58 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 a * b = a + b − 1. Find the identity element under the operation *.
Answer: C (1)
If e is the identity, a * e = a + e − 1 = a, so e − 1 = 0 and e = 1.
Express 75° in radians, leaving your answer in terms of π.
Answer: A (5π/12)
180° = π radians, so 75° = 75π/180 = 5π/12.
If log93 + 2x = 1, find x.
Answer: C (1/4)
log93 = ½ because 9½ = 3. So ½ + 2x = 1, 2x = ½ and x = ¼.
Evaluate cos(π/2 + π/3)
Answer: B (−√3/2)
cos(π/2 + π/3) = cos(π/2)cos(π/3) − sin(π/2)sin(π/3) = 0 − √3/2 = −√3/2.
Find the remainder when 5x3 + 2x2 − 7x − 5 is divided by (x − 2).
Answer: C (29)
By the remainder theorem the remainder is f(2) = 40 + 8 − 14 − 5 = 29.
A function is defined by f(x) = (3x + 1)/(x2 − 1). Find f(−3).
Answer: B (−1)
f(−3) = (−9 + 1)/(9 − 1) = −8/8 = −1.
Simplify: ∛(8/27) − (4/9)−1/2
Answer: A (−5/6)
∛(8/27) = 2/3 and (4/9)−1/2 = (9/4)1/2 = 3/2. So 2/3 − 3/2 = −5/6.
Solve 3x2 + 4x + 1 > 0.
Answer: C (x > −⅓, x < −1)
3x2 + 4x + 1 = (3x + 1)(x + 1). The product is positive when both factors have the same sign: x > −⅓ or x < −1.
The equation of a circle is 3x2 + 3y2 + 6x − 12y + 6 = 0. Find its radius.
Answer: B (√3)
Divide by 3: x2 + y2 + 2x − 4y + 2 = 0, i.e. (x + 1)2 + (y − 2)2 = 1 + 4 − 2 = 3. The radius is √3.
f(x) = p + qx, where p and q are constants. If f(1) = 7 and f(5) = 19, find f(3).
Answer: A (13)
p + q = 7 and p + 5q = 19 give 4q = 12, q = 3 and p = 4. So f(3) = 4 + 9 = 13.
The sum and product of the roots of a quadratic equation are 4/7 and 5/7 respectively. Find its equation.
Answer: B (7x2 − 4x + 5 = 0)
The equation is x2 − (sum)x + product = 0: x2 − (4/7)x + 5/7 = 0. Multiply by 7: 7x2 − 4x + 5 = 0.
f(x) = (x2 + 3)2 is defined on the set of real numbers R. Find the gradient of f(x) at x = 1/2.
Answer: B (6.5)
f'(x) = 2(x2 + 3) × 2x = 4x(x2 + 3). At x = ½: 2 × 3.25 = 6.5.
Find limx→3 (x + 3)/(x2 − x − 12)
Answer: A (−1)
The denominator is not zero at x = 3, so substitute: (3 + 3)/(9 − 3 − 12) = 6/(−6) = −1.
If y2 + xy − x = 0, find dy/dx
Answer: C ((1 − y)/(x + 2y))
Differentiate implicitly: 2y(dy/dx) + x(dy/dx) + y − 1 = 0, so (2y + x)(dy/dx) = 1 − y and dy/dx = (1 − y)/(x + 2y).
A line is perpendicular to 3x − y + 11 = 0 and passes through the point (1, −5). Find its equation.
Answer: D (3y + x + 14 = 0)
The given line has gradient 3, so the perpendicular has gradient −⅓. y + 5 = −⅓(x − 1) gives 3y + 15 = −x + 1, i.e. 3y + x + 14 = 0.
Solve 92x+1 = 813x+2.
Answer: A (−3/4)
81 = 92, so 92x+1 = 96x+4. Then 2x + 1 = 6x + 4, −4x = 3 and x = −¾.
The inverse of a function is given by f−1 : x → (x + 1)/4. Find f.
Answer: A (f : x → 4x − 1)
Let y = (x + 1)/4; then x = 4y − 1. So f is the inverse of f−1: f(x) = 4x − 1.
