, so y ≠ 5.
Simplify: (√5 + 3)/(4 − √10)
Answer: C (⅔√5 + ⅚√2 + ½√10 + 2)
Multiply by (4 + √10)/(4 + √10): (4√5 + √50 + 12 + 3√10)/6 = (4√5 + 5√2 + 12 + 3√10)/6 = ⅔√5 + ⅚√2 + ½√10 + 2.
If (6x + k)/(2x2 + 7x − 15) ≡ 4/(x + 5) − 2/(2x − 3) find the value of k.
Answer: B (−22)
6x + k = 4(2x − 3) − 2(x + 5) = 6x − 22, so k = −22.
Differentiate x/(x + 1) with respect to x.
Answer: D (1/(x + 1)2)
Quotient rule: [(x + 1)(1) − x(1)]/(x + 1)² = 1/(x + 1)².
Given that 2x + 3y − 10 = 0 and 3x = 2y − 11, calculate the value of (x − y).
Answer: D (−5)
Solving 2x + 3y = 10 and 3x − 2y = −11 gives x = −1, y = 4, so x − y = −5.
Given that f(x) = 2x and g(x) = 3x2, find the values of x for which f−1(x) = g(x).
Answer: A (x = 0 or x = 1/6)
f⁻¹(x) = x/2. x/2 = 3x² gives x(6x − 1) = 0, so x = 0 or 1/6.
If V = P logx (M + N), express N in terms of X, P, M and V.
Answer: A (N = XV/P − M)
log_x(M + N) = V/P, so M + N = x^(V/P) and N = x^(V/P) − M.
Determine the coefficient of x3 in the binomial expansion of (1 + ½x)5
Answer: C (5/4)
⁵C₃(½x)³ = 10 × x³/8, so the coefficient is 10/8 = 5/4.
Given that P = {x : 1 ≤ x ≤ 6} and Q = {x : 2 < x < 10} where x are integers, find n(P ∩ Q)
Answer: A (4)
P = {1, …, 6}, Q = {3, …, 9}. P ∩ Q = {3, 4, 5, 6}, so n = 4.
If sin X = 3/5 and cos Y = 24/25 where X and Y are acute, Find the value of cos (X + Y).
Answer: C (3/5)
cos X = 4/5 and sin Y = 7/25. cos(X + Y) = (4/5)(24/25) − (3/5)(7/25) = 96/125 − 21/125 = 75/125 = 3/5.
Find the median of the numbers 9, 7, 5, 2, 12, 9, 9, 2, 10, 10 and 18.
Answer: B (9)
In order: 2, 2, 5, 7, 9, 9, 9, 10, 10, 12, 18. The middle (6th) number is 9.
Calculate the probability that the product of two numbers selected at random with replacement from the set {−5, −2, 4, 8} is positive.
Answer: B (1/2)
The product is positive when both numbers have the same sign: 2 × 2 + 2 × 2 = 8 of the 16 pairs. P = 1/2.
Find the angle between i + 5j and 5i − j.
Answer: D (90°)
The dot product is 5 − 5 = 0, so the vectors are perpendicular: 90°.
Given that F = 3i − 12j, R = 7i + 5j and N = pi + qj are forces acting on a body, if the body is in equilibrium, find the values of p and q.
Answer: A (p = −10, q = 7)
In equilibrium F + R + N = 0, so N = −(10i − 7j) = −10i + 7j: p = −10, q = 7.
A stone was dropped from the top of a building 40m high. Find, correct to one decimal place, the time it took the stone to reach the ground. [Take g = 9.8ms−2]
Answer: A (2.9 seconds)
40 = ½ × 9.8 × t², so t² = 8.16 and t ≈ 2.86 ≈ 2.9 s.
In which of the following series can be the formula S = a/(1 − r) where a is the first term and r is the common ratio, be used to find the sum of all the terms?
Answer: D (128 + 64 + 32 + 16 + …)
The sum to infinity exists only when −1 < r < 1. Only D has such a ratio (r = ½); the others have r = 2, 5 and 3/2.
