Questions
Instructions to Candidates
 This paper consists of 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 provided 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 I (50 Marks)
Answer all the questions in this section in the spaces provided below each question
 A quadratic equation has roots as x=4 and x=2/3. Write the equation in the form ax2+bx+c=0, where a, b and c are integers. (3 marks)
 Given that 2≤p≤8 and 3≤q≤10, find the maximum value of: (2 marks)
pq
q+p  Without using a mathematical table or a calculator, write the expression below in the form a3+c, where a and c are constants (3 marks)
2
sin 90° +tan 60°  Solve for x in the equation x3+2=(8x) (3 marks)
 A truck is bought at Kshs. 1,800,000. It depreciates by 10% per annum in the first 2 years, thereafter its annual depreciation rate is 15%. Find the value of the truck after 5 years. (4 marks)
 The position vectors of points A and B are 4i5j+6k and 2i+3j+8k respectively. Calculate the magnitude of AB correct to 3 decimal places. (3 marks)
 The figure below shows the velocitytime graph of a particle that moves for 60 seconds and covered a distance of 852 metres.
Calculate the value of v (2 marks)  Make x the subject of the formula: (3 marks)
px= √(x^{2}+m/Q)  The equation of a circle is x2+y2+6x10y2=0. Determine the coordinates of the centre and the area of the circle in terms of π (3 marks)

 Expand (1+3x)^{6} in ascending powers of x up to the term in x^{3} (1 mark)
 Use your expansion to evaluate (0.997)^{6} correct to 5 decimal places. (2 marks)

 Complete the table below for the function y=x^{2}4x+5 for 1≤x≤5 (1 mark)
y 1 1.5 2 2.5 3 3.5 4 4.5 5 x 2 2 5  Use the midordinate rule with 4 strips to find the area bound by the function, the xaxis and the lines x=1 and x=5. (2 marks)
 Complete the table below for the function y=x^{2}4x+5 for 1≤x≤5 (1 mark)
 A town T lies on latitude 37°N and longitude 50°E. An airport is located on another town R whose longitude is 100W on the same latitude as T. An aeroplane leaves town T and flies westwards to R. Calculate the distance covered by the plane in km. (Take R = 6370km and π=22/7) (3 marks)
 The diagram below shows a plot of land. Shade the region R enclosed under the following conditions:
 CR≥1.5 cm
 R is more than 2 cm from line AB
 CRA≥CRB
 R is nearer to CB than CA
By construction and using a scale of 1 cm to represent 10 metres, shade the region where the borehole lies. (5 marks)
 In the figure below, O is the centre of the circle. AB is a tangent to the circle at C. AD=17cm and AO=24cm
Calculate the shaded area correct to 4 significant figures. (4 marks)  A trader makes two types of chairs; ordinary and special. The cost of each ordinary chair is Kshs. 300 while the cost of a special chair is Kshs. 700. He is prepared to spend not more than Kshs. 21, 000. It is not viable for him to make less than 20 chairs. Ordinary chairs must be less than twice the special chairs but more than 15. By taking the number of ordinary chairs as x and the special chairs as y; Write down all the inequalities representing the above information. (4 marks)
 A construction firm has two tractors; P and Q. tractor P completes a job in 4 days while tractor Q completes the work in 6 days. The two tractors start working together and after 2 days, tractor P breaks down. How long does it take Q to complete the remaining work? (3 marks)
SECTION II (50 marks)
Answer any five questions in this section
 The graph below shows a triangle ABC with vertices A2, 1, B(4, 1) and C(3, 2)
 ΔA'B'C' is the image of ABC under a transformation given by the matrix 1 2 0 1 . Determine the coordinates of A', B' and C' (2 marks)
 On the grid provided draw ∆A'B'C' and describe the transformation fully (3 marks)
 ∆A''B''C'' is the image of ∆A'B'C' under a reflection along the line y=0. Draw ∆A''B''C'' on the same pair of axes and state its coordinates (2 marks)
 Determine the matrix representing a single transformation that maps ∆ABC onto ∆ A''B''C'' (3 marks)

