57 questions. Try each one, then open the answer to check yourself and read the explanation.
Find the domain of f(x) = x/(3 − x) where x ∈ R, the set of real numbers.
Answer: A ((x : x ∈ R, x ≠ 3))
The function is undefined only when the denominator 3 − x = 0, i.e. x = 3. So the domain is all real x except 3.
Find the value of cos(60° + 45°) leaving your answer in surd form
Answer: C ((√2 − √6)/4)
cos(60° + 45°) = cos60°cos45° − sin60°sin45° = (1/2)(√2/2) − (√3/2)(√2/2) = (√2 − √6)/4.
If 5/√2 − √8/8 = m√2, where m is a constant find m.
Answer: C (2¼)
5/√2 = (5/2)√2 and √8/8 = 2√2/8 = (1/4)√2. So m = 5/2 − 1/4 = 9/4 = 2¼.
If 163x = ¼(32x−1), find the value of x.
Answer: A (−1)
Write everything as powers of 2: 212x = 2−2 × 25x−5 = 25x−7. So 12x = 5x − 7, giving x = −1.
Simplify log58 / log5√8
Answer: D (2)
log5√8 = ½ log58, so the fraction is log58 ÷ (½ log58) = 2.
The coefficient of the 7th term in the binomial expansion of (2 − x/3)10 in ascending powers of x is
Answer: C (1120/243)
The 7th term is 10C6(2)4(−x/3)6. Its coefficient is 210 × 16 × 1/729 = 3360/729 = 1120/243.
The roots of a quadratic equation are (3 − √3) and (3 + √3). Find its equation.
Answer: B (x2 − 6x + 6 = 0)
Sum of roots = 6 and product = (3 − √3)(3 + √3) = 9 − 3 = 6. The equation is x2 − (sum)x + product = 0, i.e. x2 − 6x + 6 = 0.
If (x − 3) is a factor of 2x2 − 2x + p, find the value of constant p
Answer: A (−12)
By the factor theorem, putting x = 3 gives 0: 18 − 6 + p = 0, so p = −12.
If sin x = −sin 70°, 0° < x < 360°, determine the two possible values of x.
Answer: D (250°, 290°)
sin x is negative in the 3rd and 4th quadrants with reference angle 70°: x = 180° + 70° = 250° or x = 360° − 70° = 290°.
For what values of x is (x2 − 9x + 18)/(x2 + 2x − 35) undefined?
Answer: C (−7 or 5)
A fraction is undefined when its denominator is zero: x2 + 2x − 35 = (x + 7)(x − 5) = 0, so x = −7 or 5.
Calculate, correct to one decimal place, the length of the line joining points x(3, 5) and y(5, 1).
Answer: C (4.5)
Length = √((5 − 3)2 + (1 − 5)2) = √(4 + 16) = √20 ≈ 4.47 = 4.5 to 1 d.p.
If y = 2(2x + √x)2, find dy/dx
Answer: B (4(2x + √x)(2 + 1/(2√x)))
By the chain rule, dy/dx = 2 × 2(2x + √x) × d/dx(2x + √x) = 4(2x + √x)(2 + 1/(2√x)).
Calculate, correct to one decimal place, the acute angle between the lines 3x − 4y + 5 = 0 and 2x + 3y − 1 = 0
Answer: A (70.6°)
Gradients are 3/4 and −2/3. tan θ = |(3/4 + 2/3)/(1 − ½)| = (17/12)/(1/2) = 17/6, so θ ≈ 70.6°.
Evaluate ∫12 4/x3 dx
Answer: D (1½)
∫4x−3 dx = −2x−2. From 1 to 2: (−2/4) − (−2) = −½ + 2 = 1½.
If the determinant |3 x / 2 x − 2| = −2, find the value of x. (The determinant is written row by row.)
Answer: C (4)
3(x − 2) − 2x = −2, so x − 6 = −2 and x = 4.
Given that P = {x : x is a factor of 6} if the domain of g(x) = x2 + 3x − 5, find the range of g(x).
Answer: D ((−1, 5, 13, 49))
P = {1, 2, 3, 6}. g(1) = −1, g(2) = 5, g(3) = 13, g(6) = 49, so the range is {−1, 5, 13, 49}.