If the binomial expansion of (1 + 3x)6 is used to evaluate (0.97)6, find the value of x.
Answer: C (−0.01)
1 + 3x = 0.97, so 3x = −0.03 and x = −0.01.
Find the nth term of the linear sequence (A.P) (5y + 1), (2y + 1), (1 − y), …
Answer: C ((8 − 3n)y + 1)
d = (2y + 1) − (5y + 1) = −3y. Tₙ = 5y + 1 + (n − 1)(−3y) = (8 − 3n)y + 1.
A circle with centre (5, −4) passes through the point (5, 0). Find its equation.
Answer: C (x2 + y2 − 10x + 8y + 25 = 0)
The radius is the distance from (5, −4) to (5, 0) = 4. (x − 5)² + (y + 4)² = 16 gives x² + y² − 10x + 8y + 25 = 0.
Calculate, correct to two decimal places, the area enclosed by the line 3x − 5y + 4 = 0 and the axes.
Answer: C (0.53 square units)
The line meets the axes at x = −4/3 and y = 4/5. Area = ½ × 4/3 × 4/5 = 8/15 ≈ 0.53 square units.
In how many ways can the letters of the word MEMBER be arranged?
Answer: C (180)
6 letters with M twice and E twice: 6!/(2! 2!) = 720/4 = 180.
Which of the following is not an equation of a circle?
Answer: D (x2 − y2 + 3x − 5y = 2)
A circle needs equal (and same-sign) coefficients of x² and y². D has x² − y², so it is not a circle.
A function f defined by f: x → x2 + px + q is such that f(3) = 6 and f′(3) = 0. Find the value of q.
Answer: C (15)
f′(x) = 2x + p, so f′(3) = 0 gives p = −6. Then f(3) = 9 − 18 + q = 6, so q = 15.
In what interval is the function f: x → 2x − x2 increasing?
Answer: B (x < 1)
f′(x) = 2 − 2x is positive when x < 1, so f is increasing for x < 1.
A force of 230N acts in its direction 065°. Find the its horizontal component.
Answer: A (208.5N)
065° is a bearing, measured from north. The horizontal (east) component is 230 sin 65° ≈ 208.5 N.
Calculate the variance of √2, (1 + √2) and (2 + √2)
Answer: C (2/3)
The mean is 1 + √2; deviations −1, 0, 1. Variance = (1 + 0 + 1)/3 = 2/3.
A three − digit odd number less than 500 is to be formed from 1, 2, 3, 4 and 5. If repetition of digits is allowed, in how many ways can this be done?
Answer: C (60)
First digit 1–4 (4 ways, to stay below 500), middle digit any of 5, last digit 1, 3 or 5 (3 ways): 4 × 5 × 3 = 60.
The variables x and y are such that y = 2x3 − 2x2 − 5x + 5. Calculate the corresponding change in y and x changes from 2.00 to 2.05.
Answer: A (0.58)
y(2) = 3 and y(2.05) ≈ 3.575, so the change is about 0.58.
A bag contains 5 red and 5 blue identical balls. Three balls are selected at random without replacement. Determine the probability of selecting balls alternating in color.
Answer: B (5/18)
RBR: (5/10)(5/9)(4/8) = 100/720; BRB is the same. Total 200/720 = 5/18.
The distance(s) in metres covered by a particle in motion at any time, t seconds, is given by S = 120t − 16t2. Find in metres, the distance covered by the body before coming to rest.
Answer: D (225)
v = 120 − 32t = 0 at t = 3.75 s. S = 450 − 225 = 225 m.
P(3, 4) and Q(−3, −4) are two points in a plane. Find the gradient of the line that is normal to the line PQ.
Answer: C (−3/4)
PQ has gradient 8/6 = 4/3, so the normal has gradient −3/4.
Find the unit vector in the direction opposite to the resultant of forces F1 = (−2i − 3j) and F2 = (5i − j).
