WAEC Further Mathematics 2013 past questions and answers

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

Question 2

Trigonometry

Solve Sin θ = tan θ.

  1. A.200°
  2. B.90°
  3. C.60°
  4. D.0°
Show the answer and explanation

Answer: D (0°)

Explanation

sin θ = sin θ/cos θ gives sin θ = 0 or cos θ = 1. Of the options only θ = 0° works.

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Question 3

Indices and Logarithms

Given that a5/6 × a−1/n = 1, solve for n.

  1. A.−6.00
  2. B.−1.20
  3. C.0.83
  4. D.1.20
Show the answer and explanation

Answer: D (1.20)

Explanation

a^(5/6 − 1/n) = a⁰, so 5/6 = 1/n and n = 6/5 = 1.20.

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Question 4

Indices and Logarithms

Express log ¼ + log ½ in terms of log 2.

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

Answer: C (−3 log 2)

Explanation

log ¼ + log ½ = log ⅛ = log 2⁻³ = −3 log 2.

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Question 5

Functions

If f(x) = x2 and g(x) = Sin x, find g o f.

  1. A.Sin2x
  2. B.Sin x2
  3. C.(Sin x)x2
  4. D.x Sin x
Show the answer and explanation

Answer: B (Sin x2)

Explanation

g o f(x) = g(f(x)) = g(x²) = sin x².

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Question 6

Binomial Theorem

Find the third term in the expansion of (a − b)6 in ascending powers of b.

  1. A.−15a4b2
  2. B.15a4b2
  3. C.−15a3b3
  4. D.15a3b3
Show the answer and explanation

Answer: B (15a4b2)

Explanation

Third term = ⁶C₂ a⁴(−b)² = 15a⁴b².

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Question 7

Surds

If √x + √(x + 1) = √(2x + 1), find the possible value of x.

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

Answer: D (0 and −1)

Explanation

Squaring: 2x + 1 + 2√(x(x + 1)) = 2x + 1, so x(x + 1) = 0 and x = 0 or −1. Only D contains these values (and x = 0 is the one that checks in the real numbers).

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Question 8

Polynomials

If α and β are the roots of the equation 2x2 − 6x + 5 = 0, evaluate β/α + α/β.

  1. A.24/5
  2. B.8/5
  3. C.5/8
  4. D.5/24
Show the answer and explanation

Answer: B (8/5)

Explanation

α + β = 3, αβ = 5/2. β/α + α/β = ((α + β)² − 2αβ)/αβ = (9 − 5)/(5/2) = 8/5.

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Question 9

Polynomials

Given that f(x) = 2x3 − 3x2 − 11x + 6 and f(3) = 0, factorize f(x).

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

Answer: C ((x − 3)(x + 2)(2x − 1))

Explanation

Dividing by (x − 3) gives 2x² + 3x − 2 = (2x − 1)(x + 2). So f(x) = (x − 3)(x + 2)(2x − 1).

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Question 10

Coordinate Geometry

Find the equation of the line that is perpendicular to 2y + 5x − 6 = 0 and bisects the line joining the points P(4, 3) and Q(−6, 1).

  1. A.y + 5x + 3 = 0
  2. B.2y − 5x − 9 = 0
  3. C.5y + 2x − 8 = 0
  4. D.5y − 2x − 12 = 0
Show the answer and explanation

Answer: D (5y − 2x − 12 = 0)

Explanation

The given line has gradient −5/2, so the perpendicular has gradient 2/5. Midpoint of PQ is (−1, 2): y − 2 = (2/5)(x + 1), i.e. 5y − 2x − 12 = 0.

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Question 11

Differentiation

Differentiate x2 + xy − 5 = 0 with respect to x.

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

Answer: A (−(2x + y)/x)

Explanation

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

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Question 12

Sequences and Series

The fourth term of an exponential sequence is 192 and its ninth term is 6. Find the common ratio of the sequence.

