Question 1
SurdsExpress 3/(3 − √6) in the form x + m√y.
- A.3 + 3√6
- B.3 + √6
- C.3 − √6
- D.3 − 3√6
Show the answer and explanation
Answer: B (3 + √6)
Multiply top and bottom by (3 + √6): 3(3 + √6)/(9 − 6) = 3 + √6.
53 questions. Try each one, then open the answer to check yourself and read the explanation.
Express 3/(3 − √6) in the form x + m√y.
Answer: B (3 + √6)
Multiply top and bottom by (3 + √6): 3(3 + √6)/(9 − 6) = 3 + √6.
If (1/9)2x−1 = (1/81)2−3x, find the value of x.
Answer: B (5/8)
Write both sides as powers of 3: 3^(−4x + 2) = 3^(−8 + 12x). So −4x + 2 = 12x − 8, 16x = 10 and x = 5/8.
If (x − 5) is a factor of x3 − 4x2 − 11x + 30, find the remaining factors.
Answer: C ((x + 3) and (x − 2))
Dividing by (x − 5) leaves x² + x − 6 = (x + 3)(x − 2).
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?
Answer: C (Mary walked to school because the school bus arrived late.)
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'.
Evaluate: ∫01 x(x2 − 2)2 dx.
Answer: C (1⅙)
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.
In how many ways can a committee of 3 women and 2 men be chosen from a group of 7 men and 5 women?
Answer: D (210)
⁵C₃ × ⁷C₂ = 10 × 21 = 210.
If α and β are the roots of 7x2 + 12x − 4 = 0, find the value of αβ/(α + β)2.
Answer: C (−7/36)
α + β = −12/7 and αβ = −4/7. αβ/(α + β)² = (−4/7) ÷ (144/49) = −4/7 × 49/144 = −7/36.
If f: x → 2 tan x and g: x → √(x2 + 8), find (g o f)(45°)
Answer: D (2√3)
f(45°) = 2 tan 45° = 2. Then g(2) = √(4 + 8) = √12 = 2√3.
An exponential sequence (G.P.) is given by 9/2, 3/4, 1/8, …. Find its sum to infinity
Answer: B (5⅖)
r = (3/4) ÷ (9/2) = 1/6. S∞ = a/(1 − r) = (9/2) ÷ (5/6) = 27/5 = 5⅖.
Given that M is the midpoint of T(2, 4) and Q(−8, 6), find the length MQ.
Answer: B (√26 units)
M = (−3, 5). MQ = √((−8 + 3)² + (6 − 5)²) = √(25 + 1) = √26.
Given that y2 + xy = 5, find dy/dx.
Answer: B (−y/(2y + x))
Differentiate implicitly: 2y dy/dx + x dy/dx + y = 0, so dy/dx = −y/(2y + x).
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′).
Answer: B ({x : 2 ≤ x ≤ 4})
Q′ = {x ≤ 4 or x > 12}. The part of P (2 to 8) that lies in Q′ is 2 ≤ x ≤ 4.
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.
Answer: B (13/54)
P(Atta only) = (1/6)(8/9) = 8/54; P(Tunde only) = (5/6)(1/9) = 5/54. Total 13/54.
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.
Answer: A (1)
Mean = (2 + 2k + 3 + 4 + 10)/(6 + k) = 3, so 19 + 2k = 18 + 3k and k = 1.
If 3x2 + px + 12 = 0 has equal roots, find the values of p.
Answer: D (±12)
Equal roots need p² − 4(3)(12) = 0, so p² = 144 and p = ±12.
Simplify: (log √27 − log √8)/(log 3 − log 2).
Answer: D (3/2)
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.
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.
Answer: B (3 sec.)
v = u − at: 0 = 27 − 9t, so t = 3 s.
In how many ways can four Mathematicians be selected from six?
Answer: A (15)
Order doesn't matter: ⁶C₄ = 15.
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.)
Answer: C ((4 0 / 0 4))
P = (2 1 / 0 −2). P² = (2×2 + 1×0, 2×1 + 1×(−2) / 0, (−2)(−2)) = (4 0 / 0 4).
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.
Answer: C (11.4 N)
The direction doesn't change, so a = (50 − 10)/16 = 2.5 ms⁻² and F = 4.56 × 2.5 = 11.4 N.
Find the fifth term in the binomial expansion of (q + x)7.
Answer: C (35q3x4)
The fifth term is ⁷C₄ q³x⁴ = 35q³x⁴.