The third term of geometric progression (G.P.) is 10 and the sixth term is 80. Find the common ratio.
Answer: A (2)
ar5 ÷ ar2 = 80 ÷ 10, so r3 = 8 and r = 2.
Find the axis of symmetry of the curve y = x2 − 4x − 12.
Answer: C (x = 2)
The turning point is where dy/dx = 2x − 4 = 0, i.e. x = 2. The axis of symmetry is the vertical line x = 2.
Find the equation of the tangent to the curve y = 4x2 − 12x + 7 at point (2, −1).
Answer: C (y − 4x + 9 = 0)
dy/dx = 8x − 12 = 4 at x = 2. Tangent: y + 1 = 4(x − 2), i.e. y − 4x + 9 = 0.
If h(x) = x3 − 1/x3 evaluate h(a) − h(1/a)
Answer: C (2a3 − 2/a3)
h(a) = a3 − 1/a3 and h(1/a) = 1/a3 − a3. Subtracting gives 2a3 − 2/a3.
The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.
Answer: B (12.6 years)
Total age of 15 pupils = 15 × 14.2 = 213. New total = 16 × 14.1 = 225.6. The new pupil is 225.6 − 213 = 12.6 years old.
The distance in metres of a particle from a fixed point at time t seconds is given by s = 7 + pt3 + t2, where p is a constant. If the acceleration at t = 3 seconds is 8ms−2, find the value of p.
Answer: A (1/3)
v = 3pt2 + 2t and a = 6pt + 2. At t = 3: 18p + 2 = 8, so p = 1/3.
The probabilities that a husband and wife will be alive in 15 years' time are m and n respectively. Find the probability that only one of them will be alive at that time.
Answer: C (m + n − 2mn)
Only one alive = husband alive and wife not, or wife alive and husband not: m(1 − n) + n(1 − m) = m + n − 2mn.
In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?
Answer: B (0.30)
Assuming every pupil likes at least one subject, both = 35 + 30 − 50 = 15. Probability = 15/50 = 0.30.
P, Q, R, S are points in a plane such that PQ = 8i − 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).
Answer: C ((20, 5))
PS = PQ + QR + RS = (8 + 5 + 7)i + (−5 + 7 + 3)j = 20i + 5j, so (x, y) = (20, 5).
If OA = 3i + 4j and OB = 5i − 6j where O is the origin and M is the mid-point of AB, find OM.
Answer: C (4i − j)
OM = ½(OA + OB) = ½(8i − 2j) = 4i − j.
Find the direction cosines of the vector 4i − 3j.
Answer: C (4/5, −3/5)
|4i − 3j| = √(16 + 9) = 5. The direction cosines are 4/5 and −3/5.
Yomi was asked to label four seats with letter S, R, P, Q. What is the probability that he labelled them in alphabetical order?
Answer: A (1/24)
The four letters can be arranged in 4! = 24 ways and only one is alphabetical, so the probability is 1/24.
Two forces (2i − 5j)N and (−3i + 4j)N act on a body of mass 5kg. Find, in ms−2, the magnitude of the acceleration of the body.
Answer: A (√2/5)
Resultant = (2 − 3)i + (−5 + 4)j = −i − j, of size √2 N. a = F/m = √2/5 ms−2.
Two particles are fired together along a smooth horizontal surface with velocities 4ms−1 and 5ms−1. If they move at 60° to each other, find the distance between them after 2 seconds.
Answer: C (2√21)
After 2 s they are 8 m and 10 m from the start. By the cosine rule d2 = 64 + 100 − 2(8)(10)cos 60° = 84, so d = √84 = 2√21 m.
Two forces F1 = (7i + 8j) and F2 = (3i + 4j)N act on a particle. Find the magnitude and direction of (F1 − F2).
Answer: B ((4√2N, 045°))
F1 − F2 = 4i + 4j. Its size is √32 = 4√2 N and since the i and j parts are equal it points at 45° (bearing 045°).
A stone is thrown vertically upwards and its height at any time t seconds is h = 45t − 9t2. Find the maximum height reached.
Answer: D (56.25m)
dh/dt = 45 − 18t = 0 at t = 2.5 s. h = 45(2.5) − 9(2.5)2 = 112.5 − 56.25 = 56.25 m.