Answer: B (⅕(−3i + 4j))
The resultant is 3i − 4j, of length 5. The opposite direction is −3i + 4j, so the unit vector is ⅕(−3i + 4j).
If the sum of the roots of 2x2 + 5mx + n = 0 is 5, find the value of m.
Answer: B (−2.0)
Sum of roots = −5m/2 = 5, so m = −2.
If log5(125x3/∛y) is expressed in the form p log5 x + q log5 y + k, find the values of p, q and k respectively.
Answer: D (3, −⅓, 3)
log₅ 125 + 3log₅ x − ⅓log₅ y = 3 + 3log₅ x − ⅓log₅ y. So p = 3, q = −⅓, k = 3.
Consider the statements:
x: Birds fly
y: The sky is blue
Which of the following statements can be represented as x ⇔ y?
Answer: B (Birds fly if and only if the sky is blue.)
⇔ means 'if and only if', so x ⇔ y is 'Birds fly if and only if the sky is blue'.
A binary operation * is defined on the set of real numbers, R, by p * q = p + q − pq/2, where p, q ∈ R. Find the: (a) inverse of −1 under * given that the identity element is zero. (b) truth set of m * 7 = m * 5.
(a) p * p⁻¹ = 0: p + p⁻¹ − pp⁻¹/2 = 0, so p⁻¹(1 − p/2) = −p and p⁻¹ = −2p/(2 − p).
For p = −1: p⁻¹ = 2/3 (check: −1 + 2/3 + 1/3 = 0).
(b) m + 7 − 7m/2 = m + 5 − 5m/2, so 2 = m. Truth set {2}.
Check the inverse by substituting back.
(a) Two functions p and q are defined on the set of real numbers, R, by p: y → 2y + 3 and q: y → y2 − 2. Find q o p
(b) How many four digits odd numbers greater than 4000 can be formed from 1, 7, 3, 8, 2 if repetition is allowed?
(a) q o p(y) = q(2y + 3) = (2y + 3)² − 2 = 4y² + 12y + 7.
(b) First digit 7 or 8 (2 ways, to exceed 4000); second and third digits any of 5; last digit 1, 7 or 3 (3 ways).
2 × 5 × 5 × 3 = 150.
Fill the restricted places (first and last) first.
Evaluate: ∫13 (3x − 2)5 dx
Let u = 3x − 2, du = 3 dx; u runs from 1 to 7.
∫ = (1/3)[u⁶/6] from 1 to 7 = (7⁶ − 1)/18 = 117 648/18 = 6536.
Change the limits when you substitute.
If (3x2 + 3x − 2)/((x − 1)(x + 1)) ≡ P + Q/(x − 1) + R/(x + 1), find the values of Q and R.
3x² + 3x − 2 = 3(x² − 1) + 3x + 1, so P = 3 and (3x + 1)/((x − 1)(x + 1)) = Q/(x − 1) + R/(x + 1).
3x + 1 = Q(x + 1) + R(x − 1). x = 1: 4 = 2Q, Q = 2. x = −1: −2 = −2R, R = 1.
Divide first, because the top and bottom have the same degree.
The table shows the distribution of marks scored by 64 students in a test.
Marks: 10−19, 20−29, 30−39, 40−49, 50−59, 60−69, 70−79, 80−89, 90−99
Frequency: 2, 2, 2, 8, 13, 11, 12, 10, 4
(a) Draw a histogram for the distribution. (b) Use the histogram to estimate the modal score.
(a) Class boundaries 9.5, 19.5, …, 99.5 (equal widths of 10); bar heights are the frequencies. Label axes and give a scale.
(b) Tallest bar 49.5–59.5 (13). Join its top corners to the tops of the neighbouring bars (8 and 11) and read where the lines cross: mode ≈ 49.5 + 10 × 5/(5 + 2) ≈ 56.6, about 57.
The crossing lines on the tallest bar locate the mode.
