WAEC Further Mathematics 2008 past questions and answers

57 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Surds

Simplify (√3 + √48)/√6

  1. A.3√2
  2. B.5√2
  3. C.5√2/2
  4. D.3√2/2
Show the answer and explanation

Answer: C (5√2/2)

Explanation

√48 = 4√3, so the top is 5√3. Then 5√3/√6 = 5/√2 = 5√2/2.

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Question 2

Polynomials

Find the range of values of x for which 2x2 + 7x − 15 > 0

  1. A.x < −3/2 or x > 5
  2. B.x < −5 or x > 3/2
  3. C.−3/2 < x < 5
  4. D.−5 < x < 3/2
Show the answer and explanation

Answer: B (x < −5 or x > 3/2)

Explanation

Factorise: (2x − 3)(x + 5) > 0. The product is positive outside the roots −5 and 3/2, so x < −5 or x > 3/2.

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Question 3

Functions

A function f is defined on R, the set of real numbers, by f : x → (x + 3)/(x − 1), x ≠ 2. Find f−1

  1. A.f−1 : x → (2x + 3)/(x − 1), x ≠ 1
  2. B.f−1 : x → (x + 3)/(x − 1), x ≠ −2
  3. C.f−1 : x → (x − 1)/(2x + 3), x ≠ −3/2
  4. D.f−1 : x → (x − 2)/(2x + 3), x ≠ −3
Show the answer and explanation

Answer: B (f−1 : x → (x + 3)/(x − 1), x ≠ −2)

Explanation

Let y = (x + 3)/(x − 1). Then yx − y = x + 3, so x(y − 1) = y + 3 and x = (y + 3)/(y − 1). The inverse has the same rule as f.

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Question 4

Sequences and Series

The sum of the first n terms of a linear sequence is Sn = n2 + 2n. Find the common difference of the sequence.

  1. A.5
  2. B.4
  3. C.3
  4. D.2
Show the answer and explanation

Answer: D (2)

Explanation

T1 = S1 = 3 and S2 = 8, so T2 = 5. The common difference is 5 − 3 = 2.

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Question 5

Sequences and Series

The sum of the first n terms of a linear sequence is Sn = n2 + 2n. Determine the general term of the sequence.

  1. A.n + 1
  2. B.2n + 1
  3. C.3n − 1
  4. D.4n + 1
Show the answer and explanation

Answer: B (2n + 1)

Explanation

First term 3, common difference 2, so Tn = 3 + (n − 1)2 = 2n + 1.

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Question 6

Differentiation

If f(x) = 2x2 − 3x − 1 find the value of x for which f(x) is minimum.

  1. A.3/2
  2. B.4/3
  3. C.3/4
  4. D.2/3
Show the answer and explanation

Answer: C (3/4)

Explanation

f′(x) = 4x − 3 = 0 gives x = 3/4, and f″(x) = 4 > 0 so it is a minimum.

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Question 7

Polynomials

The polynomial 2x3 + x2 − 3x + p has a remainder 20 when divided by (x − 2). Find the value of constant p.

  1. A.8
  2. B.6
  3. C.−6
  4. D.−8
Show the answer and explanation

Answer: B (6)

Explanation

By the remainder theorem f(2) = 16 + 4 − 6 + p = 14 + p = 20, so p = 6.

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Question 8

Indices and Logarithms

If 2log42 = x + 1, find the value of x.

  1. A.−2
  2. B.−1
  3. C.0
  4. D.1
Show the answer and explanation

Answer: C (0)

Explanation

log42 = 1/2 because 41/2 = 2. So 2 × 1/2 = 1 = x + 1, giving x = 0.

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Question 9

Polynomials

Which of the following quadratic curves will not intersect with the x-axis?

  1. A.y = 2 − 4x − x2
  2. B.y = x2 − 5x − 1
  3. C.y = 2x2 − x − 1
  4. D.y = 3x2 − 2x + 4
Show the answer and explanation

Answer: D (y = 3x2 − 2x + 4)

Explanation

A curve misses the x-axis when b2 − 4ac < 0. For 3x2 − 2x + 4: 4 − 48 = −44 < 0; the others have positive discriminants.

