WAEC Further Mathematics 2023 past questions and answers

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

Question 1

Surds

Express 3/(3 − √6) in the form x + m√y.

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

Answer: B (3 + √6)

Explanation

Multiply top and bottom by (3 + √6): 3(3 + √6)/(9 − 6) = 3 + √6.

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Question 2

Indices and Logarithms

If (1/9)2x−1 = (1/81)2−3x, find the value of x.

  1. A.3/4
  2. B.5/8
  3. C.−5/8
  4. D.−3/4
Show the answer and explanation

Answer: B (5/8)

Explanation

Write both sides as powers of 3: 3^(−4x + 2) = 3^(−8 + 12x). So −4x + 2 = 12x − 8, 16x = 10 and x = 5/8.

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Question 3

Polynomials

If (x − 5) is a factor of x3 − 4x2 − 11x + 30, find the remaining factors.

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

Answer: C ((x + 3) and (x − 2))

Explanation

Dividing by (x − 5) leaves x² + x − 6 = (x + 3)(x − 2).

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Question 4

Sets and Logic

Consider the statements:
x: The school bus arrived late
y: The students walked down to school.
Which of the following can be represented by y ⇒ x?

  1. A.Emmanuella did not go to school because the school bus arrived late
  2. B.The school bus arrived early and Kate ran to school
  3. C.Mary walked to school because the school bus arrived late.
  4. D.Either the school bus arrived late or Maryam walked to school
Show the answer and explanation

Answer: C (Mary walked to school because the school bus arrived late.)

Explanation

Only option C links the two statements – walking to school and the bus arriving late – as one following from the other; A, B and D involve other statements or 'or'/'and'.

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Question 5

Integration

Evaluate: ∫01 x(x2 − 2)2 dx.

  1. A.1/7
  2. B.6/7
  3. C.1⅙
  4. D.3⅙
Show the answer and explanation

Answer: C (1⅙)

Explanation

Let u = x² − 2, du = 2x dx; u goes from −2 to −1. ½∫u² du = ½[u³/3] from −2 to −1 = ½(−1/3 + 8/3) = 7/6 = 1⅙.

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Question 6

Permutations and Combinations

In how many ways can a committee of 3 women and 2 men be chosen from a group of 7 men and 5 women?

  1. A.720
  2. B.500
  3. C.350
  4. D.210
Show the answer and explanation

Answer: D (210)

Explanation

⁵C₃ × ⁷C₂ = 10 × 21 = 210.

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Question 8

Polynomials

If α and β are the roots of 7x2 + 12x − 4 = 0, find the value of αβ/(α + β)2.

  1. A.36/7
  2. B.7/36
  3. C.−7/36
  4. D.−36/7
Show the answer and explanation

Answer: C (−7/36)

Explanation

α + β = −12/7 and αβ = −4/7. αβ/(α + β)² = (−4/7) ÷ (144/49) = −4/7 × 49/144 = −7/36.

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Question 9

Functions

If f: x → 2 tan x and g: x → √(x2 + 8), find (g o f)(45°)

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

Answer: D (2√3)

Explanation

f(45°) = 2 tan 45° = 2. Then g(2) = √(4 + 8) = √12 = 2√3.

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Question 10

Sequences and Series

An exponential sequence (G.P.) is given by 9/2, 3/4, 1/8, …. Find its sum to infinity

  1. A.4⅕
  2. B.5⅖
  3. C.6¾
  4. D.13½
Show the answer and explanation

Answer: B (5⅖)

Explanation

r = (3/4) ÷ (9/2) = 1/6. S∞ = a/(1 − r) = (9/2) ÷ (5/6) = 27/5 = 5⅖.

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Question 12

Coordinate Geometry

Given that M is the midpoint of T(2, 4) and Q(−8, 6), find the length MQ.

  1. A.√24 units
  2. B.√26 units
  3. C.√28 units
  4. D.√30 units
Show the answer and explanation

Answer: B (√26 units)

Explanation

M = (−3, 5). MQ = √((−8 + 3)² + (6 − 5)²) = √(25 + 1) = √26.

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Question 13

Differentiation

Given that y2 + xy = 5, find dy/dx.

