MATHEMATICS
PAPER 2
INSTRUCTIONS TO CANDIDATES
 Write your name and Admission number in the spaces provided at the top of this page.
 This paper consists of two sections: Section I and Section II.
 Answer ALL questions from section I and ANY FIVE from section II
 All answers and workings must be written on the question paper in the spaces provided below each question.
 Show all the steps in your calculation, giving your answer at each stage in the spaces below each question.
 Non – Programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
SECTION A
 Use logarithms tables to evaluate (4mks)
³√36.72 x (0.46)²
185.4  T is a transformation represented by the matrix under T, a square of area 10cm^{2} is mapped onto a square of area 110cm^{2}. Find the value of x (3mks)
 Given that 2Cos(2x30º)=6/5 find x where 180º < x < 360º (3mks)
 Make A the subject of the formula (3mks)
T= 2m/n √^{LA}/_{3k}  A quantity P is partly constant and partly varies inversely as square of t. P = 6 when t = 6 and p = 18 when t = 3. Find t when p = 11 (3mks)

 Expand (5 + ^{x}/_{2})⁶ up to the term in x³. (2mks)
 Use your expansion to estimate the value of (^{11}/_{6})⁶ . Correct to one decimal place. (2mks)
 Solve for x in the equation. (3 Mks)
Log_{8} (x + 6) – Log8 (x – 3) = ^{2}/_{3}  Solve for x and y in the simultaneous equation below. (3 mks)
xy + 6 = 0
x – 2y = 7  The size of each interior angle of a regular polygon is five times the size of the exterior angle. Find the number of sides of the polygon. (3mks)
 If 1  2 + 2√5 = a + b√c find the value of a, b and c (3 mks)
3 √5 3 + √5  The data below shows marks scored by 8 form four students in Molo district mathematics contest 44, 32, 67, 52, 28, 39, 46, 64.Calculate the mean absolute deviation. (3 Mks)
 Steve deposited ksh.50, 000 in a financial institution in which interest is compounded quarterly. If at the end of second year he received a total amount of ksh79, 692.40. Calculate the rate of interest p.a (3 Mks)
 The points with coordinates (5,5) and (3,1) are the ends of a diameter of a circle
Centre A
Determine: The coordinates of A (1mk)
 The equation of the circle, expressing it in form x2 + y2 + ax + by + c = 0 Where a, b, and c are constants (2mks)
 Pipe x can fill an empty tank in 3 hours while pipe y can fill the same tank in 6 hours. When the tank is full, it can be emptied by pipe z in 8 hours. pipe x and y are opened at the same time when the tank is empty. If one hour later pipe z is also opened, find the total time taken to fill the tank. (3marks)
 Fatima bought maize and beans from Kami. She mixed the maize and beans in the ratio 3: 2 she bought the maize at sh.90 per kg and the beans at sh.150 per kg. If she was to make a profit of 30% what would be the selling price of 1kg of the mixture. (3mks)
 Given and find A^{‾1} B^{1} (3mks)
SECTION B

 A figure whose coordinates are A(2, 2), B(4, 1), C(4, 3) and D (2, 3) undergoes successive transformations ERS; where E, R and S are transformations represented by the matrices, On the grid provided, show the figure ABCD and its image under the successive transformations ERS. (6mks)
 Find the matrix representing the single transformation mapping the image found in (a) above back the object figure ABCD. (2mks)
Marks 1019 2029 3039 4049 5059 6069 7079 8089 9099 Frequency 2 6 10 16 24 20 12 8 2  Triangle PQR has vertices at P(2, 2), Q(4, 1) and R(6, 4). On the same grid, show the image of triangle PQR under a shear with line y = 2 invariant and point R(6, 4) is mapped onto R^{1}(2, 4). (2mks)
 The following are marks by form four students in a mathematics test.
Using an assumed mean of 54.5, calculate the Mean mark (4mks)
 Variance (4mks)
 Standard deviation (2mks)
 The diagram below represents a Cuboid ABCDEFGH in which FG =4.5cm, GH =8cm HC=6m
 Calculate the length FC (2mrks)