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Question 10

Coordinate Geometry

What is the coordinates of the centre of the circle 5x2 + 5y2 − 15x + 25y − 3 = 0?

  1. A.(15/2, −25/2)
  2. B.(3/2, −5/2)
  3. C.(−3/2, 5/2)
  4. D.(−15/2, 25/2)
Show the answer and explanation

Answer: B ((3/2, −5/2))

Explanation

Divide by 5: x2 + y2 − 3x + 5y − 3/5 = 0. The centre is (−g, −f) = (3/2, −5/2).

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Question 11

Integration

Evaluate ∫12(2 + 2x − 3x2)dx

  1. A.−2
  2. B.2
  3. C.8
  4. D.10
Show the answer and explanation

Answer: A (−2)

Explanation

The antiderivative is 2x + x2 − x3. At 2 it is 0 and at 1 it is 2, so the answer is 0 − 2 = −2.

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Question 12

Differentiation

A rectangle has a perimeter of 24m, if its area is to be a maximum, find its dimension.

  1. A.12, 12
  2. B.6, 6
  3. C.4, 8
  4. D.9, 3
Show the answer and explanation

Answer: B (6, 6)

Explanation

L + B = 12, so A = L(12 − L). dA/dL = 12 − 2L = 0 gives L = 6 and B = 6.

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Question 13

Trigonometry

Express 7π/6 radians in degrees.

  1. A.315°
  2. B.210°
  3. C.105°
  4. D.75°
Show the answer and explanation

Answer: B (210°)

Explanation

π radians = 180°, so 7π/6 = 7 × 180°/6 = 210°.

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Question 14

Matrices and Determinants

If P = (1 −2; 3 4) and Q = (−2 3; 1 0), find PQ.

  1. A.(4 1; −2 9)
  2. B.(−4 1; 2 9)
  3. C.(−4 3; −2 13)
  4. D.(−4 3; −2 9)
Show the answer and explanation

Answer: D ((−4 3; −2 9))

Explanation

Row 1: (1)(−2) + (−2)(1) = −4 and (1)(3) + (−2)(0) = 3. Row 2: (3)(−2) + (4)(1) = −2 and (3)(3) + 0 = 9.

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Question 15

Sets and Logic

Two statements are represented by p and q as follows: p: He is brilliant. q: He is regular in class. Which of the following symbols represents the statement "he is regular in class but dull"?

  1. A.q ∨ ~p
  2. B.q ∧ ~p
  3. C.~q ∧ ~p
  4. D.~q ∨ ~p
Show the answer and explanation

Answer: B (q ∧ ~p)

Explanation

'Regular in class' is q, 'dull' is not brilliant (~p), and 'but' means 'and' (∧): q ∧ ~p.

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Question 17

Binary Operations

A binary operation *, is defined on the set R, of real numbers by a * b = a2 + b + ab. Find the value of x for which 5 * x = 37.

  1. A.7
  2. B.2
  3. C.−2
  4. D.−7
Show the answer and explanation

Answer: B (2)

Explanation

5 * x = 25 + x + 5x = 25 + 6x. Setting 25 + 6x = 37 gives x = 2.

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Question 18

Differentiation

Find the derivative of 3x2 + 1/x2

  1. A.6x + 2x2
  2. B.6x + 1/(2x)
  3. C.6x − 2/x3
  4. D.6x − 1/(2x)
Show the answer and explanation

Answer: C (6x − 2/x3)

Explanation

Write 1/x2 as x−2; its derivative is −2x−3. So the answer is 6x − 2/x3.

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Question 19

Binomial Theorem

The coefficient of the 5th term in the binomial expansion of (1 + kx)8, in ascending powers of x is 35/8. Find the value of constant k.

  1. A.2
  2. B.1/2
  3. C.−1/2
  4. D.−2
Show the answer and explanation

Answer: B (1/2)

Explanation

The 5th term is 8C4(kx)4 = 70k4x4. So 70k4 = 35/8, k4 = 1/16 and k = 1/2 (k = −1/2 also fits).

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Question 20

Integration

Given that f′(x) = 3x2 − 6x + 1 and f(3) = 5, find f(x).