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

Answer: B (−y/(2y + x))

Explanation

Differentiate implicitly: 2y dy/dx + x dy/dx + y = 0, so dy/dx = −y/(2y + x).

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Question 14

Sets and Logic

Given that P = {x : 2 ≤ x ≤ 8} and Q = {x : 4 < x ≤ 12} are subsets of the universal set μ = {x : x ∈ R}, find (P ∩ Q′).

  1. A.{x : 4 ≤ x ≤ 8}
  2. B.{x : 2 ≤ x ≤ 4}
  3. C.{x : 4 < x < 8}
  4. D.{x : 2 < x ≤ 4}
Show the answer and explanation

Answer: B ({x : 2 ≤ x ≤ 4})

Explanation

Q′ = {x ≤ 4 or x > 12}. The part of P (2 to 8) that lies in Q′ is 2 ≤ x ≤ 4.

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Question 15

Probability

The probabilities that Atta and Tunde will hit a target in a shooting contest are 1/6 and 1/9 respectively. Find the probability that only one of them will hit the target.

  1. A.1/54
  2. B.13/54
  3. C.20/27
  4. D.41/54
Show the answer and explanation

Answer: B (13/54)

Explanation

P(Atta only) = (1/6)(8/9) = 8/54; P(Tunde only) = (5/6)(1/9) = 5/54. Total 13/54.

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Question 16

Statistics

The table shows the marks obtained by students in a test.
Marks: 1, 2, 3, 4, 5
Frequency: 2, k, 1, 1, 2
If the mean mark is 3, find the value of k.

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

Answer: A (1)

Explanation

Mean = (2 + 2k + 3 + 4 + 10)/(6 + k) = 3, so 19 + 2k = 18 + 3k and k = 1.

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Question 17

Polynomials

If 3x2 + px + 12 = 0 has equal roots, find the values of p.

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

Answer: D (±12)

Explanation

Equal roots need p² − 4(3)(12) = 0, so p² = 144 and p = ±12.

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Question 18

Indices and Logarithms

Simplify: (log √27 − log √8)/(log 3 − log 2).

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

Answer: D (3/2)

Explanation

log √27 = (3/2) log 3 and log √8 = (3/2) log 2, so the top is (3/2)(log 3 − log 2) and the fraction is 3/2.

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Question 19

Dynamics

A particle began to move at 27 ms−1 along a straight line with constant retardation of 9 ms−2. Calculate the time it took the particle to come to a stop.

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

Answer: B (3 sec.)

Explanation

v = u − at: 0 = 27 − 9t, so t = 3 s.

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Question 20

Permutations and Combinations

In how many ways can four Mathematicians be selected from six?

  1. A.15
  2. B.60
  3. C.90
  4. D.360
Show the answer and explanation

Answer: A (15)

Explanation

Order doesn't matter: ⁶C₄ = 15.

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Question 21

Matrices and Determinants

A linear transformation on the oxy plane is defined by P: (x, y) → (2x + y, −2y). Find P2. (Matrices written row by row; / separates the rows.)

  1. A.(4 1 / 0 4)
  2. B.(4 0 / 1 4)
  3. C.(4 0 / 0 4)
  4. D.(4 4 / 0 0)
Show the answer and explanation

Answer: C ((4 0 / 0 4))

Explanation

P = (2 1 / 0 −2). P² = (2×2 + 1×0, 2×1 + 1×(−2) / 0, (−2)(−2)) = (4 0 / 0 4).

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Question 22

Dynamics

The velocity of a body of mass 4.56 kg increases from (10 ms−1, 060°) to (50 ms−1, 060°) in 16 seconds. Calculate the magnitude of the force acting on it.

  1. A.36.5 N
  2. B.17.1 N
  3. C.11.4 N
  4. D.5.7 N
Show the answer and explanation

Answer: C (11.4 N)

Explanation

The direction doesn't change, so a = (50 − 10)/16 = 2.5 ms⁻² and F = 4.56 × 2.5 = 11.4 N.

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Question 23

Binomial Theorem

Find the fifth term in the binomial expansion of (q + x)7.