If (3 2 / 7 x)(2 / 3) = (12 / 29), find the value of x. (The 2 × 2 matrix is written row by row; the others are column vectors written top / bottom.)
Answer: A (5)
The second row gives 7 × 2 + 3x = 29, so 3x = 15 and x = 5.
The fourth term of a geometric sequence is 2 and the sixth term is 8. Find the common ratio.
Answer: B (±2)
ar3 = 2 and ar5 = 8. Dividing, r2 = 4, so r = ±2.
What percentage increase in the radius of a sphere will cause its volume to increase by 45%?
Answer: B (15%)
V = (4/3)πr3, so for small changes δV/V ≈ 3 δr/r. Then 45% = 3 × (δr/r), giving about 15%.
Evaluate 1/(1 − sin 60°) leaving your answer in surd form.
Answer: D (4 + 2√3)
1/(1 − √3/2) = 2/(2 − √3). Multiply top and bottom by 2 + √3: 2(2 + √3)/(4 − 3) = 4 + 2√3.
Find the equation of a circle with centre (−3, −8) and radius 4√6.
Answer: B (x2 + y2 + 6x + 16y − 23 = 0)
(x + 3)2 + (y + 8)2 = 96. Expanding: x2 + y2 + 6x + 16y + 73 − 96 = 0, i.e. x2 + y2 + 6x + 16y − 23 = 0.
Determine the coefficient of x2 in the expansion of (a + 3x)6.
Answer: C (135a4)
The x2 term is 6C2a4(3x)2 = 15 × 9a4x2, so the coefficient is 135a4.
The mean of 2, 5, (x + 2), 7 and 9 is 6. Find the median.
Answer: D (7.0)
The sum is 5 × 6 = 30, so 25 + x = 30 and x = 5. The numbers are 2, 5, 7, 7, 9, whose middle value is 7.
The probability that Kofi and Ama hit a target in a shooting competition are 1/6 and 1/9 respectively. What is the probability that only one of them will hit the target?
Answer: B (13/54)
P(only Kofi) + P(only Ama) = (1/6)(8/9) + (5/6)(1/9) = 8/54 + 5/54 = 13/54.
In how many ways can 3 perfects be chosen out of 8 perfects?
Answer: C (56)
Order does not matter, so the number is 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 56.
Find the standard deviation of the numbers 3, 6, 2, 1, 7 and 5.
Answer: B (2.16)
Mean = 24/6 = 4. Squared deviations: 1, 4, 4, 9, 9, 1, total 28. s.d. = √(28/6) = √4.67 ≈ 2.16.
The table shows the distribution of marks of students in a class. Find the upper class boundary of the modal class.
Marks: 5−7, 8−10, 11−13, 14−16, 17−19, 20−22
No. of students: 4, 7, 26, 41, 14, 8
Answer: C (16.5)
The modal class is 14−16 (frequency 41). Its upper class boundary is 16.5.
If 3xC2 = 15, find the value of x.
Answer: A (2)
3x(3x − 1)/2 = 15, so 9x2 − 3x − 30 = 0, i.e. 3x2 − x − 10 = (x − 2)(3x + 5) = 0. x must be a positive whole number, so x = 2.
Four doctors and two nurses are to sit round a circular table. In how many ways can this be done if the nurses are to sit together?
Answer: A (48)
Tie the two nurses together as one unit: 5 units round a table can sit in 4! = 24 ways, and the nurses can swap places in 2 ways. 24 × 2 = 48.
A basket contains 3 red and 1 white identical balls. A ball is drawn from the basket at random. Calculate the probability that it is either white or red.
Answer: D (1)
Every ball is either white or red, so the probability is 1/4 + 3/4 = 1.
A force of 200 N acting on a body of mass 20kg initially at rest causes it to move a distance of 320m along a straight line for t seconds. Find the value of t.
Answer: C (8s)
a = F/m = 200/20 = 10 ms−2. s = ½at2: 320 = 5t2, t2 = 64 and t = 8 s.
Two forces 10N and 15N act on an object at an angle of 120° to each other. Find the magnitude of the resultant force.
Answer: B (5√7)
R2 = 102 + 152 + 2(10)(15)cos 120° = 100 + 225 − 150 = 175, so R = √175 = 5√7 N.
