QUESTIONS
SECTION I (50 MARKS)
Answer all the questions from this section
 The expression (1 + ^{x}/_{2}) is taken as an approximation for . Find the percentage error in doing so if x = 0.44. (Give answer correct to 2 d.p) (3 Marks)
 Express the following in surd form and simplify by rationalizing the denominator. (3 Marks)
1
cos60º  sin45º  Make q the subject of the formula ( 3 Marks)
P= ^{3}√nqm
q  The data below shows the age of 10 pupils picked at random in a primary school 6, 11, 13, 14, 8, 7, 12, 20, P and 9 if ∑fx^{2}= 1360. Determine the value of P hence, find the standard Deviation to 3d.p (4 Marks)
 The volume, V of a cylinder varies jointly as its height, (h) and the square of its radius, (r), Calculate the percentage increase in its volume (V), when radius increases by 5% and height, h increases by 10%. (3 Marks)
 Chords AB and CD in the figure below intersect externally at Q. if AB = 5cm BQ = 6cm and DQ = 4cm, calculate the length of chord CD. (2 Marks)
 Jane can do a piece of job in 4 days while Mary can do the same piece of work in 7 days. Mary and Jane did the job together for two days before Jane fell sick. Mary was left to complete the job. How long did it take to do the job? (3 Marks)
 The sketch below represents the graph of y=x^{2}x6.Find the area bounded by the curve, xaxis, yaxis and the line x=5. (3 Marks)
 Use matrix method to determine the coordinates of the point of intersection of the two lines. 3x2y=13, 2y+x+1=0 (3 Marks)
 The figure below shows an arc of a circle through three points A, B and C.
Calculate the coordinates of the centre of the circle. (4 Marks)  A bag contains 4 white balls and 5 Red balls of similar shape and size. Two balls are picked at random without replacement. Find the probability that both balls are:
 White (1 Mark)
 Of different colour (2 Marks)
 A point P (2,3) undergoes transformation represented by the matrix (^{3}_{0} ^{3}_{1}) .Find the coordinate of the image of P. (2 Marks)
 Expand (2 + x)^{6} up to the fourth term. (2 Marks)
Hence use the above expansion to evaluate (1.96)6 correct to 4 d.p. (2 Marks)  Find the value of x in the equation 10sin^{2}x  7cos x + 2 = 0 for the range 270°≤x ≤360° (3 Marks)
 Find the value of x in (3 Marks)
Log(x2) + log (x+1) = 1 + log 4  The sum of two numbers is 9. The sum of the square of the number is 41. Find the numbers (4 Marks)
SECTION II (50 MARKS)
Answer FIVE questions ONLY from this section
 The table below shows the taxation rates for income earned.
Income in ksh pm Tax rates (%) 1 – 9680
9681 – 18800
18801 – 27920
27921 – 37040
Excess over 3704110
15
20
25
30 N.H.I.F. KSh. 320
 Cooperative society shares KSh. 6000
 Union dues KSh. 200
 Calculate Mr. Hamisi’s monthly basic salary in KSh. (7 Marks)
 Calculate his net monthly salary. (3 Marks)
 A jet leaves town P(30°S, 26°W) at 2.00 p.m on Monday and flew due \north to town Q(50°N,26°W).
 Calculate the distance covered by the jet in Km. ( take ∏=^{22}/_{7} and R=6370) (3 Marks)
 After 35 min stoppage at town Q the jet flew due East to town R a distance of 2500 nautical miles from town Q.
Find the position of town R (3 Marks)
 The local time the jet landed at R if its average speed for the whole journey is 1000km/h.
(Take 1nm=1.853km) (4 Marks)
 Use a ruler and a pair of compasses only in all constructions in this question.
 Construct the rectangle ABCD such that AB = 7.2cm and BC = 5.6cm. (3 Marks)
 Constructs on the same diagram the locus L1 of points equidistant from A and B to meet with another locus L2 of points equidistant from AB and BC at M. Measure the acute angle formed at M by L1 and L2. (3 Marks)
 Construct on the same diagram the locus of point K inside the rectangle such that K is less than 3.5cm from point M. Given that point K is nearer to B than A and also nearer to BA than BC, shade the possible region where K lies. Hence calculate the area of this region. Correct to one decimal place. (4 marks)

