Question 1
Indices and LogarithmsIf logy (1/8) = 3, find the value of y.
- A.−2
- B.−½
- C.½
- D.2
Show the answer and explanation
Answer: C (½)
y³ = 1/8, so y = ½.
52 questions. Try each one, then open the answer to check yourself and read the explanation.
If logy (1/8) = 3, find the value of y.
Answer: C (½)
y³ = 1/8, so y = ½.
If x2 − kx + 9 = 0 has equal roots, find the values of k.
Answer: D (±6)
Equal roots: k² − 36 = 0, so k = ±6.
Find the coordinates of the centre of the circle 3x2 + 3y2 − 4x + 8y − 2 = 0.
Answer: C ((2/3, −4/3))
Divide by 3 first: x² + y² − (4/3)x + (8/3)y − 2/3 = 0. The centre is (2/3, −4/3).
The function f: x → √(4 − 2x) is defined on the set of real numbers R. Find the domain of f.
Answer: B (x ≤ 2)
We need 4 − 2x ≥ 0, i.e. x ≤ 2.
Given that f(x) = (x + 1)/2, find f−1(−2).
Answer: A (−5)
f⁻¹(x) = 2x − 1, so f⁻¹(−2) = −5.
Given that (6x + m)/(2x2 + 7x − 15) = 4/(x + 5) − 2/(2x − 3), find the value of m.
Answer: D (−22)
6x + m = 4(2x − 3) − 2(x + 5) = 6x − 22, so m = −22.
Find the coefficent of x4 in the expansion of (1 − 2x)6.
Answer: C (240)
⁶C₄(−2x)⁴ = 15 × 16x⁴ = 240x⁴.
Find the 21st term of the Arithmetic Progression (AP): −4, −1.5, 1, 3.5, …
Answer: B (46)
a = −4, d = 2.5. T₂₁ = −4 + 20(2.5) = 46.
How many ways can six students be seated around a circular table?
Answer: D (120)
Around a circle: (6 − 1)! = 120.
If (2 1 / 4 3)(5 / 4) = k(17.5 / 40.0), find the value of k. (Matrices written row by row; / separates the rows.)
Answer: C (0.80)
The product is (14 / 32), and (14 / 32) = 0.8 × (17.5 / 40), so k = 0.80.
Express Cos 150° in surd form.
Answer: B (−√3/2)
cos 150° = −cos 30° = −√3/2.
A straight line parallel to 2x + 3y = 6, passes through the point (−1, 2). Find the equation of the line.
Answer: D (2x + 3y = 4)
A parallel line is 2x + 3y = c; through (−1, 2): c = −2 + 6 = 4.
α and β are the roots of the equation 2x2 − 3x + 4 = 0. Find (α + β).
Answer: C (3/2)
α + β = −b/a = 3/2.
α and β are the roots of the equation 2x2 − 3x + 4 = 0. Find (α/β + β/α).
Answer: B (−7/8)
αβ = 2. α/β + β/α = ((α + β)² − 2αβ)/αβ = (9/4 − 4)/2 = −7/8.
If B = (2 5 / 1 3), find B−1. (Matrices written row by row; / separates the rows.)
Answer: C ((3 −5 / −1 2))
|B| = 6 − 5 = 1. B⁻¹ = (3 −5 / −1 2): swap the diagonal and change the signs of both off-diagonal entries.
Given that sin x = 5/13 and sin y = 8/17, where x and y are acute, find the value of cos (x + y).
Answer: B (140/221)
cos x = 12/13, cos y = 15/17. cos(x + y) = 180/221 − 40/221 = 140/221.
Given that f(x) = 5x2 − 4x + 3, find the coordinates of the point where the gradient is 6.
Answer: C ((1, 4))
f′(x) = 10x − 4 = 6 gives x = 1, and f(1) = 5 − 4 + 3 = 4. The point is (1, 4).
If y = (1 + x)/(1 − x), find dy/dx.
Answer: A (2/(1 − x)2)
Quotient rule: [(1 − x) + (1 + x)]/(1 − x)² = 2/(1 − x)².
Evaluate ∫−10 (x + 1)(x − 2) dx
Answer: D (−7/6)
∫(x² − x − 2) dx = x³/3 − x²/2 − 2x. From −1 to 0: 0 − (−1/3 − 1/2 + 2) = −7/6.
Simplify: √128/(√32 − 2√2).
Answer: D (4)
√128 = 8√2 and √32 = 4√2, so 8√2/(2√2) = 4.
There are 7 boys in a class of 20. Find the number of ways of selecting 3 girls and 2 boys.
Answer: C (6006)
13 girls: ¹³C₃ × ⁷C₂ = 286 × 21 = 6006.
