2017 A Level H2 Maths Paper 2 Suggested Solutions CWY edited

1 2017 H2 Mathematics Paper 2 Pure Math (Suggested Solutions) 1(i) Substitute 3, 2 into 2 x y t y x t   2 32 2 3 3 t t t t              when 3, 3, 2 3 3,2 3 when 3, 3, 2 3 3, 2 3 t x y A t x y B           Length of AB =     2 2 3 3 2 3 2 3 60 2 15 units      1(ii) At 3 ,2 P p t p p         2 2 d d d 3 2 2d d d 3 y y x t x t t t            Gradient of tangent at P = 22 3 p Equation of tangent at P:   2 2 2 3 2 3 2 43 py p x p py x p            At D : let y = 0 6 6 coordinate of ,0 x D p p         At E : let x = 0  4 coordinate of 0,4 y p E p   Coordinate of F= 6 0 4 0 3 , ,2 2 2 pp pp                  

2 3Let (1) 2 (2) from (2), (3) 2 substitute (3) into (1) 6 6 x p y p yp x y xy       ⋯ ⋯ ⋯ The Cartesian equation of the curve traced by F as p varies is given by xy = 6 (or any equivalent form). 2(i) For the given arithmetic progression, 13 3, 156 a S    13 6 12 156 2 1.5 d d    (ii) For the given geometric progression, 13 3, 156 a S     13 13 13 3 1 156 1 1 52 52 52 51 0 (shown) r r r r r r         Using GC to sketch the graph of 13 52 51 y x x    and solve for the values of x- intercepts. We obtain 1.21, 1.45 or 1 r  If 1r  , all the terms of the GP will be the same resulting 13 156 S  . Hence, 1r  even though 1r  is a root of the equation. Possible values of the common ratio are 1.21, -1.45.

3 (iii) Since 0, 1.210024 r r           1 1 1 3 1.210024 100 3 1 1.5 3 1.210024 150 150 3 1.210024 150 150 0 n n n n n n                Using GC to solve, n   1 3 1.210024 150 150 n n    41 -149.9 < 0 42 991.74 > 0 43 2404.7 > 0  The smallest possible value of 42 n . 3(a) (i)  f 2 y x  results in a scaling of the graph of y = f( x) with scale factor 1 2 parallel to x-axis   f(2 ) ,0 ,0 2 x aa         coordinates of x intercept: ,0 2 a      coordinates of y intercept:  0, b (ii)  f 1 y x   results in a translation of the graph of y = f( x) by 1 unit in the positive direction of the x-axis         f( 1) f( 1) ,0 1,0 0, 1, x x a a b b        coordinates of y intercept: not possible to find (iii)  f 2 1 y x   results in a translation of the graph of y = f( x) by 1 unit in the positive direction of the x-axis, followed by scaling with scale factor 1 2 parallel to the x-axis.     f(2 1) f(2 1) 1,0 ,0 2 10, , 2 x x aa b b                 coordinates of x intercept: 1,0 2 a      coordinates of y intercept: not possible.

4 (iv)   1fy x   reflection of the graph of  fy x  about the line y x          1 1 f ( ) f ( ) ,0 0, 0, ,0 x x a a b b       coordinates of y intercept:  0, a coordinates of x intercept:  ,0 b (b) (i) 1a When 1a , the value of  g 1 will be undefined. Hence,  g x will not be a function. Thus, 1a has to be excluded from the domain of g. (b) (ii)   2 1g gg( ) g 1 g 1 1 g 1 11 1 1 11 1 , , 1 xx x x x x x x x x x x x                                        ℝ Hence g is self-inverse 1( ) ( ) g x g x   1 1g : 1 , , 1 1x x x x       ℝ . Alternatively, To find   1g x ,   1 Let g 11 1 11 1 11 1 1g : 1 , , 1 1 y x y x y x x y x x x x                 ℝ

