Mathematics Paper 2 Questions and Answers - Form 4 End Term 2 Exams 2023

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INSTRUCTIONS TO CANDIDATES

This paper consists of two sections: Section I and Section II.
Answer all questions in section I and any five questions in Section II.
Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
Marks may be given for correct working even if the answer is wrong.
KNEC Mathematical tables may be used.

SECTION I (50 Marks)

Simplify giving 4 − 3 your answer in surd form with a rational denominator. (3 marks)
5+√2 5−√2
A shopkeeper mixes 3kg of beans costing Sh. 120 per kg and 6 kg of maize costing Sh. 60 per kg. At what price must he sell the mixture so as to make profit of 30%? (3 marks)
Three quantities Q, h and r are such that Q varies jointly with h and the square of r. When h=15 and r= 3, Q =675. Find
1. An equation connecting Q, h and r (2 marks)
2. The value of Q when h=12 and r=4 (1 mark)
In the figure below, O is the centre of the circle. BOD is the diameter. AC = BC and angle BAC=25°.

Find the size of ∠AOD (2 marks)
Solve the following simultaneous equations using substitution method (4 marks)
3x + 4y = 6
4x − 6y = 1
In a certain firm there are 6 men and 4 women employees. Two employees are chosen at random to attend a seminar. Determine the probability that a man and a woman are chosen. (3 marks)
Find the equation of the tangent to the curve y = 6x² − x − 4 at the point A (1, 1). (3 marks)
The figure below shows a sector of a circle.

If the area is 30.8 cm², calculate the length of the arc AB. (2 marks)
In a math’s test the scores of eight form four students in a certain school were as follows 43,48, 54, 55,56, 57, 62, 65 .Calculate the standard deviation of the scores (4 marks)
A triangle XYZ in which XY=12.4 cm, YZ=15.6 cm and ∠XZY=60° is inscribed in a circle. Calculate the radius of the circle correct to 1 decimal place. (2 marks)
The figure below shows a cuboid.

Calculate
1. The length BE. (2 marks)
2. The angle between BE and plane ABCD correct to (4 s.f) (2 marks)
For the last 5 years the value of a car has been depreciating at a constant rate of 12 % per annum. The present value of the car is Ksh 316 640. Calculate the value of the car at the beginning of the 5 year period (3 marks)
Find scalars m and n such that (3 marks)
A transformation is represented by the matrix . This transformation maps a triangle ABC of the area 3cm² onto another triangle ’B’C’ . Find the area of the triangle A’B’C’ (3 marks)
Find the value of x given that 2log 15 − log x = log 5 + log x − 4 (4 marks)
A circle whose centre is at (1, 3) has the x – axis as its tangent. Determine the equation of the circle in the form
x²+ y²+ ax + by + c = 0 where a, b and c are integers. (4 marks)

SECTION II (50 Marks)

Instruction: Answer any five questions in this section.

Three consecutive terms in a G.P are 3^2x+1, 9^x and 81 respectively.
1. Calculate the value of x (2 marks)
2. Find the common ratio of the series. (2 marks)
3. Calculate the sum of the first 10 terms of the series. (3 marks)
4. Given that the 5^th and 7^th terms of the G.P in (a) above form the first two consecutive terms of an A.P Calculate the sum of the 1^st 20 terms of the A.P. (3 marks)
The table below shows the income tax rate for a certain year

Taxable pay per month (KSh) Tax rates (%)