 The first term of a geometric progression is 36.the sum of the first three terms is 27.
Calculate the common ratio and the value of the second term (4 Marks)  The first term of an AP is 2.the first term of a geometric sequence is also 2 and its common ratio equals the common difference of the arithmetic sequence. The square of the fifth term of arithmetic sequence exceeds the third term of the geometric sequence by 2. Find the common difference and the sum of the first 50 terms of an AP(6 Marks)
 The first term of a geometric progression is 36.the sum of the first three terms is 27.

 Draw triangle PQR whose vertices are P(1,1) Q(3,2) and R(0,3) on the grid provided. (1 Mark)
 Find the coordinates of triangle P1Q1R1 the image of triangle PQR under the transformation whose matrix is (^{3}_{1} ^{0}_{1}). Draw triangle P1Q1R1. (3 Marks)
 P1Q1R1 is then transformed onto P2Q2R2 by the transformation with matrix
Find the coordinates of P2Q2R2 and draw triangle P2Q2R2 (3 Marks)  Describe fully a single transformation which maps PQR onto P2Q2R2. Find the matrix of this transformation (3 Marks)
 Draw triangle PQR whose vertices are P(1,1) Q(3,2) and R(0,3) on the grid provided. (1 Mark)
 In the triangle PQR below, L and M are points on PQ and QR respectively such that PL: LQ=1:3 and QM:MR=1:2, PM and RL intersect at X, given that PQ = b and PR = c,
 Express the following vectors in terms of b and c
 QR (1mark)
 PM (1mark)
 RL (1mark)
 By taking PX = hPM and RX = kRL where h and k are constants find two expressions of PX in terms of h, k, b and c. Hence determine the values of the constant h and k. (6marks)
 Determine the ratio LX: XR (1mark)
 Express the following vectors in terms of b and c

 Complete the table below for the function y=2x^{2 }+ 4x 3 (2 Marks)
x 4 3 2 1 0 1 2 y 3  On the grid provided, draw the graph of the function y=2x^{2} + 4x 3 for 4≤ x ≤ 2 using the scale of 1cm to represent 1 unit on axis and 1cm to represent 2 units on yaxis. (3 Marks)
 Use the graph above to:
 Determine the roots of the equation 2x^{2} + 4x 3=0 (2 Marks)
 Solve the equation 2x^{2} +x 5 =0 (3 Marks)
 Complete the table below for the function y=2x^{2 }+ 4x 3 (2 Marks)
 The acceleration of a particle, t seconds after passing a fixed point P is given by a = 3t – 3. Given that the velocity of the particle when t = 2 is 5m/s, find;
 Its velocity when t = 4 seconds (3 Marks)
 Its displacement at the time in (a) above (3 Marks)
 The displacement during the third second (4 Marks)
MARKING SCHEME