  1. A.1/3
  2. B.1/2
  3. C.2
  4. D.3
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Answer: B (1/2)

Explanation

ar⁸ ÷ ar³ = r⁵ = 6/192 = 1/32, so r = ½.

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Question 13

Polynomials

Find the range of values of x for which x2 + 4x + 5 is less than 3x2 − x + 2.

  1. A.x > −½, x > 3
  2. B.x < −½, x > 3
  3. C.−½ ≤ x ≤ 3
  4. D.−½ < x < 3
Show the answer and explanation

Answer: B (x < −½, x > 3)

Explanation

We need 2x² − 5x − 3 > 0, i.e. (2x + 1)(x − 3) > 0, so x < −½ or x > 3.

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Question 14

Integration

Given that dy/dx = √x, find y.

  1. A.2x3/2 + c
  2. B.(2/3)x3/2 + c
  3. C.−½ ≤ x ≤ 3
  4. D.(2/3)x2 + c
Show the answer and explanation

Answer: B ((2/3)x3/2 + c)

Explanation

y = ∫x^½ dx = x^(3/2)/(3/2) + c = (2/3)x^(3/2) + c.

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Question 15

Matrices and Determinants

Given that P = (y − 2 y − 1 / y − 4 y + 2) and |P| = −23, find the value of y. (Matrix written row by row; / separates the rows.)

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

Answer: B (−3)

Explanation

|P| = (y − 2)(y + 2) − (y − 1)(y − 4) = 5y − 8 = −23, so y = −3.

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Question 16

Dynamics

An object is thrown vertically upwards from the top of a cliff with a velocity of 25 ms−1. Find the time in seconds, when it is 20 meters above the cliff. [Take g = 10 ms−2]

  1. A.0 and 1
  2. B.0 and 4
  3. C.0 and 5
  4. D.1 and 4
Show the answer and explanation

Answer: D (1 and 4)

Explanation

20 = 25t − 5t², so t² − 5t + 4 = 0 and t = 1 or 4 (going up, then coming down).

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Question 17

Integration

Evaluate ∫02 (8x − 4x2) dx.

  1. A.−16
  2. B.−16/3
  3. C.16/3
  4. D.16
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Answer: C (16/3)

Explanation

[4x² − (4/3)x³] from 0 to 2 = 16 − 32/3 = 16/3.

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Question 18

Coordinate Geometry

Find the coordinates of the point which divides the line joining P(−2, 3) and Q(4, 9) internally in the ratio 2 : 3.

  1. A.(5⅔, ⅖)
  2. B.(⅖, 5⅖)
  3. C.(⅖, 2⅖)
  4. D.(−⅖, 5⅖)
Show the answer and explanation

Answer: B ((⅖, 5⅖))

Explanation

x = (3(−2) + 2(4))/5 = 2/5, y = (3(3) + 2(9))/5 = 27/5 = 5⅖.

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Question 19

Trigonometry

The angle subtended by an arc of a circle at the centre is π/3 radians. If the radius of the circle is 12 cm, calculate the perimeter of the major sector.

  1. A.4(6 + 5π)
  2. B.4(6 + 2π)
  3. C.4(3 + 3π)
  4. D.4(3 + 5π)
Show the answer and explanation

Answer: A (4(6 + 5π))

Explanation

The major sector angle is 2π − π/3 = 5π/3. Perimeter = 2(12) + 12(5π/3) = 24 + 20π = 4(6 + 5π).

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Question 20

Functions

The function f : R → R is defined by f(x) = 3x + 2 : x > 4; 3x − 2 : x = 4; 5x − 3 : x < 4. Find f(4) − f(−3).

  1. A.28
  2. B.26
  3. C.−26
  4. D.−28
Show the answer and explanation

Answer: A (28)

Explanation

f(4) = 3(4) − 2 = 10 and f(−3) = 5(−3) − 3 = −18, so f(4) − f(−3) = 28.