1 - 9680 10

9681 - 18800 15

18801 - 27920 20

27921 - 27920 25

37041 and above 30

That year Kazembe paid a net tax of Ksh 5212 per month. His total monthly taxable allowances amounted to Ksh 15,220 and he was entitled a monthly personal relief of K.sh 1,162 .Every month the following deductions were made
- NHIF - Ksh 320
- UNION DUES - Ksh 200
- CO-OPERATIVE SHARES - Ksh 7500
  1. Calculate Kazembe’s gross tax (1 mark)
  2. Calculate Kazembe’s monthly basic salary in Ksh (6 marks)
  3. Calculate his monthly net salary (3 marks)
1. Draw a circle centre O and radius 4 cm. (1 mark)
2. From any point A on the circumference, draw two chords AB = 7 cm and AC = 7.5 cm on the opposite sides of the centre. (2 marks)
3. Join B to C and locate the orthocenter (H) of triangle ABC. Join OH and measure OH. (4 marks)
4. Use the midpoint of OH as the centre to inscribe a circle to triangle ABC and measure its radius. (3 marks)
The position of two towns X and Y are given to the nearest degrees as X(45°N , 10°E) and Y(45°N , 70°E).
1. Find the difference in longitude (1 mark)
2. The distance between the towns in:
  1. Kilometers (take the radius of the earth = 6371km and π=227) (3 marks)
  2. Nautical miles ( take 1 nautical mile to 1.853 km) (2 marks)
3. The local time at X when the local time at Y is 2.00 pm (2 marks)
4. Calculate the speed of the aeroplane moving from X to Y in; (2 marks)
  1. Km/hr
  2. Knots
A soda manufacturing company supplies two types of drinks, Fanta and Coke. The total number of crates must not be more than 400.The company must supply more crates of Fanta than Coke. However the number of crates of Fanta must not be more than 300 and the number of crates of coke must not be less than 80. Taking x to represent Fanta and y to represent Coke,
1. Write down all inequalities representing the given information (4 marks)
2. Represent the inequalities on the grid provided (4 marks)
3. The profit obtained was as follow
  - Fanta Sh. 300 per crate
  - Coke Sh. 200 per crate
    1. Use the graph to determine the number of crates of each type that should give maximum profit (1 mark)
    2. Calculate the maximum profit (1 mark)

Complete the table below for the functions y=2cos x and y=sin 2x for −180°≤x≤180° (2 marks)

x°

−180°

−150°

−120°

−90°

−60°

−30°

0°

30°

60°

90°

120

150°

180°

2x°

−360°

−300°

−60°

120°

300°

2 Cos x

−1

−2

Sin 2x

−0.87

0.87

−0.87

On the grid provided, on the same axis draw the graphs y=2cos x and y=sin 2x for −180°≤x≤180° (4 marks)
Use the graphs in (b) above to find;
1. The value of x such that for 2cos x − sin 2x=0 (1 mark)
2. State the amplitude and period of the graph y=2cos x (2 marks)
3. Find the difference in value of y when x = −45°. (1 mark)

The diagram below shows triangle OAB in which N is the mid-point of AB and M is a point on OA such that OM:MA = 2:1.

→ →
Given that OA=ǎ and OB=b̌.
1. Express in terms of ǎ and b̌.
  1. →
    AB (1 mark)
  2. →
    ON (2 marks)
  3. →
    BM (1 mark)
2. Lines ON and BM meet at X such that OX=hON and MX=kMB where h and k are constants.
  1. Express OX in terms of ǎ, b̌ and h. (1 mark)
  2. Express OX in terms of ǎ, b̌ and k. (1 mark)
  3. Hence find the value of h and k. (3 marks)
3. Find the ratio OX: XN. (1 mark)
A triangle with vertices at P(1,1) , Q(−3,2) and R (0,3) is transformed by a matrix to triangle P’Q’R’.
1. Determine the coordinates of the image (2 marks)
2. On the grid provided draw the object and the image (2 marks)
3. P’Q’R’ is then transformed by the transformation with the matrix to P’’Q’’R’’ .
  Find the coordinates of P’’Q’’R’’ and draw P’’Q’’R’’ (3 marks)
4. Find a single matrix which maps PQR onto P’’Q’’R’’ (3 marks)

MARKING SCHEME

No.