A body of mass 25kg changes its speed from 15ms−1 to 35ms−1 in 5 seconds by the action of an applied force F. Find the value of F.
Answer: A (100 N)
F = m(v − u)/t = 25 × (35 − 15)/5 = 25 × 4 = 100 N.
A particle starts from rest and moves in a straight line such that its velocity, V, at time t seconds is given by v = (3t2 − 2t)ms−1. Calculate the distance covered in the first 2 seconds.
Answer: B (4m)
s = ∫v dt = t3 − t2 (s = 0 at t = 0). At t = 2: s = 8 − 4 = 4 m.
A particle starts from rest and moves in a straight line such that its velocity, V, at time t seconds is given by v = (3t2 − 2t)ms−1. Determine the acceleration when t = 2 seconds.
Answer: D (10 ms−2)
a = dv/dt = 6t − 2. At t = 2: a = 12 − 2 = 10 ms−2.
Given that q = 9i + 6j and r = 4i − 6j, which of the following statements is true?
Answer: B (r and q are perpendicular)
q · r = 9 × 4 + 6 × (−6) = 36 − 36 = 0. A zero dot product means the vectors are perpendicular.
The functions f and g are defined on the set, R, of real numbers by f : x → x2 − x − 6 and g : x → x − 1. Find f ∘ g(3).
Answer: C (−4)
g(3) = 2, then f(2) = 4 − 2 − 6 = −4.
Find the unit vector in the direction of −5i + 12j.
Answer: C (1/13(−5i + 12j))
|−5i + 12j| = √(25 + 144) = 13, so the unit vector is (1/13)(−5i + 12j).
Find, correct to two decimal places, the acute angle between p = (13 / 14) and q = (12 / 5). (Column vectors written top / bottom.)
Answer: B (24.50°)
p · q = 156 + 70 = 226, |p| = √365, |q| = 13. cos θ = 226/(13√365) ≈ 0.9099, so θ ≈ 24.50°.
If f(x) = 6x3 + 13x2 + 2x − 5 and f(−1) = 0, find the factors of f(x).
f(−1) = 0, so (x + 1) is a factor
Divide: f(x) = (x + 1)(6x2 + 7x − 5)
6x2 + 7x − 5 = (2x − 1)(3x + 5)
Factors: (x + 1), (2x − 1) and (3x + 5)
Use the given root to find one factor, then factorise the quadratic quotient.
If α and β are the roots of the equation 2x2 − 7x + 4 = 0, find the equation whose roots are α/β and β/α.
α + β = 7/2, αβ = 2
Sum of new roots = (α2 + β2)/αβ = [(α + β)2 − 2αβ]/αβ = (49/4 − 4)/2 = 33/8
Product of new roots = (α/β)(β/α) = 1
Equation: x2 − (33/8)x + 1 = 0, i.e. 8x2 − 33x + 8 = 0
Write α² + β² as (α + β)² − 2αβ.
(a) Write down the binomial expansion of (2 − x)5 in ascending powers of x.
(b) Use your expansion in 3(a) to evaluate (1.98)5 correct to four decimal places.
(a) (2 − x)5 = 32 − 80x + 80x2 − 40x3 + 10x4 − x5
(b) 1.98 = 2 − 0.02, so x = 0.02
32 − 1.6 + 0.032 − 0.00032 + 0.0000016 − ... = 30.43168
(1.98)5 ≈ 30.4317 (4 d.p.)
Choose x so that 2 − x = 1.98, i.e. x = 0.02 (not 0.2).
Solve: 2(2y+2) − 9(2y) = −2
22y+2 = 4(2y)2; let p = 2y
4p2 − 9p + 2 = 0
(4p − 1)(p − 2) = 0, p = ¼ or p = 2
2y = 2 gives y = 1; 2y = ¼ = 2−2 gives y = −2
y = 1 or y = −2
Substitute p = 2^y to get a quadratic.
