Questions
INSTRUCTIONS TO CANDIDATES:
- This paper contains TWO sections: Section I and Section II.
- Answer ALL the questions in Section I and FIVE questions from section II.
- All answers and working MUSTbe written on the question paper in the spaces provided below each question.
- 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 I
- Use logarithm table to evaluate (4marks)
5√(75.4 x 4.832/ 0.00521) - Make b the subject of the formula given that (3 marks)
a = bd
Nb2 - d - Line PQ is the diameter of a circle such that the coordinates of P and Q are (-2, 2) and (-2,-6) respectively. Find the equation of the circle in the form . (4marks)
ax2 + ay2 + bx + cy + d = 0 - Use completing the square method to solve the equation
4 - 3x - 2x2 = 0 (3marks) - Given that P=4+2 and Q=2+2 and thatPQ=a+bc , where a, b and c are constants, find the values of a, b and c. (3 marks)
- The table below shows the temperature readings of four different solutions recorded by students to nearest during a laboratory lesson. Calculate the percentage error in P+Q/S - R to 3 d.p. (3marks)
Quantity Temperature in P 22.5 Q 19.4 R 17.3 S 26.2 - Use matrix method to solve the simultaneous equations (3marks)
2x + y =10
2x +2y =14 -
- Expand (1+2x)5 to the fourth term. (1 mark)
- Hence evaluate (1.02)5 correct to 3 decimal places. (3 marks)
- . It is known that the value of land appreciate at 7% p.a in a town. John bought a plot in the town at Ksh 500,000. Given that he plans to sell the plot after 6 years, find out how much profit he expects to get. (Give your answer correct to the nearest thousand). (3marks)
- The mass of a wire varies jointly with its length and with the square of its diameter. A section of the wire 500m long, with diameter 3mm has a mass of 31.5kg. what is the mass of 1000m of wire of diameter 2mm? (3marks)
- Mr. Gatua has a salary of sh.80000 per annum. He lives rent free in company house and is entitled to a monthly personal relief of sh.1056. Based on the tax rates given below, calculate his PAYE. (3 marks)
Taxable income (KE p.a) Rate 1 - 1500 10% 1501 - 3000 15% 3000 - 4500 25% Above 4500 35% - The third term and sixth term of a geometric series are 31/3 and 111/4 respectively. Calculate the common ratio and hence find its first term. (3marks)
- Use the figure below to answer the question that follows
Given that angle RSQ = 500, SQ= 11.83 cm and QR=12cm.A circumcirle is drawn on the triangle.Find the radius of the circle (2marks) - A Business man bought commodity A and commodity B at shs.60 and sh.72 respectively. In what ratio must he mix so that when he sells at shs.78, he makes a profit of 200%. (3 marks)
- Points A(x°N,30°E) and B(x°N, 50°E) are 1935 kilometres apart. Taking R= 6370 km and π=22/7 , find the value of x. (3marks)
- Find the gradient function of the curve y= 1/3x3 – 4x2 + 9x + 4hence,find the gradient of the curve at point (1,-4) (3marks)
- Use a scale of 1:1 in both axes to draw the graphs of y = x2 – 6x + 7 and y = x – 2 for the domain 0≤ x ≤ 6. The point of intersection of the two functions satisfy a certain quadratic equation in x. Obtain the equation in x hence calculate it’s solutions. Give answer correct to 2d.p. (10 marks)
- Points A and B are centres of two equal circles of a radius 2 cm and 10 cm apart.
- Construct the two circles in the space given below. (1mark)
- Construct the transverse common tangents to both circles. (4marks)
- Calculate the length of the transverse common tangents (Take π=22/7 ) (5marks)
- Albert, Bonny and Charles competed in a game of chess. Their probabilities of winning the game are 2/5, 3/5 and 1/10 respectively.
- Draw a probability tree diagram to show all the possible outcomes.(2 marks)
- Calculate the probability that;
- No one loses the game. (2 marks)
- Only one of them wins the game. (2 marks)
- At least one of them wins the game. (2 marks)
- At most two of them lost the game. (2marks)
- Construct rhombus ABCD such that AB=BC= 6cm and ∠ABC=600.