Working

Remarks

(4(5−√2)−3(5+√2)
(5+√2)(5−√2)

(20−4√2−15−3√2)
5²−(√2)²

5−7√2
23

M1

M1

A1

Cost price = (3×120)+(6×60) = Sh.80
9

Taking profit as 30%

130 × 80
100

=Sh.104

M1

M1

A1

Q = khr²
675 = k × 5 × 9
k = 675 = 5

15×9
Q=5hr²
Q = 5 × 12 × 4 × 4 = 960

M1

A1
B1

∠ADO=180 − 130=50°

∠AOD=150 − (2×50)=80° B1

B1
B1

x = 6−4y
3
4(6−4y) − 6y = 1

3
24 − 16y − 18y=3

24 − 3 = 34y
y = ²¹/₃₄
x = 6 − 4(²¹/₃₄)
3
x = 1³/₁₇

y = ²¹/₃₄, x = 1³/₁₇

M1

A1

M1

A1

P(M and W) or P( W and M)

(⁶/1₀×⁴/₉) + (⁴/₁₀×⁶/₉)

²⁴/₉₀ + ²⁴/₉₀ = ⁴⁸/₉₀

= ⁸/₁₅

M1

M1
A1

^dy/_dx = 12x − 1

At A, ^dy/_dx =12(1) − 1=11

Gradient of tangent = 11

y −1 =11
x − 1

y − 1=11x − 11

y =11x − 10

M1
M1

A1

Area of a sector = ^θ/₃₆₀πr²

30.8 = ⁷²/₃₆₀ × ²²/₇ × r²

r² = 49

r = 7 cm

Length of an arc =⁷²/₃₆₀ × ²²/₇ × 2 × 7 = 8.8 cm

M1

A1

Mean = 43+48+54+55+56+57+62+65 = 55
8

x	d = x − A	d²
43	−12	144
48	−7	49
54	−1	1
55	0	0
56	1	1
57	2	4
62	7	49
65	10	100
		Σd² = 348

Standard dev = 348 = 6.327
8

B1

B1 for d²

M1 A1

12.4 = 2R
sin 60
R = 12.4 = 7.2cm
2 × sin 60

M1

A1

11.

DB = √(8² + 6²) = √100 = 10 cm
BE = √(10² + 5²) = √125 = 11.18 cm
tan⁡ θ = ⁵/₁₀
θ = 26.56°

M1
A1
M1
A1

12.

316 640=P(1 − ¹²/₁₀₀)⁵

316640= P( 0.88)⁵

P = 316640
0.88⁵

P = Ksh 600001.6

M1

M1
A1

13.

4m − 3n = 5 .... (i)

3m + 2n = 8 ..... (ii)

From (i)

m = 5+3n
4

3(5+3n) + 2n = 8
4
n = 1
3m + 2(1)=8

m = 2

M1

M1

A1 for both values

14.

determinant = image area
object area

Determinant = 2 − 6 = −4
4 = area of image
3

Area of image = 3 × 4 = 12cm²

M1

M1

A1

15.

log ⁡15² − log⁡ x = log⁡ 5 + log⁡(x-4)

log⁡ (225) = log⁡(5x − 20)
x
225 = 5x − 20 ⟹ x² − 4x − 45 = 0
x

x² − 9x + 5x − 45 = 0
(x − 9)(x + 5) = 0

x = 9 or x = −5

Accept x = 9 only

M1

M1

M1

A1

16.

(x − 1)² + (y − 3)² = r²

r² = (3 − 0)² = 9

(x−1)²+ (y−3)² = 9

x² −2x + 1 + y²− 6y + 9 = 9

x² + y² − 2x − 6y + 1 = 0

M1

M1

M1

A1

17.

