56 questions. Try each one, then open the answer to check yourself and read the explanation.
Given that the straight lines kx − 5y + 6 = 0 and mx + ny − 1 = 0 are parallel, find a relationship connecting the constants m, n and k.
Answer: B (kn + 5m = 0)
The gradients are k/5 and −m/n. Parallel lines have equal gradients, so k/5 = −m/n, giving kn = −5m, i.e. kn + 5m = 0.
Given that α and β are the roots of an equation such that α + β = 3 and αβ = 2, find the equation.
Answer: A (x2 − 3x + 2 = 0)
A quadratic with roots α and β is x² − (α + β)x + αβ = 0, so x² − 3x + 2 = 0.
Which of the following is the same as sin (270 + x)?
Answer: D (− cos x)
sin(270° + x) = sin 270° cos x + cos 270° sin x = (−1)cos x + 0 = −cos x.
The sum of the first three terms of an Arithmetic Progression (A.P.) is 18. If the first term is 4, find their product
Answer: B (192)
3a + 3d = 18 with a = 4 gives d = 2, so the terms are 4, 6, 8 and their product is 192.
Two functions f and g are defined on the set R of real numbers by f : x → 2x − 1 and g : x → x2 + 1. Find the value of f−1 o g(3).
Answer: C (11/2)
g(3) = 10. The inverse of f is f−1(x) = (x + 1)/2, so f−1(10) = 11/2.
The gradient of the line passing through the points P(4, 5) and Q(x, 9) is ½. Find the value of x.
Answer: D (12)
Gradient = (9 − 5)/(x − 4) = ½, so x − 4 = 8 and x = 12.
Simplify √3/(√3 − 1) + √3/(√3 + 1)
Answer: C (3)
Over the common denominator (√3 − 1)(√3 + 1) = 2: [√3(√3 + 1) + √3(√3 − 1)]/2 = (3 + 3)/2 = 3.
Given that log2 y½ = log5 125, find the value of y.
Answer: D (64)
log₅ 125 = 3, so y½ = 2³ = 8 and y = 64.
Simplify: (tan 80° − tan 20°)/(1 + tan 80° tan 20°)
Answer: C (√3)
This is the formula for tan(A − B) with A = 80° and B = 20°, so it equals tan 60° = √3.
The binary operation * is defined on the set R, of real numbers by x * y = 3x + 3y − xy for all x, y ∈ R. Determine, in terms of x, the identity element of the operation.
Answer: A (2x/(x − 3), x ≠ 3)
The identity e satisfies x * e = x: 3x + 3e − xe = x, so e(3 − x) = −2x and e = 2x/(x − 3).
Find the fourth term of the binomial expansion of (x − k)5 in descending powers of x.
Answer: D (−10x2k3)
The fourth term is ⁵C₃ x²(−k)³ = 10x²(−k³) = −10x²k³.
The tangent to the curve y = 4x3 + kx2 − 6x + 4 at the point P(1, m) is parallel to the x-axis, where k and m are constants. Find the value of k.
Answer: D (−3)
dy/dx = 12x² + 2kx − 6 must be 0 at x = 1: 12 + 2k − 6 = 0, so k = −3.
The tangent to the curve y = 4x3 + kx2 − 6x + 4 at the point P(1, m) is parallel to the x-axis, where k and m are constants. Determine the coordinates of P
Answer: C ((1, −1))
With k = −3, y = 4 − 3 − 6 + 4 = −1 at x = 1, so P is (1, −1).
Two vectors m and n are defined by m = 3i + 4j and n = 2i − j. Find the angle between m and n.
Answer: B (79.7°)
m·n = 6 − 4 = 2, |m| = 5, |n| = √5. cos θ = 2/(5√5) ≈ 0.1789, so θ ≈ 79.7°.
Find the area of the circle whose equation is x2 + y2 − 4x + 8y + 11 = 0.
Answer: C (9π)
Centre (2, −4); r² = 2² + (−4)² − 11 = 9. Area = πr² = 9π.
Two bodies of masses 8 kg and 5 kg travelling in the same direction with speeds of x ms−1 and 2 ms−1 respectively collide. If after collision, they move together with a speed of 3.85 ms−1 find, correct to the nearest whole number the value of x.
Answer: B (5)
Momentum is conserved: 8x + 5(2) = 13 × 3.85 = 50.05, so 8x = 40.05 and x ≈ 5.
