MATHEMATICS
PAPER 2
TIME: 2 HOURS.
Instructions To Candidates
- This paper contains two sections: section I and section II
- Show all the steps in your calculations, giving your answers at each stage.
- Marks may be given for correct working even if the answer is wrong.
- Non programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
SECTION A (50 MARKS)
Answer all questions in the spaces provided
- Evaluate (3mks)
- The point (9, 4) is the image of the point (1,4) under a shear with x-axis invariant. Find the matrix of the shear.(2mks)
- In vectors , , , find x and y given that (3mks)
- Triangle ABC is such that AB =10cm, angle BAC=60° and angle ABC = 30°
- construct triangle ABC and construct the locus of P such that the area of triangle APB is 20cm2 (2mks)
- Indicate the locus of Q such that Q is equidistant from CA and CB (2mks)
- In the figure below, VABCD is a right pyramid on a rectangular base. Point O is vertically below the vertex V. AB=24cm, BC=10cm and CV=26cm Calculate the angle between the edge CV and the base ABCD (2mks)
- The escribed circle of triangle ABC touches BC at P, AB produced at Q, and AC produced at R. If AQ=10cm, AR=10cm and AO = 12cm, find the length of OQ and shaded area(4mks)
- Make K the subject at the formular(3mks)
- Solve for x if logx16 + logx32 = 18 (3mks)
- Simplify without using table leaving your answer in the form a+ b√c
1 + cos 30° sin 45°
1 - sin 60° cos 45º - Y varies inversely as the square of x. The difference between the values of y when x=6 and when x=10 is 6. Find x if y=4 (3mks)
- A ship sails due East from an island (48°N, 42°E) to another island B. The average speed of the ship is 30 knots and it takes 22 hours to reach island B. Find position of island B. (3mks)
- Solve the equation 6sin?0 - cose - 5 = 0, where 360° ≥ θ ≥ 0º (3mks)
- Calculate the standard deviation of the following data. (3mks)
No of goals 0 1 2 3 4 5 6 No. of matches 11 2 5 8 3 2 1 - The figure below shows a quadrilateral ABCD
Calculate the length of BD and AB (4mks) - Given that x=6.4 and y=5, find the correct to 1 decimal place, the percentage
error in xy (3mks)
x+y - The angle of elevation of the top of a flag post from a point A on a level ground is 23°. The angle of elevation of the top of the flag from another point B nearer the flag post and 100m from A is 40°. A, B and the bottom of the flag are collinear
Calculate- the distance from the point B to the top of the flag post(2mks)
- the height of the flag post (2mks)
SECTION B (50 MARKS)
Answer all questions in the spaces provided
- A community water tank is in the shape of a cuboid of base 6m by 5m and height of 4m. A feeder pipe of diameter 14cm supplies water to this tank at a flow of 40cm/s.
Calculate-
- the capacity of the tank is litres (2mks)
- the amount of water, in litres, delivered to this tank in one hour (3mks)
- the time taken for the tank to fill (2mks)
- The community consumes a full tank a day, with each family consuming an average of 150 litres per day. If each family pays a uniform rate of sh.350 per month, find the total amount of money due monthly (3mks)
-
- In driving to work, Jane has to pass through three sets of traffic lights. The probability that she will have to sop at any of the lights is 3/5.
- Draw a tree diagram to represent the above information(2mks)
- Using the diagram, determine the probability that on any one journey, she will have to stop at
- all the three sets(2mks)
- only one of the sets (2mks)
- only two of the sets (2mks)
- none of the sets(2mks)
-
- On the grid provided, draw the graph of the functions y = ½x2 - x+3 for 0 ≤ x ≤ 6
- Calculate the mid-ordinates for 5 strips between x=1 and x=6 and hence use the mid-ordinate rule to approximate the area under the curve between X=1, x=6 and the x-axis (3mks)
- Assuming that the area determined by integration to be the actual area. Calculate the percentage error in using the mid-ordinate rule. (4mks)
-
- Sketch the curve of y=(x-3) (2x2 - 3x + 1) (5mks)
- Given that a cone whose base radius and perpendicular height are rcm and hcm respectively. (1cm)
Determine- the radius r in terms of h if the slant height is 10cm(1mk)
- value of perpendicular height, H, which the volume of the cone is maximum and hence find the maximum possible volume (4mks)
- A carpenter takes 4 hours to make a stool and 6 hours to make a chair. It takes the carpenter at least 144 hours to make x stools and y chairs. The labour cost of making a stool is ksh.100 and that of a chair is ksh.200. The total labour cost should not exceed ksh.4,800. The carpenter must make at least 16 stools and more than 10 chairs.