- Measure BD. (1 mark)
- On the same diagram, construct the inscribed circle of triangle ACD. (3marks)
- Construct the locus of points equidistant from A and C. (3 marks)
- If x is a point on the circle in b above such that AX=XD and ∠AXD is acute, find the locus of X and make it on the diagram. (3 marks)
-
- Complete the table below (2marks)
X -180° -150° -120° -90° -60° -30° 0° 30° 60° 90° 120° 150° 180° Y = 2Cos x -1.73 1 2 1 0 Y= cos(x - 60) -0.5 0.9 0 1 -0.5 - On the same axes plot the graphs of y= cos(x-600) and y = 2 cosx (use a scale of 1unit for 300 on the x axis and 1 unit for 0.5 units on the y axis) (4mrks)
- Describe the transformation which maps y= cos(x-600) to y = 2cosx. (2marks)
- State the period and amplitude of each of the waves above. (1mark)
Amplitude Period Y = 2Cos x Y = Cos(x - 60) - Using the graph above determine the values of x for which
cos(x-60°) - 2cosx = 0 (1mark)
- Complete the table below (2marks)
-
The roof of a building is as shown in the figure above with a rectangular base ABCD. AB = 20m and AD = 8m. The ridge EF = 10m and is centrally placed. The faces ADE and BFC are equilateral triangles. Calculate- The height of E above the base ABCD (2 marks)
- The angle between the planes ABCD and ADFE (3 marks)
- The angle between the planes AED and ABCD (2 marks)
- The acute angle between lines DB and EF (3 marks)
- Kiprop has at least 50 acres of land on which he plans to plant potatoes and cabbages. Each are of potatoes requires 6 men and each are of cabbages requires 2 men. The farmer has 240men available and he must plant at least 10 acres of potatoes. The profit on potatoes is kshs.1200 per acre. If he plants x acres of potatoes and y acres of cabbages;
- Write down 3 in equalities in x and y to describe the information.(2 marks)
- Represent these in equalities graphically. (use a scale of 1:10 for both axes) (4 marks)
- Use your graph to determine the number of acres for each vegetable which will give maximum profit. (4 marks)
-
- Complete the table below for y = x2 - 3x + 5 in the range 2 < x < 8 (2marks)
x 2 3 4 5 6 7 8 y 3 9 23 33 - Use the trapezium rule with six strips to estimate the area enclosed by the curve , x- axis and the lines x=2 and x=8 (2 marks)
- Find the exact area of the region given in (b) (4 marks)
- Calculate the percentage error in the area. (2marks)
- Complete the table below for y = x2 - 3x + 5 in the range 2 < x < 8 (2marks)
Marking Scheme
No Log 75.4
4.831.8774
0.68394.83 2.5613
0.68390.00521 3.2452
3.71685.5284/5 1.276 x 101 1.1057 12.95 - a² = b²d²
Nb2 - d
Nb²a² - a²d = b²d²
Nb²a² - b²d² = a²d
b² (Na² - d²)= a²d
b² = a²d
Na² - d²
b = ± √a²d/a²N-d - (-24 -1/2 , -6+2/2)
Centre (-2,2)
radius √(-2,2)2 + (2 -6)2
NO = 16 = √16 = 4 units
(x + 2)2 + (y - 2)2 = 16
x2 + 4x + 4 + y2 - 4y + 4 = 16
x2 + y2 + 4x - 4y + 8 -16 = 0
x2 + y2 + 4x - 4y - 8 = 0 - -2x2 - 3x + 4 = 0
