SAJC_H2_MATHS_P1_Solutions

2017 Prelim Paper 1 Solutions 1 3 2 4 3 dd 4dd Vr Vr rtt     When V=20, 3 1 3 420 3 15 r r     When 1 315r   , d d V t  . 2 3 2 3 15 d4 d d d4 1 5 r t r t         Surface Area, 24 dd 8dd Ar A rrtt     When 1 315r   , 2 3d d4 1 5 r t      . 12 3 1 2 3 3 dd d dd d 15 = 8 41 5 2 15 cm /s AA r tr t            2 BC = BD + DC = tan tan3 4 hh x    BC = 3 tan tan4 1t a n t a n4 hh x x     

= 3( 1 t a n ) 31 t a n hh x x   3( 1 ) 31 hh x x   = 13 (1 )(1 )3 h hx x   = 23( 1 ) ( 2 )(1 )[1 ( 1) ) ...]32 ! h hx x x     = 23 (1 )[1 ...]3 h hx x x   = 23 (1 2 2 ...)3 h hx x   23 1 2 + 23hx x 3 (i)

(ii) 4 3 20 2 22 2 2 2 2 2 2 f( ) d 1 d cos=1 ( s i n ) d =s i n d 1c o s 2 d2 sin 2θ= θ22 22 4 a a a xx xbx a aba a ab ab ab ab ab                               4 (i)  2 2 ee axaxy bb   If 2 f( ) e xx  , then    2 fe ax ax  and so    ff fy xy a xy a x b     Hence the sequence of transformations are: 1. Scale by a factor of 1 a parallel to the x-axis, 2. Translate the resulting curve by b units in the negative y-direction.   (ii) y y=f(x) O (0,1-b) x

5 (i) Since uvw is perpendicular to  uvw ,   0 0        uvw v uvw uvw u uv vv w uw vwww uu Since 22 2 ,,  uvwuu vv ww , and ,,  vv ww wuu u v w uv , 22 2 20  uvw v w Since ,,uvw are unit vectors, 1,1, 1 uvw , 111 2 0 1 2 1cos 2 1cos 2         vw vw vw Hence, 60  5 (ii) O  V W y x O y=1/f(x) y=0

    2 Area of OVW 1 sin 602 13 1122 3 units4 OV OW      

5 (iii) Since u and vw are parallel, we have ,OV O OOU U W . Volume of OUVW     3 1 Area of OVW3 13 134 3 units12 OU   

6 (a) (i) Using integration by parts, ec o s dx nx x sin ees i n d x x nx nx xnn    O  U V W dec o s d ds i ned x x vun x x un x vxn     des i n d dc o sed x x vun x x un x vx n   

= s i n 1 ec o s ec o sed xx x nx nx nx xnn n n       = 2 sin 1 e cos 1ee c o s d x xx nx nx nx xnn n n      Rearranging, 2 11e c o s d x nx xn   = sin 1 e cose x x nx nx nn n     ec o s dx nx x = 2 2 sin 1 e cose1 x xnn x n x cnn n n              where c is a constant (ii) 2 ec o s dx nx x   = 22 2 sin 1 e cose1 x xnn x n x nn n n                  2 2 2 sin 2 1 cos 2 sin 1 cosee1 nn n n n nn n n n n n                                 For any positive integer n, 02sin n and 12cos n If n is odd, 0sin