HCI_H2_MATH_P1_ANSWERS

2015 H2 Mathematics C2 Prelim Paper 1 Solutions Qn Solutions 1(a) Let 21ux 11(1 ) 21 d 1 d 22 ux xx u u     11 1 d22 2 u uu  = 1 1 d4 uu u = 31 221 d4 uu u = 53 2212 2 45 3 uu C    = 53 2211 10 6uu C =   53 22 11 21 2110 6 x xC   1(b) Using MF15 32 PQ x  2 PQ x  6PQ x 2PQ x 28Px => 4Px 2Qx 1sin 3 sin d cos 4 cos 2 d2x xx x x x   1 sin 4 sin 2 24 2 xx C   sin 4 sin 2 84 xx C   2       fl n 1l n 1 c o s , f 0 l n 2x x x    1s i nf' ( ) , f' ( 0 ) 111 c o s xx x x          22 1c o s c o s s i n s i n1f ''( ) + 11 c o s x xx xx xx      22 22 1c o s c o ss i nf' ' ( ) + 11 c o s x xxx xx     2 11 1f ''( ) + , f '' 0 1c o s 21 x xx      32 2s i nf '''( ) + , f ''' 0 2 11 c o s xx xx    Using series formula in MF15:      23 f f 0 f '0 f ' '0 f ' ' '0 . . .2! 3! xxx x     2311f ln 2 ... 43x x x x        11c o sln ln 11 c o s xx x x        2311ln 2 ... 43xx x   

3 Let 1FGh and let 2BC h . Form equation of the ellipse: 22 22 12000 xy a   --- (1) Since the areas of ABCD and EFGH are equal: 211000 1435hh --------- (2) Substitute the point 1 1435,2Gh  into (1): 22 1 22 717.5 12000 h a  --------- (3) Substitute the point 2 1000,2Ch  into (1): 22 2 22 500 12000 h a  --------- (4) Substitute (2) into (4):  22 1 22 1.435500 12000 h a   --- (5) From (3) and (5): 2 764806.25 874.532aa   2 1749.06 1749 mmMN a  4(i) 3sloped length sin  31 3 9total area 8 3 3 24 9cosec cotsin 2 tan 2             4(ii)  2 2 924 9cosec cot 2 d9 9cot cosec cosecd2 92 c o s 1 2 sin A A          Let d 0d A   , 2cos 1  . So, 3   . 91 8 9 9 324 9cosec cot 24 24 16.2 3 s.f.32 3 2 32 3 A        3     3   3     d d A  + 0 – The maximum area is 16.2 m2 .

5(i) Since a line such as 2y  intersects the curve  fyx at 2 points, f is not one-to- one. 1 f does not exist. 5(ii) Largest possible domain for h is [3, ) . 5(iii)  2 ln 2 2yx  2 22e y x   Since 3,x 2 1 2h: 2e , , 2 x xx x       5(iv)   11hh h hx xx  which is always valid since the inverse function exists. Since domain of 1hh  is domain of  1hD2 ,    and domain of 1hh  is domain of  hD3 , . Thus the set of values is  :3xx  . 6(i) As t , sinx t and cosyt . Using trigonometric identity 22sin cos 1tt  , the Cartesian equation of C is 22 1xy . Thus the shape of C is a circle with centre at the origin and with unit length radius. 6(ii) d ec o sd tx tt  , d es i nd ty tt  . de s i n de c o s t t y t x t    At P, gradient of normal is ec o s es i n       .  equation of normal i