(a) Write the following as column vectors:
r = (10 N, 090°)
q = (8 N, 135°)
(b) use your answer in 5(a) to find (r + q)
Using bearings, a force (F, θ) = (F sin θ / F cos θ) (east component / north component)
(a) r = (10 sin 90° / 10 cos 90°) = (10 / 0)
q = (8 sin 135° / 8 cos 135°) = (4√2 / −4√2) ≈ (5.66 / −5.66)
(b) r + q = (10 + 4√2 / −4√2) ≈ (15.66 / −5.66)
Resolve each force into east and north components using its bearing; then add component by component.
(a) Evaluate 9P3/15C3 + 5C3/3P2 correct to two decimal places.
(b) A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done, if one particular tutor must be on the committee and two particular pupils must not be on the committee?
(a) 9P3 = 504, 15C3 = 455, 5C3 = 10, 3P2 = 6
504/455 + 10/6 = 1.1077 + 1.6667 = 2.77 (2 d.p.)
(b) The particular tutor is in; choose 1 more tutor from the other 5: 5 ways
Leave out the 2 particular pupils; choose 5 pupils from 8: 8C5 = 56 ways
Number of committees = 5 × 56 = 280
Remove the people whose places are already decided, then choose from those left.
A tyre manufacturing company researched into the life span of one type of their moto-cycle tyres. The results were as follows:
Distance (100km): 10 − 19, 20 − 29, 30 − 39, 40 − 49, 50 − 59, 60 − 69
Number of tyres: 30, 69, 93, 57, 36, 15
(a) Draw a histogram for the distribution.
(b) Use the histogram to estimate the mode.
(a) Class boundaries 9.5, 19.5, 29.5, 39.5, 49.5, 59.5, 69.5 on the horizontal axis; bars of equal width 10 and heights 30, 69, 93, 57, 36, 15
(b) Modal class 29.5 − 39.5 (tallest bar); join the corners of the tallest bar to the adjacent bars and read where the lines cross
Mode ≈ 29.5 + (93 − 69)/[(93 − 69) + (93 − 57)] × 10 = 29.5 + 4 = 33.5, i.e. about 3350 km
Use class boundaries so the bars touch; the mode is read from the crossing lines on the tallest bar.
A particle of mass 400g is moving under the action of two forces; F1 = (35N, 210°) F2 = (35√3N, 300°) and a resistance of 40N. Find the magnitude of the:
(a) resultant of F1 and F2;
(b) resultant force acting on the particle.
Components (east, north) from bearings: F1 = (35 sin 210°, 35 cos 210°) = (−17.5, −30.31)
F2 = (35√3 sin 300°, 35√3 cos 300°) = (−52.5, 30.31)
(a) F1 + F2 = (−70, 0); magnitude 70 N (due west). Check: the forces are at 90°, so √(352 + 3 × 352) = 70 N
(b) The resistance of 40 N opposes the motion: resultant = 70 − 40 = 30 N
The bearings differ by 90°, so Pythagoras gives the resultant quickly.
(a) Find, from first principles, the derivative of f(x) = (x + 3)2
(b) Evaluate ∫12 (x + 1)(x2 − 2x + 2)/x2 dx
(a) f(x + h) − f(x) = (x + h + 3)2 − (x + 3)2 = 2h(x + 3) + h2
[f(x + h) − f(x)]/h = 2(x + 3) + h
As h → 0: f'(x) = 2(x + 3) = 2x + 6
(b) (x + 1)(x2 − 2x + 2) = x3 − x2 + 2, so the integrand is x − 1 + 2x−2
∫ = [x2/2 − x − 2/x] from 1 to 2 = (2 − 2 − 1) − (½ − 1 − 2) = −1 + 2.5 = 3/2
For (b) multiply out the top first, then divide each term by x².
(a) If A = (−2 5 / 4 3) and B = (3 1 / 2 3) find the value of x and y such that BA = 2(3 7 / −2 x) + (y 4 / 12 −3). (Matrices are written row by row.)
(b) Two functions, f and g are defined by f : x → ½x + 1 and g : x → (5x − 1)/3. Find (i) g−1 (ii) g−1 ∘ f.
(a) BA = (3(−2) + 1(4) 3(5) + 1(3) / 2(−2) + 3(4) 2(5) + 3(3)) = (−2 18 / 8 19)
Right side = (6 + y 18 / 8 2x − 3)
6 + y = −2, so y = −8; 2x − 3 = 19, so x = 11
(b)(i) y = (5x − 1)/3 gives x = (3y + 1)/5, so g−1(x) = (3x + 1)/5
(ii) g−1(f(x)) = [3(½x + 1) + 1]/5 = (3x + 8)/10
Multiply B by A (rows of B into columns of A), then compare corresponding entries.