5 (b) (iiii)     2 1 2 g g 11 1 (1 ) 1 0 or 2 b b b b b b b         4(a) Sketch both graphs using GC, Intersections points: (1, 0) and (5.5, 2.25) Area of the plate = area enclosed   5.5 2 1 2 1 6 5 d 2 15.1875 units (exact) x x x x             (b) (i) Volume of the container =      2 1 20 1 20 2 1 22 0 1 3 2 0 d d 2 d 2 1 1 units 2 2 ( 1) 2 ( 1) y ya y y y a y y a y y a y a a a a                                               (b) (ii) 2 2 2 2 42 ( 1) 2 ( 1) 4 4 4 4 0 b b a a a a b b b b a a             

6   2 2 2 2 2 2 4 ( 4) 4(4)( ) 2(4) 4 ( 4) 4(4)( ) 2(4) 1 1 12 2 1 1 1 2 a a b a a a a a a                   Since 1, a and volume has to be formed in the same way,  1b 2 2 2 1 3 1 1 1 1 2 4 2 4 a a a a                     ∵ Since b > 1,   21 1 1 2b a a      5(i) The possible values of T are 2, 3, 4 and 5 6 5 5 P( 2) P( ) 9 8 12 T RR      6 3 5 5 P( 3) P( or ) 2! 9 8 7 14 T RYR YRR              since the first two counters can be RY or YR . P( 4) P( ) 6 3 2 3! 5 5 9 8 7 2! 6 28 T RYYR               The first 3 balls can be in any order but last one must be . R P( 5) P( ) 6 3 2 1 4! 5 1 9 8 7 6 3! 5 21 T RYYYR                The first 4 balls can be in any order but last one must be . R (ii) all E( ) P( ) 5 5 5 1 2 3 4 5 12 14 28 21 20 6 27 7 t T t T t             

7 2 2 all t ( ) ( ) E T t P T t    = 2 2 2 2 5 5 5 1 2 3 4 5 12 14 28 21        =125 14     22 2 2 Var( ) E( ) E( ) 6 125 20 75 14 7 98 T T T             (iii) 19 P( 4) P( 4) P( 5) 84 T T T       Let X denote the number of games, out of 15, that Lee takes out at least 4 counters out of the bag     19 15, 84 5 1 4 0.23791 0.238 (to 3 s.f) X B P X P X             ∼ 6 (i) Since the 4 cards in each family must be next to each other, number of ways to arrange 4 cards within each family = 5(4!) . 5Number of possible ways = 5! (4!) 955514880    (ii) Let R, B, G, Y and O be Red, Blue, Green, Yellow and Orange Let M, F, D, S be Mother, Father, Daughter, Son no. of ways for arranging the MR DR SR = 3! no. of ways for arranging the MB DB SB= 3! no. of ways for arranging the FG FY FO= 3! no. of ways for arranging the big unit and the remaining 9 cards = 10! MR DR SR FR FG FY FO FB MB DB SB X X X X X X X X X 3! Fixed 3! Fixed 3! The remaining 9 cards 2!

8 no. of ways for arranging the Red and Blue families at the ends of the big unit = 2 3Number of possible ways (3!) 10! 2 = 1567641600    (iii) No. of ways to arrange the non-fathers in a circle = (15 1)!  No. of ways to slot in the fathers = 15 5P 15 5(15 1)! P Required probability = (20 1)! 1001 or 0.258 3876      7 (i) A random sample is a sample drawn in such a w ay that each biscuit bar in the population has an equal chance of being selected as a member of the sample and each biscuit bar must be selected independently. (ii) ) Let X be the random variable” the mass of certain type of biscuit bar, in grams and be the population mean mass of biscuit ba rs, in grams”. Unbiased estimate of population mean, x = ( 32) 7.7 32 32 31.8075 40 40 x      31.8   x x x x x x x x x x x x x x x

9 Unbiased estimate of population variance,   2 2 2 2 ( 32) 1 [ ( 32) ] 39 40 1 ( 7.7) [11.05 ] 39 40 0.24533 0.245 (3 s.f) x s x          (iii) Test H0 : = 32 Against H1 : 32  at the 1 % of significance Under H 0, Since n = 40 is large, by Central Limit Theorem 0.24533 ~ (32, ) 40 X N approximately U