The 3rd and 7th terms of a Geometric Progression (G.P) are 81 and 16 respectively. Find the 5th term.
Answer: D (36)
r⁴ = 16/81, so r² = 4/9. T₅ = T₃ × r² = 81 × 4/9 = 36.
Differentiate (5x3 + x2)/x, x ≠ 0 with respect to x
Answer: A (10x + 1)
Simplify first: 5x² + x, whose derivative is 10x + 1.
A curve is given by y = 5 − x − 2x2. Find the equation of its line of symmetry.
Answer: B (x = −1/4)
The axis of symmetry passes through the turning point: dy/dx = −1 − 4x = 0, so x = −1/4.
In a class of 10 boys and 15 girls, the average score in a Biology test is 90. If the average score for the girls is x, find the average score for the boys in terms of x.
Answer: B (225 − (3/2)x)
Total = 25 × 90 = 2250. Boys' total = 2250 − 15x, so their average is 225 − (3/2)x.
A fair die is tossed twice. What is the sample size?
Answer: C (36)
6 × 6 = 36 outcomes.
Given that a = (2, 3) and b = (−1, 4) (column vectors), evaluate (2a − ¼b).
Answer: B ((17/4, 5))
2a = (4, 6) and ¼b = (−¼, 1). 2a − ¼b = (4¼, 5) = (17/4, 5).
The table shows the results of tossing a fair die 150 times.
Face: 1, 2, 3, 4, 5, 6
Frequency: 12, 18, y, 30, 2y, 45
Find the probability of obtaining a 5.
Answer: C (1/5)
105 + 3y = 150 gives y = 15, so a 5 came up 2y = 30 times. P = 30/150 = 1/5.
The table shows the results of tossing a fair die 150 times.
Face: 1, 2, 3, 4, 5, 6
Frequency: 12, 18, y, 30, 2y, 45
Find the mode.
Answer: D (6)
With y = 15 the frequencies are 12, 18, 15, 30, 30, 45. Face 6 occurs most often, so the mode is 6.
Given that a = 5i + 4j and b = 3i + 7j, evaluate (3a − 8b).
Answer: C (−9i − 44j)
3a − 8b = (15 − 24)i + (12 − 56)j = −9i − 44j.
A force (10i + 4j)N acts on a body of mass 2 kg which is at rest. Find the velocity after 3 seconds.
Answer: D ((15i + 6j) ms−1)
a = F/m = 5i + 2j; v = at = 3(5i + 2j) = 15i + 6j ms⁻¹.
Find the magnitude and direction of the vector p = (5i − 12j).
Answer: D ((13, 157.38°))
|p| = √(25 + 144) = 13. p points 5 east and 12 south, so its bearing is 90° + tan⁻¹(12/5) = 90° + 67.38° = 157.38°.
Solve: 32x − 3x+2 = 3x+1 − 27.
Answer: B (1 or 2)
Let p = 3ˣ: p² − 9p = 3p − 27, so p² − 12p + 27 = 0, p = 3 or 9. x = 1 or 2.
The velocity, V, of a particle, in ms−1, after t seconds, is V = 3t2 + 2t − 1. Find the acceleration of the particle after 2 seconds.
Answer: C (14 ms−2)
a = dV/dt = 6t + 2 = 14 ms⁻² at t = 2.
Given that f(x) = 2x2 − 3 and g(x) = x + 1, where x ∈ R, find g o f(x).
Answer: A (2(x2 − 1))
g(f(x)) = (2x² − 3) + 1 = 2x² − 2 = 2(x² − 1).
If (2x2 − x − 3) is a factor of f(x) = 2x3 − 5x2 − x + 6, find the other factor.
Answer: A ((x − 2))
(2x² − x − 3)(x − 2) = 2x³ − 5x² − x + 6, so the other factor is (x − 2).
If P = {n2 + 1 : n = 0, 2, 3} and Q = {n + 1 : n = 2, 3, 5}, find P ∩ Q.
Answer: D ({ })
P = {1, 5, 10} and Q = {3, 4, 6}. They have no common member, so P ∩ Q = { }.
(a) If f(x) = (4 − 5x)/2 and g(x) = x + 6, x ∈ R, find f o g−1
(b) P(x, y) divides the line joining (7, −5) and (−2, 7) internally in the ratio 5:4. Find the coordinates of P.
(a) g⁻¹(x) = x − 6. f(g⁻¹(x)) = (4 − 5(x − 6))/2 = (34 − 5x)/2.
(b) x = (5(−2) + 4(7))/9 = 2; y = (5(7) + 4(−5))/9 = 5/3. P(2, 5/3).
For internal division in m : n use (mx₂ + nx₁)/(m + n).