  1. A.21q4x3
  2. B.35q5x2
  3. C.35q3x4
  4. D.21q2x2
Show the answer and explanation

Answer: C (35q3x4)

Explanation

The fifth term is ⁷C₄ q³x⁴ = 35q³x⁴.

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Question 24

Matrices and Determinants

Given that P = (x 4 / 3 7), Q = (x 3 / 1 2x) and the determinant of Q is three more than that of P, find the values of x. (Matrices written row by row; / separates the rows.)

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

Answer: D (2, 3/2)

Explanation

|P| = 7x − 12 and |Q| = 2x² − 3. 2x² − 3 = 7x − 9 gives 2x² − 7x + 6 = 0, (2x − 3)(x − 2) = 0, so x = 2 or 3/2.

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Question 25

Trigonometry

Solve 6 sin 2θ tan θ = 4, where 0° < θ < 90°.

  1. A.35.26°
  2. B.30.00°
  3. C.19.47°
  4. D.18.43°
Show the answer and explanation

Answer: A (35.26°)

Explanation

sin 2θ tan θ = 2 sin θ cos θ × sin θ/cos θ = 2sin²θ. So 12 sin²θ = 4, sin θ = 1/√3 and θ ≈ 35.26°.

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Question 26

Vectors

Calculate, correct to one decimal place, the angle between 5i + 12j and −2i + 3j.

  1. A.54.8°
  2. B.56.3°
  3. C.66.4°
  4. D.76.3°
Show the answer and explanation

Answer: B (56.3°)

Explanation

Dot product = −10 + 36 = 26; magnitudes 13 and √13. cos θ = 26/(13√13) ≈ 0.5547, so θ ≈ 56.3°.

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Question 27

Dynamics

The distance S metres moved by a body in t seconds is given by S = 5t3 − (19/2)t2 + 6t − 4. Calculate the acceleration of the body after 2 seconds.

  1. A.19 ms−2
  2. B.21 ms−2
  3. C.31 ms−2
  4. D.41 ms−2
Show the answer and explanation

Answer: D (41 ms−2)

Explanation

v = 15t² − 19t + 6 and a = 30t − 19. At t = 2, a = 41 ms⁻².

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Question 28

Differentiation

Find the equation of the normal to the curve y = 3x2 + 2 at point (1, 5).

  1. A.6y − x − 29 = 0
  2. B.6y + x − 31 = 0
  3. C.y − 6x + 1 = 0
  4. D.y − 6x − 1 = 0
Show the answer and explanation

Answer: B (6y + x − 31 = 0)

Explanation

dy/dx = 6x = 6 at x = 1, so the normal has gradient −1/6. y − 5 = −(x − 1)/6 gives 6y + x − 31 = 0.

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Question 29

Statistics

Adu's scores in five subjects in an examination are 85, 84, 83, 86 and 87. Calculate the standard deviation.

  1. A.1.4
  2. B.1.6
  3. C.1.8
  4. D.2.0
Show the answer and explanation

Answer: A (1.4)

Explanation

Mean = 85. Squared deviations 0, 1, 4, 1, 4 add to 10. SD = √(10/5) = √2 ≈ 1.4.

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Question 30

Differentiation

Differentiate f(x) = 1/(1 − x2)5 with respect to x.

  1. A.10x/(1 − x2)6
  2. B.5x/(1 − x2)6
  3. C.−5x/(1 − x2)6
  4. D.−10x/(1 − x2)6
Show the answer and explanation

Answer: A (10x/(1 − x2)6)

Explanation

f(x) = (1 − x²)⁻⁵, so f′(x) = −5(1 − x²)⁻⁶ × (−2x) = 10x/(1 − x²)⁶.

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Question 31

Probability

If X and Y are two independent events such that P(X) = 1/8 and P(X ∪ Y) = 5/8, find P(Y).

  1. A.4/7
  2. B.3/7
  3. C.4/21
  4. D.1/6
Show the answer and explanation

Answer: A (4/7)

Explanation

For independent events P(X ∪ Y) = P(X) + P(Y) − P(X)P(Y): 5/8 = 1/8 + (7/8)P(Y), so P(Y) = 4/7.