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Question 21

Permutations and Combinations

A committee consists of 5 boys namely Kofi, John, Ojo, Ozo and James and 3 girls namely: Rose, Ugo and Ama. In how many ways can a subcommittee consisting of 3 boys and 2 girls be chosen, if Ozo must be on the sub-committee?

  1. A.35
  2. B.30
  3. C.18
  4. D.12
Show the answer and explanation

Answer: C (18)

Explanation

Ozo takes one place, so choose 2 of the other 4 boys (⁴C₂ = 6) and 2 of the 3 girls (³C₂ = 3): 6 × 3 = 18.

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Question 22

Statics

Forces 50N and 80N act on a body as shown in the diagram. Find, correct to the nearest whole number, the horizontal component of the resultant force.

Diagram for this question
  1. A.13N
  2. B.43N
  3. C.57N
  4. D.95N
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Answer: A (13N)

Explanation

Horizontal: 80 cos 45° − 50 cos 30° = 56.57 − 43.30 = 13.27 ≈ 13 N.

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Question 23

Statistics

The sales of five sales girls on a certain day are as follows: GH¢26.00, GH¢39.00, GH¢33.00, GH¢25.00, GH¢37.00. Calculate the standard deviation if the mean sale is GH¢32.00.

  1. A.GH¢5.65
  2. B.GH¢5.66
  3. C.GH¢6.55
  4. D.GH¢6.56
Show the answer and explanation

Answer: B (GH¢5.66)

Explanation

Deviations −6, 7, 1, −7, 5; squares add to 160. SD = √(160/5) = √32 ≈ 5.66.

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Question 24

Differentiation

A circular ink blot on a piece of paper increases its area at the rate of 4 mm2/s. Find the rate of increase of the radius of the blot when the radius is 8 mm. [Take π = 22/7].

  1. A.0.25 mm/s
  2. B.0.20 mm/s
  3. C.0.08 mm/s
  4. D.0.05 mm/s
Show the answer and explanation

Answer: C (0.08 mm/s)

Explanation

dA/dt = 2πr × dr/dt, so dr/dt = 4/(2 × 22/7 × 8) = 7/88 ≈ 0.08 mm/s.

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Question 25

Polynomials

Express (x2 + x + 4)/((1 − x)(x2 + 1)) in partial fractions.

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

Answer: B (3/(1 − x) + (2x + 1)/(x2 + 1))

Explanation

Write A/(1 − x) + (Bx + C)/(x² + 1). x = 1 gives 6 = 2A, A = 3; comparing x² terms, 1 = 3 − B so B = 2; constants, 4 = 3 + C so C = 1.

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Question 26

Dynamics

Two bodies of masses 3kg and 5kg moving with velocities 2 ms−1 and V ms−1 respectively in opposite directions collide. If they move together after collision with velocity 3.5 ms−1 in the direction of the 5kg mass, find the value of V.

  1. A.7.8 ms−1
  2. B.6.8 ms−1
  3. C.5.6 ms−1
  4. D.4.6 ms−1
Show the answer and explanation

Answer: B (6.8 ms−1)

Explanation

Taking the 3 kg body's direction as positive: 3(2) − 5V = 8(−3.5), so 6 − 5V = −28 and V = 6.8.

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Question 27

Coordinate Geometry

The equation of a circle is x2 + y2 − 8x + 9y + 15 = 0. Find its radius.

  1. A.√5
  2. B.½√15
  3. C.½√85
  4. D.√85
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Answer: C (½√85)

Explanation

Centre (4, −9/2); r² = 16 + 81/4 − 15 = 85/4, so r = ½√85.

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Question 28

Statics

A particle is acted upon by two forces 6N and 3N inclined at an angle of 120° to each other. Find the magnitude of the resultant force.

  1. A.18√3 N
  2. B.27N
  3. C.24N
  4. D.3√3 N
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Answer: D (3√3 N)

Explanation

R² = 6² + 3² + 2(6)(3) cos 120° = 45 − 18 = 27, so R = 3√3 N.

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Question 29

Vectors

If s = 3i − j and t = 2i + 3j, find (t + 3s) · (t − 3s).

