2019-1-NS-KGV

Section A[45 marks] Answer all questions in this section.

1. Sketch on the same coordinates axes, the graphs of 𝑦 = |𝑥 − 2| and 𝑦 = √𝑥 + 4, 𝑥 ≥ −4.

Hence, solve the inequality |𝑥 − 2| > √𝑥 + 4 . [5 marks]

1+2𝑥
2. Expand √ 1−𝑥 as a series in ascending powers of x, up to and including the term in 𝑥 2 . State

the range of x for which the expansion is valid. By taking 𝑥 = 0.02, use the first three non-
zero terms of the series to estimate √13 correct to four significant figures. [7 marks]

2𝑏 + 1 𝑎 𝑐−2
3. a) Determine the value of a so that matrix ( 2𝑐 𝑎2 𝑏𝑐 ) is a symmetric matrix.
8 6 2𝑐 + 1

[2 marks]
−2 1 𝑏
b) Find the value of b if matrix ( 1 3 0 ) is a singular matrix. [3 marks]
−1 11 −6

4. Express √3 + 𝑖 in the polar form. [3 marks]

12 12
Use de Moivre’s theorem, show that (√3 + 𝑖 ) + (√3 − 𝑖 ) = 213 . [4 marks]
12 12
Hence, find the roots of 2𝑧 3 − [(√3 + 𝑖 ) + (√3 − 𝑖 ) ] 𝑖 = 0 in the Cartesian form.

[3 marks]

5. The parametric equations of a conic are 𝑥 = 4 sin 𝜃 + 5 and 𝑦 = 3 cos 𝜃 − 2 and

0 ≤ 𝜃 ≤ 2𝜋.
a) Find the standard form for the equation of the conic, and identify the type of conic.
[3 marks]
b) Find the centre and foci of the conic. [3 marks]
c) Sketch the conic and indicate its centre and foci. [2 marks]
6. The planes 𝜋1 and 𝜋2 have equations 3𝑥 − 𝑦 − 𝑧 = 2 and 𝑥 + 5𝑦 + 𝑧 = 14 respectively.
Show that the point (4, 0, 10) lies in both planes. Show that the line of intersection, 𝑙1 of 𝜋1
and 𝜋2 has equation 𝒓 = 4𝒊 + 10𝒌 + (𝒊 − 𝒋 + 4𝒌). [5 marks]
The line 𝑙2 has equation 𝒓 = 3𝒋 + 9𝒌 + 𝜇(𝒊 − 3𝒋 − 𝒌).
a) Show that 𝑙1 and 𝑙2 are perpendicular. [2 marks]
b) Find the coordinates of the point of intersection Q, between 𝑙2 and 𝜋1 . [3 marks]

7. The polynomial 𝑝(𝑥) = 𝑎𝑥 3 − (1 + 6𝑎)𝑥 2 + 2(8𝑎 − 𝑏)𝑥 − 27 where a and b are real
constants, has a factor (𝑥 − 3).
a) Show that 7𝑎 − 2𝑏 = 12. [2 marks]
b) Write 𝑝(𝑥) in terms of a and x. Find the quotient in terms of a and x when 𝑝(𝑥) is
divided by (𝑥 − 3). [3 marks]
c) When 𝑝(𝑥) is divided by (𝑥 − 4), it leaves a remainder of 13, find the value of a.
[2 marks]
d) With this value of a,
i) show that 𝑝(𝑥) = 0 has only one real root. Find the set of values of x for which
𝑝(𝑥) > 0. [4 marks]
4𝑥−9
ii) express in partial fractions. [4 marks]
𝑝(𝑥)

8. The variables x, y and z satisfy the system of linear equations.

3𝑥 + 2𝑦 + 𝑧 = 1
9𝑥 + 6𝑦 + 4𝑧 = 𝑘 2
6𝑥 + 4𝑦 − 2𝑧 = 𝑘
where k is a real constant.
a) Write a matrix equation for the system of linear equations. [1 mark]
b) Reduce the augmented matrix to row-echelon form. [4 marks]
c) Determine the values of k for which the system of linear equations has infinitely many
solutions, and find the solutions in the case when k is negative. [7 marks]
d) Find the set of values of k for which the system of linear equation is inconsistent.
[3 marks]
2019-1-NS-KGV-MARKING SCHEME

1 D1
𝑦 = |𝑥 − 2|
D1

𝑦 = √𝑥 + 4

𝑥 − 2 = √𝑥 + 4 or −(𝑥 − 2) = √𝑥 + 4 M1

𝑥 = 0,5 A1

{𝑥: 𝑥 < 0, 𝑥 > 5} A1

2 1 1
1+2𝑥
√ = (1 + 2𝑥)2 (1 − 𝑥)−2
1−𝑥

1 1 1
1 (− )
(1 + 2𝑥)2 = [1 + (2𝑥) + 2 2 (2𝑥)2 + ⋯ ] OR M1
2 2!

