Mathematics Paper 2 Questions and Answers - KCSE 2022 Prediction

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INSTRUCTIONS TO CANDIDATES

This paper consists of two sections; Section I and Section II.
Answer all questions in section I and only five questions from section II.
Show all the steps in your calculations, giving the 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.
Candidates should answer the questions in English.

Questions

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

The roots of a quadratic equation are x = 1¹/₃ and x = −½. Form the quadratic equation in the form where ax² + bx + c = 0 where a,b and c are integers. (2 marks)
Find the percentage error in estimating the volume of a cuboid whose dimensions are 8 cm by 4.20 cm by 5.6 cm. (3 marks)
The cash price of a television set is sh. 25 000. A customer paid a deposit of sh. 3 750 . He paid the amount owing in 24 equal monthly installments. If he was charged simple interest at the rate of 40% p.a How much was ;
1. The carrying charge? (2 marks)
2. Each installment? (1 mark)
Without using mathematical tables or a calculator, evaluate 3 log₄(√3) − ½ log₄3 + 3log₄2 − log₄6 (3 marks)
Given that 2^−3n²(a + x)⁻ⁿ = 20^−3n² , make n the subject of the formula. (3 marks)
In the figure below, A, B, C, D and E are points on a circle. Point C is due north of point D and point E is due west of point D. Angle CAB = 27º. The angle of elevation of point B from point E is 87º.

Find the size of angle of elevation of point B from the point D. (3 marks)
Given that x = 2 − √3, find the value of (x − ¹/_x)³ (3 marks)
Expand (x − y)⁴. Hence use the expression to find the value of 1.58⁴ , giving your answer correct to 4 significant figures. (3 marks)
The diagram below shows a wedge ABCDEF.

Calculate the angle between the lines BD and BF, correct to 1 decimal place. (4 marks)
State the amplitude and the phase shift of the curve y = 2 Cos(2x+ 90°) (2 marks)
The tax authority collected the following amount of tax from different firms in a particular market.

Amount of Taxes (Ksh) 10 000 11 000 12 000 13 000 14 000

No. of firms 3 12 18 12 3

Determine the quartile deviation of the data. (3 marks)
The position of towns P(0º,51ºW) and Q(0º,37ºE).Find the distance between the two towns in kilometres, correct to one decimal place. (3 marks)
A trader bought three kinds of nuts at Ksh. 10 per kg, Ksh. 8.40 per kg and Sh. 6 per kg. He mixed them in the ratio 3: 5: 4 respectively and sold the mixed nuts making a profit of 25% in the process. At what price per kg did he sell them? (3 marks)
In the figure below, AB and CB are tangents to the circle, centre O and radius 6.5 cm. The arc AC subtends an angle of 140º at the centre of the circle.

Calculate the area of the shaded region, correct to 2 decimal places (Use π = 3.142) (4 marks)
A square QRST with vertices Q(1,1),R(3,1),S(3,3) and T(1,3) is transformed by the matrix to to Q'R'S'T' Find:
1. the are of square QRST; (2 marks)
2. the area of image Q'R'S'T' (2 marks)
Find the radius and the centre of a circle whose equation of a circle is x² + y² − 2x + 8y − 8 = 0, hence draw the circle on the grid provided below. (4 marks)

SECTION II (50 marks)
Answer any five questions from this section in the spaces provided.

1. Complete the table below for values of y = x³ − 7x² + 8x + 16 for − 2 ≤ x ≤ 6 (2 marks)
  
  x − 2 − 1 0 1 2 3 4 5 6
  
  y = x³ − 7x² + 8x + 16 − 36 18 6
2. On the grid provided, draw the graph of the function y = x³ + 7x² − 7x − 15 for − 2 ≤ x ≤ 6
  Use a scale of 2 cm to represent 1 unit on the x – axis and 1 cm represent 5 units on the y – axis. (3 marks)
3. Using the graph, determine:
  1. the average rate of change between x = 1.2 and x = 3.1 (2 marks)
  2. The instantaneous rate of change of the curve at x = − 0.5 , correct to one decimal place. (3 marks)
1. An increasing arithmetic progression (AP) has first term as a and a common difference d. The ratio of the fifth term to the twelfth term of the A.P is 6:13.Given that product of first term a and the third term is 32. Find the;
  1. Value of a and the value of d. (4 marks)
  2. Sum of the first five terms of the A.P. (2 marks)
2. Determine the least number of terms that must be added together so that the sum of the progression is more than 700. (4 marks)

The table below shows some values of the curves y = 2 cos(½x) and y = sin(½x)

Complete the table for values of y = 2 cos(½x) and y = sin(½x), correct to 1 decimal place. (2 marks)

x°	−360	−270	−180	−90	0	90	180	270	360
y = sin(½x)	0				0				0
y = 2cos(½x)	−2.0		0				0		−2.0