  1. A.f(x) = x3 − 3x + x + 20
  2. B.f(x) = x3 − 3x2 + x + 31
  3. C.f(x) = x3 − 3x2 + x + 2
  4. D.f(x) = x3 − 3x2 + x − 13
Show the answer and explanation

Answer: C (f(x) = x3 − 3x2 + x + 2)

Explanation

Integrate: f(x) = x3 − 3x2 + x + c. f(3) = 27 − 27 + 3 + c = 5, so c = 2.

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Question 21

Trigonometry

Express 1/(1 − sin 45°) in surd form

  1. A.2 + √2
  2. B.2 + √3
  3. C.2 − √2
  4. D.1 + 2√2
Show the answer and explanation

Answer: A (2 + √2)

Explanation

sin 45° = 1/√2, so the expression is √2/(√2 − 1). Multiply top and bottom by (√2 + 1): √2(√2 + 1)/1 = 2 + √2.

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Question 22

Matrices and Determinants

If |4 x; 5 3| = 32, find the value of x

  1. A.4
  2. B.2
  3. C.−2
  4. D.−4
Show the answer and explanation

Answer: D (−4)

Explanation

The determinant is 4 × 3 − 5x = 12 − 5x. 12 − 5x = 32 gives x = −4.

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Question 23

Probability

If events A and B are independent and P(A) = 7/12 and P(A∩B) = 1/2, find P(B)

  1. A.3/7
  2. B.4/7
  3. C.5/7
  4. D.6/7
Show the answer and explanation

Answer: D (6/7)

Explanation

For independent events P(A∩B) = P(A) × P(B), so P(B) = (1/2) ÷ (7/12) = 6/7.

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Question 24

Vectors

Given that AB = 5i + 3j and AC = 2i + 5j, find BC.

  1. A.−7i − 8j
  2. B.−3i + 2j
  3. C.3i − 2j
  4. D.3i + 8j
Show the answer and explanation

Answer: B (−3i + 2j)

Explanation

BC = AC − AB = (2i + 5j) − (5i + 3j) = −3i + 2j.

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Question 25

Probability

The probability of Jide, Atu and Obu solving a given problem are 1/12, 1/6 and 1/8 respectively. Calculate the probability that only one of them solves the problem.

  1. A.1/576
  2. B.55/576
  3. C.77/576
  4. D.167/576
Show the answer and explanation

Answer: D (167/576)

Explanation

Add the three cases where exactly one solves it: (1/12)(5/6)(7/8) + (11/12)(1/6)(7/8) + (11/12)(5/6)(1/8) = (35 + 77 + 55)/576 = 167/576.

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Question 26

Vectors

Two forces F1 = (10N, 020°) and F2 = (7N, 200°) act on a particle. Find the resultant force.

  1. A.(3N, 020°)
  2. B.(3N, 200°)
  3. C.(17N, 020°)
  4. D.(17N, 200°)
Show the answer and explanation

Answer: A ((3N, 020°))

Explanation

020° and 200° differ by 180°, so the forces act in opposite directions. The resultant is 10 − 7 = 3N in the direction of the larger force, 020°.

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Question 27

Statistics

The table shows the distribution of marks by some candidates in a test.
Marks: 2, 3, 4, 5, 6, 7, 8
No of students: 5, 7, 9, 6, 3, 6, 4
What is the median score?

  1. A.3.5
  2. B.4.0
  3. C.4.5
  4. D.5.0
Show the answer and explanation

Answer: B (4.0)

Explanation

There are 40 students, so the median is between the 20th and 21st scores. Cumulative frequencies are 5, 12, 21, so both are 4.

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Question 28

Statistics

The table shows the distribution of marks by some candidates in a test.
Marks: 2, 3, 4, 5, 6, 7, 8
No of students: 5, 7, 9, 6, 3, 6, 4
Find, correct to one decimal place, the mean of the distribution.

  1. A.5.5
  2. B.5.3
  3. C.5.2
  4. D.4.7
Show the answer and explanation

Answer: D (4.7)

Explanation

Σfx = 10 + 21 + 36 + 30 + 18 + 42 + 32 = 189 and Σf = 40, so the mean is 189/40 = 4.725 ≈ 4.7.