Given that dy/dx = 3x2 − 4 and y = 6 when x = 3, find y.
Answer: A (x3 − 4x − 9)
y = x3 − 4x + c. When x = 3: 27 − 12 + c = 6, so c = −9 and y = x3 − 4x − 9.
A company took delivery of 12 vehicles made up of 7 buses and 5 saloon cars for two of its departments; Personnel and General Administration. If the personnel department is to have at least 3 saloon cars, in how many ways can these vehicles be distributed equally between the Departments?
Answer: C (462)
Each department gets 6 vehicles. Personnel takes 3, 4 or 5 saloon cars: 5C37C3 + 5C47C2 + 5C57C1 = 350 + 105 + 7 = 462.
A bicycle wheel of diameter 70cm covered a distance of 350cm in 2 seconds. How many radians per second did it turn?
Answer: A (5)
Radius = 35 cm, so the angle turned is 350/35 = 10 radians in 2 seconds, i.e. 5 radians per second.
The initial velocity of an object is u = (−5 / 3) ms−1. If the acceleration of the object is a = (3 / −4) ms−2 and it moved for 3 seconds, find the final velocity. (Column vectors are written top / bottom.)
Answer: C ((4 / −9) ms−1)
v = u + at = (−5 / 3) + 3(3 / −4) = (−5 + 9 / 3 − 12) = (4 / −9) ms−1.
Find the maximum value of 2 + sin(θ + 25°)
Answer: C (3)
The largest value of any sine is 1, so the maximum is 2 + 1 = 3.
Simplify (1 + 2√3)2 − (1 − 2√3)2.
Answer: B (8√3)
(1 + 2√3)2 = 13 + 4√3 and (1 − 2√3)2 = 13 − 4√3. The difference is 8√3.
What is the angle between a = (3i − 4j) and b = (6i + 4j)?
Answer: B (87°)
a·b = 18 − 16 = 2, |a| = 5, |b| = √52. cos θ = 2/(5√52) ≈ 0.0555, so θ ≈ 86.8° ≈ 87°.
(a) Express 2√2/(√48 − √8 − √27) in the form p + q√r, where p, q and r are rational numbers.
(b) If V = A log10(M + N), express N in terms of M, V and A.
(a) √48 = 4√3, √8 = 2√2, √27 = 3√3, so the denominator is √3 − 2√2
Multiply top and bottom by √3 + 2√2: denominator 3 − 8 = −5, numerator 2√6 + 8
= −8/5 − (2/5)√6, i.e. p = −8/5, q = −2/5, r = 6
(b) log10(M + N) = V/A
M + N = 10V/A
N = 10V/A − M
Simplify each surd first, then rationalise with the conjugate.
If the quadratic equation (2x − 1) − p(x2 + 2) = 0 where p is a constant, has real roots
(a) show that 2p2 + p − 1 ≤ 0;
(b) find the value of p.
(a) Rearrange: px2 − 2x + (2p + 1) = 0
Real roots need b2 − 4ac ≥ 0: 4 − 4p(2p + 1) ≥ 0
1 − 2p2 − p ≥ 0, i.e. 2p2 + p − 1 ≤ 0
(b) (2p − 1)(p + 1) ≤ 0, so −1 ≤ p ≤ ½ (p ≠ 0 for the equation to be quadratic); the boundary values are p = −1 and p = ½
Use the discriminant condition b² − 4ac ≥ 0 for real roots.
The equation of a curve is y = x(3 − x2). Find the equation of its normal at the point where x = 2.
y = 3x − x3, dy/dx = 3 − 3x2
At x = 2: gradient of tangent = −9, y = 2(3 − 4) = −2
Gradient of normal = 1/9
y + 2 = (1/9)(x − 2)
x − 9y − 20 = 0
The normal is perpendicular to the tangent, so its gradient is −1 ÷ (tangent gradient).
If 3x2 + 2y2 + xy + x − 7 = 0, find dy/dx at the point (−2, 1).
Differentiate implicitly: 6x + 4y(dy/dx) + y + x(dy/dx) + 1 = 0
dy/dx = −(6x + y + 1)/(4y + x)
At (−2, 1): dy/dx = −(−12 + 1 + 1)/(4 − 2) = 10/2 = 5
Use the product rule on the xy term.