1 1 3
1 − (− )
(1 − 𝑥)−2 = [1 + (− ) (−𝑥) + 2 2
(−𝑥)2 + ⋯ ]
2 2!

1 1 1 3
1 (− ) 1 − (− )
= [1 + 2 (2𝑥) + 2 2
(2𝑥)2 + ⋯ ] [1 + (− 2) (−𝑥) + 2 2
(−𝑥)2 + ⋯ ] M1
2! 2!

1 1 3
= (1 + 𝑥 − 2 𝑥 2 + ⋯ ) (1 + 2 𝑥 + 8 𝑥 2 + ⋯ )

1 3 1 1
= 1 + 2 𝑥 + 8 𝑥2 + 𝑥 + 2 𝑥2 − 2 𝑥2 + ⋯

3 3 A1
= 1 + 2 𝑥 + 8 𝑥2 + ⋯
 1
for |2𝑥| < 1, |𝑥| < 2 and for |𝑥| < 1

1 A1
Expansion is valid for |𝑥| < 2

1+2(0.02) 3 3
√ 1−(0.02) ≈ 1 + 2 (0.02) + 8 (0.02)2 M1

52
√ ≈ 1.0302 M1
49

2√13 ≈ 7(1.0302)

√13 ≈ 3.6057
A1
= 3.606 (4 significant figures)
7

3a) 2𝑏 + 1 𝑎 𝑐−2
( 2𝑐 𝑎2 𝑏𝑐 )
8 6 2𝑐 + 1
M1
𝑐−2=8
A1
𝑎 = 2(10) = 20

b) −2 1 𝑏
(1 3 0)
−1 11 −6
−2 1 𝑏
|1 3 0 |=0 M1
−1 11 −6

−2[−18 − 0] − 1[−6 − 0] + 𝑏[11 − (−3)] = 0 M1

𝑏 = −3 A1

5
4
Let 𝑧 = √3 + 𝑖

2 1 𝜋 M1
|𝑧| = √(√3) + 12 = 2 and Arg z = tan−1 = M1
√3 6

𝜋 𝜋 A1
𝑧 = 2 (cos 6 + 𝑖 sin 6 )

12 𝜋 𝜋 12
(√3 + 𝑖 ) = [2 (cos 6 + 𝑖 sin 6 )] OR
12
M1
12 𝜋 𝜋
(√3 − 𝑖 ) = [2 (cos (− 6 ) + 𝑖 sin (− 6 ))]

12 12 𝜋 𝜋 12 𝜋 𝜋 12
(√3 + 𝑖 ) + (√3 − 𝑖 ) = [2 (cos 6 + 𝑖 sin 6 )] + [2 (cos (− 6 ) + 𝑖 sin (− 6 ))] M1

12𝜋 12𝜋 12𝜋 12𝜋

= 212 [cos + 𝑖 sin ] + 212 [cos (− ) + 𝑖 sin (− )] M1
6 6 6 6

= 212 [cos 2𝜋 + 𝑖 sin 2𝜋] + 212 [cos(−2𝜋) + 𝑖 sin(−2𝜋)]

= 212 [1 + 𝑖(0)] + 212 [1 + 𝑖(0)] = 213 A1

12 12
2𝑧 3 − [(√3 + 𝑖 ) + (√3 − 𝑖 ) ] 𝑖 = 0

2𝑧 3 − 213 𝑖 = 0  2𝑧 3 = 213 𝑖  𝑧 3 = 212 𝑖 M1

𝜋 𝜋
𝑧 3 = 212 (0 + 𝑖) = 212 (cos 2 + 𝑖 sin 2 )

1
1
𝜋 𝜋 3
𝑧= (212 )3 [cos ( 2 + 2𝑘𝜋) + 𝑖 sin ( 2 + 2𝑘𝜋)] M1

𝜋 2𝑘𝜋 𝜋 2𝑘𝜋
𝑧 = 16 [cos ( 6 + ) + 𝑖 sin ( 6 + )] , k = 0, 1, 2
3 3

𝜋 𝜋 √3 1
𝑘 = 0, 𝑧 = 16 [cos ( 6 ) + 𝑖 sin (6 )] = 16 ( 2 + 𝑖 2) = 8(√3 + 𝑖)

