Mathematics Paper 2 - Form 3 End Term 3 Exams 2021 Questions with Answers

121/2
Mathematics
Paper 2
2 ½ Hours

INSTRUCTIONS TO CANDIDATES

• The paper contains two sections; section I and II.
• Answer all the questions in section I and any five questions from section II.
• Non- programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise.
• Mark may be given for correct working even if the answer is wrong.

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

1. Use logarithms in all steps to evaluate. (4mks)
   2.53² x 83.45
      √0.4562
2. By using completing square method, solve for x in 4x² – 3x – 6 = 0 (3mks)
3. Make p the subject in (3mks)
   
4. If 
   
   Find the value of a and b where a and b are rational numbers. (3mks)
5.  
   1. Find the first three terms in ascending powers of x of ( 2 –x)⁵ (1mk)
   2. Hence find the value of the constant k, for which the coefficient of x in the expansion of
      ( k + x) (2- x)5 is - 8 (2mks)
6. OA = 3i + 4j – 6k and OP = i + 15k. P divides AB in the ratio 3 :–2. Write down the coordinates of B. (3mks)
7. Simplify (3 marks)
   
8. Find the relative error in the area of a parallelogram whose base is 8cm and height 5cm. (3mks)
9.  
   1. Find the inverse of the matrix (1 mark)
      
   2. Hence solve the simultaneous equation using the matrix method (2 marks)
      4x +3y = 6
      3x + 5y = 5
10. A straight line L₁ has its X intercept a = -3 and its y-intercept b = 5.
    1. Write the equation of L₁ in the formx/a+y/b=1 (1mks)
    2. Find the equation of another line L₂ which passes through (1, -2) and is perpendicular to L₁ (3mks)
11. Use reciprocals, squares and square root tables only to evaluate (3mks)
          2          _       5    
    (0.5245)²       √363.4
12. Using a ruler and a pair of compasses only construct triangle ABC such that BC=6cm, <ABC=75º and BCA=45º. Drop a perpendicular to BC from A to meet BC at O hence find the area of triangle ABC (3mks)
13. A two digit number is such that the difference between the ones digit and the tens digit is 2. If the two digits are interchanged, the sum of the new and the original number is 132. Find the original number (3mks)
14. A quantity P varies partly as the cube of Q and partly varies inversely as the square of Q. when Q = 2, P = 108 and when Q = 3, P = 259. Find the value of P when Q = 6. (3mks)
15. Solve for y in the following equation below: (4mks)
    log₄y + log_y4 = 2
16. Obtain the values of x for which the matrix is singular (3mks)
    

SECTION II (50 MARKS)
Answer any five questions in this sections in the spaces provided.

17. The table below shows income tax rates.
    

    monthly taxable income	Rate of tax ( Ksh/ £)
    1- 435	2
    436- 870	3
    871-1305	4
    1306-1740	5
    Excess over 1740	6

    
    An employee earns a monthly basic salary of sh. 30,000 and is also entitled to taxable allowances amounting to Ksh. 10,480.
    1. Calculate the gross income tax (4mks)
    2. The employee is entitled to a personal tax relief of Ksh. 800 per month. Determine the net tax.(2mks)
    3. If the employee received a 50% increase in his total income, calculate the parentage increase on the income tax. (4mks)
18. In the figure below, O is the centre of the circle. PQ and PR are tangents to the circle at Q and R respectively. <PQS=40º and <PRS = 30º. RTU is a straight line.
    