Five students are to be selected from a large population. If 60% of them are boys and the rest are girls, find the probability that:
(a) exactly 3 of them are boys;
(b) at least 3 of them are girls.
P(boy) = 3/5, P(girl) = 2/5 (binomial, n = 5)
(a) 5C3(3/5)3(2/5)2 = 10 × 27/125 × 4/25 = 216/625 (0.3456)
(b) P(3, 4 or 5 girls) = 5C3(2/5)3(3/5)2 + 5C4(2/5)4(3/5) + (2/5)5
= 720/3125 + 240/3125 + 32/3125 = 992/3125 (≈ 0.317)
Each selection is a binomial trial with p = 0.6 for a boy.
The table shows the marks obtained by groups of students in a class test.
Marks: 40−44, 45−49, 50−54, 55−59, 60−64, 65−69
No. of Students: 4, 9, 18, 23, 10, 6
(a) Draw a histogram for the distribution.
(b) Use your histogram to estimate the median of the distribution.
(a) Class boundaries 39.5, 44.5, 49.5, 54.5, 59.5, 64.5, 69.5 on the x-axis; bars of heights 4, 9, 18, 23, 10, 6, all of width 5
(b) N = 70, so the median is the 35th mark; cumulative frequencies 4, 13, 31, 54 put it in the 54.5 − 59.5 bar
Median ≈ 54.5 + (35 − 31)/23 × 5 ≈ 55.4 (an estimate of about 55 to 56 from the histogram)
Use class boundaries so the bars touch; the median splits the total area in half.
The position vector of a particle of mass 3 kg moving along a space curve is given by r = (4t3 − t2)i − (2t2 − t)j at any time t seconds. Find the force acting on it at t = 2 seconds.
v = dr/dt = (12t2 − 2t)i − (4t − 1)j
a = dv/dt = (24t − 2)i − 4j
At t = 2: a = 46i − 4j
F = ma = 3(46i − 4j) = (138i − 12j) N (magnitude ≈ 138.5 N)
Differentiate twice to get acceleration, then use F = ma.
A uniform plank PQ of length 8m and mass 10 Kg is supported horizontally at the end P and at point R, 3 metres from Q. A boy of mass 20 kg walks along the plank starting from P. If the plank is in equilibrium, calculate the (a) reactions at P and R when he walked 1.5 metres; (b) distance he had walked when the two reactions are equal; (c) distance he walked before the plank tips over. [Take g = 10ms−2].
Weights: plank 100 N at 4 m from P, boy 200 N; R is 5 m from P; RP + RR = 300 N
(a) Moments about P: 5RR = 100 × 4 + 200 × 1.5 = 700, RR = 140 N, RP = 160 N
(b) Equal reactions: 150 N each; 150 × 5 = 400 + 200x, x = 1.75 m
(c) About to tip about R when RP = 0: 200(x − 5) = 100 × 1, x = 5.5 m from P
Take moments about a support to remove one unknown reaction.
(a) The nth term of a sequence is given by Tn = 4Tn−1 − 3. If twice the third term is five times the second term, find the first three terms of the sequence.
(b) Given that the matrix (2 0 1 / 5 −3 1 / 0 4 6) multiplied by the column (1 / m / r) equals the column (k / 2 / 26), find the values of constants k, m and r. (Matrices are written row by row.)
(a) T2 = 4T1 − 3, T3 = 4T2 − 3 = 16T1 − 15
2T3 = 5T2: 32T1 − 30 = 20T1 − 15, T1 = 5/4
First three terms: 1¼, 2, 5
(b) 2 + r = k; 5 − 3m + r = 2; 4m + 6r = 26
From the second, r = 3m − 3; then 4m + 18m − 18 = 26, m = 2
r = 3, k = 5
Express T₂ and T₃ in terms of T₁; multiply out the matrix row by row.
