RI 2015 P1 A-Level H2 Math Solution

1 | P a g e Question 1 No. Suggested Solution Remarks for Student (i) 2 ay bx cx   3 d 2 d y a bx x  At x = 1, d 2,d y x we have 2 2 ...(1)a b   At  1.6, 2.4 ,  we have   2 1.6 2.4 ...(2)1.6 a b c   At  0.7,3.6 ,  we have   2 0.7 3.6 ...(3)0.7 a b c   Using GC to solve (1), (2) and (3): 3.59345 3.593, 5.18691 5.187, 7.30274 7.303 a b c       Can key in   2 1 1.6 (ii) Using GC, x = –0.589 (3 d.p.) (iii) 5.187 7.303 (3 d.p.)y x   Raffles Institution H2 Mathematics Solution for 2015 A-Level Paper 1

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (i) Label asymptotes: y = 1 and x = 1 (ii) From graphs and GC, the graphs intersect at 1.73, 0.414, 1.73x For the inequality to hold, we have 1.73 0.414 or 1.73x x    y = 1 x = 1 1 O 1 2

3 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) Note that 1 1 2 3f f f ... f n n n n n n                             is the sum of area of the n rectangles in the diagram. As n, the sum will approach the exact area under the curve. Hence   1 0 1 1 2 3lim f f f ... f f dn n x xn n n n n                               Since f is any continuous function, we can use y = x2 for convenience. (ii) 143 3 3 13 3 3 0 0 1 1 2 ... 3 3lim d 4 4n n x x xn n                  

4 | P a g e Question 4 No. Suggested Solution Remarks for Student Length of rectangle 2 2x y  Length of semi-circle 2 2d x y   But length of semi-circle also    1 2 2 2 22 x x x x x        Thus, 2 2 2 2 2 2 1 1 22 2 d x y x x y d x x y d x x                 Total area, A 21 2xy x  21 1 1 22 2 2x d x x x        21 22xd x   1 42x d x  This is an quadratic expression with max value attained when 8 dx , that is the mid-point of the 2 roots 0 and .4 d Thus, max value of A 21 1 42 8 8 32 d dd d               m2 So, 1 32k

5 | P a g e Question 5 No. Suggested Solution Remarks for Student (i)     2 2 14 2 replace by 3 3 replace by 1 34 y x x x y x y y y x         Translate 3 units in the positive x directions followed by scaling of factor 1 4 parallel to y-axis. (ii) (iii) 1 1 O 1 3 4 x

6 | P a g e 1 O 2 4 x y 2 5/4 1

7 | P a g e Question 6 No. Suggested Solution Remarks for Student (i)     2 3 2 32 2 8ln 1 2 2 ... 2 22 3 3 x xx x x x x        (ii)        2 3 2 3 2 3 4 1 1 21 1 ... 2! 3! 1 1 2 ...2 6 c c c c c cax bx ax bcx bx bx ab c c ab c c cax abcx x x    