x2 + 3/2x - 2 = 0
x2 + 3/2x - 2 = 0
x2 + 3/2 - 2
x2 + 3/2x + 9/16 = 2 + 9/16 = 41/16
(x + 3/4)2 = 41/16
x + 3/4 = ± √41/4
x = 0.8508 or -2.3508 - 4 + √2 x 2 + √2 = 4(2 - √2) + √2(2 - √2)
2 +√2 2 - √2 4-2
8 - 4√2 + 2√2 - 2
2
6 - 2√2 = 3 - √2
2
a = 3
b = -1
c = 2 - Max 22.55 + 19.45 = 105 = 4.77
26.15 - 17.35 22
Min 22.45 + 19.35 = 209 = 4.64
26.25 - 17.25 25
Actual 22.5 + 19.4 = 419 = 4.71
26.20 - 17.30
4.77 - 4.65
2 x 100
4.71
11/200 ÷ 4.7 x 100 = 1.168 % -
-
-
1 + 5 (22) + 10(2x)2 + 10(2x)3
1 + 10x + 40x2 + 80x3 - (1.02)5 : (1 + 0.02)5
2x = 0.02
x = 0.01
1 + 10 x 0.01 + 40x (0.01)2 + 80 x (0.01)2
1 + 0.1 + 0.004 + 0.00008
1.10408
-
- L = P (1 + r/100)n
500,000 (1 + 0.07)6
500,000 (1.07)6
750,365.1759
750, 365.1759 - 500,000
250,365.1759 ≈ 250,000 - Mass → M
Length → diameter
M = kld2 where 1l is a constant
31.5 = 300 x 3x 3k
k = 31.5 = 7/1000
500 x 9
M = 7/1000 x 1000 x 2 x 2
= 28kg - 80,000 x 1.15 = 4600
20
(150 + 225 + 375 + 35) x 20 = 1308.33
12
1308.33 - 105.6
= 252.33
≈ 252 - ar2 = 10/3
ar5 = 45/4
a = 10/3 / r2
a = 45/4 / r5
10/3 x 1/r2 = 45/4 x 1/r5
10/3r2 = 45/4r5
40r5 = 135r2
r3 = 135/40
r = 3/2
a x 9/4 = 10/3
a = 40/27 - a/SinA = b/Sin B= c/SinC = 2R
12?Sin 30 = 2R
R= 12
2Sin 300
=7.8324 - 60A + 72B = 68
A+B
60A + 72B = 68A + 68B
-8A = -4B
A/B = 4/8 = 1/2
A:B = 1:2 - 20/360 x 22/7 x 2 6370Cos s = 1935
Cos x = 1935 x 7 x 360 = 0.8699
20 x 22 x 2 x 6370
x = Cos-1 0.8699
= 29.55 - dy/dx = x2 - 8x + 9
(1)2 - 8(1) + 9
1 + 9 - 8 = 2 -
x2 - 6x + 7 = x -2
x2 - 7x + 9 = 0
x = 7 ± √49 - 4x 1 x 9
2
x = 7 ± √13
2
x = 5.30 or x 1.70 -
Length of belt
√52 - 22 = √21 = 4.5826
Cos-1 2/5 = 66.42°
180 - 66.42 = 113.58
113.58 x 2 = 227.16
2(22.7/360 x 22/7 x 4) + (4.5826 x 4)
15.85 + 18.33
34.18 -
-
-
- 2/5 x 3/15 x 1/10
3/125 - 2/5 x 2/5 x 9/10 or 3/5 x 3/5 x 9/10 or 3/5 x 2/5 x 1/10
18/125 + 81/250 + 3/125
= 125/250 - 1 - (3/5 x 2/5 x 7/10)
1 - 27/125
= 98/125 - 1 - 123/250
127/250
- 2/5 x 3/15 x 1/10
-
-
- 11.3cm
-
-
-
X -180° -150° -120° -90° -60° -30° 0° 30° 60° 90° 120° 150° 180° Y = 2Cos x -2 -1.73 -1 0 1 1.73 2 1.73 1 0 -1 -1.73 -2 Y= cos(x - 60) -0.5 -0.7 -1 0.9 -0.5 0 0.5 0.9 1 0.9 0.5 0 -0.5 -
- Translation by translation vector (-60) followed by a stretch , stretch 2 x- axis invariant
-
Amplitude Period Y = 2Cos x 2 360 ° Y = Cos(x - 60) 1 360° - x= -120° and 60°
-
-
- Height √48 - 25
√23 = 4.7958 - tan θ = √23/15
θ tan-1 = √23/15 = 17.73° - Cos θ = 5/√48
θ = Cos-1 5/5/√48 = 43.81 - α = Tan-1 4/15 = 14.93°
- Height √48 - 25
-
- x + y > 50
x > 10
3x + y < 120 -
- (10, 90)
118000
- x + y > 50
-
-
x 2 3 4 5 6 7 8 y 3 5 9 15 23 33 45 - h = 8 - 2 = 1
6
1/2 x 1 {(3 + 45 ) + 2 (5 + 9 + 15 + 23 + 33)}
1/2 {48 + 170}
1/2 {218}
= 109 - 82∫(x2 - 3x + 5)dx = [x3/3 - 3x2/2 + 5x + c]82
[512/3 - 192/2 + 40 + c] - [8/3 - 12/2 + 10 + c]
108 sq units - 109 - 108 = 1
1/108 x 100
= 0.9259%
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