 The size of the angle between the lines FC and FH (2mks)
 Size of the angle between the line AB and FH. (2mks)
 The size of the angle between the planes ABHE and the plane FGHE.(2mks)
 The total surface area of the cuboid (closed) (2mks)

 Complete the table below, giving all your values correct to 2 d. p. for the functions y = cos x and y = 2cos (x + 30)º
xº 0º 60º 120º 180º 240º 300º 360º 420º 480º 540º Cos x 1.00 1.00 0.50 2Cos(x+30) 1.73 1.73 0.00  For the function y = 2cos (x + 30)0
State: The period (1mk)
 Phase angle (1mk)
 On the same axes draw the waves of the functionsy = cos x and y = 2cos (x + 30)º for 0º ≤ x ≤ 540.Use the scale 1cm rep 30º horizontally and 2cm rep 1 unit vertically. (4mks)
 Use your graph above to solve the inequality 2cos (x + 30o) ≤ cos x (2mks)
 Complete the table below, giving all your values correct to 2 d. p. for the functions y = cos x and y = 2cos (x + 30)º
 A teacher had 5 red, 6 black and 9 blue pens in a box. The pens were all identical except for the colour.
 If one pen is picked from the box, what is the probability that it is
 Red. (1mk)
 Not black. (1mk)
 The teacher asked a student to pick two pens from the box, one at a time, without replacement. Find the probability that
 Both pens are of the same colour. (3mks)
 They are of different colours. (2mks)
 If the first student was allowed to take away two blue pens and another student was asked to pick two pens without replacement. What is the probability that the second student picked pens of same colour? (3mks)
 If one pen is picked from the box, what is the probability that it is
 In the figure below, PQR is the tangent to the circle at Q. TS is a diameter and TSR and QUV are straight lines. QS is parallel to TV. Angle SQR = 35° and TQV = 60°.
 Find the following angles, giving reasons for each answer.
 QTS. (2mks)
 QRS. (2mks)
 QVT. (2mks)
 UTV. (2mks)
 QUT. (2mks)
 Find the following angles, giving reasons for each answer.
 Use ruler and a pair of compasses only in this question
 Construct triangle ABC such that AB = 6cm, AC=BC and angle ACB = 135° 4mks
 On one side only construct the locus of P such that:
 < APB = 67.5° 1mk
 area of triangle , APB = 9cm² 3mks

 Locate P1 and P2 the two possible positions of P which satisfy the two conditions above 1mk
 Measure the distance between P1 and P2. 1mk
 An arithmetic progression has the first term a and the common difference d.
 Write down the third, ninth and twenty – fifth terms of the progression. (3 Mks)
 The progression is increasing and the third, ninth and twentyfifth terms form the first three Consecutive terms of a geometric progression. If the sum of the seventh term and twice the sixth term of the arithmetic progression is 78.
Calculate The first term and the common difference (5 Mks)
 The sum of the first nine terms of the arithmetic progression (2 Mks)
MARKING SCHEME