  1. A.−77
  2. B.−71
  3. C.−53
  4. D.−41
Show the answer and explanation

Answer: A (−77)

Explanation

t + 3s = 11i + 0j and t − 3s = −7i + 6j. Dot product = 11(−7) + 0(6) = −77.

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Question 30

Trigonometry

If 2sin2 θ = 1 + Cos θ, 0° ≤ θ ≤ 90°, find θ.

  1. A.30°
  2. B.45°
  3. C.60°
  4. D.90°
Show the answer and explanation

Answer: C (60°)

Explanation

2(1 − cos²θ) = 1 + cos θ gives 2cos²θ + cos θ − 1 = 0, so cos θ = ½ (or −1). θ = 60°.

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Question 31

Statistics

Find the upper quartile of the following scores: 41, 29, 17, 2, 12, 33, 45, 18, 21, 43 and 5.

  1. A.45
  2. B.41
  3. C.33
  4. D.21
Show the answer and explanation

Answer: B (41)

Explanation

In order: 2, 5, 12, 17, 18, 21, 29, 33, 41, 43, 45. The upper quartile is the ¾(11 + 1) = 9th score, 41.

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Question 32

Matrices and Determinants

Given that P = (3 4 / 2 x), Q = (1 3 / −2 4), R = (−5 25 / −8 26) and PQ = R, find the value of x. (Matrices written row by row; / separates the rows.)

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

Answer: D (5)

Explanation

PQ = (−5 25 / 2 − 2x 6 + 4x). So 2 − 2x = −8 (and 6 + 4x = 26), giving x = 5.

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Question 33

Probability

Two out of 10 tickets on sale for a raffle draw are winning tickets. If a guest bought two tickets, what is the probability that both tickets are winning ticket?

  1. A.1/90
  2. B.1/45
  3. C.1/20
  4. D.1/10
Show the answer and explanation

Answer: B (1/45)

Explanation

P = 2/10 × 1/9 = 1/45.

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Question 34

Vectors

P and Q are the points (3, 1) and (7, 4) respectively. Find the unit vector along PQ. (Column vectors written as (x, y).)

  1. A.(4, 3)
  2. B.(0.6, 0.8)
  3. C.(0.8, 0.6)
  4. D.(−0.8, −0.6)
Show the answer and explanation

Answer: C ((0.8, 0.6))

Explanation

PQ = (7 − 3, 4 − 1) = (4, 3) and |PQ| = 5, so the unit vector is (0.8, 0.6).

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Question 35

Functions

If g(x) = (x + 1)/(x + 2), (x ≠ −2), find g−1(2).

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

Answer: D (−3)

Explanation

Solve (x + 1)/(x + 2) = 2: x + 1 = 2x + 4, so x = −3.

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Question 36

Statistics

Calculate the mean deviation of 1, 2, 3, 4, 5, 5, 6, 7, 8, 9.

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

Answer: A (2)

Explanation

Mean = 50/10 = 5. The absolute deviations 4, 3, 2, 1, 0, 0, 1, 2, 3, 4 add to 20, so MD = 20/10 = 2.

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Question 37

Vectors

If V = (−2, 4) and U = (−1, 5) (column vectors), find |U + V|.

  1. A.3√10
  2. B.√82
  3. C.√46
  4. D.2√5
Show the answer and explanation

Answer: A (3√10)

Explanation

U + V = (−3, 9), |U + V| = √(9 + 81) = √90 = 3√10.

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Question 38

Coordinate Geometry

Find the equation of the straight line that passes through (2, −3) and perpendicular to the line 3x − 2y + 4 = 0.

  1. A.2y − 3x = 0
  2. B.3y − 2x + 5 = 0
  3. C.3y + 2x + 5 = 0
  4. D.2y − 3x − 5 = 0
Show the answer and explanation

Answer: C (3y + 2x + 5 = 0)

Explanation

The given line has gradient 3/2, so the perpendicular has gradient −2/3: 3(y + 3) = −2(x − 2), i.e. 3y + 2x + 5 = 0.

