Mathematics Paper 2 Questions and Answers - Lainaku II Joint Mock Examination 2023

INSTRUCTIONS TO CANDIDATES

• This paper contains TWO sections: section I and section II
• Answer ALL the questions in Section I and only 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.

SECTION I (50 MARKS)
Answer ALL the questions in this section.

1. Without using logarithms tables or calculator: Evaluate. (3mks)
   Log₁₀96 + 34 Log₁₀625 − Log₁₀12
2. Find the value of p if the expression px^2 − 60x + 25 is a perfect square, given that p is a constant. (2mks)
3. The price of a new car is shs. 800,000. If it depreciates at a constant rate to shs. 550,000 within 4 years, find the annual rate of depreciation to 2d.p. (3mks)
4. Mwangi truncated ⁷/₉ to 3 decimal places. Calculate th e percentage error resulting from the truncating. (3mks)
5. The position vector of A and B are D is a point on AB such that AD:DB is 2:1. Find the co-ordinates of D. (3mks)
6. A quantity y varies partly as x² and partly as x. When y = 6, x = 1 when y = 30, x = 3. Find y when x = − 3.  (3mks)
7. The coordinates of the end points of diameter are A(2,4) and B(2,6). Find the equation of a circle in the form ax² + by² +cx + dy + e = 0 (3mks)
8. A bag M contains 8 balls of which 3 are red and 5 are white. Another bag N contains 7 balls of which 4 are red and 3 are white. A bag is chosen at random and two balls picked at random without replacement. Using a tree diagram, Find the probability that the two balls chosen are red in colour. (4mks)
9. Find the equation of tangent to the curve below at the indicated point y = x³ + 3x² − 3 at x = 2 giving your answer in form of y = mx + c (3mks)
10. Solve the equation 8 sin²x = 7 − 2cos x for 0°≤ x ≤360° (4mks)
11. Chord WX and YZ intersect externally at Q. The secant WQ= 11cm and QX= 6 cm while ZQ= 4cm.
                                               
    1. Calculate the length of chord YZ. (2mks)
    2. Using the answer in (a) above, find the length of the tangent SQ to 2 d.p. (2mks)
12. Find the possible values of x given that x-1 x+1 3x x is a singular matrix. (3mks)
13. Determine the semi interquartile range for the following set of numbers. (3mks)
    4, 9, 5, 4, 7, 6, 2, 1, 6, 7, 8.
14. A carrot patch is in the shape of trapezium. There are 13 carrot plants in the first row, 15 in the second row and 17 in the next and so on. If there are 47 plants in the last row, how many rows are there and how many plants. (3mks)
15. Two brands of coffee cost Kshs 120 and Kshs 144 per kilogram respectively. A wholesaler blends the two and sells the blend at Kshs 168 thereby making a profit of 20%. Find the ratio in which he blends them. (3mks)
16. Make x the subject of the formula (3mks)
                             

SECTION II (50 MARKS)
Answer any FIVE questions in this section.

17.  
    1. Complete the table below, giving the values correct to 2 decimal places (2mks)
       

       X°	0°	15°	30°	45°	60°	75°	90°	105°	120°	135°	150°	165°	180°
       Cos 2x°	1.00		0.50	0.00	−0.50		−1.00	−0.87		0.00		0.87	1.00
       Sin(x°+30°)	0.50		0.87	0.97	1.00		0.87	0.71	0.50	0.26	0.00		−0.50

    2. Using the grid provided draw on the same axes the graph of y=Cos 2x0 and y = Sin (xo+300) for 00≤x≤1800 (5mks)
    3. Find the period of the curve y = Cos 2x° (1mk)
    4. Using the graph, estimate the solutions to the following questions to the nearest degree;
       1. Sin(x° + 30°) − Cos 2x°= 0 (1mks)
       2. Cos 2x° = 0.50 (1mk)
18. The figure below shows points on the earth’s surface.
                                          
    1. State the positions of A, B, C and D in coordinate form. (2mks)
    2. An aircraft flies from A to B along latitude 40°N, B to C along longitude 30°E, C to D along latitude 40°S and then flew back to A along longitude 300W. Calculate to 2 d.p the total distance it covered .(Take radius of the earth = 6370 km and π = ²²/₇) (5mks)
    3. If the aircraft leaves A at 8.00am at a speed of 720km/h to B. At what local time is it expected at B? Give your answer to the nearest minute. (3mks)
19.  
    1. AB and CD are chords of a circle. Construct the circle with centre O and measure its radius (4mks)
                                                 