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Question 29

Probability

The table shows the distribution of marks by some candidates in a test.
Marks: 2, 3, 4, 5, 6, 7, 8
No of students: 5, 7, 9, 6, 3, 6, 4
If a student is selected at random, what is the probability that she scored at least 6 marks?

  1. A.3/40
  2. B.1/4
  3. C.13/40
  4. D.27/40
Show the answer and explanation

Answer: C (13/40)

Explanation

Students scoring 6, 7 or 8: 3 + 6 + 4 = 13 out of 40, so the probability is 13/40.

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Question 30

Vectors

Express r = (12, 210°) in the form ai + bj

  1. A.6(−i + √3j)
  2. B.6(−i − √3j)
  3. C.6(i − √3j)
  4. D.6(i + √3j)
Show the answer and explanation

Answer: B (6(−i − √3j))

Explanation

Read 210° as a bearing (clockwise from north): a = 12 sin 210° = −6 and b = 12 cos 210° = −6√3, so r = −6i − 6√3j = 6(−i − √3j).

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Question 31

Permutations and Combinations

A test consists of 12 questions out of which candidates are to answer 10. If the first 6 questions are compulsory, in how many ways can each candidate select her questions?

  1. A.40
  2. B.25
  3. C.15
  4. D.10
Show the answer and explanation

Answer: C (15)

Explanation

After the 6 compulsory questions she chooses 4 more from the remaining 6: 6C4 = 15.

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Question 32

Dynamics

A body starts from rest and moves in a straight line with a uniform acceleration of 5ms−2. How far, in m, does it go in 10 seconds?

  1. A.50
  2. B.250
  3. C.350
  4. D.500
Show the answer and explanation

Answer: B (250)

Explanation

s = ut + ½at2 = 0 + ½ × 5 × 100 = 250 m.

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Question 33

Permutations and Combinations

If n items are arranged two at a time, the number obtained is 20. Find the value of n.

  1. A.5
  2. B.10
  3. C.15
  4. D.40
Show the answer and explanation

Answer: A (5)

Explanation

nP2 = n(n − 1) = 20, so n2 − n − 20 = 0, (n − 5)(n + 4) = 0 and n = 5.

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Question 34

Vectors

If p = (2, −2) and q = (3, 4), find |q − ½p|

  1. A.2√2
  2. B.√13
  3. C.5
  4. D.√29
Show the answer and explanation

Answer: D (√29)

Explanation

q − ½p = (3 − 1, 4 + 1) = (2, 5). Its magnitude is √(4 + 25) = √29.

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Question 35

Vectors

Find the value of constant k for which a = 4i − kj and b = 3i + 8j are perpendicular.

  1. A.2/3
  2. B.3/2
  3. C.2
  4. D.3
Show the answer and explanation

Answer: B (3/2)

Explanation

Perpendicular vectors have a zero dot product: 12 − 8k = 0, so k = 3/2.

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Question 36

Probability

A pack of cards has 6 of them with a ⊗ written on them, 5 with ⊕ and 9 with ∅. What is probability that single card drawn at random is not a ⊕?

  1. A.1/4
  2. B.2/5
  3. C.2/3
  4. D.3/4
Show the answer and explanation

Answer: D (3/4)

Explanation

There are 20 cards and 5 have ⊕, so 15 do not. The probability is 15/20 = 3/4.

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Question 37

Dynamics

The initial and final velocities of an object of mass 5kg are u = (1, 3) and v = (4, 7) respectively. Find the magnitude of its change in momentum.

  1. A.25
  2. B.15
  3. C.3√7
  4. D.√10
Show the answer and explanation

Answer: A (25)

Explanation

Change in velocity = (3, 4), so change in momentum = 5(3, 4) = (15, 20). Its magnitude is √(225 + 400) = 25.

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Question 38

Polynomials

If y = x2 − 6x + 11 is written in the form y = a(x − h)2 + k, find the value of (a + h + k)

  1. A.−4
  2. B.−3
  3. C.0
  4. D.6
Show the answer and explanation

Answer: D (6)

Explanation

Completing the square: y = (x − 3)2 + 2, so a = 1, h = 3, k = 2 and a + h + k = 6.