 Write down the first three terms of the sequence whose nth term is given by Tn=5n2 (1 mark)
 The third and the sixth terms of a geometric sequence are 18 and 486 respectively. Find the first term and the common ratio of the sequence. (3 marks)
 The first and the last terms of an arithmetic progression are 8 and 190 respectively. If the sum of the first n terms of this arithmetic progression is 3094, find the number of terms in the progression (2 marks)
 The second, fourth and seventh terms of an arithmetic progression are the first three terms of a geometric progression. Find the common ratio of the geometric progression if the first term of the arithmetic progression is 2 (4 marks)

 Three variables P, Q and R are such that P varies partly as the square of Q and partly inversely as the square root of R. Determine:
 The relationship between P, Q and R given that when P=11^{1}/_{3}, Q=2 and R=9 and also when P=14.75, Q=5 and R=64 (4 marks)
 Q when P=145^{11}/_{18} and R=1.44 (2 marks)
 Four quantities A, B, C and D are such that A varies jointly with B, the square root of C and inversely as the square of D. Find the percentage change in A if B increases by 21%, C decreases by 36% and D increases by 10% (4 marks)
 Three variables P, Q and R are such that P varies partly as the square of Q and partly inversely as the square root of R. Determine:
 A particle moves along a straight line such that its displacement S metres from a given point S=t^{3}5t^{2}+4 where t is time in seconds. Calculate:
 The displacement of the particle at t=5 (2 marks)
 The velocity of the particle when t=5 (3 marks)
 The values of t when the particle is momentarily at rest (3 marks)
 The acceleration of the particle t=2 (2 marks)
 The figure below shows a right pyramid standing on a rectangular base ABCD. AB=8 cm, BC=15 cm and each slant edge is 12 cm long. N is the midpoint of BC
Calculate to two decimal places The vertical height of the pyramid (3 marks)
 The volume of the pyramid. (1 mark)
 The obtuse angle between the planes VBC and VAD of the pyramid (4 marks)
 The angle between line VD and the base (2 marks)

 Complete the table below giving the values correct to 2 decimal places. (2 marks)
x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° y= sin (x + 30°) 0.50 0.87 0.00 0.87 0.50 0.50 y= 2 cos (x + 3) 1.73 0.00 1.73 1.73 0.00 2.00 1.73  On the same set of axes, draw the graphs of y=sin (x+30°) and y=2cos (x+30°) for 0°≤x≤360°.
Use the scales xaxis; 1 cm to represent 30° and yaxis; 2 cm to represent 1 unit. (5 marks)  Use you graphs to solve the equation 2cos x+30°sin (x+30°) =0 (2 marks)
 State the amplitude of y=2cos (x+30°) (1 mark)
 Complete the table below giving the values correct to 2 decimal places. (2 marks)
 A triangle OPQ, R and S are points on OP and OQ respectively, such that OR: RP=2:3 and OS: SQ=2: 1. PS and QR intersect at T. Given that OP=p and OQ = q.
 Express in terms p and q
 QR (1 mark)
 PS (1 mark)
 Given that QT=hQR and PT= kPS, express OT in terms of
 h,p and q (2 marks)
 k,p and q (2 marks)
 Find the values of h and k (4 marks)
 Express in terms p and q
 The test scores obtained by 40 students were recorded as shown in the table below
Marks No. of students 61  65
66  70
71  75
76  80
81  85
86  904
5
9
8
8
6 Using a working mean of 73, calculate
 the mean mark (4 marks)
 the standard deviation (3 marks)