Evaluate limx→3 ((x2 − 2x − 3)/(x − 3))
Answer: A (4)
x² − 2x − 3 = (x − 3)(x + 1). Cancel (x − 3) to get x + 1, which is 4 at x = 3.
A bag contains 2 red and 4 green sweets of the same size and shape. Two boys pick a sweet each from the box, one after the other, without replacement. What is the probability that at least a sweet with a green wrapper is picked?
Answer: D (14/15)
P(no green) = P(both red) = 2/6 × 1/5 = 1/15. So P(at least one green) = 1 − 1/15 = 14/15.
If f(x) = mx2 − 6x − 3 and f′(1) = 12, find the value of constant m.
Answer: A (9)
f′(x) = 2mx − 6, so f′(1) = 2m − 6 = 12 and m = 9.
A body is acted upon by two forces F1 = (5N, 060°) and F2 = (10N, 180°). Find the magnitude of the resultant force.
Answer: D (8.66N)
The angle between the forces is 120°. R² = 5² + 10² + 2(5)(10)cos 120° = 125 − 50 = 75, so R = √75 ≈ 8.66 N.
The equation of a curve is given by y = 2x2 − 5x + k. If the curve has two intercepts on the x-axis, find the value(s) of constant k.
Answer: C (k < 25/8)
Two x-intercepts means two distinct real roots: b² − 4ac > 0, so 25 − 8k > 0 and k < 25/8.
Find the value of p which will make (x2 − x + p) a perfect square.
Answer: B (¼)
x² − x + ¼ = (x − ½)², so p = ¼ (half the x-coefficient, squared).
The polynomial g(x) = 2x3 + 3x2 + qx − 1, has the same remainder when divided by (x + 2) and (x − 1). Find the value of constant q.
Answer: C (−3)
g(−2) = −16 + 12 − 2q − 1 = −5 − 2q and g(1) = 2 + 3 + q − 1 = 4 + q. Setting them equal: −5 − 2q = 4 + q, so q = −3.
The table shows the marks obtained by two pupils in a test.
Marks: 5 − 7, 8 − 10, 11 − 13, 14 − 16, 17 − 19, 20 − 22
Freq.: 4, 7, 26, 41, 14, 8
Find the upper class boundary of the class containing the third quartile.
Answer: C (16.5)
Total frequency is 100, so Q₃ is the 75th mark. Cumulative frequencies are 4, 11, 37, 78, …, so it lies in 14 − 16, whose upper class boundary is 16.5.
The table shows the marks obtained by two pupils in a test.
Marks: 5 − 7, 8 − 10, 11 − 13, 14 − 16, 17 − 19, 20 − 22
Freq.: 4, 7, 26, 41, 14, 8
Find the probability that a pupil selected at random had a mark of at least 14.
Answer: D (0.63)
Marks of 14 or more: 41 + 14 + 8 = 63 out of 100, so the probability is 0.63.
How many ways can 12 people be divided into three groups of 2, 7 and 3 in that order?
Answer: A (7,920)
Choose 2 from 12, then 7 from the remaining 10, then the last 3: ¹²C₂ × ¹⁰C₇ × ³C₃ = 66 × 120 × 1 = 7920.
Given that P = (4 9) and Q = (−1 −2; 3 2), which of the following operations is possible?
Answer: D (PQ)
P is 1 × 2 and Q is 2 × 2. PQ needs columns of P (2) = rows of Q (2), so it works; QP, P² and Q²P do not.
Given that P = (4 9) and Q = (−1 −2; 3 2), evaluate |Q| P.
Answer: A ((16 36))
|Q| = (−1)(2) − (−2)(3) = −2 + 6 = 4, so |Q|P = 4(4 9) = (16 36).
The equation of a circle is given by x2 + y2 − 4x + 2y − 3 = 0. Find the radius and the coordinates of its centre.
Answer: B (2√2, (2, −1))
Complete the squares: (x − 2)² + (y + 1)² = 3 + 4 + 1 = 8. Centre (2, −1), radius √8 = 2√2.
Simplify nP3/nC2 + nP0
Answer: D (2n − 3)
ⁿP₃/ⁿC₂ = n(n − 1)(n − 2) ÷ [n(n − 1)/2] = 2(n − 2) = 2n − 4, and ⁿP₀ = 1, giving 2n − 3.
X and Y are two independent events. If P(X) = ⅕, and P(X ∩ Y) = 2/15, find P(Y).
