Questions
INSTRUCTIONS TO CANDIDATES
 The paper contains two sections. Section I and Section II
 Answer all the questions in section I and any five questions from 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.
 Nonprogrammable silent electronic calculators and KNEC Mathematical tables may be used except where stated otherwise.
SECTION 1 (50 MARKS)
Answer all questions in this section
 Use logarithm tables to evaluate to 4 significant figures (4mks)
 Simplify ^{3}/_{2+√2} + ^{4√2}/_{2√2.} Write your answer in the form a+ b√C (3mks)
 Expand (p  3q )^{5} (1mks)
hence state Coefficient of p^{4}q (1mks)
 Fourth term in the expansion (1mk)
 Make c the subject of the formula b = √(kac) , hence find the value of c when k= 1, a=4 and b= 2 (3mk
 Given that Find value of x such that A is a singular matrix. (3mks)
 The dimensions of a rectangle are 40cm and 45cm. If there is an error of 5 % in the dimensions find the percentage error in calculating area of the rectangle. (3 mks)
 Solve the equation
Log _{2} (2 +3x) + 3log_{2} 2 = 2+log_{2} (2x+6) (3mks)  The cash price of a TV set is Ksh 13800. A customer opts to buy the set on hire purchase terms by paying a deposit of Ksh. 2280. If the simple interest of 20% p.a is charged on the balance and customer is required to pay 24 equal monthly instalments calculate the amount of each instalment. (2mks)
 Chords PQ and RS intersect internally at point T. Given that PT = 3.2 cm, TQ= 4.7cm and TS = 5.2cm, find the length of chord RS. (3mks
 On the line AB below show by shading the region R above the line such that
 R is nearer A than B
 R is not more than 3.0 cm from A (4mks)
 < ARB ≥900
 Determine the radius and centre of a circle whose equation is
3x^{2} +3y^{2}18x+12y9=0 (3mks)  Grade A coffee costs sh.100 per kg while grade B costs sh150 per kg. Find the ratio in which the two grades should be mixed so that by selling the mixture at sh.147 per kg a 5% profit is realised. (3mks)
 The following table shows income tax rates
Income Ksh per month Rate in ksh per every sh.20 18400
840118000
1800130000
Above 300002
3
4
5  In a transformation, an object A of area 4cm^{2} is mapped into B of area 48cm2 by a transformation whose matrix is determine possible values of y. (3mks)
 The figure below shows a triangle ABC not drawn to scale. Calculate the length marked b given that AB= 240cm <BAC = 300 and <ACB =450 (3mks)
 Two variables R and V are such that R= kvn where k and n are constants. The table below shows values of logR and logV to 2d.p.
Log V 0.48 0.60 0.70 0.78 0.85 0.90 Log R 1.43 1.68 1.88 2.03 2.16 2.28
SECTION II (50 MARKS)
Answer only five questions in this section

 Complete the table below for values of y for the curve
Y= x^{3}5x^{2}+2x+9 for 2≤ x ≤5 (2mks)X 2 1.5 1 0 1 2 3 4 5 Y  Draw a graph of y= x^{3}5x^{2}+2x+9 for 2 ≤ x ≤ 5 (3mk
 Use your graph to solve the equations
 X^{3}5x^{2}+2x+9 =0 (2mks)
 X^{3}5x^{2}+6x = 5 (3 mks)
 Complete the table below for values of y for the curve
 The cost Y of producing a number of items varies partly as X and partly inversely as X. To produce 2 items it costs sh. 135 and to produce 3 items it costs sh.140.
 find Law connecting Y and X. (5mks)
 Cost of producing 10 items. (2mks)
 Number of items produced at a cost of sh.180 (3mks)
 The first, fourth and thirteenth terms of an AP correspond to the first three consecutive terms of an increasing Geometric progression.
Given that the first term of the AP is a and common difference is d Write down the first three terms of the GP in terms of a and d. (1mk)
 The sum of the third and eleventh terms of the AP is 30.
Calculate; The first term and common difference of the AP (5mks)
 Common ratio of the GP (2mks)
 Sum of the first 10 terms of the GP (2mks)