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Question 32

Binary Operations

The table shows the operation * on the set {x, y, z, w}.
Row x: x * x = y, x * y = z, x * z = x, x * w = w
Row y: y * x = z, y * y = w, y * z = y, y * w = x
Row z: z * x = x, z * y = y, z * z = z, z * w = w
Row w: w * x = w, w * y = x, w * z = w, w * w = z
Find the identity element.

  1. A.x
  2. B.y
  3. C.z
  4. D.w
Show the answer and explanation

Answer: C (z)

Explanation

Row z and column z reproduce the other element every time (z * a = a * z = a), so z is the identity.

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Question 33

Coordinate Geometry

Find the radius of the circle 2x2 + 2y2 − 4x + 5y + 1 = 0

  1. A.√5/6
  2. B.5/6
  3. C.33/4
  4. D.√33/4
Show the answer and explanation

Answer: D (√33/4)

Explanation

Divide by 2: x² + y² − 2x + (5/2)y + ½ = 0. Centre (1, −5/4); r² = 1 + 25/16 − ½ = 33/16, so r = √33/4.

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Question 34

Integration

Evaluate: ∫(2x + 1)3 dx.

  1. A.8(2x + 1)2 + k
  2. B.6(2x + 1)2 + k
  3. C.⅙(2x + 1)4 + k
  4. D.⅛(2x + 1)4 + k
Show the answer and explanation

Answer: D (⅛(2x + 1)4 + k)

Explanation

∫(2x + 1)³ dx = (2x + 1)⁴/(4 × 2) + k = ⅛(2x + 1)⁴ + k.

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Question 35

Statics

A uniform beam PQ of length 80 cm and weight 60 N rests on a support at X where |PX| = 30 cm. If the body is kept in equilibrium by a mass m kg which is placed at P, calculate the value of m. [Take g = 10 ms−2]

  1. A.2.0
  2. B.2.5
  3. C.3.0
  4. D.4.0
Show the answer and explanation

Answer: A (2.0)

Explanation

The weight acts at the centre, 40 cm from P, i.e. 10 cm beyond X. Moments about X: 10m × 30 = 60 × 10, so m = 2.0 kg.

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Question 36

Polynomials

If m and (m + 4) are the roots of 4x2 − 4x − 15 = 0, find the equation whose roots are 2m and (2m + 8).

  1. A.x2 + 8x + 15 = 0
  2. B.x2 − 8x − 15 = 0
  3. C.x2 + 2x + 15 = 0
  4. D.x2 − 2x − 15 = 0
Show the answer and explanation

Answer: D (x2 − 2x − 15 = 0)

Explanation

The sum of the roots is 1, so 2m + 4 = 1 and m = −3/2. New roots −3 and 5: sum 2, product −15, giving x² − 2x − 15 = 0.

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Question 37

Binomial Theorem

Find the coefficient of the 6th term in the binomial expansion of (1 − 2x/3)10 in ascending powers of x.

  1. A.−896x6/9
  2. B.−896x6/27
  3. C.−896x5/9
  4. D.−896x5/27
Show the answer and explanation

Answer: D (−896x5/27)

Explanation

The 6th term is ¹⁰C₅(−2x/3)⁵ = 252 × (−32/243)x⁵ = −896x⁵/27.

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Question 38

Sequences and Series

An exponential sequence (G.P.) is given by 8√2, 16√2, 32√2, …. Find the nth term of the sequence

  1. A.8n√2
  2. B.8√2n
  3. C.√(2(n + 3))
  4. D.2(n + 2)√2
Show the answer and explanation

Answer: D (2(n + 2)√2)

Explanation

a = 8√2, r = 2. Tₙ = 8√2 × 2ⁿ⁻¹ = 2³ × 2ⁿ⁻¹ × √2 = 2ⁿ⁺²√2.

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Question 39

Vectors

Given that r = (10 N, 200°) and n = (16 N, 020°), find (3r − 2n).

  1. A.(62 N, 200°)
  2. B.(62 N, 020°)
  3. C.(62 N, 240°)
  4. D.(62 N, 280°)
Show the answer and explanation

Answer: A ((62 N, 200°))

Explanation

3r = (30 N, 200°). −2n is 32 N in the opposite direction to 020°, i.e. (32 N, 200°). Both point along 200°, so 3r − 2n = (62 N, 200°).