Answer: A (2/3)
For independent events P(X ∩ Y) = P(X) × P(Y), so P(Y) = (2/15) ÷ (1/5) = 2/3.
Given that p = 4i + 3j, find the unit vector in the direction of p.
Answer: D (⅕(4i + 3j))
|p| = √(4² + 3²) = 5, so the unit vector is p/|p| = ⅕(4i + 3j).
A particle is projected vertically upwards with a speed of 40 ms−1. At what times will it be 35m above its point of projection? [Take g = 10ms−2]
Answer: A (1 sec. and 7 sec.)
35 = 40t − 5t², so t² − 8t + 7 = 0, (t − 1)(t − 7) = 0: at 1 s (going up) and 7 s (coming down).
Three students are working independently on a Mathematics problem. Their respective probabilities of solving the problem are 0.6, 0.7 and 0.8. What is the probability that at least one of them solves the problem?
Answer: D (0.976)
P(none solves) = 0.4 × 0.3 × 0.2 = 0.024, so P(at least one) = 1 − 0.024 = 0.976.
Given that R = (4, 180°) and S = (3, 300°), find the dot product R.S.
Answer: B (−6)
The angle between the vectors is 300° − 180° = 120°, so R·S = 4 × 3 × cos 120° = 12 × (−½) = −6.
Calculate, correct to one decimal place, the standard deviation of the numbers −1, 5, 0, 2 and 9.
Answer: C (3.6)
Mean = 15/5 = 3. Squared deviations 16, 4, 9, 1, 36 sum to 66, so SD = √(66/5) = √13.2 ≈ 3.6.
A group of 5 boys and 4 girls is to be chosen from a class of 8 boys and 6 girls. In how many ways can this be done?
Answer: A (840)
⁸C₅ × ⁶C₄ = 56 × 15 = 840.
A force of 30 N acts at an angle of 60° on a body of mass 6 kg initially at rest on a smooth horizontal plane. Find the distance covered in 10 seconds.
Answer: B (125 m)
Horizontal component = 30 cos 60° = 15 N, so a = 15/6 = 2.5 ms⁻². s = ½ × 2.5 × 10² = 125 m.
Three forces F1 = (8N, 300°), F2 = (6N, 090°) and F3 = (4N, 180°) act on a particle. Find the vertical component of the resultant force.
Answer: D ((6 − 4√3)N)
Vertical components (angles from the positive x-axis): 8 sin 300° = −4√3, 6 sin 90° = 6, 4 sin 180° = 0. Total = 6 − 4√3 N.
The second term of a geometric progression is 3. If its sum to infinity is 25/2, find the value of the common ratio.
ar = 3 and a/(1 − r) = 25/2
a = 3/r, so 3/(r(1 − r)) = 25/2
25r − 25r² = 6, 25r² − 25r + 6 = 0
(5r − 3)(5r − 2) = 0
r = 3/5 or r = 2/5 (both have |r| < 1, so both are valid)
Eliminate a between the two equations to get a quadratic in r.
Solve x2/3 − 5x1/3 + 6 = 0
Let y = x1/3: y² − 5y + 6 = 0
(y − 2)(y − 3) = 0, y = 2 or y = 3
x = 2³ = 8 or x = 3³ = 27
Substitute y = x1/3 to get a quadratic, then cube the answers.
The normal to the curve y = 2x2 + x − 3 at the point (2, 7) meets the x-axis at point P. Find the coordinates of P.
dy/dx = 4x + 1 = 9 at x = 2 (gradient of tangent)
Gradient of normal = −1/9
Normal: y − 7 = −(1/9)(x − 2), i.e. x + 9y − 65 = 0
On the x-axis y = 0, so x = 65
P is (65, 0)
The normal is perpendicular to the tangent, so its gradient is −1/(dy/dx).
(a) Write down the first four terms of the binomial expansion of (2 − ½x)5 in ascending powers of x.
(b) Use your expansion in (a) to find, correct to two decimal places, the value of (1.99)5
(a) 2⁵ + 5(2⁴)(−½x) + 10(2³)(−½x)² + 10(2²)(−½x)³
= 32 − 40x + 20x² − 5x³
(b) 2 − ½x = 1.99 gives x = 0.02
32 − 40(0.02) + 20(0.02)² − 5(0.02)³ = 32 − 0.8 + 0.008 − 0.00004 = 31.20796
(1.99)⁵ ≈ 31.21
Use ⁵Cᵣ coefficients with powers of 2 decreasing; then choose x so that 2 − ½x = 1.99.