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Question 39

Coordinate Geometry

The distance between P(x, 7) and Q(6, 19) is 13 units. Find the values of x

  1. A.1 or −7
  2. B.1 or 7
  3. C.1 or 11
  4. D.5 or −5
Show the answer and explanation

Answer: C (1 or 11)

Explanation

(6 − x)2 + 122 = 169, so (6 − x)2 = 25 and 6 − x = ±5. Hence x = 1 or 11.

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Question 40

Statics

In the diagram below, forces P, Q and 50N are acting on a body at M. If the system is in equilibrium, calculate, in terms of θ, the magnitude of P.

Diagram for this question
  1. A.50cosθ/sin(θ + 45)°
  2. B.50cosθ/cos(θ + 45)°
  3. C.50sinθ/cos(θ + 45)°
  4. D.50sinθ/sin(θ + 45)°
Show the answer and explanation

Answer: D (50sinθ/sin(θ + 45)°)

Explanation

By Lami's theorem P/sin(angle between Q and 50N) = 50/sin(angle between P and Q). The angle between Q and the weight is 180° − θ and between P and Q is θ + 45°, so P = 50 sinθ/sin(θ + 45°).

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Theory question 101

Functions

A function f is defined on the set R, of real numbers by f : x → px2 + qx + 2, where p and q are constants. If f(−2) = 0 and f(1) = 3, find f(−4)

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Model answer

f(−2) = 4p − 2q + 2 = 0, so 2p − q = −1
f(1) = p + q + 2 = 3, so p + q = 1
Adding: 3p = 0, so p = 0 and q = 1
f(x) = x + 2, so f(−4) = −2

Explanation

Form two equations from the given values, solve for p and q, then substitute x = −4.

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Theory question 102

Trigonometry

If sin A = 3/5 and cos B = 15/17, where A is obtuse and B is acute, find the value of cos(A + B).

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Model answer

A is obtuse, so cos A = −4/5
B is acute, so sin B = 8/17
cos(A + B) = cos A cos B − sin A sin B
= (−4/5)(15/17) − (3/5)(8/17) = −60/85 − 24/85
= −84/85

Explanation

Use right-angled triangles (3, 4, 5 and 8, 15, 17) and remember cosine is negative for an obtuse angle.

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Theory question 103

Differentiation

Differentiate, with respect to x, x3 − 2x from the first principle

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Model answer

y = x3 − 2x; y + δy = (x + δx)3 − 2(x + δx)
δy = 3x2δx + 3x(δx)2 + (δx)3 − 2δx
δy/δx = 3x2 + 3xδx + (δx)2 − 2
As δx → 0: dy/dx = 3x2 − 2

Explanation

Give x a small increase δx, find δy, divide by δx and let δx tend to 0.

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Theory question 104

Coordinate Geometry

A straight line passes through the point P(−1, 3). Another line which passes through Q(−4, 4) intersects the first line at a point R(k, 5), where k is constant. If ∠PRQ = 90° find the values of K.

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Model answer

Gradient of PR = (5 − 3)/(k + 1) = 2/(k + 1)
Gradient of QR = (5 − 4)/(k + 4) = 1/(k + 4)
Perpendicular: product = −1, so 2 = −(k + 1)(k + 4)
k2 + 5k + 6 = 0, (k + 2)(k + 3) = 0
k = −2 or k = −3

Explanation

Lines at right angles have gradients whose product is −1.

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Theory question 105

Probability

In a hotel the breakfast menu is a choice between yam (Y) or plantain (P) or both. The Venn diagram shows the choices made by 25 guests of the Hotel. (a) Find the value of x. (b) What is the probability that a guest chosen at random chose only one of the two?

Diagram for this question
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Model answer

(a) (2x + 1) + x + (x − 2)2 = 25
x2 − x − 20 = 0, (x − 5)(x + 4) = 0
x = 5 (x cannot be negative)
(b) P only = 11, Y only = 9, so 20 guests chose only one
Probability = 20/25 = 4/5

Explanation

The three regions of the Venn diagram add up to the 25 guests.