Evaluate ∫13 (x − 1)/(x + 1)2 dx.
(x − 1)/(x + 1)² = 1/(x + 1) − 2/(x + 1)².
∫ = [ln(x + 1) + 2/(x + 1)] from 1 to 3 = (ln 4 + ½) − (ln 2 + 1) = ln 2 − ½ ≈ 0.193.
Split the fraction first (write x − 1 as (x + 1) − 2).
(a) Given that log10 p = a, log10 q = b and log10 s = c, express log10(p1/3q4/s2) in terms of a, b and c.
(b) The radius of a circle is 6 cm. If the area of the circle is increasing at the rate of 20 cm2s−1, find, leaving the answer in terms of π, the rate at which the radius is increasing.
(a) ⅓log p + 4 log q − 2 log s = a/3 + 4b − 2c.
(b) A = πr², dA/dt = 2πr dr/dt. 20 = 12π dr/dt, so dr/dt = 5/(3π) cm s⁻¹.
Connect the rates with the chain rule.
If (x + 1) and (x − 2) are factors of the polynomial g(x) = x4 + ax3 + bx2 − 16x − 12, find the values of a and b.
g(−1) = 1 − a + b + 16 − 12 = 0, so b − a = −5.
g(2) = 16 + 8a + 4b − 32 − 12 = 0, so 2a + b = 7.
Subtracting: 3a = 12, a = 4, b = −1.
Use the factor theorem twice.
Bottles of the same sizes produced in a factory are packed in boxes. Each box contains 10 bottles. If 8% of the bottles produced are defective, find correct to two decimal places, the probability that a box chosen at random contains at least three defective bottles.
Binomial: n = 10, p = 0.08, q = 0.92.
P(0) = 0.4344, P(1) = 0.3777, P(2) = 0.1478.
P(X ≥ 3) = 1 − 0.9599 ≈ 0.04.
'At least three' = 1 − P(0, 1 or 2).
The table shows the heights in cm of some seedlings in a certain garden:
Height (cm): 36 − 40, 41 − 45, 46 − 50, 51 − 55, 56 − 60
Frequency: 3, 9, 21, 12, 5
(a) Draw a cumulative frequency curve for the distribution (b) Using the curve in 6(a), find the semi-interquartile range.
(a) Upper boundaries 40.5, 45.5, 50.5, 55.5, 60.5; cumulative frequencies 3, 12, 33, 45, 50. Plot (start at (35.5, 0)) and join with a smooth curve.
(b) Q₁ (12.5th) ≈ 45.6 cm; Q₃ (37.5th) ≈ 52.4 cm.
Semi-interquartile range = ½(52.4 − 45.6) ≈ 3.4 cm (values read from a graph may differ slightly).
Read the quartiles at a quarter and three quarters of the total frequency.
A parallelogram MNQR has vertices M(4, −6), N(10, 2), Q(8, 16) and R(x, y). Find the coordinates of R.
Diagonals MQ and NR bisect each other: M + Q = N + R.
R = (4 + 8 − 10, −6 + 16 − 2) = (2, 8).
The diagonals of a parallelogram share a midpoint.
Forces F1 (18 N, 330°), F2 (10 N, 090°) and F3 (25 N, 180°) act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.
Components (east, north) = (F sin θ, F cos θ):
F₁ = (−9, 15.59); F₂ = (10, 0); F₃ = (0, −25).
Resultant = (1, −9.41); magnitude ≈ 9.5 N.
Direction: bearing 180° − tan⁻¹(1/9.41) ≈ 173.9°.
Resolve each force into east and north components.
(a) Simplify: 1/(1 − cos θ) + 1/(1 + cos θ) and leave the answer in terms of sin θ.
(b) Find the equation of the line joining the stationary points of y = x2(x − 3) and the distance between them.
(a) (1 + cos θ + 1 − cos θ)/(1 − cos²θ) = 2/sin²θ.
(b) y = x³ − 3x²; dy/dx = 3x² − 6x = 0 at x = 0 and 2. Points (0, 0) and (2, −4).
Line: y = −2x. Distance = √(4 + 16) = 2√5 ≈ 4.47.
1 − cos²θ = sin²θ.
(a) If f(x) = (2x − 3)/((x2 − 1)(x − 2)), (i) find the values of x for which f(x) is undefined. (ii) express f(x) in partial fractions.
(b) A circle with centre (−3, 1) passes through the point (3, 1). Find its equation.
(a)(i) x = 1, −1 or 2.
(ii) A/(x − 1) + B/(x + 1) + C/(x − 2): x = 1 gives A = ½; x = −1 gives B = −5/6; x = 2 gives C = ⅓.
f(x) = 1/(2(x − 1)) − 5/(6(x + 1)) + 1/(3(x − 2)).