The images of points (2, 3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the:
(a) matrix A;
(b) inverse of A;
(c) point whose image is (−1, 1).
(a) Let A = (a b / c d) (row by row): 2a + 3b = 3, 4a + 5b = 5 give b = 1, a = 0
2c + 3d = 4, 4c + 5d = 6 give d = 2, c = −1
A = (0 1 / −1 2)
(b) |A| = 0 × 2 − 1 × (−1) = 1; A−1 = (2 −1 / 1 0)
(c) A−1(−1 / 1) = (2(−1) − 1(1) / 1(−1) + 0) = (−3 / −1); the point is (−3, −1)
Set up two pairs of simultaneous equations from the images; use the inverse to go backwards.
(a) Find the equation of the tangent to curve x2/4 + y2 = 1 at point (1, √3/2).
(b) Express (3x + 2)/(x2 + x − 2) in partial fractions.
(a) x/2 + 2y(dy/dx) = 0, so dy/dx = −x/(4y)
At (1, √3/2): dy/dx = −1/(2√3) = −√3/6
y − √3/2 = −(√3/6)(x − 1)
√3x + 6y − 4√3 = 0 (or x + 2√3y − 4 = 0)
(b) x2 + x − 2 = (x − 1)(x + 2); 3x + 2 = A(x + 2) + B(x − 1)
x = 1: 5 = 3A, A = 5/3; x = −2: −4 = −3B, B = 4/3
(3x + 2)/(x2 + x − 2) = 5/[3(x − 1)] + 4/[3(x + 2)]
Differentiate implicitly for the gradient; factorise the denominator for the partial fractions.
A survey conducted revealed that four out of every twenty taxi drivers do not have valid driving license. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that: (a) exactly 2; (b) more than 3; (c) at least 5 have valid driving license.
P(valid) = 16/20 = 0.8, P(not valid) = 0.2; binomial with n = 6
(a) 6C2(0.8)2(0.2)4 = 15 × 0.64 × 0.0016 = 0.015
(b) P(4) + P(5) + P(6) = 0.24576 + 0.393216 + 0.262144 = 0.901
(c) P(5) + P(6) = 0.393216 + 0.262144 = 0.655
Four out of twenty have no licence, so the probability of a valid licence is 0.8.
The table shows the frequency distribution of marks scored by some candidates in an examination.
Marks: 0 − 9, 10 − 19, 20 − 29, 30 − 39, 40 − 49, 50 − 59, 60 − 69, 70 − 79, 80 − 89, 90 − 99
Frequency: 2, 5, 8, 18, 20, 15, 5, 4, 2, 1
(a) Draw the cumulative frequency curve of the distribution.
(b) Use your graph to estimate the: (i) Semi − interquartile range of the distribution; (ii) percentage of candidates who passed with distinction if the least mark for distinction was 72.
(a) Upper class boundaries 9.5, 19.5, ..., 99.5 against cumulative frequencies 2, 7, 15, 33, 53, 68, 73, 77, 79, 80; join with a smooth curve
(b)(i) N = 80: Q1 = 20th mark ≈ 29.5 + (5/18) × 10 ≈ 32.3; Q3 = 60th mark ≈ 49.5 + (7/15) × 10 ≈ 54.2
Semi-interquartile range = (54.2 − 32.3)/2 ≈ 10.9 (accept about 10.5 to 11 from the graph)
(ii) Read the cumulative frequency at 71.5: about 73.8, so about 80 − 73.8 ≈ 6 candidates scored 72 or more
Percentage ≈ 6.2/80 × 100 ≈ 7.8% (about 8%)
Quartiles are at N/4 and 3N/4 on the cumulative frequency axis.
A bag contains 4 red, 6 blue and 8 green identical marbles.
(a) If three marbles are drawn at random, without replacement, calculate the probability that: (i) all will be green; (ii) all will have the same colour.