 On the grid provided, draw an ogive to represent the information in the table (3 marks)
 Use the ogive to estimate the marks scored by the 25th student. (1 mark)
 Using a working mean of 73, calculate
Marking Scheme
 x + 4 = 0
x  2/3 = 0 → 3x  2 = 0
x(3x  2) + 4(3x  2)=0
3x^{2}  2x + 12x  8 =0
3x^{2} + 10x  8= 0  max p  min q = 8  3
min q + min p 3+ 2
2
1 + √3  2(1  √3)
(1 + √3)(1 √3)
2  2√3 = 2  2√3 = 2  2√3 = √3 1
1 (√3)^{2 } 13 2
Sin 90°  log_{2}(8x)  log_{2}(x3) = log_{2}2^{2}
Log_{2}(8x/x3)= log24
8  x = 3
x  3
4(8  x) = x 3
32  4x = x 3
32 + 3 = x + 4x → 5x = 35
x = 7
2log_{2}2  1458000
Amount = 1458000 (1  15/100)3
895394.25
AB= OB  OA =  AB = √(6)^{2} + 8^{2} + 2^{2}
√104
= 10.198 units
AB = 6i + 8j  2k  v(71) = 852
v =852/17 = 24m/s
(px)^{2} = x^{2} + m/Q  p^{2}x^{2}  x^{2} = m/Q
x^{2}(p^{2}  1)= m/Q
x^{2} = m/Q x 1/p^{2}1
x = ± √(m/Q(p^{2}1))
x^{2} + 6x + 3^{2} + y^{2}  10y + (5)^{2} = 2 + 3^{2} + (5)^{2}  (x + 3)^{2} + (y 5)^{2} = 36
(x + 3)^{2} + (y  5)^{2} = 6^{2}
→(3,5)
Centre 6
Radius = (x  a)^{2} + (y  b)^{2} = r^{2} units
(x  a)^{2} + (y  b)^{2} = r^{2} 
 Binomial coeffecients
(1 + 3x)^{6} = 1 + 6(3x) + 15(3x)^{2} + 20(3x)^{3} + ...
(1 + 3x)^{6} = 1 + 18x + 135x^{2} + 540x^{3} + ...
1 + 3x = 0.0997 → 3x = 0.997  1 = 0.003  x = 0.001
 Binomial coeffecients

1
1.5 2 2.5 3 3.5 4 4.5 5 x 1.25 1 1.25 2 3.25 5 7.25 10
=10° + 50° = 60°
RT = ^{60}/_{360} x 2 x ^{22}/_{7} x 6370Cos37°  Cos θ = 7/24 → Cos1(7/24)= 73.04°
Change in longitude
AOB = 2 x 73.04° = 146.08°
Area of ΔAOB = 1/2 x 24 x 24 x sin 146.08° 

= 160.7140cm^{2}Area of sector = 146.0/360 x 22/7 x 7x7 = 62.4898cm^{2}
160.7140  62.4898 = 98.2242 = 98.22cm^{2}  Shaded area 300x + 700y < 21,000
x + y > 20
x < 2y
x>15
P and Q in 1 day → 1/4 + 1/6 = 5/12
P and Q in 2 days → 2 x 5/12 = 5/6  Remainder = 1  5/6 = 1/6
Q does 1/6 in 1 day 

 A''B''C''

 Let the matrix be 4p  2q = 4
7p = 11 → 11p = 11/7
T_{1} = 5 x 1  3 = 3
T_{2} = 5 x 2  2 = 8
T_{3} = 5 x 3  2 = 13
T_{3} = ar^{31} = ar^{2} = 18
T_{6} = ar^{61} = ar^{5} = 486


 ar^{5} = 486
ar^{2} 18
r^{3} = 27
r = ³√27
= 3   3094 = n/2(8 + (190))
3094 x 2 (n  190)
n = 3094 x 2
(8  190)
n= 34
T_{2} = a + d , T_{4} = a + 3d , T_{7} = a + 6d  Number of terms
2 + d , 2 + 3d , 2 + 6d
2 + 6d = 2 + 3d
2 + 3d 2 +d
(2 + 6d)(2 + d) = (2+ 3d)(2 + 3d)
2(2+d)+ 6d(2 + d)= 2(2 + 3d)+ 3d (2 + 3d)  4 + 14d + 6d^{2} = 4 + 12d + 9d^{2}
14d  12d = 9d^{2}  6d^{2}
2d = 3d^{2} → 2 = 3d
d= 2/3
r = 2 + 6 x 2/3 = 3/2 = 1 ½
2 + 3 x 2/3
³√27 = 2/3
P= kQ2 + c/√R
11 ^{1}/_{3} = 4k + c/√9 → 34/3 = 4k + c/3 → 34 = 12k + c ...(i)
14¾ = 5k + c/√64 → 59/4 = 5k + c/8 → 118 = 40k + c (ii)
 ar^{5} = 486