(a) A bag contains 10 red and 8 green identical balls. Two balls are drawn at random from the bag, one after the other, without replacement. Find the probablity that one is red and the other is green.
(b) There are 20% defective bulbs in a large box. If 12 bulbs are selected randomly from the box, calculate the probablity that between two and five are defective.
(a) RG or GR: 2 × (10/18)(8/17) = 160/306 = 80/153.
(b) Binomial n = 12, p = 0.2. 'Between two and five' (3 or 4): ¹²C₃(0.2)³(0.8)⁹ + ¹²C₄(0.2)⁴(0.8)⁸ = 0.2362 + 0.1329 ≈ 0.3691.
(If 2 and 5 are included: P(2) + … + P(5) ≈ 0.7057.)
Two orders (red first or green first) both count.
Forces F1 (10N, 090°), F2 (20N, 210°) and F3 (4N, 330°) act on a particle. Find, correct to one decimal place, the magnitude of the resultant force.
Components (east, north) = (F sin θ, F cos θ):
F₁ = (10, 0); F₂ = (−10, −17.32); F₃ = (−2, 3.46).
Resultant = (−2, −13.86); magnitude = √(4 + 192) = √196 = 14.0 N.
Resolve each bearing into east and north parts, then add.
Given that w = 8i + 3j, x = 6i − 5j, y = 2i + 3j and |z| = 41, find z in the direction of w + x − 2y.
w + x − 2y = (8 + 6 − 4)i + (3 − 5 − 6)j = 10i − 8j; its length is √164 = 2√41.
Unit vector = (5i − 4j)/√41.
z = 41 × (5i − 4j)/√41 = √41(5i − 4j) = 5√41 i − 4√41 j.
(If |z| = √41 is meant, z = 5i − 4j.)
A vector of given length in a given direction = length × unit vector.
(a) If (x + 2) is a factor of g(x) = 2x3 + 11x2 − x − 30, find the zeros of g(x).
(b) Solve 3(2x) + 3y−2 = 25 and 2x − 3y+1 = −19 simultaneously.
(a) g(x) = (x + 2)(2x² + 7x − 15) = (x + 2)(2x − 3)(x + 5). Zeros: −2, 3/2, −5.
(b) Let a = 2ˣ, b = 3ʸ: 3a + b/9 = 25 and a − 3b = −19.
a = 3b − 19, so 9b − 57 + b/9 = 25, 82b/9 = 82, b = 9 and a = 8.
x = 3, y = 2.
Substitute a = 2ˣ and b = 3ʸ to get ordinary simultaneous equations.
(a) Find the derivative of y = x2(1 + x)3/2 with respect to x.
(b) The centre of a circle lies on the line 2y − x = 3. If the circle passes through P(2, 3) and Q(6, 7), find its equation.
(a) Product rule: dy/dx = 2x(1 + x)^(3/2) + (3/2)x²(1 + x)^(1/2) = ½x(1 + x)^(1/2)(4 + 7x).
(b) The centre is on the perpendicular bisector of PQ: midpoint (4, 5), gradient −1, so x + y = 9.
With 2y − x = 3: y = 4, x = 5. Centre (5, 4); r² = (5 − 2)² + (4 − 3)² = 10.
(x − 5)² + (y − 4)² = 10, i.e. x² + y² − 10x − 8y + 31 = 0.
The centre is the same distance from P and Q, so it lies on their perpendicular bisector.
(a) Simplify: (log2 8 + log2 16 − 4 log2 2)/log4 16
(b) The first, third and seventh terms of an Arithmetic Progression (A.P) from three consecutive terms of a Geometric Progression (G.P). If the sum of the first two terms of the A.P is 6, find its: (i) first term; (ii) common difference.
(a) (3 + 4 − 4)/2 = 3/2.
(b) (a + 2d)² = a(a + 6d) gives 4d² = 2ad, so a = 2d (d ≠ 0).
a + (a + d) = 6 gives 5d = 6.
