NAME: Mr DULK Marking Scheme

MATOPO HIGH SCHOOL(BICC)

General Certificate of Education Ordinary Level

PURE MATHEMATICS 4027/1

Paper 1

Holiday Assignment 2020 SESSION 2hours 30 minutes

Candidates answer on the question paper.

Additional materials: Data booklet MF7

Mathematical tables/ electronic calculators

Allow candidates 5 minutes to count pages before the examination

TIME 2hours 30 minutes

INSTRUCTIONS TO CANDIDATES

Answer all questions in this paper

Check that all pages are in the booklet and ask the invigilator for a replacement if there are duplicate or
missing pages.

Write your answers in the spaces provided on the question paper using black or blue pens.

If working is needed for any question it must be shown in the space below that question.

Omission of essential working will result in loss of marks.

Decimal answers which are not exact should be given correct to three significant figure unless stated
otherwise. Decimal answers in degrees should be given to one decimal place.

INFORMATION FOR CANDIDATES

The number of marks is given in brackets [] at the end of each question or part question.

Copyright: DULKtec Exam Board, 2020

1. Simplify
1 49
(a) (3 16)3/4 = (16)3/4 1

4
√49 3
= (4 )
√16

√7
= ( 2 )3 1

7√7
= 1
8

1 1 1+ √2
(b) = 1− × 1+ 1
1− √2 √2 √2

1+ √2
= 1− 2

1+ √2
= 1
−1

= −1(1 + √2) or −1 − √2 1

2. a. Find the value of A and the value of B if,

A(x + 3) + B(x + 2) ≡ 4x + 9

Let x = − 3; B(−3 + 2) = − 12 + 9

B=3 2 marks for any correct value

Let x = − 2; A(−2 + 3) = − 8 + 9 1 mark for the other value

A=1
 4𝑥 + 9
(b). Hence or otherwise, express in partial fractions.
(𝑥 + 3)(𝑥 + 2)

Since we can re arrange the identity in question 2(a) above and rewrite it as,
𝐀 𝐁 4𝑥 + 9
+ (𝑥 + 3) ≡ (𝑥 + 2)(𝑥 + 3) , by simply dividing everything by (𝑥 + 2)(𝑥 + 3)
(𝑥 + 2)

Since we have already calculated the values of A and B in 2(a), then we are simply substituting
and get the final answer,
4𝑥 + 9 𝟏 𝟑
≡ (𝑥 + 2) + (𝑥 + 3) 3marks
(𝑥 + 2)(𝑥 + 3)

3a. Show that (2x− 1) is a factor of f(x) = 2x3 + x2 + x – 1 ,

2x− 1 is a factor → 0.5 is a root.

∴ If 2x− 1 is a factor of f(x) the f(0.5) must be zero

f(0.5) = 2(0.5)3 + 0.52 + 0.5 – 1 1

f(0.5) = 0.25 + 0.25 + 0.5 – 1

f(0.5) =0 hence 2x− 1 is a factor of f(x) since f(0.5) = 0. 1

(b) Hence find the quadratic factor of f(x). x2 + x + 1

2x− 1 2x3 + x2 + x – 1
2x3 − x2
2x2 + x
2marks
2x2 − x
2x − 1
2x − 1
0 0
∴ f(x) = (2x− 1)( x2 + x + 1). Hence the quadratic factor is x2 + x + 1 1mark
 4a. Express 3x2 + 12x +5 , in the form a(x + b)2 + c, where a, b and c are constants to be
found.

3x2 + 12x +5 ≡ 3(x2 + 4) + 5

≡ 3(x2 + 4 + 22 - 22) + 5 1marks for method

≡ 3(x + 2) 2 + 5 – 12 accept any other correct method

≡ 3(x + 2) 2 – 7 1 for the answer

(b) Hence find the coordinates of the turning point of 3x2 + 12x +5

At the turning point, 3(x + 2) 2 = o

x+2=0 →x=−2 1mark

y - coordinate = − 7

∴ Turning point is ( − 2 ; − 7) 1mark

(c) State with a reason the general shape of the graph of 3x2 + 12x +5

It is a U – shaped graph because the coefficient of x2 is positive or is greater than zero

1mark for shape

1mark for the reason

5a. Solve the simultaneous equations, y = 4x

y = x2 + 3,

using the substitution method, 4x= x2 + 3 1mark

rearranging and solving for x, x2 - 4x+ 3 = 0

x2 − x – 3x + 3= 0

(x – 1)(x – 3) = 0

x = 1 or 3, but y = 4x substituting the values of x, 2marks for all correct values of x

 when x = 1, y = 4(1) =4

when x = 3, y = 4(3) = 12 2marks, each correct value of y carries 1mark

when x = 1, y = 4

when x = 3, y = 12

(b) Solve the equation, Sin 2x = 0,4751, for 0o<x≤ 360o.

