QUESTIONS
SECTION I (50 MARKS)
Answer all the questions from this section
 Use Logarithms correct to four significant figures to evaluate. (4marks)
∛(24.36 × 0.066547)
1.48^{2 }  Find the value of x given that the matrix is singular (3 marks)
 In the figure below QT is a tangent to the circle at Q. PXRT and QXS are straight lines. PX = 6cm, RT = 8cm, QX = 4.8cm and XS = 5cm.
Find the length of QT (3 marks)  Use the trapezium rule with seven ordinates to find the area bounded by the curve y= x^{2 }+ 1 lines x = 2, x = 4 and x – axis (3 marks)
 Given that x=√ tp make p the subject of the formula (3 marks)
2m+p 
 Construct triangle PQR such that PQ = 7cm, QR = 5cm and <PQR=30° (2 marks)
 Construct the focus L1 of points equidistant from P and Q to meet the locus L2 of points equidistant from Q and R (2 marks)
 The points (5, 5) and (3, 1) are ends of a diameter of a circle centre A. Determine:
 The coordinates of A. (1 mark)
 The equation of a circle expressing it in form x^{2} + y^{2 }+ ax + by + c = 0 (2 marks)
 A transformation is represented by the matrix . This transformation maps a triangle ABC of the area 12.5cm^{2} onto another triangle A′B′C′. Find the area of triangle A′B′C′. (3marks)
 Pipe A can fill a tank in 2 hours, pipes B and C can empty the tank in 5 hours and 6 hours respectively. How long would it take
 To fill the tank if A and B are left open and C closed (2 Marks)
 To fill the tank with all the pipes open (2 Marks)

 Expand and simplify (13x)^{5} upto the term in x^{3} (2 marks)
 Hence use your expansion to estimate (0.97)5 correct to 4d.p. (2 marks)
 Solve for x in the equation:
2cos4x= 1 for 0º < x <180º (3 marks)  Wanjiku pays for a car on hire purchase in 15 monthly instalments. The cash price of the car is Ksh.300, 000 and the interest rate is 15%p.a. A deposit of Ksh.75, 000 is made. Calculate her monthly repayments. (3 marks)
 The gradient function of a curve is given ^{dy}/_{dx} = 3x^{2} – 8x + 2. If the curve passes through the point, (2, –2), find its equation. (3 marks)
 Simplify the following surds leaving your answer in the form a+ b√c (3marks)
√5 + √2
2√2 √5 2√2 √5  The sum of two numbers is 24. The difference of their squares is 144. What are the two numbers? (3marks)
 The data below represents the marks scored by 9 form 4 students in an exam:
40, 37, 39, 40,41,43,44, 37,44
Calculate the interquartile range of the above data (3 marks)
SECTION II (50 MARKS)
Answer five questions only from this section
 The following table shows the rate at which income tax was charged during the year 2021
Monthly taxable income in Ksh. Tax rate % 0 – 9860
9861 – 19720
19721 – 29580
29581 – 39440
39441 – 49300
49301 – 59160
over 5916010
15
20
25
30
35
40 His monthly Taxable income (2 marks)
 Calculate his net monthly tax (6 marks)
 Maina’s monthly salary (2 marks)
The above diagram represents a wooden prism. ABCD is a rectangle. Points E and F are directly below C and B respectively. M is the midpoint of CD. AB = 8 cm, BC = 10 cm and CE = 4.5 cm. Calculate the size of angle CDE (2 marks)
 Calculate the
 Length of AC (2 marks)
 Angle AC makes with the plane ADEF (2 marks)
 Find the:
 Length of MB (2 marks)
 Angle CBM (2 marks)
 An aeroplane left town P(6ºN,15ºE) to another town Q(65ºN ,165ºW) at a speed of 200 knots using the shortest route. (Take π = ^{22}/_{7} ,R = 6370km)
 Find
 The shortest distance travelled in nautical miles. (3 marks)
 The time taken from P to Q in hours. (2 marks)
 Another plane left P at 1.30 p.m local time and travelled to T (65ºN, 60ºE) along the parallel of latitude. Find
 The distance between P and Q to the nearest kilometres. (3 marks)
 The local time of arrival at T if the plane flew at 470km/hr. (2 marks)
 Find
 The probability that a student goes to school by a bodaboda is ^{2}/_{3} and by a matatu is ¼ . If he uses a bodaboda the probability that he is late is ^{2}/_{5} and if he uses matatu the probability of being late is ^{3}/_{10}. If he uses other means of transport the probability of being late is ^{3}/_{20}.
 Draw a tree diagram to represent this information. (3marks)
 Find the probability that he will be late for school. (3marks)
 Find the probability that he will be late for school if he does not use a matatu. (2marks)
 What is the probability that he will not be late to school? (2marks)
 A farmer has 50 acres of land. He has a capital Shs. 2,400 to grow carrots and potatoes as cash crops. The cost of growing carrots is Shs.40 per acre and that of growing potatoes is Shs.60 per acre. He estimates that the respective profits per acre are Shs.30 (on carrots) and Shs. 40 (on potatoes). By letting x and y to represent the acres of carrots and potatoes respectively:
 Form suitable inequalities to represent this information. (4marks)
 By representing this information on a graph, determine on how many acres he should grow each crop for maximum profit (4marks)
 Find the maximum profit. (2marks)
 The 2nd and 5th terms of an arithmetic progression are 8 and 17 respectively. The 2nd, 10th and 42nd terms of the A.P. form the first three terms of a geometric progression. Find
 the 1st term and the common difference. (3marks)
 the first three terms of the G.P and the 10th term of the G.P. (4marks)
 The sum of the first 10 terms of the G.P. (3marks)
 In the triangle PQR below L and M are points on PQ and QR respectively such that PL:LQ=1:3 and QM:MR=1:2, PM and RL intersect at X, given that PQ = b and PR = c
 Express the following vectors in terms of b and c
 QR (1mark)
 PM (1mark)
 RL (1mark)
 By taking PX = hPM and RX = kRL where h and k are constants find two expressions of PX in terms of h, k, b and c. Hence determine the values of the constant h and k. (6marks)
 Determine the ratio LX:XR (1mark)
 Express the following vectors in terms of b and c
 Given that y = 2sin 2x and y = 3cos (x + 45)º
 Complete the table below. (2mks)
X 0º 20º 40º 60º 80º 100º 120º 140º 160º 180º 2sin 2x 0 1.97 0.68 0.68 1.73 1.29 0.00 3cos (x + 45º) 2.12 1.27 0.78 2.46 2.72 2.12  Use the data to draw the graphs of y = 2 sin 2x and y = 3 cos (x + 45º) for 0º ≤ x ≤ 180º on the same axes. (4marks)
 State the amplitude and period of each curve. (2marks)
 Use the graph to solve the equation 2 sin 2x – 3cos (x + 45º) = 0 for 0º ≤ x ≤ 180º (2marks)
 Complete the table below. (2mks)
MARKING SCHEME
No. Log 24.36
0.066547
1.48
10^{1 }x 9.045
0.90451.3867
2.8231
0.2098
0.1703
X 2
0.3406
0.2098
0.3406
1.8692 x ^{1}/_{3}
^{3}/_{3} ̅+2.8692
3
=1.9564 x(x + 7) + 12 = 0
x^2 + 7x + 12 = 0
x^2 + 3x + 4x + 12 = 0
x(x + 3) + 4(x+3) = 0
(x + 3 ) (x + 4) = 0
x = 3 or 4 M1  X R = 4.8 × 5 =4
6
QT^{2} = PT × RT
QT^{2 }= 18 × 8
QT = √144
QT = 12cm X 2 1 0 1 2 3 4 y 5 2 1 2 5 10 17
= 31 sq. Units x = √ tp
2m+p
x^{2} = √ tp
2m+p
2mx^{2}+px^{2}=tp
2mx^{2}= tppx^{2}
2mx^{2}= p(tx^{2})
p = 2mx^{2} t  x^{2} 