- Write down inequalities to represent the above information(3mks)
- Draw the inequalities in (a) above on the grid provided(4mks)
- The carpenter makes a profit of ksh.40 on a stool and ksh100 on a chair. Use the graph to determine the maximum profit the carpenter can make. (3mks)
MARKING SCHEME
SECTION A
(x4 - x3 + x2 -2x)2
=[16-8 +4-4]- [16+8+4-4]
= 16-
1 + 4k = 9
4k=8
k= 2
x = 16 = x=4
-3x - y = 11 =y=-23-
-
- Locus θ labeled
Bisector of AC and BC
-
- Diagonal = √242 - 102 = 26
Cost = 13/26 -0 = 600 - OQ=√122 - 102
= √44
= 6.633
Sinθ = 10/12
θ = 56.44
Area Δ AOQ = ½ x 6.633 x 10 = 33.165cm2
Area sector OQP = 56.44 x 22/7 x 6.635
360
= 21.68cm2
Shaded area = 2(33.165-21.68)
= 22.97cm2 - R2 = k2 + a2/Hg
K2 = hgR2 - a2
K =±√hgR2 - a2 - 4logx2 + 5logx2 = 18
9 logx2= 18
Logx2 = 2
x2 = 2
x=√2 = 1.4142
=4+√6 x 4
4 4-√6
4+√6(4+√6) = 16+8√6 + 6
4-√6 (4+√6) 10
=11 + 4√6
5
- Y =k
x2
k - k = 6
36 100
64k =6
3600
K-337.5
4 = 337.5
X2
X = √337.5
4
= 9.186
D = SXT
22 x 30 = 660nm
60θ = 660
θ = 110
B (48°N, 53°E)- 6(1-сos2 θ) - cos θ -5 = 0
6 - 6cos2θ - cos θ -5 =0
6cos2 θ + cosθ -1 = 0
6cos2θ +3 cosθ - 2cos θ-1 = 0
3 cosθ(2cosθ+1)-1(2cosθ+1) = 0
(3cosθ-1) (2cosθ+1)=0
cosθ =1/3 or cosθ = -1/2
θ= 150°, 210°, 70.53°, 289.47° -
X F Fx d=x-m d2f 0 11 0 -2 0 1 2 2 -1 1 2 5 10 0 0 3 8 24 1 3 4 3 12 2 16 5 2 10 3 45 6 1 6 4 96 32 64 161
fx
= 64/32
=2
s.d=√Σd2f
Σf
=√5.031 = 2.243 - BD2 - 152 +102 -2(15)(10) cos40
BD2 = 95.19
BD = 9.756
12 = 9.756
sinB sin50
sinB = 12 sin50
9.756
B = 70.43; D = 59.57
AB = 9.756
Sins9.57 sin50
AB = 10.98cm - Vw =6.4(5) = 2.8070
6.4+5
Vmax = 6.45(5.5) = 3.2696
4.5+6.35
Vmin = 6.35(4.5) = 2.3912
6.45+5.5
A.E = 3.2696 - 2.3912 = 0.4392
2
%E=0.4392 x 100 - 15.6%
2.8070 -
tan40 = h/x
tan23= h/100+x
xtan40=(100+x) tan23
0.8391x -0.4245 = 42.45
x =102.38m
PB = 102.382 +85.912 =133.65m
or cos40-102.35/PB = PB = 133.65m- Height = xtan40 = 102.38 x tan40 = 85.91m
SECTION B
-
-
- Capacity = 6 x 5 x 4 x 1000
= 120000L - Amount = (22/7 x 72 x 40 x 60 x 60)/100
=22176Litres - Time = 120000
22176
=5.411 hours
- Capacity = 6 x 5 x 4 x 1000
- No. of families = 120000/150 = 800
Amount = 800 x 350
= 280,000
-
-
-
-
- P(3S) = 3/5 x 3/5 x 3/5 =9/125
- P (Is only)
3/5 x 2/5 x 2/5 + 2/5 x 3/5 x 2/5 + 2/5 x 2/5 x 3/5
=36/125 - P (2s only)
3/5 x 3/5 x 2/5 + 3/5 x 2/5 x 3/5 + 2/5 x 3/5 x 3/5
=54/125 - P (none) = 2/5 x 2/5 x 2/5 =8/125
-
-
x 0 1 2 3 4 5 6 y 3 2.5 3 4.5 7 10.5 15 - y1=2.6, y2=3.6, y3=5.6, y4 =8.6, y5=12.6
Area = 1 [2.6+3.6+5.6_8.6+12.6]
= 33 sq. units
= 36 - 22/3
= 331/3
% error = 1/3 x 100
100
= 1.01%
-
- x-intercept = when y=o
(x-3)(2x2 - 3x + 1)=0
X=3 or 2x - 2x -x+1=0
(2x - 1)(x-1)=0
X= ½ or 1
(3,0), (½,0), (1, 0) y-intercept = when x=0
y=-3 (0, -3)
St. pts=y=2x3 - 9x2 + 10x -3
dy/dx = 6x2 - 18x + 10
6x2 - 18x + 10 = 0
X = 2.264 or 0.7362
St pts. (0.7362, 0.2821)
And
(2.264, -3.282)
Nature of st. pts
d2y/dx2 = 12x - 18
when x=0.7362
d2y=-ve
dx2
(0.7362, 0.2821) = Max st. pt
When x = 2.264
d2y=+ve
dx2
(2.264, -3.282) = Min st. pt. -
- r2 = 100-h2 = r = √100 - h2
- V=1/3π(100 - h2)h = v= 100πh1/3
3
dv = 100π - πr2 = dv=0
dh 3 dh
h= √100/3 - 5.773
- x-intercept = when y=o
-
- 4x + 6y ≥ 144
100x + 200y ≤ 4800
X ≥ 16 and y > 10 -
- Max point (16,16)
Objective function: z=40x + 100y
Profit = 40(16) + 100 (16)
=sh 2240
- 4x + 6y ≥ 144
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