    Calculate by giving reasons
    1. <QRS (2mks)
    2. <RTQ (2mks)
    3. <RPQ (2mks)
    4. Reflex <QOR (2mks)
    5. <TRO given that TR = TQ (2mks)
19. Three darts players Jane, Kelly and Brony are playing in a completion the probability that Jane, Kelly and Brony hit the bull’s eyes is 1⁄5 , 2⁄5 and 3⁄10 respectively.
    1. Draw a probability tree diagram to show all the possible outcomes for the players. (4mks)
    2. Calculate the probability that :
       1. Jane or Brony hit the bull’s eye. (2mks)
       2. All the three fail to hit the bull’s eye. (2mks)
       3. Only two fails to hit the bull’s eye. (2mks)
20. Three towns X, Y and Z are such that X is on a bearing of 120º and 20km from Y. Town Z is on a bearing of 220º and 12cm from X
    1. Using a scale of 1cm to represent 2km, show the relative position of the places (3mks)
    2. Find;
       1. The distance between Y and Z (2mks)
       2. The bearing of X from Z (1mk) (iii) The bearing of Z from Y (1mk)
       3. The area of the figure bounded by XYZ (3mks)
21. The fourth, seventh and sixteenth term of an arithmetic progression are in geometric progression. The sum of the first six terms of the arithmetic progression is 12.
    Determine the:
    1. First term and the common difference of the arithmetic progression. (6mks)
    2. Common ratio of the geometric progression. (2mks)
    3. Sum of the first six terms of the geometric progression. (2mks)
    1. Draw the graph of y = 2x² - 3x - 5 taking the values of xin the interval -2 ≤ x ≤ 4. (5mks)
    2. Use the graph in to solve the equation 2x2 - 3x - 5 = 0 (2mks)
    3. Using a suitable straight line, solve the equation 2x2 - 5x - 3 = 0 (3mks)
22. Draw the quadrilateral with vertices at A (-6,-1) B (-6,-4) C(3,-7) and D (3,2).
    1. On the same grid draw the image of ABCD under enlargement centre (0,-1) scale factor 1/3 label the image A^IB^IC^ID^I (3mks)
    2. Draw A^IIB^IIC^IID^II the image of A^IB^IC^ID^I under rotation of +90º about (1,0) (2mks)
    3. Draw A^IIIB^IIIC^IIID^III the image of A^IIB^IIC^IID^II under a reflection in the line y-x = 0 (2mks)
    4. Draw A^IVB^IVC^IVD^IV the image of A^IIIB^IIIC^IIID^III under translation (² ₃) and write the co-ordinate of the final image. (3mks)
23. The volume of two similar solid cylinders are 4096cm³ and 1728cm³.
    1. If the curved surface area of the smaller one is 112cm². Find the height of the larger cylinder if the radius is 7cm. (4mks)
    2. The diagram below represents a solid made up of a hemisphere mounted on a cone. The radius of the hemisphere and cone are each 6cm, and the height of the cone is 9cm.
       
    3. Calculate the volume of the solid. Take π = ²²/₇ (6mks)                                                                 
                                                                      

MARKING SCHEME

SECTION I

1.  
   
2.  
   
3.  (T-R)³= P² + n
                   m² 
   (T-R)³ m²= p²+n
   [(T-R)³ m² ] - n=p² 
   
   √[(T-R)³ m² ] - n=p
4.  
   
5. 102⁵(-x)⁰ + 5.2⁴(-X)¹ + 10.2³(-X)² 
   = 32-80X + 80X² 
   (k=x)(32 - 80x + 80x²) = -8x
   -80kx + 32x = -8x
   (-80k + 32)x = -8x
   -80k= -8-32
   k= 1/2
6.  
   
7. 5-4(1-sin²x) = 4sin x
   5-4+4sin²x-4sin x=0
   4sin² x-4sinx +1=0
   4sin² x- 2sin x - 2sin x +1=0
   2sin x(2sin x -1) -1( 2sin x -1)=0
   (2 sin x - 1) (2 sin x - 1)= 0
   sin x = 1/2
   x = 30º
8.  
   maximum area = 8.5 x 5.5= 46.75cm²
   minimum area  = 7.5 x 4.5= 33.75cm² 
   relative error    = 0.5 + 0.5  = 0.1625
                               8       5
9.  
   
   x = 1 4/11, y = 2/11
10.  
    1.  x/3 - y/5 = 1
    2.  for perpendicular lines, m₁ x m₂ = -1
       m₁ = Δy = 5-0= 5 
                Δx     0-3  -3
       m₂= 3/5
       (1, -2)
       y--2  = 3
       x -1     5
       (y+2)5 = (x-1) 3
       5y+10=3x-3
       5y=3x-13
       y=3/5x -13/5
11.  
           2       _     5      
    (0.5245)²     √363.4
           2            _              5      
    (5.245 x 10^-1)²     √3.634 x 100
           2            _              5      
    (27.510 x 1/100)     (1.9063 x 10)
           2     _      5      
    0.27510      19.063 
    2 x      1       _  5 x     1      
          0.27510           19.063
    2 x            1         _  5 x          1         
          10^-1 x 2.7510           10¹ x 1.9063
    (2 x 10 x 0.3635) - ( 5 x 10^-1 x 0.5247)
    7.27 - 0.26235
    7.00765
12. Let the original number be xy
    y - x = 2
    (10y + x) + (10 + y) = 132
    11x + 11y = 132 → x + y = 12
    y - x = 2
    y + x=  
    2y = 14 → y = 7
                     x = 5
    xy = 57
13.   
    p = KQ³ + m 
                    Q²
    8k + m = 108
            4
    32K + m = 432
    27k + m = 259
              9
    243k + m = 2331
    211k = 1899
    k=9, m = 144
    p = 9 x 6³ + 144
                         6²
    =1948
14.  
    log ₄ y +   1      = 2
                 log ₄ y 
    let x = log ₄ y  
    x +  1   = 2
           x
    x² + 1 = 2x
    2² - 2x + x + 1 = 0
    ( x - 1)( x - 1) = 0
    x = 1 = log ₄ y
    log ₄ y = 1
    y = 4
15.  
    