On the grid provided, draw the graph of y = 2 cos(½x) and y = sin(½x) for −360° ≤ x ≤ 360° on the same axes. (5 marks)
Use your graph in (b) above to find the values of;
1. x in 2cos(½x) − sin(½x) = 0 (2 marks)
2. y when 2cos(½x) = sin(½x) (1 mark)

The runway in Wilson airport is 3.22 km long and it is wide enough to accommodate the landing of two aircrafts simultaneously, An aircraft x landed at the runway with a speed of 144 km/hr and maintained this speed till the end of the runway when it took off again, Another aircraft y landed at runway half a minute later with a speed of 252 km/hr and followed x on the parallel runway throughout to the end of the runway when it also took off again.
1. How far from starting point was x at the instant y overtook x. (4 marks)
2. Find the distance between the two planes at the instant when y is beginning to take off.(3 marks)
3. Selina took a train from her home to the Nairobi. At a railway station she met another train travelling in the opposite direction at 25 km/hr. If the train she boarded was travelling at a speed of 15 km/hr and it took 10.8 seconds to completely pass the other train. Calculate the length of each train, if the train Selina boarded was twice as long as the train moving in the opposite direction. (3 marks)
The charge C in shillings per person for a certain seminar is partly fixed and partly inversely proportional to the total number of N of people.
1. Write down an expression for C in terms of N. (2 marks)
2. When 100 people attended the charge was Ksh 8,700 per person. While for 35 people charge is Ksh 10,000 per person. Calculate the fixed charges and hence state the equation connecting charge C. (4 marks)
3. If a person had paid the full amount and doesn’t attend, the fixed charge is refunded. A group of people paid but 10% of them did not attend. After the refund the organizers remained with Ksh574,000.Find the number of people initially in the group. (4 marks)
In certain game, ²/₃ of the players are boys while the rest are girls. 80% of the boys and 90% of the girls are right handed. The probability that a right handed player will break the rules of the game is 0.1.Otherwise it is 0.3, these probabilities being independent of the sex of the player.
1. Draw a tree diagram to represent this information. (2 marks)
2. Use your tree diagram in (a) above determine;
  1. The probability that a player chosen at random from the game is left handed. (2 marks)
  2. The probability that a game rule is broken is a session. (2 marks)
  3. The probability of a girl breaking a rule of the game. (2 marks)
  4. The probability that a boy who is right handed doesn’t break the rules and a girl who is left handed break rules of the game. (2 marks)
In the diagram drawn below, point D and E are such that AD = ¹/₃OB and OE = ¹/₃OA
1. Given that vectors OA = a and OB = b, express in terms of a and b:
  1. OD (1 mark)
  2. BE (1 mark)
  3. BD (2 marks)
2. Line OD and BE intersect at X such that OX = kOD and BX = hBE and where k and h , express OX in two different ways (2 marks)
3. Using the expressions in (b) above, determine.
  1. the values of k and h. (3 marks)
  2. Hence state the ratio in which k divides OD. (1 mark)
A plan for a national stadium is in the form a pentagon ABCDE inscribed in a circle, a hockey pitch is to be constructed outside the stadium. The centre of the pitch is fixed at a point X. Given that the angle DAC = 36º, angle ACB = 42° AE = ED, BC = AB and that the tangent at C meets ED produced at X.
1. Calculate the angles of the pentagon. (4 marks)
2. Giving reasons, calculate ∠DXC. (3 marks)
3. Given that DX = 20m, calculate the area of triangle CDX. (3 marks)

Marking Scheme

(x − ⁴/₃)(x + ½) = 0
x² − ⁵/₆x− ²/₃ = 0
6x² − 5x − 4 = 0
Actual volume = (8 × 4.20 × 5.6) = 188.16cm³
Maximum volume = (8.5 × 4.205 × 5.65) = 201.945125cm³
Minimu volume = (7.5 × 4.195 × 5.55) = 174.616875cm³
Absolute error = 201.945125 − 1740616875
2
= 13.664125
Percentage error = (13.664125) × 100%
188.16
= 7.262%
Alternative method
(0.5 + 0.005 + 0.005) × 100%
8 4.20 5.6
= 7.261906%
1. Sh.(25 000 – 3730) = Sh. 21 250
  Carrying charge = (21250 × ⁴⁰/₁₀₀ × 2)
  = Sh. 17 000
2. 21250 + 17000 = Sh. 1593.75
  24
3 log₄(√3) − ½ log₄3 + 3log₄2 − log₄6