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Question 40

Trigonometry

Given that sin x = 4/5 and cos y = 12/13, where x is an obtuse angle and y is an acute angle, find the value of sin (x − y)

  1. A.16/65
  2. B.48/65
  3. C.56/65
  4. D.63/65
Show the answer and explanation

Answer: D (63/65)

Explanation

x obtuse: cos x = −3/5; y acute: sin y = 5/13. sin(x − y) = (4/5)(12/13) − (−3/5)(5/13) = 48/65 + 15/65 = 63/65.

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Theory question 101

Permutations and Combinations

If 9Cx = 4[7Cx−1], find the values of x.

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Model answer

9!/(x!(9 − x)!) = 4 × 7!/((x − 1)!(8 − x)!).
Cancel 7!, (x − 1)! and (8 − x)!: 72/(x(9 − x)) = 4.
x(9 − x) = 18, so x² − 9x + 18 = 0, (x − 3)(x − 6) = 0.
x = 3 or x = 6.

Explanation

Write both sides in factorials and cancel the common parts.

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Theory question 102

Differentiation

The volume of a cube is increasing at the rate of 3½ cm3s−1. Find the rate of change of the side of the base when its length is 6 cm.

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Model answer

V = s³, so dV/dt = 3s² ds/dt.
3.5 = 3(36) ds/dt.
ds/dt = 3.5/108 = 7/216 ≈ 0.0324 cm s⁻¹.

Explanation

Connect the two rates with the chain rule.

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Theory question 103

Functions

The inverse of a function f is given by f−1(x) = (5x − 6)/(4 − x), x ≠ 4. Find the: (a) function, f(x) (b) value of x for which f(x) = 5.

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Model answer

(a) f is the inverse of f⁻¹. Let y = (5x − 6)/(4 − x): 4y − xy = 5x − 6, x(5 + y) = 4y + 6.
f(x) = (4x + 6)/(x + 5), x ≠ −5.
(b) (4x + 6)/(x + 5) = 5 gives 4x + 6 = 5x + 25, x = −19 (check: f⁻¹(5) = 19/(−1) = −19).

Explanation

The inverse of f⁻¹ is f itself.

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Theory question 104

Sequences and Series

The first term of an Arithmetic Progression is −8, the last term is 52 and the sum of terms is 286. Find the: (a) number of terms in the series; (b) common difference.

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Model answer

(a) S = n(a + l)/2: 286 = n(−8 + 52)/2 = 22n, so n = 13.
(b) l = a + (n − 1)d: 52 = −8 + 12d, so d = 5.

Explanation

Use S = n(a + l)/2 when the first and last terms are known.

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Theory question 105

Statistics

The table shows the distribution of the heights (cm) of 60 seedlings in a vegetable garden.
Heights (cm): 0.1 − 0.3, 0.4 − 0.6, 0.7 − 0.9, 1.0 − 1.4, 1.5 − 1.9, 2.0 − 2.2, 2.3 − 2.5
Frequency: 6, 9, 12, 15, 3, 6, 9
(a) Draw a histogram for the distribution. (b) Use the histogram to estimate the modal height of the seedlings.

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Model answer

(a) Class boundaries (widths): 0.05–0.35 (0.3), 0.35–0.65 (0.3), 0.65–0.95 (0.3), 0.95–1.45 (0.5), 1.45–1.95 (0.5), 1.95–2.25 (0.3), 2.25–2.55 (0.3).
Widths differ, so plot frequency density (frequency per 0.1 cm): 2, 3, 4, 3, 0.6, 2, 3. Draw touching bars on the boundaries, label axes, state scale.
(b) Tallest bar: 0.65–0.95. Join its top corners to the adjacent bars' tops (both height 3) and read where the lines cross: mode ≈ 0.65 + ½ × 0.3 = 0.80 cm.

Explanation

With unequal widths, plot frequency density, not frequency.

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Theory question 106

Probability

There are 6 boys and 8 girls in a class. If five students are selected from the class, find the probability that more girls than boys are selected.