    

SECTION II

17.  
    1. Taxable income 30000 + 10480 = 40480
       in $= 2024
       435 x 2= 870
       435 x 3= 1305
       435 x 4= 1740
       435 x 5= 2175
       284 x 6= 1704
       Gross tax = 7794
    2.  Net tax = 7794 - 800
       = 6994
    3.  870+1305+1740+2175+ (1296  x 6)
       New net tax = 13866 – 80= 13066
       Increase in tax = 13066 – 6994= 6072
       %tage increase =  6072 x 100
                                   6994
       = 86.82%
       
       
18.  
    
    

    1. QRS = 40º angles in alternate segment

    2. < RTQ = ½ (180º – 40º) = 70º
       angle at the centre is twice that subtended at the circumference.

    3. <RPQ180º – (70+70) = 40º angle sum of a triangle

    4. Reflex <QOR = 360º – 140º = 220º angles at a point

    5. < TRO = (180º – 70º) = 55º
                             2
       Base angles of isosceles triangle

19.  
    1.  
       
    2.  
       1.   
          
          
          
       2.  all fair= 4 x 3 x 7  =   42
                       5    5   10     125
       3. only two fail:
          
20. 1.  
       
    2.  

       1. 
          YZ=9.9 + 0.1 or 9.9 - 0.1
                9.9 x 2=19.8 
                 19.8 + 0.2
                  19.8 - 0.2

       2. 40 +/ - 1º

       3. 158 +/- 1º

       4. Area = ½ x10x6xsin80^o

          = 29.544cm3
          1cm rep 2km
          1cm² rep 4km²
          29.544cm² rep
          (29.544x4)
          = 18.176km² 

21.  
    1.   a+6d = a+14d
         3+3d     a+6d
       a²+12ad+36d² = a²+15ad+ad+45d²
       12ad + 36d² = 18ad+45d² 
       -6a=ad
       sn6 = 3(2a+5d) = 12
       2a+5d = 4
       a= -3, d=2
    2.  a+6d = a 
        3+3d    3
       =3
    3.  
       sn = 3⁶ - 1/ 3-1
            = 364
       
       
22. 
    

    x	-3	-2	-1	0	1	2	3	4
    x2	9	4	1	0	1	4	9	16
    2x2	18	8	2	0	2	8	18	32
    -3x	9	6	3	0	-3	-6	-9	-12
    -5	-5	-5	-5	-5	-5	-5	-5	-5
    7	22	9	0	-5	-6	-3	4	15

    
    
    1.   
       y = 2x2 – 5x – 5
       0 = 2x2 – 5x -5
       Y=0
       X=-1 or x = 2.5
    2. 
       Y = 2x-2
       X = -0.5 or x = 3
23.  
    
24.  

    V.S.F = 4096
                1728
    
    L.S.F = ₃√4096 = 16 = 4
                  1728      12    3
    
    A.S.F = (4/3)² = 16/9

    1.  
       16/9 x x/112
       
       Curved S.A of larger container 9x = 16 x 112
        16 x 112 = 119 1
              9                9
       1792
          9
       22 x 14 x H = 1792
        7                      9
       H = 1792 x    7   = 4.525
                9       308         
    2. Volume of hemisphere = 4 x  1 x 22 x 6³ = 19008 = 452.57
                                           3     2     7                42
        1 πr²h
        3
        1 x 22 x 6² x 9 = 339.429
        3     7
       Total volume of solid = 452.57 + 339.429= 791.999cm3