(a) Copy and complete the table for the relation y = 2cos x + 3sin x for 0° ≤ x ≤ 360°
x: 0°, 30°, 60°, 90°, 120°, 150°, 180°, 210°, 240°, 270°, 300°, 330°, 360°
y: 2.00, 3.23, __, __, 1.6, __, __, −3.2, __, __, __, __, __
(b) Using a scale of 2cm to 30° on the x-axis and 2cm to one unit on the y-axis, draw the graph of y = 2cos x + 3sin x for 0° ≤ x ≤ 360°
(c) From your graph, find the (i) maximum value of y, correct to two decimal places; (ii) solution of the equation (2/3)cos x + sin x = 5/6
(a) y = 2.00, 3.23, 3.60, 3.00, 1.60, −0.23, −2.00, −3.23, −3.60, −3.00, −1.60, 0.23, 2.00
(b) Smooth sine-shaped curve through these points
(c)(i) Maximum y ≈ 3.61 (exact value √13 ≈ 3.606, near x ≈ 56°)
(ii) Multiply by 3: 2cos x + 3sin x = 2.5; read where the curve meets y = 2.5: x ≈ 10° and x ≈ 102°
For (c)(ii) turn the equation into y = 2.5 by multiplying by 3.
(a) If y = (2x + 3)7 + (x − 1)/(2x − 1), find the value of dy/dx when x = −1.
(b) Using the substitution, u = x + 2, evaluate ∫12 (x − 1)/(x + 2)4 dx
(a) dy/dx = 14(2x + 3)6 + [(2x − 1) − 2(x − 1)]/(2x − 1)2 = 14(2x + 3)6 + 1/(2x − 1)2
At x = −1: 14 + 1/9 = 127/9 (14 1/9)
(b) x − 1 = u − 3, dx = du, limits u = 3 to 4
∫(u−3 − 3u−4)du = [−1/(2u2) + 1/u3] from 3 to 4
= (−1/32 + 1/64) − (−1/18 + 1/27) = −1/64 + 1/54 = 5/1728
Chain rule for the power, quotient rule for the fraction; change the limits when you substitute.
The images of (3, 2) and (−1, 4) under a linear transformation T are (−1, 4) and (7, 11) respectively. P is another transformation where P : (x, y) → (x + y, x + 2y)
(a) Find the matrices T and P of the linear transformations T and P.
(b) Calculate TP.
(c) Find the image of the point x(4, 3) under TP.
(a) Let T = (a b / c d): 3a + 2b = −1, −a + 4b = 7 give a = −9/7, b = 10/7
3c + 2d = 4, −c + 4d = 11 give c = −3/7, d = 37/14
T = (−9/7 10/7 / −3/7 37/14); P = (1 1 / 1 2) (matrices row by row)
(b) TP = (1/7 11/7 / 31/14 34/7)
(c) TP(4, 3): (4/7 + 33/7, 62/7 + 102/7) = (37/7, 164/7)
Write the images as matrix equations and solve two pairs of simultaneous equations.
The table gives the distribution of marks of 60 candidates in a test.
Marks: 23−25, 26−28, 29−31, 32−34, 35−37, 38−40
frequency: 3, 7, 15, 21, 10, 4
(a) Draw a cumulative frequency curve of the distribution.
(b) From your curve, estimate the (i) 80th percentile; (ii) median (iii) semi-interquartile range.
(a) Upper class boundaries 25.5, 28.5, 31.5, 34.5, 37.5, 40.5 with cumulative frequencies 3, 10, 25, 46, 56, 60; plot and join with a smooth curve
(b)(i) 80th percentile = 48th mark ≈ 34.5 + (2/10) × 3 ≈ 35.1
(ii) Median = 30th mark ≈ 31.5 + (5/21) × 3 ≈ 32.2
(iii) Q1 = 15th mark ≈ 29.5, Q3 = 45th mark ≈ 34.4; semi-interquartile range ≈ (34.4 − 29.5)/2 ≈ 2.4 (accept readings from the curve)
Plot cumulative frequency against upper class boundaries.
(a) Eight coins are tossed once. Find, correct to three decimal places, the probability of obtaining (i) exactly 8 heads; (ii) at least 5 heads; (iii) at most 1 head.
(b) In how many ways can four letters from the word SHEEP be arranged (i) without any restriction; (ii) with only one E.
(a) Binomial with n = 8, p = ½; each outcome count over 256
(i) 1/256 ≈ 0.004
(ii) (56 + 28 + 8 + 1)/256 = 93/256 ≈ 0.363
(iii) (1 + 8)/256 = 9/256 ≈ 0.035
(b)(i) Four different letters S, H, E, P: 4! = 24; two E's with two of S, H, P: 3 × 4!/2! = 36; total 60
(ii) One E with S, H, P: 4! = 24
Count arrangements with repeated letters by dividing by the factorial of the repeats.