9^x = 81
3^2x+1 9^x
9^x+x = 3^4+2x+1 3^4x = 3^2x+5 4x=2x+5
2x=5 ⟹x=2.5
Common ratio r = 81 = 81
9^2.5 243
=¹/₃
a = 3^{2(2.5) +1} = 3⁶ = 729
S₄ = 729 (1−(¹/₃)⁴)
(1− ¹/₃)
= 729 1−(¹/₈₁)
(1− ¹/₃)
= 729 × ⁸⁰/₈₁ ×³/₂= 1080
For the A.P, a = 9,d = −6
S₂₀ = ²⁰/₂ {(2×9) + (20−1) −6}
=10(18 − 114)
= −960

M1

A1
M1
A1

M1

M1
A1

B1

M1
A1

18.

5212 + 1162 = 6374
Tax due
1^st band = 0.1 × 9680 = Sh 968
2^nd band = 0.15 × 9120 = Sh 1368
3^rd band = 0.2 × 9120 = Sh 1824
4^th band = 0.25 × k = 2214
k=Sh.8856
Taxable income = 9680 + 9120 + 9120 + 8856
= Ksh.36 776
Basic salary =Ksh.(36776 − 15220) = Ksh 21 556
Total deductions = 320 + 200 + 7500 + 5212 = 13 232
Net salary = Ksh 36 776 − 13 232

=Ksh 23 544

B1

M1
M1
M1
A1

M1
M1
A1

19.

OH = 1.4 cm
Radius = 2.0 cm B1

B1 for circle r = 4 cm
B2 for AC and AB
B1 for BC
B2 for perpendicular bisectors
B2 for circle touching points A,B and C only

B1
B1

20.

70 − 10 = 60°
1. Distance = ⁶⁰/₃₆₀ × ²²/₇ × 2 × 6371 × cos ⁡45
  = (60 × 22 × 2 × 6371 × cos⁡45)
  2520
  = 4719.5 km
2. distance = 4719.5
  1.853
  = 2546.9 nm
Time difference = 4×60=24 mins or 4 hours
2.00 p.m. + 4 hrs = 6.00 p.m
1. Speed = 4719.5 km = 1179.875 km/h
  4 hrs
2. speed = 2546.9 nm = 636.74 knots
  4 hrs

B1

M1
M1

A1
M1
A1

B1
B1

B1
B1

21.

x + y ≤ 400
x > y
x ≤ 300
y ≥ 80
B1 for each of the inequalities total 4 marks
1. Fanta = 300 crates
  Coke = 100 crates
2. (300×300) + (200×100)=Ksh.110 000

B1
B1
B1
B1

B1

B1

22.

x°

−180°

−150°

−120°

−90°

−60°

−30°

0°

30°

60°

90°

120

150°

180°

2x°

−360°

−300°

−240°

−180°

−120°

−60°

0°

60°

120°

180°

240°

300°

360°

2Cos x

−2

−1.73

−1

1.73

−1

−1.73

−2

Sin 2x

0.87

−0.87

−0.87

0.87

0.87

−0.87

−0.87

b)
MathF42023ET2P1Ans21

S & A =1

P = 2

C =1

x = 90° and x = −90° B1

Amplitude = 2 B1
Period = 360° B1

y = sin⁡2x, y = −1
y = 2 cos x, y=1.42
Difference in y ⟹ 1.425−(−1) = 2.425 B1

1. − ǎ +b ̌
2. ǎ + ½ (− ǎ + b̌)
  =½ ǎ + ½ b̌
3. ²/₃ ǎ − b̌
1. OX = h(½ ǎ + ½b̌) = ½hǎ + ½ hb̌
2. OX = OM + MX = ²/₃ ǎ + k( −²/₃ ǎ + b̌ )
  =(²/₃ − ²/₃ k) ǎ + kb̌
3. ½hǎ + 1½hb̌ = (²/₃ − ²/₃k) ǎ + kb̌
  ^h/₂ = ²/₃ − ²/₃ k ................ (i)
  ^h/₂ = k ⟹ h = 2k ..... (ii)
  ^2k/₂ = ²/₃ − ²/₃ k
  6k = 4 − 4k
  k = ²/₅
  h= 2 × ²/₅ = ⁴/₅
4:5