    2. Construct the loci of a points x which are equidistant from line AB and CD (1mk)
    3. Construct the loci of a points Z which are 2cm from the circumference of the circle. (2mks)
    4. A point P moves such that CP DP, It is not more than 2cm from the circumference of the circle and its distance from line CD is not more than its distance from AB. Show the region P by shading it. (3mks)
20. The diagram below represents a cuboid ABCDEFGH in which FG = 6 cm, GH = 12 cm and HC = 8 cm
                                               
    Calculate to 2 decimal places;
    1. The length of FC (2mks)
    2.  
       1. The size of the angle between the lines FC and FH (2mks)
       2. The size of the angle between the lines AB and FH (2mks)
    3. The size of the angle between the planes ABHE and the plane FGHE (2mks)
    4. The space occupied by the Cuboid. (2mks)
21. The points A (1,4), B(-2,0) and C (4,−2) of a triangle are mapped onto A¹(7,4), B¹(x,y) and C¹ (10,16) by a transformation N = a b c d . Find
    1. Matrix N of the transformation (4mks)
    2. Coordinates of B¹ (2mks)
    3. A^IIB^IIC^II are the image of A¹B¹C¹ under transformation represented by matrix M = (2 −1 1 0) Write down the co-ordinates of A^IIB^IIC^II (2mks)
    4. A transformation N followed by M can be represented by a single transformation K.
       Determine K (2mks)
22. A particle moves along a straight line such that its displacement S metres, from a given point is
    S = t³ – 5t² + 3t + 4. Where t is time in seconds find;
    1. The displacement of the particle at t = 5 (2mks)
    2. The velocity of the particle when t =5 (2mks)
    3. The values of t when the particle is momentarily at rest. (3mks)
    4. The acceleration of the particle when t = 2. (3mks)
23. In the figure below ,O is the center of the circle. PQR is a tangent to the circle at Q. Angle PQS = 28°,angle UTQ = 54° and UT = TQ.
                                                         
    Giving reasons, determine the size of
    1. Angle STQ. (2mks)
    2. Angle TQU. (2mks)
    3. Angle TQS (2mks)
    4. Reflex angle UOQ . (2mks)
    5. Angle TQR. (2mks)
24. A company wishes to buy two types of squash machines; Electric and manual. A manual machine requires four attendants whereas an electric one also requires four. An electric machine fills 300 packets per hour; a manual one can fill 200 packets per hour. The numbers of packets to be filled are not more than 3000 per hour and the number of attendants should be at least 40. By letting x be the number of electric Machines and y be the number of manual Machines answer the following questions.
    1. Form all the possible inequalities which will represent the above information. (4mks)
    2. On the grid provided draw the inequalities and shade the unwanted region. (3mks)
    3. If for every hour it is used, an electric machine brings a profit of shs.200 and a manual one shs.500, determine the number of machines of each type that should be installed in order to maximize profit per hour. Find the maximum profit (3mks)

MARKING SCHEME.