5𝜋 5𝜋 √3 1
𝑘 = 1, 𝑧 = 16 [cos ( 6 ) + 𝑖 sin ( 6 )] = 16 (− + 𝑖 2) = 8(−√3 + 𝑖)
2

3𝜋 3𝜋
𝑘 = 2, 𝑧 = 16 [cos ( 2 ) + 𝑖 sin ( 2 )] = 16(0 − 𝑖) = −16𝑖

The roots are 8(√3 + 𝑖), 8(−√3 + 𝑖) and −16𝑖. A1

10
5a)
𝑥 = 4 sin 𝜃 + 5 and 𝑦 = 3 cos 𝜃 − 2
𝑥−5 𝑦+2 M1
sin 𝜃 = or cos 𝜃 =
4 3

Using sin2 𝜃 + cos 2 𝜃 = 1

𝑥−5 2 𝑦+2 2
( ) +( ) =1
4 3

(𝑥−5)2 (𝑦+2)2 A1
+ =1
16 9

This is an ellipse. A1
b
𝑎 = 4, 𝑏 = 3

𝑐 2 = 42 − 32  𝑐 = √7, −√7 M1
A1
Centre (5, −2)
A1
Foci (5 + √7, −2) and (5 − √7, −2)
c)

Label
– D1
𝐶(5, −2)
Shape
– D1
𝐹(5 − √7, −2) 𝐹(5 + √7, −2)

8
6 When x = 4, y = 0, z = 10

3𝑥 − 𝑦 − 𝑧 = 3(4) − 0 − 10 = 2 or 𝑥 + 5𝑦 + 𝑧 = 4 + 5(0) + 10 = 14 M1

Hence, the point (4, 0, 10) lies in both planes A1

3 1
𝒏1 = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝜋1 = (−1) or 𝒏2 = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝜋2 = (5)
−1 1
𝒊 𝒋 𝒌
𝒏1 × 𝒏2 = |3 −1 −1| = (−1 + 5)𝒊 − (3 + 1)𝒋 + (15 + 1)𝒌 M1
1 5 1
= 4𝒊 − 4𝒋 + 16𝒌

 Direction of line of intersection is 𝒊 − 𝒋 + 4𝒌 M1

Since point (4, 0, 10) lies in both planes, therefore (4, 0, 10) lies on the line of M1
intersection.

Equation of line of intersection between the two planes is 𝒓 = 4𝒊 + 10𝒌 + (𝒊 − A1

𝒋 + 4𝒌) [Shown]

a) 𝑙1 : 𝒓 = 4𝒊 + 10𝒌 + (𝒊 − 𝒋 + 4𝒌) and 𝑙2 : 𝒓 = 3𝒊 + 9𝒌 + (𝒊 − 3𝒋 − 𝒌)

1 1
(−1) ∙ (−3) = 1 + 3 − 4 = 0 M1
4 −1
Hence, 𝑙1 and 𝑙2 are perpendicular. A1

b) 0 1 3
𝑙2 : 𝒓 = (3) + 𝜇 (−3) and 𝜋1 : 𝒓 ∙ (−1) = 2
9 −1 −1
𝜇 3
( 3 − 3𝜇 ) ∙ (−1) = 2 M1
9−𝜇 −1

3𝜇 − 3 + 3𝜇 − 9 + 𝜇 = 2  𝜇 = 2 M1

2
(3 − 3(2)) Coordinates of Q = (2, -3, 7) A1
9−2
10
7a)
𝑝(3) = 𝑎(3)3 − (1 + 6𝑎)(3)2 + 2(8𝑎 − 𝑏)(3) − 27 = 0 M1

27𝑎 − 9 − 54𝑎 + 48𝑎 − 6𝑏 − 27 = 0

7𝑎 − 2𝑏 = 12 A1
b)
𝑝(𝑥) = 𝑎𝑥 3 − (1 + 6𝑎)𝑥 2 + 2(8𝑎 − 𝑏)𝑥 − 27

𝑝(𝑥) = 𝑎𝑥 3 − (1 + 6𝑎)𝑥 2 + (16𝑎 + 12 − 7𝑎)𝑥 − 27

A1
𝑝(𝑥) = 𝑎𝑥 3 − (1 + 6𝑎)𝑥 2 + 3(3𝑎 + 4)𝑥 − 27

𝑎𝑥 3 − (1 + 6𝑎)𝑥 2 + 3(3𝑎 + 4)𝑥 − 27 = (𝑥 − 3)(𝑎𝑥 2 + 𝑞𝑥 + 9)