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Question 39

Permutations and Combinations

If nC3/nP2 = 1, find the value of n.

  1. A.8
  2. B.7
  3. C.6
  4. D.5
Show the answer and explanation

Answer: A (8)

Explanation

ⁿC₃ ÷ ⁿP₂ = [n!/(3!(n − 3)!)] × [(n − 2)!/n!] = (n − 2)/6 = 1, so n = 8.

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Question 40

Statics

A body is kept at rest by three forces F1 = (10N, 030°), F2 = (10N, 150°) and F3. Find F3.

  1. A.(12N, 090°)
  2. B.(10 N, 270°)
  3. C.(10 N, 180°)
  4. D.(10N, 120°)
Show the answer and explanation

Answer: B ((10 N, 270°))

Explanation

F₃ must equal −(F₁ + F₂). As bearings, F₁ + F₂ = 10 N due east, so F₃ = (10 N, 270°). (Measuring angles from the x-axis instead, F₁ + F₂ = 10 N at 90° and F₃ is again at 270°.)

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Theory question 101

Binomial Theorem

(a) If the coefficient of x2 and x3 in the expansion of (p + qx)7 are equal, express q in terms of p.
(b) A man makes a weekly contribution into a fund. In the first week, he paid ₦180.00, second week ₦260.00, third week ₦340.00 and so on. How much would he have contributed in 16 weeks?

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Model answer

(a) Coefficient of x² = ⁷C₂p⁵q² = 21p⁵q²; of x³ = ⁷C₃p⁴q³ = 35p⁴q³.
21p⁵q² = 35p⁴q³ gives q = 21p/35 = (3/5)p.
(b) A.P. with a = 180, d = 80, n = 16.
S₁₆ = (16/2)[2(180) + 15(80)] = 8 × 1560 = ₦12,480.00.

Explanation

The general term of (p + qx)ⁿ is ⁿCᵣ pⁿ⁻ʳ(qx)ʳ.

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Theory question 102

Differentiation

Calculate the gradient of the curve x3 + y3 − 2xy = 11 at the point (2, −1).

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Model answer

Differentiate implicitly: 3x² + 3y²(dy/dx) − 2y − 2x(dy/dx) = 0.
dy/dx = (2y − 3x²)/(3y² − 2x).
At (2, −1): (−2 − 12)/(3 − 4) = 14.

Explanation

Differentiate xy with the product rule.

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Theory question 103

Differentiation

A side of a rectangle is three times the other. If the perimeter increases by 2%, find the percentage increase in the area of the rectangle.

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Model answer

Sides x and 3x: perimeter 8x, area 3x².
A 2% increase in perimeter (same shape) makes each side 1.02 times as long.
New area = 3(1.02x)² = 1.0404 × 3x².
Percentage increase in area = 4.04%.

Explanation

Area scales as the square of the length factor.

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Theory question 104

Coordinate Geometry

The line 2y = x + 3 meets the circle x2 + y2 − 2x + 6y − 15 = 0 at point M and N, where N is in the first quadrant. Find the coordinates of M and N.

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Model answer

Substitute x = 2y − 3: (2y − 3)² + y² − 2(2y − 3) + 6y − 15 = 0.
5y² − 10y = 0, so y = 0 or y = 2.
y = 0: x = −3; y = 2: x = 1.
N = (1, 2) (first quadrant), M = (−3, 0).

Explanation

Solve the line and circle simultaneously.

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Theory question 105

Probability

Three school prefects are to be chosen from four girls and five boys. What is the probability that: (a) Only boys will be chosen; (b) more girls than boys will be chosen?

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Model answer

Total selections ⁹C₃ = 84.
(a) ⁵C₃ = 10, P = 10/84 = 5/42.
(b) 3 girls: ⁴C₃ = 4; 2 girls and 1 boy: ⁴C₂ × ⁵C₁ = 30. P = 34/84 = 17/42.