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Theory question 106

Statistics

The table shows the distribution of marks obtained by some candidates in a test
Marks: 10 − 14, 15 − 24, 25 − 29, 30 − 39, 40 − 44, 45 − 49
No. of candidates: 14, 30, 22, 18, 12, 4
Draw a histogram for the distribution.

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Model answer

Class boundaries: 9.5–14.5, 14.5–24.5, 24.5–29.5, 29.5–39.5, 39.5–44.5, 44.5–49.5
Class widths 5, 10, 5, 10, 5, 5 are unequal, so bar heights must be frequency densities
Heights per 5 marks: 14, 15, 22, 9, 12, 4 (frequency ÷ width × 5)
Bars drawn touching on the class boundaries, with a stated scale and labelled axes

Explanation

The classes have different widths, so the area (not the height) of each bar must show the frequency.

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Theory question 107

Dynamics

A car travelling at a velocity of 50 km h−1, covers a distance of 20km. If it was accelerating at 6km h−2, calculate, correct to one decimal place, the time the car took to cover the distance.

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Model answer

s = ut + ½at2: 20 = 50t + 3t2
3t2 + 50t − 20 = 0
t = (−50 + √2740)/6 ≈ 0.39
t ≈ 0.4 hours (about 23 minutes)
(Taking 50 km h−1 as the final speed instead also gives 0.4 h.)

Explanation

Use s = ut + ½at² and solve the quadratic in t, keeping the positive root.

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Theory question 108

Vectors

The magnitude of a force xi + 15j is 17N. Find the: (a) possible value of x; (b) directions of the forces, correct to the nearest degree.

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Model answer

(a) x2 + 152 = 172, x2 = 64, x = ±8
(b) 8i + 15j: tan θ = 15/8, θ = 62° to the positive x-axis (bearing 028°)
−8i + 15j: 180° − 62° = 118° to the positive x-axis (bearing 332°)

Explanation

The magnitude of ai + bj is √(a² + b²); there are two forces, one for each value of x.

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Theory question 109

Sequences and Series

(a) The sum of the first n terms of a sequence is given by Sn = 5n2/2 + 5n/2. Write down the first four terms of the sequence and find an expression for the nth term.
(b) The equation of a circle is given by x2 + y2 − 10x − 8y + 25 = 0. (i) Show that the circle touches the X-axis. (ii) Find the coordinates of the point of contact.

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Model answer

(a) S1 = 5, S2 = 15, S3 = 30, S4 = 50
Terms: 5, 10, 15, 20
Tn = Sn − Sn−1 = 5n
(b)(i) Centre (5, 4), radius √(25 + 16 − 25) = 4; distance from centre to the x-axis = 4 = radius, so the circle touches it (or: y = 0 gives (x − 5)2 = 0, equal roots)
(ii) Point of contact (5, 0)

Explanation

Each term is the difference of two consecutive sums; a circle touches a line when the line is exactly one radius from the centre.

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Theory question 110

Polynomials

(a) If α and β are the roots of the equation 2x2 + 5x − 6 = 0, find the equation whose roots are (α − 2) and (β − 2).
(b) Given that ∫0k(x2 − 2x)dx = 4/3, find the values of k.

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Model answer

(a) α + β = −5/2, αβ = −3
Sum of new roots = α + β − 4 = −13/2
Product = αβ − 2(α + β) + 4 = −3 + 5 + 4 = 6
x2 + (13/2)x + 6 = 0, i.e. 2x2 + 13x + 12 = 0
(b) [x3/3 − x2] from 0 to k = k3/3 − k2 = 4/3
k3 − 3k2 − 4 = 0
This cubic has only one real root: k ≈ 3.36 (by trial or iteration)

Explanation

Find the sum and product of the new roots from those of the old ones; for (b) integrate, substitute the limits and solve for k.

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Theory question 111

Polynomials

(a) Express (2x2 − 5x + 1)/(x3 − 4x2 + 3x) in partial fractions.
(b) If |x − 3 −4 3|
|5 2 2|
|2 −4 6 − x| = −24, find the value of x.