= log₄2³
Let m = log₄2³ ⇒ 2^2m = 2³
2m = 3
m = 1½
∴ log₄2³ = 1½
2^−3n²(a + x)⁻ⁿ = 20^−3n²
(a + x)⁻ⁿ = (10³ⁿ)⁻ⁿ
a + x = 10³ⁿ
log(a + x) = 3n log 10
n = ¹/₃ log(a + x)
The required angle is BDE
Now ∠EBC = 90° angles in a semicircle in triangle EBC
∠BEC = 27°, hence;
∠ECB = 90 − 27 = 63°(angles in a traingle)
But ∠BDE = ∠BCE, angles in the same segment.
Hence ∠BDE = 63°
1 = 1 × 2 + √3 = 2 + √3 = 2 + √3
x 2 − √3 2 + √3 1
x − ¹/_x = (2 − √3) − (2 + √3) = −2√3
(x − ¹/_x)³ = (−2√3)³
=(−2)³ (√3)²
= −24√3
(x  − y)⁴ = x⁴ + 4x³ ( −y) + 6x²  − y² + 4x( −y)³ + ( −y)⁴
(1.58)⁴ = (2  − 0.42)²
2⁴  − 4(2)² (0.42) + 6(2)²  − 4(2)(0.42)³ + (0.42)⁴
= 16  − 13.44 + 4.2336  − 0.592704 + 0.013069123
= 6.213965123
≅ 6.214
BF = √(4² + 2²)
= 2√5
DE = √(5² + 2²)
= √29
Cos β = 41 + 20 − 29
2 × √41 × √20
= 56.0°
y = 2Cos(2x + 90°)
Amplitude = 2
Phase shift = − 180° = 45°(left of 45°)
4
Q₁ = ⁴⁸/₄ = 12^th item = 11,000
Q₃ = ¾ × 48 = 36^th item
= 13,000
Q.D = 13,000 − 11,000
2
= 1,000
Latitude difference = 51° + 37° = 88°
Distance in kilometres; ⁸⁸/₃₆₀ × ²²/₇ × 2 × 6370
= 9787.6km
1. Side of QRST = √((1 − 1)² + (3 − 1)²)
  = 2
  ∴ Area = 2 × 2 = 4
2. The area of image Q'R'S'T'
  = det × 4
  = 6 × 4
  =24sq
AB = 6.5 tan 70°
= 17.86cm
Area of quadrilateral OABC = 2(½ × 6.5² tan 70°)
= 116.08cm²
Area of sector OAC = ¹⁴⁰/₃₆₀ × 3.142 × 6.5²
= 51.62cm²
Shaded area = 116.08 − 51.62
= 64.47cm²
C fraction in 1 minute = ¹/₃₀ − ¹/₄₀ = ¹/₁₂₀
C takes 120 minutes
B fraction in 1 minute = ¹/₃₀ − ¹/₄₅ = ¹/₉₀
B takes 90 minutes
A fraction in 1 minute = ¹/₃₀ − ¹/₉₀ = ⁵/₃₆₀
A takes = ³⁶⁰/₅ = 72 minutes
x² − 2x + ( ⁻²/₂) + y² + 8y + (⁸/₂)² = 8 + 1 + 16
(x − 1)² + (y + 4)² = 5²
Centre (1, −4)
Radius 5 units
1. x − 2 − 1 0 1 2 3 4 5 6
  
  y = x³ − 7x² + 8x + 16 − 36 0 16 18 12 4 0 6 28
3. Average rate of change;
  = 3 − 17
  3.1−1.2
  = − 7.37
4. Tangent at x = −0.5
  Gradient = −6 − 30
  −1.5 − 0.8
  = 15.7
1. 1. a + 4d = 6 ⇒ a = 2d .........(i)
    a + 11d 13
    a(a + 2d) = 32 ...................(ii)
    2d(2d + 2d) = 32
    8d² = 32
    d² = 4
    d = ±2
    d = 2
    a = 2d
    = 2 × 2
    = 4
  2. S₅ = ⁵/₂{2 × 4 + (5 − 1)2}
    = ⁵/₂(8 + 8)
    =40
2. ⁿ/₂ { 2 × 4 + (n − 1)2} > 700
  ⁿ/₂ {6 + 2n} > 700
  2n² + 6n − 1400 > 0
  n² + 3n − 700 > 0
  n² + 28n − 25n − 700 > 0
  n(n + 28) − 25(n + 28) > 0
  n − 25) (n + 28) > 0
  ∴ n − 25 > 0
  n > 25
  n = 25 + 1
  = 26

x°	−360	−270	−180	−90	0	90	180	270	360
y = sin(½x)	0	−0.7	− 1.0	− 0.7	0	−0.7	1.0	0.7	0
y = 2cos(½x)	−2.0	−1.4	0	1.4	2.0	1.4	0	− 1.4	−2.0