Explanation

Count favourable selections with combinations.

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Theory question 106

Statistics

The table shows the distribution of ages of 22 students in a school.
Age (years): 12 − 14, 15 − 17, 18 − 20, 21 − 23, 24 − 26
Freq.: 6, 10, 3, 2, 1
Using an assumed mean of 19, calculate, correct to three significant figures, the: (a) mean age; (b) standard deviation of the distribution.

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Model answer

Class marks 13, 16, 19, 22, 25; d = x − 19 = −6, −3, 0, 3, 6.
Σfd = −36 − 30 + 0 + 6 + 6 = −54; Σfd² = 216 + 90 + 0 + 18 + 36 = 360.
(a) Mean = 19 + (−54/22) = 16.5 years (16.545).
(b) SD = √(360/22 − (54/22)²) = √(16.364 − 6.025) = √10.339 = 3.22.

Explanation

Use class marks and the coded deviations d = x − 19.

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Theory question 107

Dynamics

The initial velocity of a particle of mass 0.1 kg is 40 ms−1 in the direction of the unit vector j. The velocity of the particle changed to 30 ms−1 in the direction of the unit vector i. Find the change in momentum.

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Model answer

Initial momentum = 0.1(40j) = 4j; final = 0.1(30i) = 3i.
Change = 3i − 4j (N s).
Magnitude = √(9 + 16) = 5 kg m s⁻¹.

Explanation

Momentum is a vector: subtract the initial from the final.

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Theory question 108

Dynamics

A stone is dropped vertically downwards from the top of a tower of height 45m with a speed of 20 ms−1. Find the: (a) time it takes to reach the ground; (b) speed with which it hits the ground. [Take g = 10 ms−2]

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Model answer

(a) s = ut + ½gt²: 45 = 20t + 5t², so t² + 4t − 9 = 0.
t = −2 + √13 ≈ 1.61 s.
(b) v² = u² + 2gs = 400 + 900 = 1300, v ≈ 36.06 ms⁻¹.

Explanation

Take downwards as positive; u = 20 and g = 10 both act downwards.

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Theory question 109

Differentiation

(a) Differentiate (x2 + 1)/(x + 1)2 with respect to x.
(b)(i) Evaluate the determinant (1 2 −1 / 2 3 −1 / −1 1 3) (rows separated by /).
(ii) Using the answer in 9(b)(i), solve the system of equations: x + 2y − z = 4; 2x + 3y − z = 2; −x + y + 3z = −1.

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Model answer

(a) Quotient rule: [2x(x + 1)² − 2(x + 1)(x² + 1)]/(x + 1)⁴ = [2x(x + 1) − 2(x² + 1)]/(x + 1)³ = 2(x − 1)/(x + 1)³.
(b)(i) Δ = 1(9 + 1) − 2(6 − 1) − 1(2 + 3) = −5.
(ii) Cramer's rule: Δₓ = 25, Δᵧ = −15, Δ_z = 15.
x = 25/−5 = −5, y = −15/−5 = 3, z = 15/−5 = −3.

Explanation

Check the solution in all three equations.

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Theory question 110

Trigonometry

(a) Using a scale of 2 cm to 30° on the x-axis and 2 cm to 0.2 units on the y-axis, on the same graph sheet, draw the graphs of y = sin 2x and y = cos x for 0° ≤ x ≤ 210° at intervals of 30°.
(b) Using the graphs in 10(a), find the truth set of: (i) sin 2x = 0; (ii) sin 2x − cos x = 0.

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Model answer

(a) x: 0, 30, 60, 90, 120, 150, 180, 210
sin 2x: 0, 0.87, 0.87, 0, −0.87, −0.87, 0, 0.87
cos x: 1, 0.87, 0.5, 0, −0.5, −0.87, −1, −0.87
Plot both smooth curves on the same axes.
(b)(i) Where y = sin 2x crosses the x-axis: x = 0°, 90°, 180°.
(ii) Where the curves meet: x = 30°, 90°, 150°.