No Log 36.72
(0.46)2 ⇒ 2 (‾1.6628)
3.474 x 10^{1}
=0.3474  A.S.F = det
5x2 + 6 = 110/10
5x2 + 6 11 = 0
5x2 = 5
x = ± 1  2 cos(2x – 30) = 6/5
Cos(2x 30) =  0.6
Cos –ve in 3rd quad
2x – 30 = 53.13
2x – 30 = 233.13
2x = 263.10
x = 131.57°
2x30=54053.13
X=258.44
2x  30°= 360 + 233.13°
2x = 623.13
x = 311.57
x=131.57,258.44,311.57  ^{nT}/_{2m} = √ ^{LA}/_{3k } n²T² = L  A
4m² 3K
4m^{2} (L – A) = 3k (n^{2} T^{2})
4Lm^{2} – 4Am^{2} = 3kn^{2}T^{2}A = 4LM²  3kn²T²
4m²  p=c + ^{k}/_{t²}
6= c + ^{k}/_{36}
18=c + ^{k}/_{9}
216 = 36c + k  (i)
162 = 9 c + k (ii)
54 = 27c
^{54}/_{27} = c = 2 k = 144
when p = 11
11 = 2 + ^{144}/_{t²}
9t^{2} = 144
t = 4 
 Coefficients: 1 6 15 20
Terms: 5^{6}, 5^{5} (^{x}/_{2}), 5^{4}(^{x}/_{2})^{2}, 5^{3} (^{x}/_{2})^{3}15625, ^{3125x}/_{2}, ^{625x²}/_{4}, ^{125x³}/_{8}( 5 + x/2)⁶ = 15625 + ^{3125}/_{3x} + ^{9375x³}/_{4} + ^{625x³}/_{2} + ...
x= 1 Therefore, (11/2)^{6} = 15625 + ^{3125}/_{3} + ^{9375}/_{4} +^{ 625}/_{2}  = 15625 + 1041.667 + 2343.75 + 312.5
=19322.9
 Coefficients: 1 6 15 20
 Log8 (^{x+6}/_{x3}) = Log_{8}8^{²/₃}
^{x+6}/_{x3} = 4
x + 6 = 4x – 12
3x = 18
x = 6  8 x = 7 + 2y
( 7 + 2y) y + 6 = 0
7y + 2y^{2} + 6= 0
2y^{2} + 7y + 6 = 0
2y^{2} + 4y + 3y + 60 = 0
2y(y + 2) +3 (y + 2) = 0
(2y + 3) ( y + 2) = 0
When y =2, x = 3
y =^{3}/_{21}, x = 4  X+5X=180
6X=180
X=30
360/30
12 sides  1( 3 + √5)  ( 2 + 2√5)(3 √5)
(3  √5)(3 + √5)
3 + √5  [62√5 + 6√5 10]
95
3+√5  4√5 + 4
4
7  3√5 = 7  3 √5
4 4 4
hence a = ^{7}/_{4}, b= ^{3}/_{4} and c =5  Mean = 44 + 32 + 67 + 52 +28 + 39 + 46 + 64
372/8 = 46.5x 44 32 67 52 28 39 46 64 d 2.5 14.5 20.5 5.5 18.5 7.5 0.5 17.5 (d) 2.5 14.5 20.5 5.5 18.5 7.5 0.5 17.5
Σxd = 87 = 10.875
N 8  ^{50,000}/_{50,000} ( 1 + r/100)⁸ = ^{79692.40}/_{50,000}
⁸√(1 + r/1000) = ⁸√1.593848
1 +^{ r}/_{100} = 1.06
^{r}/_{100} = 0.06
R = 6% per interest period
Per annum = 6 x 4 = 24% 
 Co – ordinate of
A ( ^{53}/_{2} ^{51}/_{2})
A (^{2}/_{2} ^{4}_{/2})
A ( 1,2)  r = √(51)^{2} + (52)^{2}
=√16+9
=5 units
( x – 1)^{2} + ( y – 2) ^{2} = 5^{2}x^{2} – 2x+ 1 + y^{2} 4y + 4 = 25
x^{2} + y^{2} – 2x  4y + 5 = 25
x^{2} y^{2} – 2x – 4y – 20 = 0
 Co – ordinate of
 x fills ^{1}/_{3} of tank in 1 hour
y fills ^{1}/_{6} of tank in 1 hour
z empties ^{1}/_{8} of tank in 1 hour
x and y open will fill ^{1}/_{3} + ^{1}/_{6} = ^{1}/_{2} of tank in 1 hour
remaining fraction = 1  ^{1}/_{2} = ^{1}/_{2}
x, y and z both will fill ^{1}/_{3} + ^{1}/_{6}  ^{1}/_{8} = ^{3}/_{8} of tank in 1 hour
time taken by x, y and z to fill ½ of remaining tank
= ^{1}/_{2} ÷ ^{3}/_{8}=1 ^{1}/_{3} hours
Total time = 1 + 1 ^{1}/_{3} = 2 ^{1}/_{3} hours  M:B
3:2
3 x 90 x 150 x 2 = cost price
3+ 5
CP= 270 + 300
8
S.P = ^{130}/_{100} + 71.25
= Ksh 92.625
=Ksh 92.60 
Det2120 =1 Det 1211=1 

