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.
- Non-programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
Questions
SECTION A (50 MARKS)
Answer all the questions in this section in the spaces provides
- A school bursar wishes to obtain the sum of the following amounts of money paid in as school fees for three students by the CDF: Ksh 20 760, Ksh 49 105 and Ksh 17 352.The bursar estimates the sum by first rounding each of the amounts to 3 significant figures.
- Determine the estimated sum (1 mark)
- Determine the percentage error in this estimate (2 marks)
- Solve for x in : log5(3x - 2) + log5(2x - 1) = 0 without using mathematical tables or calculator. (3 marks)
- In the figure below, ABX is a tangent. Angle CAB = 36° and angle ACB = 42°.
Calculate the size of angle BDC (2 marks) - Make P the subject of the formula in (3 marks)
- Given that 0 tan15° = 2 - √3 = − , simplify (2 marks)
1
tan15°. - Solve the equation 5cos2θ + 2 = 3sin2θ - 2cosθ for the range 0 ≤ θ ≤ 360(4 marks)
- Solve the simultaneous equation (4 marks)
2x - y =3
x2 - xy = -4 - Use binomial expansion to simplify(√2 + √5)4(√2 - √5 )4(4 marks)
- The life expectancy in hours of 40 bulbs are shown in the table below.
Expectancy
(hours)90 - 94 95 - 99 100 - 104 105 - 109 110 - 114 115 - 119 Frequency (f) 3 10 12 9 4 2 - Given that y is inversely proportional to xn and k is the constant of proportionality and that x = 2 , when y = 4½ , and x = 3, when y =11/3 . Find the values of n and k . (4 marks)
- A poultry farmer vaccinated 640 of his 700 chicken against a disease. Two months later 15% of the vaccinated and 70% of the unvaccinated chicken contracted the disease. Calculate the probability that the chicken chosen at random contracted the disease. (3 marks)
- A pilot leaves point T (60°S ,15°W ) and flies due East for a distance of 1260nm to a point U . Determine the position of U . (3 marks)
- The equation of a circle is given as 1/3x2 + 1/3y2 - 11/3x + 2y - 1=0 . Draw the circle on the grid provided. (4 marks)
- Pipes S and T can fill a tank in 2 hours and 3 hours respectively. Pipe U can empty the full tank in 4 hours. How long will it take to fill the tank with all the pipes running? (2 marks)
- Table below is part of tax table for annual income for the year 2020.
Taxable income in K£ p.a. Rate in Kshs. per K£ Under K£4201 2 From K£4201 but under K£8401 3 From K£8401 but under K£12601 4 - The position vectors of points A, B and C are a, b and c respectively. Point C divides AB in the ratio 5 : 2. − Express c in terms of a and b. (2 marks)
SECTION B (50 MARKS)
Answer only five questions in this section in the spaces provided
- The Hire Purchase (H.P) price of a public address system was Ksh 448 000. A deposit of Ksh 112 000 was paid followed by 24 equal monthly instalments. The cash price of the public address system was 15% less than the H.P price.
- Calculate :
- The monthly instalment. (2 marks)
- The cash price. (2 marks)
- A customer decided to buy the system in cash and was allowed an 8% discount on the cash price. He took a bank loan to buy the system in cash. The bank charged compound interest on the loan at rate of 16% p.a. compounded quarterly. The loan was repaid in 12 2 years. Calculate the amount repaid to the bank by the end of the 12 2 years. (3 marks)
- Express as a percentage of the Hire Purchase price, the difference between the amounts repaid to the bank and the Hire Purchase price. (3 marks)
- Calculate :
-
- Complete the table below for the function y= cos x° and y=2Cos½x° , for < x < 360° (2 marks)
x° 0 30 60 90 120 150 180 210 240 270 300 330 360 Cos x° 1 0.87 0.50 -0.87 -1 -0.50 0.50 1 2 Cos½ x° 2 1.93 1.41 0.52 0 -1 -1.73 -2 - On the grid provided, draw the graph of y= cos x° and y=2Cos½x° using a scale of: 1cm to represent 300 on x-axis and 4 cm to represent 1 unit on y-axis (5 marks)
- Use your graph to determine:
- The period of the function y=2Cos½x° (1 mark)
- The transformation that maps the function y= cos x° onto y=2Cos½x° (1 mark)
- The values of x for which cos½x° - ½cos x° = 0 (1 mark)
- Complete the table below for the function y= cos x° and y=2Cos½x° , for < x < 360° (2 marks)
- The figure ABCDEF below represents a roof of a house. AB = DC = 16 m, BC = AD = 9 m, AE = BF = CF = DE =7.5 m and EF = 12 m.