2x = Sin – 1 0,4751

2x = 28.4 using 180on +(-1)n ∝, where ∝= 28.4 to 1d.p and n = 0,1,2,3,…

2x = 28.4o; 151.6o; 388.4o; 511.6o; 748.4o, then dividing throughout by 2.

x = 14.2 o; 75.8o; 194.2o; 255.8 1 mark for each correct value of x = 4marks

6a. Find the coordinates of the mid-point of a line AB if A is (2; - 5) and B is (4;1).
2+4 − 5+1
mid-point = ( ; ) 1
2 2

= (3 ; − 2) 1

(b) Find the gradient of line AB

−5−1
Gradient = 1
2−4

−6
= −2

Gradient = 3 1

(c). Find the gradient of the line perpendicular to AB

Since the line is perpendicular to line AB then its gradient is a negative reciprocal of the gradient
−1
of line AB, ie its gradient = 3
2marks
 6d. Hence find of the equation of the perpendicular bisector of the line AB

Perpendicular bisector of line AB passes through the midpoint of AB which is (3 ; − 2)

−1
ant it has gradient therefore we are now introducing a second point (x;y) and calculate the
3
gradient of the line as if we don’t know the gradient,
𝑦− −2 −1
Gr = , but gr =
𝑥−3 3

−1 𝑦+ 2
∴ = , simplifying, 1
3 𝑥−3

3y = - x + 3 – 6

3y = - x – 3 or 3y + x = – 3 or 3y + x + 3 = 0 1

(e) Find the length of line AB

|𝐴𝐵| = √(1— −5)2 + (4 − 2)2 1

= √(6)2 + (2)2

= √36 + 4

=√40 1

|𝐴𝐵| = 2√10 units

(c). Solve the equation, 3(2x – 1) = 9 – x

3(2x – 1) = 3 –2 x 1

2x – 1 = –2 x 1

4x = 1
1
x = 0.25 or 4 1
 7a. Solve the equations, (2x – 1)(x2 – x – 12) = 0

(2x – 1)(x – 4)(x+3) = 0

x = 0.5 or 4 or – 3 1mark for each value = 3marks

(b) Hence solve the inequality, (2x – 1)(x2 – x – 12) ≥ 0

Critical values are 0.5 or 4 or – 3,

𝑥 ≤ −3 −3 ≤ 𝑥 ≤ 0.5 0.5 ≤ 𝑥 ≤ 4 𝑥≥4

2x – 1 − − + +
x+3 − + + +
x-4 − − − +
f(x) − + − +
2marks

Solution, −3 ≤ 𝑥 ≤ 0.5 ∪ 𝑥 ≥ 4 2marks

8. The third and 7th terms of an arithmetic progression are71 and 55 respectively.

(a) Find (i) the first term and the common difference.

T3 = a + (3 – 1)d, but T3 = 71

∴ 71 = a + 2d equation 1 1mark

T7 = a + (7 – 1)d, but T7 = 55

∴ 55 = a + 6d equation 2 1mark

Solving equation 1 and equation 2 simultaneously,

71 = a + 2d 1mark

55 = a + 6d

Subtracting equation 2 from 1 we get 16 = − 4d →d=−4 1mark

Substituting – 4 for d in any of the two equations, 1 or 2, eg

71 = a + 2( − 4 ) → a = 71 + 8 = 79 1mark

First term (a) = 79

Common difference (d) = − 4

 (ii) the sum of the first 45 terms
45
S45 = [2(79) + (45– 1)( − 4 )] 2marks
2

45
= [158 − 176]
2

S45 = − 405 1mark

(c). Find the fifth term of the progression

T5 = 79 + (5– 1)( − 4 ) 1mark

= 79 – 16 1mark

T5 = 63 1mark

9. (a) Differentiate y = 2x3 + 3x2 - 12x + 6 with respect to x.