 Two towns on latitude 300 N are 3000km apart. Find the longitude difference of the two towns. (Take π = 22/7 and radius of earth to be 6370km) (2mks)
 The position of the airport P and Q are P (600N, 450W) and Q (600N, K0E)
It takes a plane 5 hrs to travel due East from P to Q at an average speed of 600 knots. Calculate the value of K (3mks)
 The local time at P is 10.45 am when is the local time at Q when the plane reached there? (3mks)
 Calculate the shortest distance between A(300S, 360E) and B (300S, 1440W) (2mks)
 The probability that Andrew goes to bed on time is 2⁄3. If he goes to bed on time the probability that he wakes up early is 3⁄5 otherwise it is 1⁄7. If Andrew wakes up late, the probability that he will be punctual for class is1⁄4 otherwise its is( 2)⁄7.
 Draw a tree diagram to represent above the information. (2mks)
 Determine the probability that;
 He will wake up late (2mks)
 He will wake up early and arrive in class late (2mks)
 He will go to bed late but arrive class early (2mks)
 He will be late for class. (2mks)

 A shear parallel to xaxis (xaxis invariant) maps point (3,1) onto ( 5,1). If S is the transformation find the matrix that defines S (3mks)
 A transformation X maps points (1,3) and 2,3) onto (2,4) and (3, 1) respectively.
Determine the matrix of transformation ` (4mks)  Transformations R and T are represented by matrices and respectively, point P has coordinates ( 3,2)
 Find coordinates of RT(P) (3mks)
 A transport company runs a fleet of two types of buses operating between Meru and Nairobi. Coach buses and Minibuses. A coach bus carries 52 passengers and 200kg of luggage while a minibus carries 32 passengers and 300kg of luggage. On one Saturday, there were 500 passengers with 3500 kg of luggage to be transported, the company could only use a maximum of 15 buses all together.
 if the company uses x coach buses and y minibuses write down all inequalities that satisfy the given conditions. (4mks)
 Represent the inequalities graphically in the grid provide
(use a scale of 1cm to represent 1 unit) (3mks)  if the cost of running one coach bus is sh.7200 and that of running one minibus is sh. 6000 use the graph above to determine the minimum cost of running the vehicles (3 mks)
 The velocity of a particle after t seconds is given by V= t2 – 4t+4.
 Find displacement of the particles during the third second (4mks)
 Determine the time when the particle is momentarily at rest (3mks)
 The acceleration of the particle after 2 seconds (3mks)
Marking Scheme

No. Log 1.000
0.03506
28.5 
+
90.35
118.87
4.917×100
4.9170.0000
2.5448
1.4552
2.0751÷3
0.6917  3 + (4√2)/
(2+√2) (2√2)
= 3(2√(2)+(4√2)(2+√2)
(2+√2)(2√2)
=3√2 + 8 + 4√22√22
42
= 12√2
2  (p3q)^{5} = p^{5}+5p^{4}(3q)10p^{3}(3q)^{2}+10p^{2}(3q)^{3}
= p5 +15p^{4}q +90p^{3}q^{2} 270p^{2}q^{3} Coefficient =15
 4th term =270p^{2}q^{3}
 b=√kac
b^{2}=kc
c=kb^{2} a
c=1²2² =  3/4
4  D = 3x(2x2) 6(x6)=0
=6x^{2}6x6x+6=0
=6x^{2}12x+6=0
=6x^{2}6x6x+6 =0
6x(x1) 6(x1)=0
(x1)(6x6)=0
X=1  Error = maxmin × 100
2
actual
= ½(2041.21573.2) × 100
40×45
= 13%  Log_{2}8(2+3x) = log_{2}4(2x+6)
16 +24x = 8x + 24
16x = 8
X= ½  5.2(RT) = 3.2x4.7
RT = 3.2 x 4.7
5.2
= 2.59
RS = 5.2 + 2.89
= 8.09 
 3X^{2} + 3Y^{2} – 18X + 12Y – 9 =0
X^{2} + Y^{2} 6X +4Y =0
X^{2}  6X + 9 + Y^{2} +4Y + 4 = 3+ 9+4
(X  3)^{2} + (Y + 2)^{2} = 16
Centre (3,2)
Radius = 4  147 ≫ 105%
Cp = 100 x 147 = sh 140
105
Let the ratio b 1⋮ n
= (100x1+150n)140
1+n
100 +105n = 140 + 140n
10n = 40
n =4
ratio 1⋮4  Total tax= 3038+1162= 4200
Tax calc
^{8400}/_{20} x2 = 840
^{9600}/_{20} x3 = 1440
^{x}/_{20} x 4 = 1920
total = 4200
x= ^{1920x20}/_{4} = 9600
Income = (18000+9600) x ^{100}/_{115}
= sh 24000 M1  x (x+3) 12=0
x^{2} + 3x12 = 0
x^{2}+ 4xx 12 =0
x(x+4)1(x+4)=0
(x1)(x+4)= 0
X=1 x=4 M1  240 = b
sin45 sin105
b = ^{240}/_{sin45} x sin105
327.8461 B1 