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.)
Answer: D (2, 3/2)
|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.
Solve 6 sin 2θ tan θ = 4, where 0° < θ < 90°.
Answer: A (35.26°)
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°.
Calculate, correct to one decimal place, the angle between 5i + 12j and −2i + 3j.
Answer: B (56.3°)
Dot product = −10 + 36 = 26; magnitudes 13 and √13. cos θ = 26/(13√13) ≈ 0.5547, so θ ≈ 56.3°.
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.
Answer: D (41 ms−2)
v = 15t² − 19t + 6 and a = 30t − 19. At t = 2, a = 41 ms⁻².
Find the equation of the normal to the curve y = 3x2 + 2 at point (1, 5).
Answer: B (6y + x − 31 = 0)
dy/dx = 6x = 6 at x = 1, so the normal has gradient −1/6. y − 5 = −(x − 1)/6 gives 6y + x − 31 = 0.
Adu's scores in five subjects in an examination are 85, 84, 83, 86 and 87. Calculate the standard deviation.
Answer: A (1.4)
Mean = 85. Squared deviations 0, 1, 4, 1, 4 add to 10. SD = √(10/5) = √2 ≈ 1.4.
Differentiate f(x) = 1/(1 − x2)5 with respect to x.
Answer: A (10x/(1 − x2)6)
f(x) = (1 − x²)⁻⁵, so f′(x) = −5(1 − x²)⁻⁶ × (−2x) = 10x/(1 − x²)⁶.
If X and Y are two independent events such that P(X) = 1/8 and P(X ∪ Y) = 5/8, find P(Y).
Answer: A (4/7)
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.
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.
Answer: C (z)
Row z and column z reproduce the other element every time (z * a = a * z = a), so z is the identity.
Find the radius of the circle 2x2 + 2y2 − 4x + 5y + 1 = 0
Answer: D (√33/4)
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.
Evaluate: ∫(2x + 1)3 dx.
Answer: D (⅛(2x + 1)4 + k)
∫(2x + 1)³ dx = (2x + 1)⁴/(4 × 2) + k = ⅛(2x + 1)⁴ + k.
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]
Answer: A (2.0)
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.
If m and (m + 4) are the roots of 4x2 − 4x − 15 = 0, find the equation whose roots are 2m and (2m + 8).
Answer: D (x2 − 2x − 15 = 0)
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.
Find the coefficient of the 6th term in the binomial expansion of (1 − 2x/3)10 in ascending powers of x.
Answer: D (−896x5/27)
The 6th term is ¹⁰C₅(−2x/3)⁵ = 252 × (−32/243)x⁵ = −896x⁵/27.
An exponential sequence (G.P.) is given by 8√2, 16√2, 32√2, …. Find the nth term of the sequence
Answer: D (2(n + 2)√2)
a = 8√2, r = 2. Tₙ = 8√2 × 2ⁿ⁻¹ = 2³ × 2ⁿ⁻¹ × √2 = 2ⁿ⁺²√2.
Given that r = (10 N, 200°) and n = (16 N, 020°), find (3r − 2n).
Answer: A ((62 N, 200°))
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°).
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)
Answer: D (63/65)
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.
If 9Cx = 4[7Cx−1], find the values of x.
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.
Write both sides in factorials and cancel the common parts.
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.
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.
Connect the two rates with the chain rule.
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.
(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).
The inverse of f⁻¹ is f itself.
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.
(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.
Use S = n(a + l)/2 when the first and last terms are known.
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.
(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.
With unequal widths, plot frequency density, not frequency.
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.
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.
'More girls than boys' in a group of 5 means 3, 4 or 5 girls.
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.
(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.
Choose the equation of motion that contains the quantities you know.
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.)
P is the midpoint of NO, so MP = ½(MN + MO).
MP = ½(22i − 2j) = 11i − j.
The position of a midpoint is the average of the two end vectors.
(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).
(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).
From first principles: find δy, divide by δx, then let δx → 0.
(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.
(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.
Use the remainder theorem: the remainder on dividing by (x − c) is g(c).
(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.
(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.
A repeated factor needs a fraction for each power of it.
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.
(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.
Plot cumulative frequencies at the upper class boundaries.
(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.
(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.
Deviations from the mean always sum to zero.
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.
(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.
Work with the i and j parts together as vectors.
(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.
(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.
Balance clockwise and anticlockwise moments; friction opposes the motion about to happen.