 Relationship
 28k = 84
k = 84 = 3
28
12 x 3 + c = 34 → c = 34  12 x 3
c = 2
P = 3Q^{2}  2/√R
P = 145 ^{11}/_{18}
R= 1.44
P = 3Q^{2}  2/√R → 145 ^{11}/_{18} = 3Q2  2/√1.44
3Q^{2} = 145 ^{11}/_{18} + 2/1.2
Hence
3Q^{2} = 2651
18  Q^{2} = 2651/18 x 1/3 = 2651/54
Q = √2651/54 = 7.007
A_{α} kB√C → A = kB√C
D^{2} D^{2}
B_{1} = 1.21B, C1 = √0.64C = 0.8√C
D_{1} = (1.1D)2 = 1.21D2
A_{1} = k (1.21B)0.8√C = 1.21 x 0.8 (kB)√C
1.21D^{2} 1.21D^{2}
A_{1} = 0.8A
Percentage change in A = 0.8 1 x 100
1
 Relationship
 A, B , C and D = 20%
A decrease of 20%


 t(3t  10)= 0
t =0  Velocity at 3t = 10 → t = 3 ^{1}/_{3}
Hence t = 3 ^{1}/_{3}
t = 2  Time at dv/dt = 6t  10
a = 6 x 2  10 = 2m/s^{2}
t= 0
VN = √(12^{2}  7.5^{2})  VN = √87.75 = 9.367cm at VO = √(9.3672  42) = 8.46998 ≅ 8.47
Volume = 1/3 x 8 x 15 x 8.47 = 338.80
 t(3t  10)= 0

 Obtuse VMN = 180°  50.56° = 129.44°
Sin θ = 8.47/12
θ sin1(8.47/12)= 44.90°
0°  Volume of pyramid= 30°
 Consider ΔVMN where M is the midpoint o f AD
60°, 90°, 120°, 150°, 180°  Consider ΔVDO
210°, 240°
 Obtuse VMN = 180°  50.56° = 129.44°

 Table values
x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° y= sin (x + 30°) 0.50 0.87 1 0.87 0.87 0.00 0.5 0.87 1.00 0.87 0.50 0.00 0.50 y= 2 cos (x + 3) 1.73 1.00 0.00 1.00 1.73 2.00 1.73 1.00 0.00 1.00 1.73 2.00 1.73 
S!  given scales used
P1  plotting
C1  drawing
P1  plotting
C1  drawing
 Table values


 QR = QO + OR
QR = q + ^{2}/_{5}p  PS = PO + OS
PS= p + ^{2}/_{3}q
 QR = QO + OR
 QT = hQR → QT = h(q + ^{2}/_{5} p)
QT = ^{2h}/_{5} p  hq
PT = kPS → PT = k (p + ^{2}/_{3}q)
PT = ^{2k}/_{3 }q  kp OT = OQ + QT = q  hq + ^{2h}/_{5} p
OT = (1h)q + ^{2h}/_{5} p
OT = OP + PT → p  kp + ^{2k}/_{3} q
OT = (1k)p + ^{2k}/_{3}q
 OT = OQ + QT = q  hq + ^{2h}/_{5} p
 h and 2h/5 = 1  k → 2h + 5k = 5 ...(i) x 3
1  h = 2k/3 → 3h + 2k = 3...(ii) x 2
6h + 15k = 15
6j + 4k = 6
11k = 9 → k = 9/11
2h + 5 x 9/11 = 5
2h = 5  45/11
2h = 10/11 → h = 10/11 ÷ 2 = 5/11
k = 9/11
h = 5/11


 d = xA
classes f d d^{2} fd^{2} cf 61  65 4 10 63 100 400 4 65.5 66  70 5 5 68 25 125 9 70.5 71  75 9 0 73 0 0 18 75.5 76  80 8 5 78 25 200 26 80.5 81  85 8 10 83 100 800 34 85.5 86  90 6 15 88 225 1350 40 90.5 40 2875  Mean (x) = A + Σfd/Σf = 73 + 145/40  M1
Mean= 76.625  Standard deviation
s= 7.664
 Mean (x) = A + Σfd/Σf = 73 + 145/40  M1

 student should draw curve
 Student should be able to estimate.
 d = xA
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