 A [5 + 3 , 5 + 1]
2 2
A (1, 2)  (x  a)^{2} + (y  b)^{2} = r^{2}
(5  1)^{2}+ (5  2)^{2} = r^{2}
4^{2}+ 3^{2} = 52
radius 5 units
(x  1)^{2} + (y  2)^{2}= 52
x^{2}  2x + 1 + y^{2}  4y + 4 = 25
x^{2}  2x + y^{2}  4y  20 = 0
 A [5 + 3 , 5 + 1]
 Determinmant = 2 – 12 = –10
A.S.F= 10 = 10
A=10×12.5 = 125 cm^{2} 
 ½  ^{1}/_{5} = ^{3}/_{10}
Required time ^{10}/_{3}
3^{1}/_{3} or 3 hrs 20 min  ½  ^{1}/_{5}  ^{1}/_{6} = ^{4}/_{30}
Required time 30/4
7½ hrs
 ½  ^{1}/_{5} = ^{3}/_{10}

 1^{5}  5(3x) + 1^{3} x 10 (3x)^{2}  12 x 10 (3x)^{3}
1  15x + 90x^{2}  270x^{3}  (0.97)^{5} = (1  0.03)^{5}
3x = 0.03
x = 0.01
(0.97)^{5} = 1  15(0.01) + 90(0.01)^{2}  270(0.01)^{3}
= 0.8587
 1^{5}  5(3x) + 1^{3} x 10 (3x)^{2}  12 x 10 (3x)^{3}
 cos 4x = ½
cos^{1}  ½ = 60º
x = 30º, 127.5º, 150º  P = 300,000  75000= 225,000
A = 225,000 × 1.15^{1.25}
15
=225000 × 1.190 = 267950
15 15
= Ksh.17863.38  ^{dy}/_{dx} = 3x^{2}8x+2
y = x^{3}4x^{2}+2x+c
At x =2 y=2
 2 = 816+4+c
C=2
y = x^{3} 4x^{2} + 2x+2  (√5 (2√2+√(5)+√2 (2√2√(5)
(2√2√(5) (2√2+√(5)
2√10+5+4√10
85
9+√10
3  x + y = 24
x^{2} + y^{2} = 144
x^{2} – (24 –x )2 = 144
x^{2} – [576 48x + x^{2}] = 144
x^{2} 576 + 48x – x^{2} = 144
48x = 720
x =15
y = 24 15
=9
The two numbers are 9 and 15  37,37,39,40,40,41,43,44,44
Q1 = 37+39 =38
2
Q3 = 43+44=43.5
2
Interquartile range = (43.538)
= 5.5 
 taxable income
42000+1300042000 x 10/100 = 42000+130003150
= sh. 51850  9860 x ^{10}/_{100} = 986
9860 x ^{15}/_{100} = 1479
9860 x ^{20}/_{100}= 1972
9860 x ^{25}/_{100}= 2465
9860 x ^{30}/_{100} = 2958
2550 x ^{35}/_{100} = 892.50
=10752.50
Total less relief 1062
sh.8352.50 pm  Net salary = 518508352.50
sh.43497.50
 taxable income