1 + 0.44
2
1 + 0.22 = 1.22
√1 + 0.44
= 1.2
Error = 122  1.2
= 0.02
0.02 x 100
1.2
= 1.67% 
1 = 22
½  ^{1}/_{12}
1 = 2√2
√22 √22
2√2
2√2 x √2+2
√22 √2+2
4 + 4√2
2
= 22√2 
p^{3} = nq  m
q
p^{3}q = nq  m
p^{3}q  nq = m
q(p^{3}  n) = m
p^{3}  n p^{3}  m
q = m
p^{3}n 
36 + 49 + 64 + 81 + 100 + 121 + 144 + 169 + 196 + 400 + p2
1260 + p^{2} = 1360
p2 = 100
p = 10
s.d = √∑fx^{2}^{ } (∑fx)^{2}
∑f ∑f
√1360  (10)^{2}
10 10
= √136  121
= √15
= 3.873 
v x πr^{2}h
v = πr^{2}h
v_{1} = πr_{1}^{2}h_{1}
= π(1.05)^{2} x 1.1πr^{2}h
V_{1} = 1.1025 x πr^{2}h
= 1.21275v
= 1.21275v  v
v
= 21.275% 
AQ x QB = CQ x QD
11 x 6 = (x + 4)4
66 = 4x + 16
4x = 50
x = 12.5 
(¼ x ^{1}/_{7})^{2}
^{11}/_{28} x 2
= ^{11}/_{4} work donw
= ^{3}/_{14}
1 day = ^{3}/_{14}
=^{3}/_{14} x ^{7}/_{1} = 1½
1½ + 2 = 3½ days 
(^{27}/_{3}  ^{9}/_{2}  18)(0) + (^{125}/_{3}  ^{25}/_{2}  30)  (^{27}/_{3 } ^{9}/_{2}  18)
(13.5)t  ^{5}/_{6} + 13½
= 13½ + 12^{2}/_{3}
= 26^{1}/_{6} sq units 
3x  2y = 13
x + 2y = 1 
(2+3 , 5+8) = (2.5 , 6.5)
2 2
(3+4 , 8+10) = (3.5 , 9)
2 2
gradient AB = 85 = 3
32
y  6.5 = ^{1}/_{3}
x  2.5
3y  19.5 = x + 2.5
3y + x = 22
gradient BC = 108 = 2
43
y9 = ½
x3.5
2y  18 = x + 3.5
2y + x = 21.5
3y + x = 22
2y + x = 21.5
y = 0.5
2(0.5) + x = 21.5
1 + x = 21.5
x = 20.5
(20.5 , 0.5) 

^{4}/_{9} x ^{3}/_{8} = ^{1}/_{6}

(^{4}/_{9} x ^{5}/_{8}) + (^{5}/_{9} x ^{4}/_{8})
= ^{5}/_{18} + ^{5}/_{18}
= ^{5}/_{9}


(^{3}_{0} ^{0}_{1})(^{2}_{3}) = (^{6}_{3})
p1(6,3) 

2^{6} + 6 (2^{3})x + 15(2)^{4}x^{2} + 20(2)^{3}x^{3}
= 64 + 102x + 240x^{2} + 160x^{3} 
(2+x)^{6} = (20.04)^{6}
x = 0.04
64 + 192(0.04) + 240(0.04)^{2} + 160(0.04)^{3}
64  7.68 + 0.384  0.01024
= 56.6938


10(1  cos^{2}x)  7cosx + 2 = 0
10  10cos^{2}x  7cosx + 2 = 0
10cos^{2}x + 7cosx  12 = 0
10y2 + 15y  8y  12 = 0
5y (2y + 3)4(2y + 3) = 0
(5y  4)(2y + 3) = 0
y = ^{4}/_{5} or 2y = 3
y = 0.8
cos x = 0.8
x = 36.87
360  36.87
= 323.13 
logx 2 + logx+1 = log10 + log4
log((x2)(x1)) = log(10x4)
(x2)(x1) = 40
x2  x  42 = 0
x(x7) +6(x7)=0
x=7
log(x2)+log(x+1)  log4=1
log(x2)(x+1) = 1
4
(x2)(x+1)=10
4
(x2)(x+1)=40
x^{2}x42=0
x =7 
x+y = 9
x^{2} + y^{2} = 41
y = 9x
x^{2} + (9x)^{2} = 41
x^{2}  9x + 20 = 0
x(x5) 4(x5) = 0
(x4)(x5) = 0
x = 4 or x = 5
when x = 4
y = 5
when x = 5
y = 4 

gross tax = 5512 + 162 + 150
= 5824
9680 x 0.1 = 968
9120 x 0.15 = 1368
9120 x 0.2 = 1824
y x 0.25 = 1664
y = 6656
27920 + 6656
= 34576
34576  (10000+2400+2820)
= 19356 
34576  (5512 + 320 + 6000 + 200 + 1500)
= 34576  13532
= 21,044