S/NO	SOLUTION	MARKS	COMMENT
1.	Log1096 + Log1062534 − Log1012   6253/4 = 125   Log10{96x125} = Log101000                12   Log101000 = 3	M1  M1  A1	Working 62534 = 125  For Log10 {96x125}                       12  For answer = 3
		3
2.	If ax2 + bx + c then ac = (b/2)2   25p = (60/2)2   25p = 900    p = 36	M1  A1	For 25 p = (60/2)2
		2
3.	550,000 = 800,000(1− r/100)4   550,000 = (1 − r/100)4   4√0.6875 = (1 − r/100)   0.91058 = 1− r/100   r/100 = 1 − 0.91058   r = 0.08942x100   r = 8.94%	M1  M1  A1	For correct substitution.  For 4√0.6875 = 0.91058  For 8.94% in 2 d.p.
		3
4.	79=0.777777777777   Truncated value = 0.777       = 0.1 %	M1  M1  A1	For truncated value  For acual value – truncated x100                actual value
		3
5.	Coordinate of D(8, 4, 6)	M1  M1  A1	For ratio theorem  For correct working.  Penalize if brackets missing.
		3
6.	Y ᾳ x2+x   Y =a x2+ bx   6 = a+b   30 = 9a+3b   a = 2 and b=4   equation of Y = 2x2 + 4x   When x = −3   Y = 2(−3)2 + 4(−3)= 18−12 = 6	M1  M1  A1	For the two equations  For the equation  Y =2x2 + 4x  For y = 6
		3
7.	r = √02+12   r = 1  Equation  (x − 2)2 +(y − 5)2 = 12  X2 + y2 − 4x −10y + 28 = 0	M1  M1  A1	For centre (2,5)  For radius = 1
		3
8.	3/56 + 1/7 = 11/56	B1    B1          M1  A1	For inclusion of 12 on tree diagram  For the rest of the tree diagram.  For method
		4
9.	Gradient= dydx = 3x2 + 6x   M = 3(4) + 6(2)2   M = 12 + 12= 24   Y = 23 + 3(22) −3 = 17   Equation at (2, 17)   y − 17 = 24   x − 2        1   Working out   Y = 24x − 31	M1  M1  A1	For M = 24  For y = 17  For Y = 24x − 31
		3
10.	8(1−cos2x) + 2cos x −7 = 0   8 − 8cos2x + 2cos x − 7 = 0   8cos2x − 2cos x − 1 = 0   Letting cos x = y   8y2 − 2y−1 = 0   Solving;   8y2 − 4y + 2y − 1 = 0   4y(2y −1) + 1(2y −1) = 0   (4y + 1) (2y −1) =0   Y= −¼ or y= ½   Cos x = −¼ or Cos x = ½   Therefore   For Cos x = −¼ x = 104.48° or 255.52°   For Cos x = ½ x = 60° or 300°	M1      M1        A1  A1	For 8cos2x − 2cos x −1 = 0  For solving quadratic equation to get  cos x = −¼or Cos x = ½  For x = 104.48° and 255.52°  For x = 60° and 300°
		4
11	QX.QW = QZ.QY 6 x 11 = 4X (4 + YZ) 66 = 16 + 4YZ YZ = 12.5   SQ2 = QZ x QY SQ2 = 4 x 16.5 SQ2 = 66 SQ = √66 SQ = 8.12 cm	M1  A1  M1  A1	For correct substitution  For correct substitution  Answer must be in 2 d.p
		4
12.	For Determinant X(x − 1)−3x(x + 1)=0           X2 − x −3x2 −3x = 0          −2x2 − 4x = 0          −2x(x + 2) = 0          −2x = 0 or x + 2 = 0           X = 0 or x = −2	M1  M1  A1	For determinant = 0  Solving for x
		3
13.	Lower Quartile(Q1) = 4   Upper Quartile(Q3) = 7   Interquartile range 7−4 = 3                                    2      2     = 1.5	M1  M1  A1
		3
14.	13, 15, 17,…. Is an AP  13+(n −1)2 = 47  13 + 2n − 2 = 47  2n = 36  n = 18 ,rows = 18  S18 = 18/2(26 + 34)       = 9 x 60       = 540 i.e 540 plants	M1  A1  A1	Substituting in AP formula  For rows = 18.  For 540 plants.
		3
15.	Ratio  x   y   x + y   Cost  120  144     Total  120x  144y   120x + 144y   Marked price = 168 x 100 = 140                                 120  120x + 144y = 140         x + y   120x + 144y = 140x + 140y   4y = 20x   X:y = 1 : 5	M1      M1        A1	For marked price=140  Accept any method used
		3
16.	P2 = x + 2W          4x+3R  P2(4x + 3R) = x + 2W  4P2x – x = 2W − 3p2R  x(4P2 − 1)= 2W − 3P2R  x = 2W − 3P2R             4P2 − 1	M1  M1  A1	For p2 = x + 2W sq both sides.                 4x+3R
		3