(b) If each marble is replaced before another is drawn, calculate the probability that all will have the same colour.
Total 18 marbles
(a)(i) (8/18)(7/17)(6/16) = 336/4896 = 7/102
(ii) P(GGG) + P(BBB) + P(RRR) = 7/102 + (6 × 5 × 4)/4896 + (4 × 3 × 2)/4896 = (336 + 120 + 24)/4896 = 480/4896 = 5/51
(b) (8/18)3 + (6/18)3 + (4/18)3 = (512 + 216 + 64)/5832 = 792/5832 = 11/81
Without replacement the numbers on top and bottom both drop by one each draw.
(a) An object P of mass 6.5kg is suspended by two light inextensible strings AP and BP. The strings make angles 50° and 60° respectively with the downward vertical. (i) Express the forces acting on P in component form. (ii) If P is at rest, write down the vector equation connecting all the forces. (iii) Calculate, correct to one decimal place, the tensions in the strings.
(b) A particle of mass 5kg moves with initial velocity (−1 / 2)ms−1 and final velocity (3 / 4)ms−1. Find the magnitude of its change in momentum. (Column vectors are written top / bottom.)
(a) Take g = 10 ms−2: weight W = 65 N. Let T1 (in AP) and T2 (in BP)
(i) T1 = (−T1 sin 50° / T1 cos 50°), T2 = (T2 sin 60° / T2 cos 60°), W = (0 / −65)
(ii) T1 + T2 + W = 0
(iii) Horizontal: T1 sin 50° = T2 sin 60°; vertical: T1 cos 50° + T2 cos 60° = 65
Solving (or Lami's theorem, angle between strings 110°): T1 = 65 sin 60°/sin 110° ≈ 59.9 N, T2 = 65 sin 50°/sin 110° ≈ 53.0 N
(with g = 9.8: 58.7 N and 51.9 N)
(b) Change in momentum = m(v − u) = 5[(3 / 4) − (−1 / 2)] = 5(4 / 2) = (20 / 10) Ns
Magnitude = √(400 + 100) = √500 = 10√5 ≈ 22.4 Ns
Resolve horizontally and vertically; the horizontal parts of the two tensions must balance.
(a) m(2 / 1) + n(−1 / 2) = (5 / −1) where m and n are scalars. Find the value of (m + n). (Column vectors are written top / bottom.)
(b) A(−1, 3), B(2, −1) and C(5, 3) are the vertices of ABC. (i) Express in column notation, the unit vectors parallel to AB and AC. (ii) Use a dot product to calculate BÂC, correct to the nearest degree.
(a) 2m − n = 5 and m + 2n = −1
Double the first and add: 5m = 9, m = 9/5; n = 2m − 5 = −7/5
m + n = 2/5
(b)(i) AB = (3 / −4), |AB| = 5, unit vector (3/5 / −4/5)
AC = (6 / 0), |AC| = 6, unit vector (1 / 0)
(ii) cos BÂC = (AB · AC)/(|AB||AC|) = 18/30 = 0.6
BÂC ≈ 53°
Use AB = B − A and AC = C − A, not the position vectors themselves.
(a) Forces F1 = (2N, 210°), F2 = (4N, 120°) act on a particle of mass 7kg which is at rest. Calculate the: (i) acceleration of the particle; (ii) velocity of the particle after 3 seconds
(b) F1 = (2i + 3j)N, F2 = (−5j)N and F3 = (6i − 4j)N act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.
(a) Bearings 210° and 120° differ by 90°, so R = √(22 + 42) = √20 = 2√5 ≈ 4.47 N
(components: F1 = (−1, −1.732), F2 = (3.464, −2), sum (2.464, −3.732))
(i) a = R/m = 4.47/7 ≈ 0.64 ms−2
(ii) v = u + at = 0 + 0.639 × 3 ≈ 1.92 ms−1
(b) F1 + F2 + F3 = 8i − 6j, so F4 = −8i + 6j
|F4| = √(64 + 36) = 10 N
Direction: tan−1(6/8) ≈ 36.9° above the negative x-axis, i.e. 143.1° to the positive x-axis (bearing ≈ 306.9°, N53.1°W)
In equilibrium the fourth force is minus the sum of the other three.