 Sin < CDE = ^{4.5}/_{8}
<CDE = 34.23º 
 AC = √(10^{2}+ 8^{2})
= √164 = 12.81  DE = 8^{2 } 4.5^{2} = 6.61 cm
AE = √(6.61^{2}+ 10^{2}) = √143.75
= 11.99
Tan <CAE = 4.5/11.99
<CAE = 20.57º
 AC = √(10^{2}+ 8^{2})

 MB = √(10^{2 }+ 4^{2} ) = √116
= 10.77 cm  Sin <CBM = ^{4}/_{10.77}
<CBM = 21.8º
 MB = √(10^{2 }+ 4^{2} ) = √116
 Sin < CDE = ^{4.5}/_{8}


longitudinal difference=25×2=50º
distance=60×50
=3000 nm t=^{3000}/_{200}
=15 hr

 Lat.diff= 65  15 = 500
PT= ^{50}/_{360} ×2π× 6370 Cos65
= 2350 km  t = d/s = ^{2350}/_{470}= 5hr
time difference btn longitude=50 × 4 min = 200min
= 3hr 20min
1.30 pm+3hr 20 min+5h
9.50pm
 Lat.diff= 65  15 = 500

 ^{ }

 ^{ }^{2}/_{3} x ^{2}/_{5} + ¼ x ^{3}/_{10}^{ 1}/_{12} x ^{3}/_{20}^{4}/_{15} + ^{3}/_{40} + ^{1}/_{80}
= ^{17}/_{48}  P(BL) or P(OL)
^{2}/_{3} x ^{2}/_{5} +^{ 1}/_{12} x ^{3}/_{20}^{4}/_{15} + ^{1}/_{80}
^{67}/_{240}  P(Not late to school) = 1 – P(Late to school)
= 1  ^{17}/_{48 }= ^{31}/_{48}


 Let the carrots be x, potatoes y and the total profit be p.
The inequalities that represents this information are:
x + y = ≤ 50
40x + 60y ≤ 2400
x ≥ 0. and y ≥ 0
maximum profit
P= 30x + 40y = 30( 30) + 40 ( 20)
= sh 1700
 Let the carrots be x, potatoes y and the total profit be p.

 a + d = 8
a + 4d = 17
3d = a
d = 3
a = 5  2nd = 8
10th = 5 + 9 x 3 = 32
42nd = 5 + 41 x 3 = 128
GP is 8, 32, 128,    
a = 8, r = 4
nth term of G.P = arn – 1
10th term = 8(4)^{9}
= 2097152  =  ^{8}/_{3} x 1048575
= 2796200
 a + d = 8


 QR = QP + PR
=  b + c  PM = PQ + QM
= b + ^{1}/_{3} (b+c)
= ^{2}/_{3} b +^{1}/_{3} c
RL = RQ + QL
= b – c – ¾b
= ¼ bc
 QR = QP + PR
 PX = h (^{2}/_{3} b+ ^{1}/_{3} c)
^{2}/_{3} hb+ ^{1}/_{3} hc …….. (i)
Also PX = PR + RX
= c + k (¼b – c)
= (1 – k) c +¼kb……………(ii)
Equating equations (i) to (ii) and comparing
coefficients of vectors b and c,
K = ^{8}/_{3} h and ^{1}/_{3} h=1k
Solving the above eqtns simultaneously
^{1}/_{3} h=1  ^{8}/_{3}h
h =^{1}/_{3}
h = ^{8}/_{3 }×^{1}/_{3} = ^{8}/_{9}
RX = ^{8}/_{9}
LX:XR = 1:8


X 0º 20º 40º 60º 80º 100º 120º 140º 160º 180º 2sin 2x 0 1.28 1.97 1.73 0.68 0.68 1.73 1.97 1.29 0.00 3cos (x + 45º) 2.12 1.27 0.17 0.78 1.72 2.46 2.90 2.99 2.72 2.12  y = 2 sin 2x
Amplitude = 2
Period = 180º  y = 3 cos(x + 45º)
Amplitude = 3
Period = 360º
2 sin 2x – 3 cos ( x + 45º) = 0
X = 20º
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