Explanation

The truth set in (ii) is where the two curves intersect.

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Theory question 111

Sequences and Series

(a) The sum of the first three terms of a decreasing exponential sequence (GP) is equal to 7 and the product of these three terms is equal to 8. Find the: (i) common ratio; (ii) first three terms of the sequence.
(b) Using the trapezium rule with ordinates at x = 1, 2, 3, 4 and 5, calculate, correct to two decimal places, the value of ∫15 (x + 2/x2) dx.

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(a) Let the terms be a/r, a, ar. Product a³ = 8, so a = 2.
2/r + 2 + 2r = 7 gives 2r² − 5r + 2 = 0, r = ½ or 2. Decreasing: (i) r = ½.
(ii) 4, 2, 1.
(b) y = 3, 2.5, 3.2222, 4.125, 5.08 at x = 1 to 5; h = 1.
Area ≈ ½(3 + 5.08) + 2.5 + 3.2222 + 4.125 = 13.89.

Explanation

Writing the terms as a/r, a, ar makes the product easy.

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Theory question 112

Differentiation

(a) Find the maximum and minimum points of the curve y = 2x3 − 3x2 − 12x + 4.
(b) Sketch the curve in 12(a) above.

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(a) dy/dx = 6x² − 6x − 12 = 6(x − 2)(x + 1) = 0, so x = 2 or −1.
d²y/dx² = 12x − 6: at x = −1 it is −18 (maximum), at x = 2 it is 18 (minimum).
Maximum point (−1, 11); minimum point (2, −16).
(b) Cubic rising from the left to (−1, 11), down through (0, 4) to (2, −16), then rising again.

Explanation

Use the second derivative to tell the maximum from the minimum.

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Theory question 113

Statistics

The table gives the relationship between the height, in meters, of a plant and the number of days it is left to grow.
Number of days (x): 10, 20, 30, 40, 50, 60, 70, 80
Height (y): 1.0, 1.1, 1.2, 1.4, 1.6, 1.8, 2.0, 2.3
(a) Using a scale of 2cm to represent 0.5 units on the y-axis and 2 cm to represent 10 units on the x-axis, draw a scatter diagram for the information. (b) Find x̄ the mean of x, and ȳ the mean of y, and plot (x̄, ȳ) on the diagram. (c) Draw the line of best fit to pass through (x̄, ȳ) and (10, 1). (d) From the graph, find the: (i) equation of the line of best fit; (ii) height of plant in 75 days.

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(a) Plot the eight points (10, 1.0) … (80, 2.3).
(b) x̄ = 360/8 = 45; ȳ = 12.4/8 = 1.55. Plot (45, 1.55).
(c) Straight line through (10, 1) and (45, 1.55).
(d)(i) Gradient = 0.55/35 ≈ 0.0157; y = 0.0157x + 0.843 (about y = 0.016x + 0.84).
(ii) At x = 75: y ≈ 2.02 m.

Explanation

Read values from your line, or use its equation.

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Theory question 114

Probability

(a) A bag contains 5 blue, 4 green and 3 yellow balls. All the balls are identical except for colour. Three balls are drawn at random without replacement. Find the probability that: (i) all three balls have the same colour; (ii) two balls have the same colour.
(b) The table shows the ranks of the marks scored by 7 candidates in Physics and Chemistry tests.
Physics: 6, 5, 4, 3, 2, 7, 1
Chemistry: 7, 6, 2, 4, 1, 5, 3
Calculate the Spearman's rank correlation coefficient.

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(a) ¹²C₃ = 220 selections.
(i) ⁵C₃ + ⁴C₃ + ³C₃ = 10 + 4 + 1 = 15; P = 15/220 = 3/44.
(ii) Exactly two the same: ⁵C₂ × 7 + ⁴C₂ × 8 + ³C₂ × 9 = 70 + 48 + 27 = 145; P = 145/220 = 29/44.
(b) d = −1, −1, 2, −1, 1, 2, −2; Σd² = 16.
ρ = 1 − 6(16)/(7(49 − 1)) = 1 − 96/336 = 0.714.