Five digit numbers are formed from digits 4, 5, 6, 7 and 8. (a) How many of such numbers can be formed if repetition of digits is (i) allowed (ii) not allowed? (b) How many of the numbers are odd, if repetition of digits is not allowed?
(a)(i) Each of the 5 places can take any of 5 digits: 5⁵ = 3125
(ii) 5 × 4 × 3 × 2 × 1 = 5! = 120
(b) The last digit must be 5 or 7 (2 ways); the other four places: 4! = 24
Odd numbers = 2 × 24 = 48
Fill the units place first when the number must be odd.
Four boys participated in a competition in which their respective chances of winning prizes are 1/5, 1/4, 1/3 and 1/2. What is the probability that at most two of them win prizes?
P(win): 1/5, 1/4, 1/3, 1/2; P(not win): 4/5, 3/4, 2/3, 1/2
P(all four win) = 1/5 × 1/4 × 1/3 × 1/2 = 1/120
P(exactly three win) = (4/5)(1/4)(1/3)(1/2) + (1/5)(3/4)(1/3)(1/2) + (1/5)(1/4)(2/3)(1/2) + (1/5)(1/4)(1/3)(1/2) = (4 + 3 + 2 + 1)/120 = 10/120
P(at most two) = 1 − 1/120 − 10/120 = 109/120 ≈ 0.908
It is quicker to subtract P(three win) and P(four win) from 1.
An object is projected vertically upwards, its height, h m, at time t seconds is given by h = 20t − (3/2)t2 − (2/3)t3. Find (a) the time at which it is momentarily at rest; (b) correct to two decimal places, the maximum height reached by the object
(a) v = dh/dt = 20 − 3t − 2t² = 0
2t² + 3t − 20 = 0, (2t − 5)(t + 4) = 0
t = 2.5 s (t = −4 rejected)
(b) h = 20(2.5) − 1.5(2.5)² − (2/3)(2.5)³ = 50 − 9.375 − 10.417 ≈ 30.21 m
Velocity is dh/dt; the object is at rest (at the top) when the velocity is zero.
(a) The roots of the equation x2 + mx + 11 = 0 are α and β, where m is a constant. If α2 + β2 = 27, find the value of m.
(b) The line 2x + 3y = 1 intersects the circle 2x2 + 2y2 + 4x + 9y − 9 = 0 at points P and Q where Q lies in the fourth quadrant. Find the coordinates of P and Q.
(a) α + β = −m, αβ = 11
α² + β² = (α + β)² − 2αβ = m² − 22 = 27
m² = 49, m = ±7
(b) y = (1 − 2x)/3; substitute and multiply by 9:
18x² + 2(1 − 2x)² + 36x + 27(1 − 2x) − 81 = 0
26x² − 26x − 52 = 0, x² − x − 2 = 0, x = 2 or x = −1
x = 2: y = −1 (fourth quadrant), so Q(2, −1)
x = −1: y = 1, so P(−1, 1)
Use sum and product of roots in (a); in (b) substitute the line into the circle.
(a) Solve the equation √(4x − 3) − √(2x − 5) = 2
(b) Find the finite area enclosed by the curve y2 = 4x and the line y + x = 0.
(a) √(4x − 3) = 2 + √(2x − 5); square: 4x − 3 = 4 + 4√(2x − 5) + 2x − 5
2x − 2 = 4√(2x − 5), so x − 1 = 2√(2x − 5)
Square: x² − 2x + 1 = 8x − 20, x² − 10x + 21 = 0, x = 3 or x = 7
Check: x = 3 gives 3 − 1 = 2; x = 7 gives 5 − 3 = 2; both are solutions
(b) Intersections: x = −y, y² = −4y, so y = 0 or y = −4; points (0, 0) and (4, −4)
Area = ∫−40 (−y − y²/4) dy = [−y²/2 − y³/12] from −4 to 0 = 0 − (−8 + 16/3) = 8/3 square units
Square twice and check both answers in the original equation; for the area integrate with respect to y.
(a) Given that |5 2 −3; −1 k 6; 3 9 (k + 2)| = −207, find the values of the constant k.