Angle diff = 30 + 50 = 80º
PQ = 80 x 2 x ^{22}/_{7} x 6370
360
= 8898km 

x x 60cos50 = 2500
x = 2500
60cos50
= 64.82º
64.82º  26º = 38.82
R(50ºN , 38.82ºE) 
Total distance = (8898 + (2500 x 1.853))
= 13530.5km
Time taken = 13530.5
1000
= 13.5305hrs
Local time at R = 2.00 + 13.5305hrs + 35min + (4x64.82)
60
8.26amTues





45º

^{45}/_{360} x ^{22}/_{7} x 3.5^{2} = 4.8125cm^{2}



a + ar + ar^{2} = 27
36 + 36r + 36r^{2} = 27
36 + 36r + 9 = 0
4r^{2} + 4r + 1 = 0
2r + 1 = 0
r = ½
2nd term = ½ x 36
= 18 
a = 2
r = d
(a+4d)^{2} = 2r^{2} + 2
(2+4d)2 = 2d + 2
4 + 16d + 16d^{2} = 2d^{2} + 2
14d^{2} + 16d + 2 = 0
7d^{2} + 8d + 1 = 0
7d^{2} + 7d + d + 1 = 0
7d(d+1)+1(d+1)=0
(7d+1)(d+1)=0
d = ^{1}/_{7} or d = 1
S50 = ^{50}/_{2}{4 + (501)1}
= 25(449)
= 1125
or
^{50}/_{2}{4+(501)^{1}/_{7}}
= 25(3)
= 75




P^{1}(3,2)
Q(9,1)
R(0,3) 
P^{2}(2,2)
Q^{2}(6,4)
R^{2}(0,6) 
Enlargement centre (0,0) scale factor 2




b+c

b + ^{1}/_{3}(b+c)
= ^{2}/_{3}b + ^{1}/_{3}c 
bc¾b
=¼bc


px = h(^{2}/_{3}b + ^{1}/_{3}c)
= ^{2}/_{3}hb + ^{1}/_{3}hc
px = c + kkb
^{2}/_{3}hb + ^{1}/_{3}hc = (1k)c + kb
^{2}/_{3}h = k = k = ^{8}/_{3}h
^{1}/_{3}h = 1k
^{1}/_{3}h = 1  ^{8}/_{3}h
^{3h}/_{3} = ^{1}/_{3}
h = ^{1}/_{3 }
k =^{ 8}/_{9} 
1:8


x
4
3
2
1
0
1
2
y
13
3
3
5
3
3
13


 x = 2.6
x = 0.6  y = 2x^{2} + 4x  3
0 = 2x^{2} + x  5
y =3x + 2
x = 1.8
x = 1.4
 x = 2.6


 a = 3t  3
v = 3t^{2}  3t + c
2
5 = ^{3}/_{2}(2)^{2}  3(2) + c
v = ^{3}/_{2}t^{2}  3t + 5
v = ^{3}/_{2}(4)^{2}  3(4) + 5
= 24  12 + 5
= 17m/s  S = t ^{3}  ^{3}/_{2}t^{2} + 5t +c
2
s = 0 t = 0
c = 0
S = t ^{3}  ^{3}/_{2}t^{2} + 5t
2
4 ^{3}  ^{3}/_{2}(4)^{2} + 5(4)
2
= 32  24 + 20
= 28m 
( 3 ^{3}  ^{3}/_{2}(3)^{2} + 5(3))  ( 2 ^{2}  ^{3}/_{2}(2)^{2} + 5 (2))
2 2
= (^{27}/_{2}  ^{27}/_{2} + 15)  (2  6 + 10)
= 15  6
=9M
 a = 3t  3
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