1. −234° ±1° and 126°±1°
2. values of y
  y₁ = − 0.9
  y₂ = 0.9

1. Distance in 1 minute 144 × 30 = 1.2km
  3600
  D_252km/h = 1.2 + x D_144km/h = x
  S = 252 km/h S = 144km/h
  T = 1.2 + x T = x
  252 144
  1.2 + x = x
  252 144
  1.75x − x = 1.2
  0.75x = 1.2
  x = 1.6 km
  D_252km/h = 1.2 + x
  = 1.2 + 1.6
  = 2.8 km
  Alternatively;
  Relative speed = 252 − 144
  = 108 km/h
  T = 1.2 = 1 hrs
  10.8 90
  D = ¹/₉₀ × 144
  = 2.8km
2. Remaining distance = 3.22 – 2.8 = 0.42 km
  Distance = 144 × 0.42 = 0.24km
  252
  (0.42 − 0.24) = 0.18km
3. Distance = 2x + x = 3x
  RS = (25 + 15) = 40km/h
  3x = 10.83
  40 3600
  x = 40 × 10.8
  3 × 3600
  = 0.04km = 40m
  Train 1 = 40m
  Train 2; 2 × 40 = 80m
1. C = k + ^m/_N
2. 8,700 = k + ^m/₁₀₀
  10,000 = k + ^m/₃₅8700 − ^m/₁₀₀ = 10000 − ^m/₃₅
  100m − 35m = 4,550,000
  m = 4550000
  65
  = 70,000
  k = 85,000 − ^m/₁₀₀
  = 85,000 − 70000
  100
  = 8000
  C = 8000 + 70000
  N
3. Let the number of people be x; P = ⁹⁰/₁₀₀x = 0.9x
  Amount per person = 57400
  0.9x
  57400 = 800 + 70000
  0.9x 0.9x
  ¹/_0.9x(574000 − 70000) = 8000
  x = 504000
  0.9 × 8000
  = 70
2. 1. P(BL) or P(GL)
    P(BL) or P(GL) = (²/₃ × ²/₁₀) + (¹/₃ × ¹/₁₀)
    = ¹/₆
  2. P(BRB) or P(BLB) or P(GRB) or P(GLB)
    = (²/₃ × ⁸/₁₀ × ¹/₁₀) + (²/₃ × ²/₁₀ × ³/₁₀) + (¹/₃ × ⁹/₁₀ × ¹/₉) + (¹/₃ × ¹/₁₀ × ³/₁₀)
    = ⁴/₇₅ + ¹/₂₅ + ¹/₃₀ + ¹/₁₀₀
    = ⁴¹/₃₀₀
  3. P(GRB) or P(GLB)
    P(GRB) or P(GLB) = (¹/₃ × ⁹/₁₀ × ¹/₁₀) + (¹/₃ × ¹/₁₀ × ³/₁₀)
    = ⁹/₃₀₀ + ¹/₁₀₀
    = ¹/₂₅
  4. P(BRB') or P(GLB')
    P(BRB') or P(GLB') = (²/₃ × ⁸/₁₀ × ⁹/₁₀) + (¹/₃ × ¹/₁₀ × ³/₁₀)
    = ¹⁴⁴/₃₀₀ × ³/₃₀₀
    = ³/₆₂₅
1. 1. OD = a + ¹/₃b
  2. BE = −b + ¹/₃a = ¹/₃a − b
  3. BD = BO + OA + AD
    = −b + a + ¹/₃b
    = a − ²/₃b
2. OX = k(a + ¹/₃b) = ka + ¹/₃kb
  OX = OB + hBE
  = b + h(¹/₃a − b) = b(1 − h)+¹/₃ha
3. 1. ka = ¹/₃kb = b(1 − h) + ¹/₃ha
    k = ¹/₃h
    ¹/₃k = 1 − h ⇒ k = 3(1 − h)
    ¹/₃h = 3 − 3h
    3¹/₃h = 3
    h = ⁹/₁₀
    k = ¹/₃ × ⁹/₁₀
    = ³/₁₀
  2. OX:XD = 3:7
1. ∠ABC = 180° − 84° = 96°
  ∠ACD = 180° − (84 + 36)° = 60°
  ∠DEA = 180° − 60° = 120°
  ∠EDA = ∠EAD = 180 − 120 = 30°
  2
  ∠BCD = ∠ACD + ∠ABC
  = 60+ 42 = 102°
  ∠CDE = ∠CDA + ∠EDA
  = 84 + 30
  =114°
  ∠EDA = 120°
  ∠EAB = ∠EAD + ∠DAC ∠ CAB
  = 30 + 36 + 42
  = 108°
2. ∠CAB = ∠BCY = 42° Alternative angles are equal
  ∠XDC = 180 − (30 + 84) = 66°, Angles on straight line are supplementary/ add up to 180°
  ∠DXC = 180 − (66 + 36) = 78°, sum of angles in a triangle add up to 180°
3. DC = sin 78 ( 28 ) = 33.28m
  sin 36
  A = ½ × 33.28 × 20 sin 66°
  = 305.05m²