Explanation

Spearman: ρ = 1 − 6Σd²/(n(n² − 1)).

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Theory question 115

Probability

(a) The probability that a man wins a race is 0.8. In four different races, what is the probability that he wins: (i) all races; (ii) no race; (iii) at most 3 races?
(b) A class consists of 5 girls and 10 boys. If a committee of 5 is chosen at random from the class find the probability that: (i) 3 boys are selected; (ii) at least one girl is selected.

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(a) p = 0.8, q = 0.2, n = 4.
(i) 0.8⁴ = 0.4096. (ii) 0.2⁴ = 0.0016. (iii) 1 − 0.4096 = 0.5904.
(b) ¹⁵C₅ = 3003.
(i) ¹⁰C₃ × ⁵C₂ = 120 × 10 = 1200; P = 1200/3003 = 400/1001 ≈ 0.400.
(ii) No girl: ¹⁰C₅ = 252; P(at least one girl) = 1 − 252/3003 = 131/143 ≈ 0.916.

Explanation

'At least one' is 1 minus 'none'.

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Theory question 116

Vectors

(a) Three vectors a, b and c are (8, 3), (6, −5) and (2, 3) (column vectors) respectively. Find the vector d such that |d| = √41 and d is in the direction of a + b − 2c.
(b) The coordinates of A and B are (3, 4) and (3, n) respectively. If O is the origin and ∠AOB = 30°, find, correct to 2 decimal places, the values of n.

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(a) a + b − 2c = (10, −8), magnitude √164 = 2√41.
d = √41 × (10, −8)/(2√41) = (5, −4).
(b) OA·OB = 9 + 4n = |OA||OB| cos 30° = 5√(9 + n²)(√3/2).
Square: 4(9 + 4n)² = 75(9 + n²) gives 11n² − 288n + 351 = 0.
n = (288 ± √67500)/22: n = 24.90 or n = 1.28.

Explanation

Use the dot product for the angle between two vectors.

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Theory question 117

Statics

(a) A particle is under the action of forces P = (4N, 030°) and R = (10N, 300°). Find the force that will keep the particle in equilibrium.
(b) A train travelling at 45 ms−1 is brought to rest after covering a distance of 1500m. Find the: (i) time taken to come to rest; (ii) uniform retardation.

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(a) East/north components: P = (2, 3.464), R = (−8.660, 5).
Resultant = (−6.660, 8.464). The equilibrant is (6.660, −8.464).
Magnitude = √(44.36 + 71.64) = 10.77 N; bearing = 180° − tan⁻¹(6.660/8.464) = 141.8°.
Force = (10.77 N, 141.8°).
(b)(i) s = ½(u + v)t: 1500 = 22.5t, t = 66.7 s.
(ii) 0 = 45² + 2a(1500), a = −0.675; retardation 0.675 ms⁻².

Explanation

The equilibrant is equal and opposite to the resultant.

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Theory question 118

Dynamics

The displacement S metres of the particle from a fixed point O at time t seconds is given by S = t2 − 6t + 5.
(a) On a graph sheet, draw a displacement-time graph for the interval 0 ≤ t ≤ 6.
(b) From the graph, find the: (i) time at which the velocity is zero; (ii) average velocity over the interval 0 ≤ t ≤ 4; (iii) total distance covered in the interval 0 ≤ t ≤ 5.

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(a) t: 0, 1, 2, 3, 4, 5, 6; S: 5, 0, −3, −4, −3, 0, 5. Plot a smooth U-shaped curve.
(b)(i) Velocity is zero at the lowest point, t = 3 s.
(ii) (S(4) − S(0))/4 = (−3 − 5)/4 = −2 ms⁻¹.
(iii) From S = 5 to −4 (9 m) then back to 0 (4 m): 13 m.

Explanation

Distance counts both directions; displacement does not.

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