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Model answer

Total selections: ¹⁴C₅ = 2002.
3 girls, 2 boys: ⁸C₃ × ⁶C₂ = 56 × 15 = 840; 4 girls, 1 boy: 70 × 6 = 420; 5 girls: ⁸C₅ = 56.
Favourable = 1316. P = 1316/2002 = 94/143 ≈ 0.657.

Explanation

'More girls than boys' in a group of 5 means 3, 4 or 5 girls.

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Theory question 107

Dynamics

A bus travels with a velocity of 6 ms−1. It then accelerates uniformly and travels a distance of 70 m. If the final velocity is 20 ms−1, find, correct to one decimal place, the: (a) acceleration; (b) time to travel this distance.

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Model answer

(a) v² = u² + 2as: 400 = 36 + 140a, so a = 364/140 = 2.6 ms⁻².
(b) v = u + at: 20 = 6 + 2.6t, so t = 14/2.6 ≈ 5.4 s.

Explanation

Choose the equation of motion that contains the quantities you know.

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Theory question 108

Vectors

P is the mid-point of NO and equidistant from MN and MO. If MN = 8i + 3j and MO = 14i − 5j; find MP. (MN, MO and MP are vectors.)

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Model answer

P is the midpoint of NO, so MP = ½(MN + MO).
MP = ½(22i − 2j) = 11i − j.

Explanation

The position of a midpoint is the average of the two end vectors.

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Theory question 109

Differentiation

(a) Find the derivative of 4x − 7/x2 with respect to x, from first principle.
(b) Given that tan P = 3/(x − 1) and tan Q = 2/(x − 1), find tan (P − Q).

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Model answer

(a) y + δy = 4(x + δx) − 7/(x + δx)²; δy = 4δx + 7[1/x² − 1/(x + δx)²] = 4δx + 7(2xδx + δx²)/(x²(x + δx)²).
δy/δx = 4 + 7(2x + δx)/(x²(x + δx)²). As δx → 0: dy/dx = 4 + 14x/x⁴ = 4 + 14/x³.
(b) tan(P − Q) = (3/(x − 1) − 2/(x − 1)) ÷ (1 + 6/(x − 1)²) = (1/(x − 1)) × (x − 1)²/((x − 1)² + 6) = (x − 1)/(x² − 2x + 7).

Explanation

From first principles: find δy, divide by δx, then let δx → 0.

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Theory question 110

Polynomials

(a) A quadratic polynomial, g(x) has (2x + 1) as a factor. If g(x) is divided by (x + 1) and (x − 2), the remainders are −6 and −5 respectively. Find: (i) g(x); (ii) the zeros of g(x).
(b) Find the third term when (x/2 − 1)8 is expanded in descending powers of x.

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Model answer

(a)(i) Let g(x) = (2x + 1)(ax + b). g(−1) = −(−a + b) = a − b = −6; g(2) = 5(2a + b) = −5, so 2a + b = −1.
a = −7/3, b = 11/3: g(x) = (2x + 1)(11 − 7x)/3 = (−14x² + 15x + 11)/3.
(ii) Zeros: x = −½ and x = 11/7.
(b) Third term = ⁸C₂(x/2)⁶(−1)² = 28x⁶/64 = 7x⁶/16.

Explanation

Use the remainder theorem: the remainder on dividing by (x − c) is g(c).

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Theory question 111

Polynomials

(a) Express (8x2 + 8x + 9)/((x − 1)(2x + 3)2) in partial fractions.
(b) The coordinates of the centre and circumference of a circle are (−2, 5) and 6π units respectively. Find the equation of the circle.

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(a) Let it equal A/(x − 1) + B/(2x + 3) + C/(2x + 3)²; 8x² + 8x + 9 = A(2x + 3)² + B(x − 1)(2x + 3) + C(x − 1).
x = 1: 25 = 25A, A = 1. x = −3/2: 15 = −(5/2)C, C = −6. x² terms: 8 = 4A + 2B, B = 2.
Answer: 1/(x − 1) + 2/(2x + 3) − 6/(2x + 3)².
(b) 2πr = 6π gives r = 3. (x + 2)² + (y − 5)² = 9, i.e. x² + y² + 4x − 10y + 20 = 0.

Explanation

A repeated factor needs a fraction for each power of it.