17.	X°  0°  15°  30°  45°   60°  75°  90°   105°  120° 135° 150° 165° 180°  Cos 2x°  1.00  0.87   0.50  0.00  −0.50   −0.87   −1.00   −0.87   −0.50   0.00  0.50   0.87  1.00  Sin(x°+30°)  0.50  0.71  0.87   0.97   1.00   0.97  0.87  0.71  0.50  0.26  0.00  −0.26  −0.50 Marking points in bold on the table For all correct-2mks, For 4 to 6 correct 1 mk, For 3 and below 0 mk  Graph – distribution of the marks on the graph.               Period of curve y = cos 2x° 180°   Solution to Sin(x°+ 300) −Cos 2x° = 0 Sin(x° + 300) = Cos 2x° 21° ± 1° or 141° ± 1°   cos 2x° = 0.50 30° or 150°	B1  B1      B1	For 180°   For both 21°±1° and 141° ± 1° For both 30° or 150°
		10
18.	A(40°N, 30°W)  B (40°N, 30°E)  C (40°S, 30°E)  D(40°S, 30°W) AB = 60 x 2 x 22 x 6370cos 40° = 5112.069917         360          7 AB = 80 x 2 x 22 x 6370 = 8897.777777         360          7 CD = 60 x 2 x 22 x 6370cos 40°=  5112.069917          360         7 DA = 80 x 2 x 22 x 6370 = 8897.777777         360          7 Total = 28,019.70 Km   Time taken = 5112.069917 = 7hrs 06 min                                720 Local time at B Difference in longitude = 60° Time difference = 60° x 4 = 240 min = 4 hrs  Local time at B   8 : 00 am                            +4: 00                              12:00 noon Local time aircraft arrived at B  12:00 noon + 7:06    19:06 hrs or 7:06 P.M	B1 B1 M1 M1 M1 M1 A1 M1 M1 A1	For both A and B For both C and D For 28,019.70 Km  to 2 d.p.   For 7hrs 06 min For local time at B 12:00 noon For 19:06 hrs or 7:06 P.M
		10
19.
20.	(i) Tan Ɵ =  8                    13.4164     Tan Ɵ = 0.5962            Ɵ = 30.800 ii)Tan Ɵ = 12/6    Tan Ɵ = 2            Ɵ = Tan−2            Ɵ = 63.43° Tan Ɵ = 8/12            = Tan− 0.6667            = 33.69° Space occupied is volume Volume = 6cm x 12cm x 8cm              = 576 cm2	M1  A1        M1  A1        M1  A1    M1  A1       M1   A1	For FH  For FC      Tan Ɵ =        8         13.4164    For Tan Ɵ   = 12/6    Tan Ɵ =   8/12    For expression of volume
		10
21.	a + 4b = 7 4a − 2b = 10 Solving simultaneous equation gives; a = 3 and b = 1 c + 4d = 4 4c − 2d = 16 Solving simultaneous equation gives; c = 4 and d = 0 The matrix become N =     B1 (x,y) = (−6, −8)    A”(10,7) B”(-4,-6) C”(4,10)   N followed by M is MN	M1  M1  M1  A1    M1  A1    M1  A1    M1  A1	For expression to find the matrix For a=3 and b= 1 c=4 and d= 0 For matrix N. Correct and in bracket.   For working Correct and in bracket.  For correct working
22.	S = 53 − 5(52) + 3(5) + 4 =125 −125 + 15 + 4 = 19m   ds/dt = 3t2 − 10t +3 =3(5)2 − 10(5) +3 = 75 − 50 +3 = 28 m/s   3t2 −10t +3 = 0 Solving 3t2 − 9t − t + 3 = 0 3t(t − 3)−1(t − 3)=0 (3t − 1)(t − 3)=0 t = 1/3 s or t = 3s   Acceleration = dv/dt =6t-10 = 6(2) − 10 = 12 − 10 = 2m/s	M1  A1  M1  A1  M1  M1  A1  M1  M1  A1	Correct substitution   For correct gradient function   For velocity expression=0 For working out quadratic equation  For expression of acceleration. For correct substitution
		10
23.	Angle STQ = 28° Angles in alternate segment are equal   Angle TQU = 63° Base angles of isosceles triangle are equal   Angle TQS = 35° Angles in a triangle adds up to 180°   Reflex angle UOQ = 252° Angles at a point add up to 360°.   Angle TQR = 117° Angles in alternate segment are equal	B1  B1  B1  B1  B1  B1  B1  B1  B1  B1	For reason For reason     For reason     For reason     For reason
		10
24.	Inequalties 3x + 2y < 30 x + y ≥ 20 x > 0 (Attendants are non-negative) y > 0                       Objective function 200 x + 500y To Maximize (0,15) Electric = 0 Manual = 15 Machines Max Profit 200 (0) + 500 (15) = 7,500 = sh 7,500/-	B1  B1    B1  B1      B1      B1    B1  B1    B1  B1  B1	Correct drawing and shading of x > 0 Correct drawing and shading of y > 0 Correct drawing and shading of x + y ≥10 Correct drawing and shading of 3x + 2y < 30
		10