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Model answer

(a) x3 − 4x2 + 3x = x(x − 1)(x − 3)
A/x + B/(x − 1) + C/(x − 3): A = 1/3, B = 1, C = 2/3
= 1/(3x) + 1/(x − 1) + 2/(3(x − 3))
(b) Expanding along the first row: (x − 3)(20 − 2x) + 4(26 − 5x) + 3(−24) = −2x2 + 6x − 28
−2x2 + 6x − 28 = −24, x2 − 3x + 2 = 0
x = 1 or x = 2

Explanation

Factorise the denominator fully before splitting; expand the determinant along the first row.

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Theory question 112

Matrices and Determinants

(a) The point P(3, −5) is rotated through an angle 60° anticlockwise about the origin. (i) Obtain the matrix for the rotation. (ii) Find the image P′ of point P under the rotation.
(b) A linear transformation is given by N: (x, y) → (2x + 3y, 3x − y). (i) Write down the matrix N of the transformation. (ii) If N2 + aN + bI = O, where a, b ∈ R, I is the 2 × 2 unit matrix and O is the 2 × 2 null matrix, find the values of a and b.

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Model answer

(a)(i) (cos 60° −sin 60°; sin 60° cos 60°) = (1/2 −√3/2; √3/2 1/2)
(ii) x′ = 3/2 + 5√3/2 = (3 + 5√3)/2, y′ = 3√3/2 − 5/2 = (3√3 − 5)/2
P′((3 + 5√3)/2, (3√3 − 5)/2) ≈ (5.83, 0.10)
(b)(i) N = (2 3; 3 −1)
(ii) N2 = (13 3; 3 10)
13 + 2a + b = 0 and 3 + 3a = 0
a = −1, b = −11 (check: 10 − a·1 + b = 10 + 1 − 11 = 0)

Explanation

Matrices are written (row 1; row 2). Compare corresponding entries of N² + aN + bI with zero.

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Theory question 113

Probability

A survey indicated that 65% of the families in an area have cars. Find, correct to three decimal places, the probability that among 7 families selected at random in the area (a) exactly 5; (b) 3 or 4; (c) at most 2 of them have cars.

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Model answer

Binomial with n = 7, p = 0.65, q = 0.35
(a) 7C5(0.65)5(0.35)2 = 0.298
(b) P(3) + P(4) = 0.1442 + 0.2679 = 0.412
(c) P(0) + P(1) + P(2) = 0.0006 + 0.0084 + 0.0466 = 0.056

Explanation

Use P(r) = ⁿCᵣ pʳ qⁿ⁻ʳ and add the cases needed.

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Theory question 114

Statistics

The following table shows the distribution of marks obtained by some students in an examination.
Marks: 0 − 9, 10 − 19, 20 − 29, 30 − 39, 40 − 49, 50 − 59, 60 − 69, 70 − 79, 80 − 89, 90 − 99
No. of students: 50, 50, 40, 60, 100, 100, 50, 25, 15, 10
(a) Construct a cumulative frequency table for the distribution. (b) Draw an ogive for the distribution. (c) Use your graph in (b) to determine the (i) semi-interquartile range; (ii) number of students who failed, if the pass mark for the examination is 37; (iii) probability that a student selected at random scored between 20% and 60%.

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Model answer

(a) Cumulative frequencies: 50, 100, 140, 200, 300, 400, 450, 475, 490, 500
(b) Plot cumulative frequency against upper class boundaries 9.5, 19.5, …, 99.5 and join with a smooth curve
(c)(i) Q1 (125th) ≈ 25.8, Q3 (375th) ≈ 57; semi-interquartile range ≈ (57 − 25.8)/2 ≈ 15.6 (about 15–16 from the graph)
(ii) Reading at 36.5: about 182 students failed (about 180)
(iii) (400 − 100)/500 = 3/5

Explanation

Read quartiles at the 125th and 375th positions of 500 on the ogive.