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Theory question 112

Statistics

The table shows the distribution of marks scored by some candidates in an examination
Marks: 11 − 20, 21 − 30, 31 − 40, 41 − 50, 51 − 60, 61 − 70, 71 − 80, 81 − 90, 91 − 100
Number of candidates: 5, 39, 14, 40, 57, 25, 11, 8, 1
(a) Construct a cumulative frequency table for the distribution. (b) Draw a cumulative frequency curve for the distribution. (c) Use the curve to estimate the: (i) number of candidates who scored marks between 24 and 58; (ii) lowest mark for distinction, if 12% of the candidates passed with distinction.

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(a) Upper class boundaries 20.5, 30.5, …, 100.5; cumulative frequencies 5, 44, 58, 98, 155, 180, 191, 199, 200 (total 200).
(b) Plot cumulative frequency against upper boundary (start at (10.5, 0)) and join with a smooth S-shaped curve.
(c)(i) From the curve: about 19 scored up to 24 and about 141 up to 58, so about 122 candidates.
(ii) 12% of 200 = 24 candidates, i.e. cumulative frequency 176: reading across gives a mark of about 69.

Explanation

Plot cumulative frequencies at the upper class boundaries.

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Theory question 113

Probability

(a) A bag contains 16 identical balls of which 4 are green. A boy picks a ball at random from the bag and replaces it. If this is repeated 5 times, what is the probability that he: (i) did not pick a green ball; (ii) picked a green ball at least three times?
(b) The deviations from a mean of values from a set of data are −2, (m − 1), (m2 + 1), −1, 2, (2m − 1) and −2. Find the possible values of m.

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Model answer

(a) p(green) = 1/4, q = 3/4, n = 5.
(i) (3/4)⁵ = 243/1024 ≈ 0.237.
(ii) P(3) + P(4) + P(5) = 10(1/4)³(3/4)² + 5(1/4)⁴(3/4) + (1/4)⁵ = (90 + 15 + 1)/1024 = 53/512 ≈ 0.104.
(b) Deviations from the mean add up to 0: m² + 3m − 4 = 0, (m + 4)(m − 1) = 0, so m = 1 or m = −4.

Explanation

Deviations from the mean always sum to zero.

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Theory question 114

Dynamics

A particle of mass 2 kg moves under the action of a constant force, F N, with an initial velocity (3i + 2j) ms−1 and a velocity of (15i − 4j) ms−1 after 4 seconds. Find the: (a) acceleration of the particle; (b) magnitude of the force F; (c) magnitude of the velocity of the particle after 8 seconds, correct to three decimal places.

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(a) a = (v − u)/t = (12i − 6j)/4 = 3i − 1.5j ms⁻².
(b) F = ma = 6i − 3j; |F| = √45 ≈ 6.71 N.
(c) v = u + at = (3i + 2j) + 8(3i − 1.5j) = 27i − 10j; |v| = √829 ≈ 28.792 ms⁻¹.

Explanation

Work with the i and j parts together as vectors.

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Theory question 115

Statics

(a) A see-saw pivoted at the middle is kept in balance by weights of Richard, John and Philip such that only Richard whose mass is 60 kg sits on one side. If they sit at distances 2m, 3m and 4m respectively from the pivot and Philip is 15 kg, find the mass of John.
(b) A body of mass 12 kg rests on a rough plane inclined at an angle of 30° to the horizontal. The coeffient of friction between the body and the plane is 2/3. A force of magnitude P Newtons acts on the body along the inclined plane. Find the value of P, if the body is at the point of moving: (i) down the plane; (ii) up the plane.

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(a) Moments about the pivot: 60 × 2 = J × 3 + 15 × 4, so 120 = 3J + 60 and John's mass J = 20 kg.
(b) (g = 10) R = 120 cos 30° ≈ 103.9 N; limiting friction = ⅔ × 103.9 ≈ 69.3 N; weight component down the plane = 120 sin 30° = 60 N.
(i) About to move down (friction up the plane): P + 60 = 69.3, P ≈ 9.3 N.
(ii) About to move up (friction down the plane): P = 60 + 69.3 ≈ 129.3 N.

Explanation

Balance clockwise and anticlockwise moments; friction opposes the motion about to happen.

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