(b) The equation of a curve is x(y2 + 1) − y(x2 + 1) + 4 = 0. Find the (i) gradient of the curve at any point (x, y); (ii) equation of the tangent to the curve at the point (−1, −3)
(a) 5[k(k + 2) − 54] − 2[−(k + 2) − 18] − 3[−9 − 3k] = −207
5k² + 10k − 270 + 2k + 40 + 27 + 9k = −207
5k² + 21k + 4 = 0, (5k + 1)(k + 4) = 0
k = −1/5 or k = −4
(b)(i) Differentiate implicitly: (y² + 1) + 2xy dy/dx − (x² + 1)dy/dx − 2xy = 0
dy/dx = (2xy − y² − 1)/(2xy − x² − 1)
(ii) At (−1, −3): dy/dx = (6 − 9 − 1)/(6 − 1 − 1) = −4/4 = −1
Tangent: y + 3 = −(x + 1), i.e. x + y + 4 = 0
Expand the determinant along the first row; in (b) differentiate each term with respect to x, treating y as a function of x.
(a) Using the trapezium rule with seven ordinates, evaluate ∫03 dx/(x2 + 1), correct to two decimal places.
(b) Using matrix method, solve −2x + y = 3; −x + 4y = 1
(a) h = 0.5; x = 0, 0.5, 1, 1.5, 2, 2.5, 3
y = 1, 0.8, 0.5, 0.3077, 0.2, 0.1379, 0.1
Area ≈ (h/2)[y₀ + y₆ + 2(y₁ + … + y₅)] = 0.25[1.1 + 2(1.9456)] = 0.25 × 4.9912 ≈ 1.25
(b) (−2 1; −1 4)(x; y) = (3; 1); determinant = −8 + 1 = −7
Inverse = −(1/7)(4 −1; 1 −2)
x = −(1/7)(12 − 1) = −11/7, y = −(1/7)(3 − 2) = −1/7
Seven ordinates means six strips of width 0.5; for (b) multiply by the inverse matrix.
(a) If 18Cr = 18Cr+2, find rC5
(b) In a community, 10% of the people tested positive to the HIV virus. If 6 persons from the community are selected at random, one after the other with replacement, calculate correct to four decimal places, the probability that: (i) exactly 5; (ii) none; (iii) at most 2: tested positive to the virus.
(a) ⁿCᵣ = ⁿCₛ with r ≠ s means r + s = n: r + (r + 2) = 18, r = 8
⁸C₅ = 56
(b) Binomial, n = 6, p = 0.1, q = 0.9
(i) P(5) = 6(0.1)⁵(0.9) = 0.000054 ≈ 0.0001
(ii) P(0) = 0.9⁶ = 0.531441 ≈ 0.5314
(iii) P(≤ 2) = 0.531441 + 6(0.1)(0.9)⁵ + 15(0.1)²(0.9)⁴ = 0.531441 + 0.354294 + 0.098415 = 0.98415 ≈ 0.9842
Use ⁿCᵣ = ⁿCₙ₋ᵣ; then the binomial formula ⁶Cᵣ(0.1)ʳ(0.9)⁶⁻ʳ.
The table shows the distribution of ages of workers in a company
Age in (years): 17 − 21, 22 − 26, 27 − 31, 32 − 36, 37 − 41, 42 − 46, 47 − 51, 52 − 56
No of workers: 12, 24, 30, 37, 45, 25, 10, 7
(a) Using an assumed mean of 39, calculate the (i) mean; (ii) standard deviation; of the distribution.
(b) If a worker is selected at random from the company for an award, what is the probability that he is at most 36 years old?
Midpoints 19, 24, 29, 34, 39, 44, 49, 54; d = x − 39: −20, −15, −10, −5, 0, 5, 10, 15
Σf = 190, Σfd = −755, Σfd² = 17325
(a)(i) Mean = 39 + (−755/190) = 39 − 3.97 ≈ 35.03 years
(ii) SD = √(17325/190 − (755/190)²) = √(91.18 − 15.79) = √75.39 ≈ 8.68
(b) Workers aged at most 36: 12 + 24 + 30 + 37 = 103
Probability = 103/190 ≈ 0.542
Work with deviations from the assumed mean: mean = A + Σfd/Σf, SD = √(Σfd²/Σf − (Σfd/Σf)²).
The table shows the marks obtained by some candidates in Physics (y) and Mathematics (x) tests.
Mathematics: 43, 46, 48, 39, 30, 60, 8, 45, 40
Physics: 54, 53, 63, 30, 44, 75, 20, 33, 49
(a) (i) Represent this information on a scatter diagram. (ii) Find x̄ and ȳ, the mean of x and y respectively. (iii) Draw the line of best fit to pass through (x̄, ȳ)
(b) Find the equation of the line in (a)(iii).