X 2 1.5 1 0 1 2 3 4 5 Y 23 8.625 1 9 7 1 3 1 19 
 Y = nX + ^{m}/_{X}
Y = nX + ^{m}/_{X}
135 = 2n + ^{m}/_{2}
140 = 3n + ^{m}/_{3}
4m + n = 270
9m + n = 420
5m = 150
M = 30
n = 270120 =150  y= 30x + ^{150}/_{x}
y = 30x10 +^{ 150}/_{10} = 315  30x + ^{150}/_{x} = 180
30 x^{2} +150 =180x
30 x^{2} – 180x +150 = 0
(x5)(x1)= 0
X=5 or x=1
 Y = nX + ^{m}/_{X}

 a a+3d a+ 12d

 a+ 2d + a+ 10d =30
2a + 12a =30
(a+3d)/a = (a+12d)/(a+3d)
(a+3d )^{2} = a^{2} + 12ad
9d^{2} 6ad + 9d^{2} = a^{2} +12ad
9d^{2} – 6ad =0
9d^{2} 90d(156d) = 0
9d^{2} 90d +36d2 = 0
45 d^{2} + 90d = 0
9d (5d  10)= 0
d = 0 or d= 2
a = 156d
= 1512
= 3  r = (a+ 3d)/a = (3+6)/3
= 3  S10 = 3(310 1) / 2 = 888572
 a+ 2d + a+ 10d =30

 Dist AB = ^{α}/_{360} x 2π R Cosθ
3000 = ^{α}/_{360} x 2π R Cos30
Α = 31.150  60α cos θ = 5x600
α = 50/cos60 = 1000
K = 100 – 45 = 550
Long diff = 45^{0} + 55^{0} =90^{0}
Time diff = 100 x4 / 60
= 6hrs 40mins
Time at Q = 10.45am + 6hrs 40 mins
17 25 HRS
Time when the plane reached 17 25hrs + 5 hrs
22 25 HRS
 Dist AB = ^{α}/_{360} x 2π R Cosθ

 P(wakes up early) = p(BE’ or B’ E’)
= 2/3 x 2/5 + 1/3 x 6/7 = 11/14  P(wakes up early but late for class)= P(BEC’ or B’ EC’)
= 2/3 x 2/5 x 5/7 + 1/3 x 1/7 x 5/7
= 26/147  P(bed late but early for class)= P(B’EC or B’ E’C)
= 1/3 x 1/7 x 2/7 + 1/3 x 6/7 x 1/4
= 25/294  P(late) = P(BEC ‘or B E’C’ or B’EC’ or B’E’C’)
= 2/3 x 3/5 x 5/7 + 1/3 x 1/7 x 5/7 + 2/3 x 2/5 x 3/4 + 1/3 x 6/7 x 3/4
=1289/1470
 P(wakes up early) = p(BE’ or B’ E’)

 1 k 3 = 5
0 1 1 1
3 + k = 5
K = 2 
a + 3b = 2
2a + 3b = 3
3a = 5 a = 5/3
3b = 2 – 5/3
3b = 1/3 b= 1/9
c + 3d = 4
2c + 3d = 1
3c = 5 c = 5/3
d = (4 – 5/3)/ 3 d= 7/9 
(15,6)
 1 k 3 = 5

 52x + 32y ≥ 500
13x + 8y ≥ 125………………………………………1
200x + 300y ≥ 3500
2x + 3y ≥ 35…………………………………………..2
x + y ≤ 15 ………………………………………………3
x ≥ 0 , y ≥ 0
 52x + 32y ≥ 500

 ∫^{³}_{₂}(t2 – 2t + 4) dt
[t^{3}/3 – t2 + 4t]^{³}_{₂}(27/3 – 9+12) – (8/3 4 +8)
12 – 4  8/3
8 – 8/3
(248)/3
16/3 or 5 1/3  t2 – 4t + 4 = 0
t = 4 ± √(164(4)
2
4 ± 0 = 2
2  a = dv/dt
= 2t – t (at t = 2)
a= 2(2) – 2
a= 2 m/s^{2}
 ∫^{³}_{₂}(t2 – 2t + 4) dt
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