Comparing coefficients of 𝑥 2 : −(1 + 6𝑎) = 𝑞 − 3𝑎 M1

𝑞 = −(3𝑎 + 1)
A1
The quotient = 𝑎𝑥 2 − (3𝑎 + 1)𝑥 + 9
c)
Remainder theorem 𝑝(4) = 13

𝑎(4)3 − (1 + 6𝑎)(4)2 + 3(3𝑎 + 4)(4) − 27 = 13 M1

64𝑎 − 16 − 96𝑎 + 36𝑎 + 48 − 27 = 13

𝑎=2 A1
di)
𝑝(𝑥) = (𝑥 − 3)(2𝑥 2 − 7𝑥 + 9) = 0

𝑥 − 3 = 0  𝑥 = 3 (real root)

2𝑥 2 − 7𝑥 + 9 = 0

𝑏 2 − 4𝑎𝑐 = (−7)2 − 4(2)(9) = −23 < 0 (no real roots) M1

Therefore 𝑝(𝑥) = 0 has only one real root (shown)

(𝑥 − 3)(2𝑥 2 − 7𝑥 + 9) > 0
7 9
(𝑥 − 3) [2 (𝑥 2 − 𝑥 + )] > 0
2 2

7 2 49 9 7 2 23 M1
(𝑥 − 3) [2 ((𝑥 − ) − + )] > 0 or (𝑥 − 3) [2 (𝑥 − ) + ]>0
4 16 2 4 8

7 2 7 2 23
M1
Since (𝑥 − 4) ≥ 0, 2 (𝑥 − 4) + > 0 for 𝑥 ∈ 𝑅
8
 Therefore (𝑥 − 3) > 0

The set of values of x : 𝑥 ∈ (3, ∞) A1

ii) 4𝑥−9 4𝑥−9 𝐴 𝐵𝑥+𝐶
= (𝑥−3)(2𝑥 2 −7𝑥+9) ≡ 𝑥−3 + 2𝑥 2 −7𝑥+9 M1
𝑝(𝑥)

4𝑥 − 9 = 𝐴(2𝑥 2 − 7𝑥 + 9) + (𝐵𝑥 + 𝐶)(𝑥 − 3) M1

When 𝑥 = 3, 4(3) − 9 = 𝐴[2(3)2 − 7(3) + 9]

1 9 A1
𝐴 = 2 , 𝐵 = −1, 𝐶 = 2

1 9
4𝑥−9 −𝑥+
2 2
= + 2𝑥 2 −7𝑥+9
𝑝(𝑥) 𝑥−3

1 9−2𝑥 A1
= 2(𝑥−3) + 2(2𝑥 2 −7𝑥+9)

15
8a) 3 2 1 𝑥 1
(9 6 4 ) ( 𝑦 ) = (𝑘 2 ) A1
6 4 −2 𝑧 𝑘
b) 3 2 1 1
(9 6 4 |𝑘 2 ) M1
6 4 −2 𝑘
3 2 1 1 M1
𝑅2 −3𝑅1 →𝑅2
𝑅3 −2𝑅1 →𝑅3
(0 0 1 |𝑘 2 − 3) M1
0 0 −4 𝑘 − 2 A1
c 3 2 1 1
𝑅3 + 4𝑅2 → 𝑅3 (0 0 1| 𝑘 2 − 3 ) M1
0 0 0 4𝑘 2 + 𝑘 − 14
The system of linear equations has infinitely many solutions when
M1
4𝑘 2 + 𝑘 − 14 = 0

(4𝑘 − 7)(𝑘 + 2) = 0 M1
7 A1
𝑘=4 or 𝑘 = −2

3 2 11
When 𝑘 = −2, (0 0 1|1) M1
0 0 00
𝑧 = 1 and 3𝑥 + 2𝑦 + 𝑧 = 1 M1

When 𝑧 = 1, 3𝑥 + 2𝑦 = 0
3
If 𝑥 = 𝑡, 𝑦 = − 2 𝑡

3 A1
The solutions are 𝑥 = 𝑡, 𝑦 = − 2 𝑡, 𝑧 = 1 where 𝑡 ∈ 𝑅
d The system of linear equations is inconsistent when
4𝑘 2 + 𝑘 − 14 ≠ 0 M1
(4𝑘 − 7)(𝑘 + 2) ≠ 0 M1
7
The set of values of k is {𝑘: 𝑘 ≠ 4 , 𝑘 ≠ 2, 𝑘 ∈ 𝑅} A1

15