(a) A manufacturer produces light bulbs which are tested in the following way; A batch is accepted in either of the following cases: (i) a first sample of 5 bulbs contains no faulty bulbs; (ii) a first sample of 5 bulbs contains at least one faulty bulb but a second sample of size 5 contains no faulty bulb. If 10% of the bulbs are faulty, what is the probability that the batch is accepted?
(b) A bag contains 15 identical marbles of which 3 are black. Keshi picks a marble at random from the bag and replaces it. If this is repeated 10 times, what is the probability that he (i) did not pick a black ball? (ii) picked a black ball at most three times?
(a) P(sample of 5 has no faulty bulb) = 0.95 = 0.5905
P(accepted) = 0.5905 + (1 − 0.5905)(0.5905) = 0.5905 + 0.2418 ≈ 0.832
(b) P(black) = 1/5, P(not black) = 4/5
(i) (4/5)10 ≈ 0.107
(ii) Σk=03 10Ck(1/5)k(4/5)10−k = 0.1074 + 0.2684 + 0.3020 + 0.2013 ≈ 0.879
Add the probabilities of the two separate ways a batch can be accepted.
(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6j) and (2i + 3j) and the [distances] of the ships after t seconds are in metres, find (i) MN (ii) |MN|, when t = 3 seconds.
(b) A particle is acted upon by forces F1 = 5i + pj, F2 = qi + j, F3 = 2pi + 3j and F4 = −4i + qj, where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the value of p and q.
(a) Position at time t: M = (3 + 5t)i + (7 + 6t)j, N = (4 + 2t)i + (5 + 3t)j
(i) MN = N − M = (1 − 3t)i − (2 + 3t)j; at t = 3, MN = −8i − 11j
(ii) |MN| = √(64 + 121) = √185 ≈ 13.6 m
(b) Sum of forces = 0
i: 5 + q + 2p − 4 = 0, so 2p + q = −1
j: p + 1 + 3 + q = 0, so p + q = −4
p = 3, q = −7
Position at time t = starting position + t × velocity; in equilibrium the i and j parts each add to zero.
A particle moves from point O along a straight line such that its acceleration at any time, t seconds is a = (4 − 2t)ms−2. At t = 0, its distance from O is 18 metres while its velocity is 5ms−1,
(a) At what time will the velocity be greatest?
(b) calculate the; (i) time; (ii) distance of the particle from O when the particle is momentarily at rest.
v = ∫(4 − 2t)dt = 4t − t2 + 5 (v = 5 at t = 0)
s = 2t2 − t3/3 + 5t + 18 (s = 18 at t = 0)
(a) v is greatest when a = 0: t = 2 s (v = 9 ms−1)
(b)(i) v = 0: t2 − 4t − 5 = 0, (t − 5)(t + 1) = 0, t = 5 s
(ii) s = 50 − 125/3 + 25 + 18 = 51⅓ m
Integrate acceleration to get velocity and again to get distance, using the starting values for the constants.
(a) The position vectors of the points X and Y are x = (−2i + 5j) and y = (i − 7j) respectively. Find (i) (3x + 2y); (ii) (y − 2x); (iii) the angle between x and y; (iv) the unit vector in the direction of (x + y).
(b) A bullet of mass 0.084kg is fired horizontally into a 20 kg block of wood at rest on a smooth floor. If they both move at a velocity of 0.24ms−1 after impact, calculate, correct to two decimal places, the initial velocity of the bullet.
(a)(i) 3x + 2y = (−6i + 15j) + (2i − 14j) = −4i + j
(ii) y − 2x = (i − 7j) − (−4i + 10j) = 5i − 17j
(iii) x·y = −2 − 35 = −37; |x| = √29, |y| = √50; cos θ = −37/√1450 ≈ −0.9717, θ ≈ 166.3°
(iv) x + y = −i − 2j, |x + y| = √5; unit vector = (−i − 2j)/√5
(b) Momentum: 0.084u = (20 + 0.084)(0.24) = 4.82016
u ≈ 57.38 ms−1
Use the dot product for the angle and conservation of momentum for the bullet.