(b) r = 6: (x + 3)² + (y − 1)² = 36, i.e. x² + y² + 6x − 2y − 26 = 0.
Factorise x² − 1 first.
(a) If f(x) = ∫(4x − x2) dx and f(3) = 21, find f(x).
(b) The second, fourth and eight terms of an Arithmetic Progression (A.P.) form the first three consecutive terms of a Geometric Progression (G.P.). The sum of the third and fifth terms of the A.P. is 20. Find the: (i) first four terms of the A.P., (ii) sum of the first ten terms of the A.P.
(a) f(x) = 2x² − x³/3 + c; f(3) = 18 − 9 + c = 21, c = 12. f(x) = 2x² − x³/3 + 12.
(b) (a + 3d)² = (a + d)(a + 7d) gives 2d² = 2ad, so a = d.
(a + 2d) + (a + 4d) = 8d = 20, d = 2.5 = a.
(i) 2.5, 5, 7.5, 10.
(ii) S₁₀ = 5(5 + 22.5) = 137.5.
For G.P. terms, (middle)² = first × third.
(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, correct to two decimal places, the probability that: (i) at least 3 students passed the test: (ii) between 3 and 6 students failed the test. (b) A fair die is thrown five times, find the probability of obtaining a six three times.
P(pass) = 0.8, P(fail) = 0.2, n = 7.
(a)(i) 1 − P(0, 1 or 2 pass) = 1 − 0.0047 ≈ 1.00 (0.9953).
(ii) Between 3 and 6 (4 or 5 failed): ⁷C₄(0.2)⁴(0.8)³ + ⁷C₅(0.2)⁵(0.8)² ≈ 0.0287 + 0.0043 = 0.03. (Including 3 and 6: ≈ 0.15.)
(b) ⁵C₃(1/6)³(5/6)² = 250/7776 = 125/3888 ≈ 0.032.
Use the binomial distribution with the right probability for each event.
The table shows the frequency distribution of the ages of patients in a clinic.
Age (years): 17 − 19, 20 − 22, 23 − 28, 29 − 34, 35 − 43
No of patients: 6, 9, 12, 18, 18
(a) Draw a histogram for the distribution. (b) Find, correct to two decimal places, the mean age of the patients.
(a) Boundaries 16.5, 19.5, 22.5, 28.5, 34.5, 43.5 (widths 3, 3, 6, 6, 9). Plot frequency density (per 3 years): 6, 9, 6, 9, 6, as touching bars.
(b) Class marks 18, 21, 25.5, 31.5, 39; Σf = 63; Σfx = 108 + 189 + 306 + 567 + 702 = 1872.
Mean = 1872/63 ≈ 29.71 years.
With unequal class widths, plot frequency densities.
(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° respectively to the horizontal and the tension in AB is 48 N. If the system is in equilibrium: (i) sketch a diagram to represent the information; (ii) calculate the tension in DB, (iii) find the value of q. [Take g = 10 ms−2]
(b) A ball is thrown vertically upwards with a velocity of 20 ms−1. Find, correct to two decimal places the: (i) maximum height reached by the ball; (ii) time taken to reach the maximum height. [Take g = 10 ms−2]
(a)(i) Sketch: B below the support line, BA at 30° and BD at 60° to the horizontal on opposite sides, weight 10q N straight down.
(ii) Horizontally: 48 cos 30° = T cos 60°, so T = 83.14 N.
(iii) Vertically: 48 sin 30° + 83.14 sin 60° = 24 + 72 = 96 = 10q, so q = 9.6.
(b)(i) v² = u² − 2gh: h = 400/20 = 20.00 m.
(ii) t = u/g = 2.00 s.
Resolve the tensions horizontally and vertically.
(a) Given that m = (6i + 8j) and n = (−8i + (7/3)j), find the: (i) magnitudes and directions of m and n, (ii) angle between m and n.
(b) The position vectors of points P, Q, R, S are (−2, 3), (10, 4), (3, 12) and (4, 0) (column vectors) respectively. Show that PQ is pependicular to RS
(a)(i) |m| = 10, at tan⁻¹(8/6) = 53.13° to the positive x-axis (bearing 036.87°).
|n| = √(64 + 49/9) = 25/3 ≈ 8.33, at 180° − tan⁻¹(7/24) = 163.74° to the positive x-axis (bearing 286.26°).
(ii) m·n = −48 + 56/3 = −88/3; cos θ = (−88/3)/(10 × 25/3) = −0.352, θ ≈ 110.61°.
(b) PQ = (12, 1), RS = (1, −12); PQ·RS = 12 − 12 = 0, so PQ ⊥ RS.
Two vectors are perpendicular when their dot product is zero.