𝑑𝑦
= 6x2 + 6x − 12 2marks
𝑑𝑥

(b) Hence find the coordinates of the points on the curve y = 2x3 + 3x2 − 12x + 6 for which y
has a stationary value.
𝑑𝑦
At the point where y has stationary points, =0
𝑑𝑥

∴ 6x2 + 6x − 12 = 0

x2 + x − 2 = 0, by dividing throughout by 6.

x2 + 2x – x − 2 = 0

(x – 1)(x + 2) = 0 → x = 1 or – 2 2marks

When x = 1, y = 2(1)3 + 3(1)2 − 12(1) + 6 → y=−1 1mark

When x = – 2, y = 2(– 2)3 + 3(– 2)2 − 12(– 2) + 6 → y = 26 1mark

∴ coordinates are (1 ; − 1) and ( − 2 ; 26)

 (c). For each of the points in (b) determine whether the value of y is maximum or
minimum.

d2y/dx2 = 12x + 6

at (1 ; − 1), d2y/dx2 = 12(1) +6

= 18 (+ve) 1mark

d2y/dx2 > 0

∴ at (1 ; − 1), y has a minimum value. 1mark

at ( − 2 ; 26) d2y/dx2 = 12(- 2 ) +6

= − 18 (−ve) 1mark

d2y/dx2 < 0

∴ at ( − 2 ; 26), y has a maximum value. 1mark

10. (a) Find the integral of ∫(4𝑥 + 5)dx

4
∫(4𝑥 + 5)dx = 2 x2 +5x + k 1mark

= 2x2 +5x + k 1mark

2
(b) Evaluate, ∫0 3𝑥 2 dx
2
2
∫0 3𝑥 2 dx = 𝑥 2 1mark
0
= (2)2 – (0)2 1mark

= (4) – 0
2
∫0 3𝑥 2 dx = 4 1mark

11. (a). The first term of a Geometric series is 3 and the common ratio is 0.8. Find the sum of the
first 24 terms, giving your answer to 3 significant figures.
3 3
S24 = 1−0.8 [1 – 0.824] = 0.2 [1 – 0.824] = 15[1 – 0.824] = 15(0.995277633)

= 14.9291645 ≈ 14.9 to 3s.f 4marks

 2
11. (b) Given that f: x→x2 – x − 6 and g: x→ , find
𝑥−2

2
(i)gf = 𝑥 2 – 𝑥 − 6 1mark
−2

2
= 𝑥2 – 𝑥 − 8 1mark

(ii) g – 1 , the inverse of g.

2
Let y = 𝑥−2 , making x the subject

y(x – 2) = 2

xy – 2y = 2 1mark

xy = 2 + 2y
2+2𝑦
x= 1mark
𝑦

2+2𝑥
∴ g–1 = 1mark
𝑥

11. (c). Sketch f showing clearly the axes intercepts and the turning point.

f: x→x2 – x − 6 , we can express f as f: x→(x+2)(x – 3)

∴ x – axis intercepts = ( −2 ; 0 ) and (3;0). And when x = 0, y = - 6 hence (0 , - 6) - y – intercept.

𝑑𝑦 1 −25 1 −25
= 2x – 1 then x = 2 Hence y = → turning point = (2 ; )
𝑑𝑥 4 4
y

y = x2 – x − 6

x
−2 3
[4]

−6
1 −25
(2 ; )
4
 12. Given that OP = 3i + 4k, OQ = i + pj + k and OR = 2i + 3j + 3k. Find

(a) → = OQ - OP
𝑃𝑄

1 3
= (𝑝) – (0) 1mark
1 4
−2
=( 𝑝 ) 1mark
−3

(b) the unit vector parallel to →

𝑃𝑄

|𝑃𝑄| = √(−2)2 + 𝑝2 + (−3)2 1mark

= √13 + 𝑝2 1mark
−2𝑖+𝑝𝑗−3𝑘
Therefore unit vector parallel to PQ = or any other correct vector form. 1mark
√13+𝑝2

(c). the value of p for which OQ is perpendicular to PQ.

OQ PQ = 0

-2 + p2 – 3 = 0

p2 = 5

p = ±√5 4marks