RI 9740/02 2013

1 | Page   A x x y y Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i) We have f {1}\D   and g .R   Since gf ,RD the composite function fg does not exist. (ii) We have gf ( ) 1 2f ( ) 212 1 312 1 1 63. 1 xx x x x x          If 1(gf ) (5),x  then 5g f ( ) ,x so 5. 63 1 4x x   Therefore 1(gf ) (5) 4.  It is also fine to write this as 3( 1) .1 x x   The form given in the answer makes the next part easier. Question 2 No. Suggested Solution Remarks for Student (i) Consider one of the kite shapes: We have tan .6 3x y yx  The base of the prism is an equilateral triangle with side 22 3 .ay ax  The area of this base, S, is    2 (base) (height) 1 23 1 23 s i n23 3 23 . 2 4 ax ax a S x        Raffles Institution H2 Mathematics Solution for 2013 A-Level Paper 2

2 | Page   Therefore   21 32 . 34 aVS x x x  (ii) We have     2 23 23d3 43d4 3 63 .24 3 ax axV xx axax       At stationary values,   3 23 63 04 or (reject). 63 3 d 2 0d ax ax aaxx V x      We have      2 2 d3 23 6 3 63 2 34d 32 6 3 . V ax axx ax       At , 63 ax 2 2 d 30 ,d V ax   so this gives a maximum value of 2 31 32 3 .45 463 63 aa aVa         Product rule. 23 ax is rejected as it is too large, giving a “base” with zero length. For “real life” problems, you should use the second derivative test. Question 3 No. Suggested Solution Remarks for Student (i) Let f( ) l n( 1 2s i n ) ,yx x so e1 2 s i n .y x Differentiating implicitly gives 22 2 332 32 de2 c o sd ddee 2 s i n dd dd dde3 e e 2 c o sdddd y yy yy y y xx yy xxx yy yy xxxxx         When 0,x 0e, 1y y and 23 23 dd d 2, 4, 14.d dd yy y x x x   Therefore (0) 2, f (f0 ) 4(0) 0, f     and (0) 4.f1  A direct differentiation approach gets messy real fast. For such problems, use implicit differentiation as far as possible.

3 | Page   Maclaurin’s Theorem gives 2 3 3 2 (0) (ff ff( ) f( 0 ) 1! 2 0) (0) 7 ! 2 3! 2 3 xx x xx x x            You can use the standard series from MF15 to verify this answer if you have time. A faster, but less reliable way to check is to pick a small value of x, say 0.1, and verify that the value from this expression agrees with the value of f(0.1) from the GC. (ii) We have 23 3 23 23 () () ()es i n 1 2! 3! 3! 26 , ax ax ax nxnx ax nx an nnx anx x                     so comparing with 23 7f( ) 2 2 3xx x x    gives 2n and 12. aan   The next term in the series of es i nax nx is 23 33(1 )2 2 1 .26 3 x x    Thankfully this is non-zero, or we will have to find the next term. Question 4 No. Suggested