- Calculate, correct to 4 significant figures, the perpendicular distance of EF from the plane ABCD. (3 marks)
- Calculate, correct to 2 decimal places, the angle between:
- The planes ADE and ABCD. (2 marks)
- The line AE and the plane ABCD. (2 marks)
- The planes ABFE and DCFE. (3 marks)
- In an experiment involving two variables T and Q , the following results were obtained.
T 0.45 1.35 1.80 2.55 3.50 4.50 5.40 6.00 Q 8.6 7.2 6.3 5.3 4.0 2.5 1.2 0.3 - On the grid provided, plot a graph of Q against T and draw the line of best fit for the data. (4 marks)
- The variables T and Q are connected by the equation Q aT b = + where a and b are constants.
Determine:- The values of a and b (3 marks)
- The equation of the line of best fit. (1 mark)
- The value of T when Q = 0 (2 marks)
- In the figure below, the shaded region is bounded by the lines y x = 3 , y x = and the curve y xx = − 8 .The two straight lines intersect the curve at points P and Q .
- Determine the coordinates of point P and Q . (2 marks)
- Calculate the exact area of the:
- Region bounded by the curve and the line y = x . (3 marks)
- Region bounded by the curve and the line y = 3x . (3 marks)
- Shaded region. (2 marks)
- The vertices of a rectangle ABCD are A (-1,1), B(1,1), C(1,4) and D(-1,4). The vertices of its image under transformation T are A'(1,1) B'(3,1) C'(9,4) and D'(7,4).
-
- Draw on the grid provided rectangle ABCD and its image A'B'C'D' under the transformationT . (2 marks)
- Describe fully the transformationT . ( 3 marks)
- Determine the matrix of transformation T (2 marks)
- On the same grid as in (a), draw rectangle A"B"C"D" the image of rectangle ABCD under a stretch with line y =1 invariant and stretch factor 2. State the coordinates of A"B"C"D" (3 marks)
-
- The figure below shows a rectangle ABCD with AB = 96 m and diagonal AC = 120 m.
- Construct the locus L1 of points equidistant from A and B and locus L2 of points equidistant from BC and BA. If L1 and L2 meet at N inside the rectangle, locate point N. (3 marks)
- A point X is to be located inside the rectangle such that it is nearer B than A and also nearer AB than BC. If it is not greater than 45 m from N shade the region where the points could be located. (4 marks)
- Calculate the area of the region X (3 marks)
-
- The first term of an arithmetic progression (AP) is −11 .The sum of the first 8 terms of AP is 52.
- Find the common difference of AP. (2 marks)
- Given that the sum of the first n terms of the AP is greater than 920. Find the least value of n . (3 marks)
- The 3rd, 7th, and 17th terms of another AP form the first three terms of a geometric progression (GP). If the common difference of the AP is 3.
Find:- The first term of GP (3 marks)
- The sum of the first 7 terms of the GP. (2 marks)
- The first term of an arithmetic progression (AP) is −11 .The sum of the first 8 terms of AP is 52.
Marking Scheme
-
- 20 800 + 49 100 + 17 400 = 87 300
- % error = 87300 - 87217 x 100%
87217
8300 % or 0.09516%
87217
- 20 800 + 49 100 + 17 400 = 87 300
- Log 5(3x-2) + Log (2x-1) = 0
Log5[(3x-2)(2x-1)] = Log51
⇒6x2 - 7x + 2=1
6x2 - 7x + 1=0
(6x+1)(x-1)=0
x= -1/6 or x=1
x=1 - <ABD = 42°(Alternate segemnt theorem)