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Theory question 115

Permutations and Combinations

(a) Five (5) female and seven (7) male teachers applied for 4 vacancies in a Junior High School. The teachers are equally qualified. Find the number of ways of employing the 4 teachers. If (i) there is no restriction; (ii) at least 2 of them are females.
(b) The table shows the positions awarded to 7 contestants by Judges x and y in a competition.
Contestant: P, Q, R, S, T, U, V
Judges X: 2, 7, 1, 3, 6, 5, 4
Judges Y: 4, 6, 2, 3, 7, 1, 5
(i) Calculate, correct to one decimal place, the Spearman's rank correlation coefficient. (ii) interprete your answer in (b)(i) above.

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Model answer

(a)(i) 12C4 = 495
(ii) 5C2×7C2 + 5C3×7C1 + 5C4 = 210 + 70 + 5 = 285
(b)(i) d = −2, 1, −1, 0, −1, 4, −1; Σd2 = 24
ρ = 1 − 6(24)/(7(49 − 1)) = 1 − 144/336 = 0.57 ≈ 0.6
(ii) There is a moderate positive agreement between the two judges' rankings

Explanation

'At least 2 females' means 2, 3 or 4 females; add the three cases.

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Theory question 116

Vectors

The position vectors of points 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 position vector of M; (b) |PM| and |PQ|; (c) the acute angle between PM and PQ, correct to one decimal place; (d) the area of PQRM.

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Model answer

(a) OM = OP + OR − OQ = −2i − 6j
(b) PM = −6i − j, |PM| = √37; PQ = −3i + 8j, |PQ| = √73
(c) PM·PQ = 18 − 8 = 10; cos θ = 10/(√37 √73) = 0.1924; θ = 78.9°
(d) Area = |PM||PQ| sin θ = √37 × √73 × sin 78.9° = 51 square units

Explanation

In parallelogram PQRM, PM = QR, so M = P + R − Q.

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Theory question 117

Dynamics

(a) Forces F1 = (−5, 4)N, F2 = (2, 5)N, F3 = (2, −1)N, F4 = (3, −5)N, act on a body, Find the: (i) resultant of these forces; (ii) fifth force that will keep the system in equilibrium.
(b) A body, moving at 20ms−1 accelerates uniformly at 2.5ms−2 for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest t seconds with a uniform retardation. If the ratio of acceleration to retardation is 3:4 (i) sketch the velocity - time graph of the journey; (ii) find t, (iii) find the total distance of the journey.

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Model answer

(a)(i) R = (−5 + 2 + 2 + 3, 4 + 5 − 1 − 5) = (2, 3)N (magnitude √13 N)
(ii) Fifth force = −R = (−2, −3)N
(b)(i) Graph: line from 20 to 30 ms−1 over 0–4 s, horizontal at 30 ms−1 from 4 to 12 s, then down to 0
Speed after 4 s: 20 + 2.5 × 4 = 30 ms−1
(ii) Retardation = 2.5 × 4/3 = 10/3 ms−2; t = 30 ÷ 10/3 = 9 s
(iii) Distance = area: ½(20 + 30)(4) + 30 × 8 + ½(30)(9) = 100 + 240 + 135 = 475 m

Explanation

The equilibrant is equal and opposite to the resultant; distance is the area under the velocity–time graph.

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Theory question 118

Dynamics

(a) An object is thrown up a smooth plane inclined at an angle of 30° to the horizontal. If the plane is 15m long and the object comes to rest at the top, find the (i) initial speed of the object (ii) time taken to reach the top.
(b) Force of magnitudes 5N, 5√3 N, 10N, 5√3 and 5N act on a body P, of mass 5kg as shown in the diagram. Find the (i) magnitude of the resultant force; (ii) acceleration of the body.

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Model answer

(a) Retardation = g sin 30° = 10 × ½ = 5 ms−2 (taking g = 10)
(i) 0 = u2 − 2(5)(15), u = √150 ≈ 12.2 ms−1
(ii) t = u/5 = 2.45 s
(b) Forces at 60°, 30°, 0°, −30°, −60° to the 10N direction: components across cancel
Along: 10 + 2(5√3 cos 30°) + 2(5 cos 60°) = 10 + 15 + 5 = 30 N
(i) Resultant = 30 N along the 10N force
(ii) a = F/m = 30/5 = 6 ms−2

Explanation

On a smooth slope the only retarding force is the weight component mg sin θ.

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