 S1 R1 E1 (A”’B ”’C ”’D ”’) = ABCD
 ΔPQR – For object figure plotted correctly
PQR – For image figure plotted correctly


marks f Mp(x) d(xA) fd d2 fd2 1019
2029
3039
4049
5059
6069
7079
8089
90992
6
10
16
24
20
12
8
214.5
24.5
34.5
44.5
54.5
64.5
74.5
84.5
94.540
30
20
10
0
10
20
30
4080
180
200
160
0
200
240
240
801600
900
400
100
0
100
400
900
16003200
5400
4000
1600
0
2000
4800
7200
3200140 31400  Mean = A + ^{∑fd}/_{∑f}= 54.5 + ^{140}/_{100} = 55.9
 Variance = ^{31400}/_{100} – (^{140}/_{100})^{2}
= 312.04  standard deviation = √variance
= √312.04 = 17.66
 Mean = A + ^{∑fd}/_{∑f}= 54.5 + ^{140}/_{100} = 55.9

 FC = √6^{2} + 4.5^{2} + 8^{2}= 10.97

 Sin θ = 6
0.97
θ = 33.16 
Tan θ = 8 ÷ 4.5
= 1.778
θ = 60.64
 Sin θ = 6

Tan θ = 6 ÷8
θ = 36. 87  Total surface area
2 (8 x 6) + 2 (4.5 x 8) + 2(4.5 x 6)
= 222cm^{2}


xº 0º 60º 120º 180º 240º 300º 360º 420º 480º 540º Cos x 1.00 0.50 0.50 1.00 0.5 0.50 1.00 0.50 0.50 1.00 2Cos(x+30) 1.73 0.00 1.73 1.73 0.00 1.73 1.73 0.00 1.73 1.73 
 Period = 360º
 phase angle = 30º

 37.5º ≤ x ≤ 217.5º
397.5º ≤ x ≤ 540º



 ^{5}/_{20} = ^{1}/_{4}
 ^{5}/_{20} + ^{9}/_{20}
=^{14}/_{20} = ^{7}/_{10}

 ( ^{5}/_{20} X ^{4}/_{19})+(^{6}/_{20} X ^{5}/_{19)} + (^{9}/_{20} X ^{8}/_{19})
= 20/380 + 30/380 + 72/380
= ^{122}/_{380} = ^{61}/_{190= }1 ^{61}_{/90= }^{29}/_{90}
 ( ^{5}/_{20} X ^{4}/_{19})+(^{6}/_{20} X ^{5}/_{19)} + (^{9}/_{20} X ^{8}/_{19})


 <QTS = <SQR = 35° B1
Angles in alternate segments  <QST = 55° B1
<QRS = 55  35° B1
= 20°  <QVT = <SQV = 30°
 90 – (30° + 35) = 25°
 180 – (60° + 65°) = 55°
 <QTS = <SQR = 35° B1
 ½ x 6x h = 9✓
h= 3✓
P_{1}p_{2} = 5.6 cm 
 T_{3} = a + 2d
T_{9} = a + 8d
T_{25} = a + 24d 
 ^{a+8d}/_{a+2d =} ^{a+24d}/_{a+8d}
a^{2} + 2bad + 48d^{2} = a^{2} + 16ad + 64d^{2}
10ad = 16d^{2}
a = ^{16d}/_{10} or ^{8}/_{5d}
a + 6d + 2(a + 5d) = 78
3a + 16d = 78
24d + 80d = 390
104d = 390
d = 3.75
a = ^{8}/_{5} x 3.75
a = 6  S_{9} = 4.5 (12 + (8 X 3.75)
4.5 (12 + 30)
= 4.5 (42)
= 189
(10 Marks)
 ^{a+8d}/_{a+2d =} ^{a+24d}/_{a+8d}
 T_{3} = a + 2d
Download Mathematics Paper 2 Questions and Answers  Kapsabet Pre Mock 2021/2022.
Tap Here to Download for 50/
Get on WhatsApp for 50/
Why download?
 ✔ To read offline at any time.
 ✔ To Print at your convenience
 ✔ Share Easily with Friends / Students
Join our whatsapp group for latest updates