<BDC = 42° + 36° = 78° - L² = x - PT
π² Py
L²Py = xπ² - PTπ²
L²Py + PTπ² = xπ²
P = xπ²
L²y + Tπ² - 1 = 1
tan 15° 2 -√3
1(2+√3)
(2-√3)(2+√3)
2+√3
4-3
2+√3 - 5Cos2θ + 2= 3(1-Cos2θ) - 2cosθ
8Cosθ + 2Cosθ -1 = 0
(4Cosθ - 1)(2Cos θ+1)=0
cosθ =¼ or cos θ = -½
θ = 75.52°. 120°, 240°, 288.48° - x2 - x(2x-3)=-4
x2 - 2x2 + 3x=-4
x2 - 3x - 4=0
(x+1)(x-4)=0
x=-1 or x=4
y=-5 or y=5 - (√2 + √5)4 - (√2-√5)4
(√2 - √5)4 = (√2)4 + 4(√2)3(√5) + 6(√2)2(√5)2 + 4(√2)(√5)3 + (√5)4
= 4 + 8√10 x 60 + 20√10 = 25 = 89 + 28√10
(√2 - √5)4 = 4 - 8√10 + 60 - 20√10 + 25 = 89 - 28√10
(√2 + √5)4 - (√2 - √5)4 = (89 +28√10) - (89 - 28√10)
56√10 -
Cumulative frequency 3 13 25 34 38 40
Q3 = 104.5 + (30-25/9) x 5= 107.28
quartile deviation
= 107.28 - 98.0
2
= 4.639 - y = k/xn ⇒ 4½ = k/2n and 11/3 = k/3n
4½ x 2n = 11/3 x 3n
(2/3)n = (2/3)3
n=3
k= 4/3 x 27 =36 - p( contracted disease) = (⁶⁴⁰/₇₀₀ x ¹⁵/₁₀₀ ) + (⁶⁰/₇₀₀ x ⁷⁰/₁₀₀)
24/175 + 3/50
69/350 - 60θCos60° = 1260
θ= 42°
U(60°S, 27°E) - 1/3x2+1/3y2-11/3x+2y-1=0
x2-4x+4+y2+6y+9=3+4+9
(x-2)2+(y+3)2=42
Centre(2,-3) radius = 4 units - In 1h 1/2 + 1/3 - 1/4 = 7/12 is filled
time taken to fill tank k=12/7
=15/7 - 1st slab: 4200 x 2 = Ksh 8,400
2nd slab: 4200 x 3 = Ksh 12,600
3rd slab: y x 4 = (37000 - 21000)
y=4000
annual income = 4200 + 4200 + 4000 = 12400 - OC = 5/3b + -2/3a
= 5/3b - 2/3a -
-
- 448000 - 112000
24
Ksh14000 - 85 x 448000
100
Ksh 380,800
- 448000 - 112000
- 92/100 x 380800
350336 = (1 + 4/100)10
Ksh 518,582 - 518582.86 - 448000 x 100%
448000
70582.86 x 100
448000
15.76%
-
-
-
-
x° 60 90 120 210 270 330 Cos x 0.5 1 -0.5 -0.87 0 0.87 2cos½x° 1.73 1.41 1 -0.52 -1.41 -1.93 -
- Period = 720°
- Stretch along y - axis, stretch factor 2 followed by stretch along x-axis, stretch factor 2
- 225° ± 2°
-
-
- height of traingle = √7.52 - 4.52 = 6m
vertical height = √62 - 22
= 5.657m -
- Cos θ = 2/6
θ = 70.53° - Sin α = 5.657/7.5
α = 48.96° - Slant length= √7.52 - 22
=7.228m
Sin½β= 4.5/ 7.228
β= 2 x 38.50
β= 77.00°
- Cos θ = 2/6
- height of traingle = √7.52 - 4.52 = 6m
-
-
-
- a=gradient = 7.0 - 4.0
1.5 - 3.5
a= -1.5
b= y - intercept = 9.2 ± 0.1 - Q= -1.5T + 9.2
- 0=-1.5T + 9.2
1.5T=9.2
T=6.133
or when Q=0
T= x-intercept
T = 6.15 ± 0.04
- a=gradient = 7.0 - 4.0
-
-
- 8x-x2=3x
5x-x2
x(5-x)=0
x=0 or x=5
P(5,15)
8x-x2=x
7x-x2=0
x97-x)=0
x=0 or x=7
Q(7,7) -
- ∫70 (8x-x2)dx - ∫70 xdx
[4x2 - x³/3 + c]70 -[x²/2+c]70
[4 x 72 - 7³/3] - [7²/2]
812/3 - 241/2
=571/6 - ∫50 (8x-x2)dx - ∫50 3dx
[4x2 - x³/3 + c]50 -[x²/2+c]50
[4 x 52 - 5³/3] - [3 x 5²/2]
581/3 - 371/2
=205/6 - Shaded area = 571/6 - 201/6
361/3 sq units
- ∫70 (8x-x2)dx - ∫70 xdx
- 8x-x2=3x
-
-
-
- Transformation is a shear
linear x- axis in variant
point B(1,1) mapped onto point B'(3,1) -
-
- A*(-1,1), B*(1,1), C*(1,7), D*(-1,7)
-
-
-
- θ= 45 ± 1°
Area of region x = 45/360 x 22/7 x 452
=795.54m2
-
-
-
- 8/2[2x-11+(8-1)d]=52
-88+28d=52
d=5 - n/2[2x=11+(n-1)5]>920
-22n+5n2-5n>1840
5n2-27n-1840>0
n= 27 ± √(-27)2-4x5x-1840
2 x 5
n= 27 ± √37529
10
n= 22.072 or -16.672
Hence least volume of n=23 terms
- 8/2[2x-11+(8-1)d]=52
-
- a+18 = a+48
a+6 a+18
a2+36a+324 = a2 + 54a + 288
18a=36
a=2
1st term of G.P = 2+2 x 3=8 - S7 = 8(2.5⁷ ‾ ¹)
2.5 -1
S7 = 13021/12
- a+18 = a+48
-
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