(i) a = 12/5; (ii) d = 6/5. (Check: 12/5, 24/5, 48/5 form a G.P. with ratio 2.)
For three G.P. terms, (middle)² = first × last.
(a) An association is made up of 6 farmers and 8 traders. If an executive body of 4 members is to be formed, find the probability that it will consist of at least two farmers.
(b) The probability of an accident occuring in a given month in factories X, Y, and Z are 1/5, 1/12 and 1/6 respectively. Find the probablity that the accident will occur in: (i) none of the factories; (ii) all the factories; (iii) at least one factory.
(a) ¹⁴C₄ = 1001. 2 farmers: 15 × 28 = 420; 3: 20 × 8 = 160; 4: 15. P = 595/1001 = 85/143 ≈ 0.594.
(b)(i) (4/5)(11/12)(5/6) = 11/18.
(ii) (1/5)(1/12)(1/6) = 1/360.
(iii) 1 − 11/18 = 7/18.
'At least one' = 1 − P(none).
The essays of 10 candidates were ranked by three examiners as shown in the table.
Candidate: A, B, C, D, E, F, G, H, I, J
Examiner I: 1st, 3rd, 6th, 2nd, 10th, 9th, 7th, 4th, 8th, 5th
Examiner II: 2nd, 1st, 3rd, 9th, 7th, 4th, 8th, 10th, 5th, 6th
Examiner III: 3rd, 2nd, 1st, 6th, 9th, 8th, 7th, 5th, 4th, 10th
(a) Calculate the Spearman's rank correlation coefficient of the ranks assigned by: (i) Examiners I and II; (ii) Examiners I and III (iii) Examiners II and III. (b) Using the results in 13(a), state which two examiners agree most.
ρ = 1 − 6Σd²/(n(n² − 1)), n = 10, so n(n² − 1) = 990.
(a)(i) I and II: Σd² = 144, ρ = 1 − 864/990 ≈ 0.13.
(ii) I and III: Σd² = 90, ρ = 1 − 540/990 ≈ 0.45.
(iii) II and III: Σd² = 78, ρ = 1 − 468/990 ≈ 0.53.
(b) Examiners II and III agree most (largest coefficient).
The bigger the coefficient, the closer the agreement.
(a) A car is moving with a velocity of 10ms−1. It then accelerates at 0.2ms−2 for 100m. Find, correct to two decimal places the time taken by the car to cover the distance.
(b) A particle moves along a straight line such that its distance S metres from a fixed point O is given by S = t2 − 5t + 6, where t is the time in seconds. Find its: (i) initial velocity; (ii) distance when it is momentarily at rest
(a) 100 = 10t + 0.1t², so t² + 100t − 1000 = 0 and t = (−100 + √14 000)/2 ≈ 9.16 s.
(b)(i) v = 2t − 5; at t = 0, v = −5 ms⁻¹ (5 ms⁻¹ towards O).
(ii) v = 0 at t = 2.5 s: S = 6.25 − 12.5 + 6 = −0.25, i.e. 0.25 m from O on the other side.
Momentarily at rest means v = 0.
The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal level. A body of mass 800g is hung on the string at a point O, 2.4m from Y. If the system is kept in equilibrium by a horizontal force P acting on the body and the tensions are equal, calculate: (a) ∠XOY; (b) the magnitude of the force P; (c) the tension T in the string.
OX = 1.8 m, OY = 2.4 m, XY = 3 m.
(a) 1.8² + 2.4² = 9 = 3², so ∠XOY = 90°.
OX makes 53.1° with the horizontal (cos = 0.6) and OY makes 36.9° (cos = 0.8). Weight = 0.8 × 10 = 8 N.
(c) Vertically: T sin 53.1° + T sin 36.9° = 8, so 1.4T = 8 and T ≈ 5.71 N.
(b) Horizontally: P = T cos 36.9° − T cos 53.1° = 0.2T ≈ 1.14 N.
Resolve the forces at O horizontally and vertically; take g = 10 ms⁻².