(c) Use your equation in (b) to find, correct to one decimal place, the mark in Physics for a candidate who scored 28 in mathematics
(a)(i) Plot the 9 points (x, y) with Mathematics on the horizontal axis
(ii) x̄ = 359/9 ≈ 39.9; ȳ = 421/9 ≈ 46.8
(iii) Draw a straight line through (39.9, 46.8) with roughly equal numbers of points on each side
(b) Least squares: Σxy = 18379, Σx² = 15979
b = (18379 − 9 × 39.89 × 46.78)/(15979 − 9 × 39.89²) ≈ 1585.6/1658.9 ≈ 0.956
y − 46.8 = 0.956(x − 39.9), i.e. y ≈ 0.956x + 8.65 (a line drawn by eye gives a similar equation)
(c) x = 28: y ≈ 0.956(28) + 8.65 ≈ 35.4
The line of best fit must pass through the mean point (x̄, ȳ); read or calculate its gradient.
(a) Find, correct to one decimal place, the angle between p = (3; −1) and q = (3; 4).
(b) ABCD is a square with vertices at A (0,0), B(2,0), C(2, 2) and D (0, 2). Forces of magnitudes 10N, 15N, 20N and 5N act along BA, BC, DC and AD respectively. Find the (i) magnitude; (ii) direction; of their resultant.
(a) p·q = 9 − 4 = 5; |p| = √10, |q| = 5
cos θ = 5/(5√10) = 0.3162, θ ≈ 71.6°
(b) BA: (−10, 0); BC: (0, 15); DC: (20, 0); AD: (0, 5)
Resultant = (10, 20)
(i) |R| = √(10² + 20²) = √500 ≈ 22.4 N
(ii) tan θ = 20/10 = 2, θ ≈ 63.4° to AB (the positive x-direction), i.e. a bearing of about 026.6°
Write each force as a vector using the direction of the side it acts along, then add.
A body of mass 20kg is suspended at point A by two inextensible strings to points X and Y, where |AX| = 1.7m, |AY| = 2.1m and |XY| = 2.9m. If the body remains in equilibrium, calculate, correct to three significant figures, the tension in each string. [Take g = 10ms−2]
Assume X and Y are at the same level. Weight = 200 N
Cosine rule in triangle AXY: angle X = cos⁻¹[(1.7² + 2.9² − 2.1²)/(2 × 1.7 × 2.9)] ≈ 45.67°; angle Y ≈ 35.39°; angle A ≈ 98.94°
Horizontally: TX cos 45.67° = TY cos 35.39°
Vertically: TX sin 45.67° + TY sin 35.39° = 200
TX = 200 cos 35.39°/sin 81.06° ≈ 165 N (string AX)
TY = 200 cos 45.67°/sin 81.06° ≈ 141 N (string AY)
Find the angles of the triangle with the cosine rule, then resolve forces horizontally and vertically.
(a) The position vectors of points P, Q and R are 2i − j, i + kj and 3i + 2j respectively, where k is a constant. If the area of triangle PQR is 4 square units, find the value of k.
(b) A train moving along a straight track accelerates uniformly from rest at 300 km h−2. When its speed reaches 100 km h−1, it moves with that velocity for 40 minutes. It then decelerates uniformly to a stop in 20 minutes. (i) Sketch the velocity time graph for the journey. (ii) Determine the magnitude of the retardation. (iii) Find the total distance covered by the train. (iv) What is the total time it travelled before stopping?
(a) P(2, −1), Q(1, k), R(3, 2)
Area = ½|2(k − 2) + 1(2 + 1) + 3(−1 − k)| = ½|−k − 4| = 4
|k + 4| = 8, so k = 4 or k = −12
(b) Time to reach 100 km/h = 100/300 h = 20 min
(i) Trapezium-shaped graph: rises from 0 to 100 km/h in 20 min, level for 40 min, falls to 0 in 20 min
(ii) Retardation = 100 ÷ (1/3) = 300 km h⁻²
(iii) Distance = area = ½(1/3)(100) + (2/3)(100) + ½(1/3)(100) = 16⅔ + 66⅔ + 16⅔ = 100 km
(iv) Total time = 20 + 40 + 20 = 80 minutes (1 h 20 min)
Use the area-of-triangle formula from coordinates; on a velocity–time graph distance is the area under it.