EJC_H2_2020_Prelim_P1_Solution

Eunoia Junior College 2020 JC 2 Prelim Exam H2 Mathematics Paper 1 Suggested solution 2 2 ( 3)x y y x   ----------(1) Differentiating throughout with respect to x, we have d d2 2 ( 3) d d y yx y y x x x    .  d 2 3 2d y y x y xx     d 2 d 2 3 y y x x y x     For tangent to be parallel to y-axis, gradient must be undefined. So 2 3 0y x   . 2 3x y   Sub into (1): 2 2(2 3) (2 )y y y y   23 12 9 0 ( 3)( 1) 0 y y y y         3y  or 1 The points at which the tangents are parallel to the y-axis are ( 3, 3)  and (1, 1) . Alternatively: 2 3 0y x   3 2 xy   Sub into (1): 2 2 3 3 ( 3)2 2 x xx x              2 2 2 6 9 3 6 9 04 x xx x x           2 2 3 0x x    1 or 3x   1 or 3y    The points at which the tangents are parallel to the y-axis are ( 3, 3)  and (1, 1)

Suggested solution   2 2 2d tan 2 secd x x xx    3 2 2 2 2 2 2 2 2 2 1sec d 2 sec d2 d 2d d 2 sec tand x x x x x x uu x x x v x x v xx          Therefore,   3 2 2 2 2 2 2 2 2 1 1sec d tan 2 tan d tan ln sec 2 2 where is an arbitrary constant        x x x x x x x x x x c c 2 Suggested solution (i) d d d d d d r r V t V t  3 24 d 43 d VV r r r    dr d d d d d r V t V t  Given d 12d V t  (constant) 2 2 dr 12 3 d 4t r r    At r= 5, 2 d 3 3 0.0382 cm/mind (5) 25 r t     (ii) 24A r  d 8d A rr    2 d d d d d 3 248      A A dr t r t r r r In 10 min, the volume of the balloon 12 10V    120 3cm . 3 3 4 1203 90 3.0598 V r r       At t=10 min, 3d 24 24 7.84 cm/mind 90 3.0598 A t   

2020 JC2 H2 Mathematics Preliminary Examination 3 Suggested solution (i) (ii)Multiplying b to 1 ax a x   gives 1 abb x a x   ( 0b ) Hence from the graph, 0 1 , 11 bx a x a xx       4 Suggested solution (i) de e d x x yy x         2 2 2 2 1 1 2 d d 2 11 1 ln 1 1 02 1 ln e 12 x y yF y y y yy y y d y d                       where d is an arbitrary constant. (ii) 1 1e (e e ) (e e )2 2 x x x x x        1 (e e ) (e e ) 1 e e d 1 d2 e e 2 e e 1 ln(e e ) e e 02 x x x x x x x x x x x x x x F x x x c                               where c is an arbitrary constant. (0,ab) (a,0) O (a+1, b) x =1 x y y =0

(iii) From (ii), 2 2 2 2 1 1 1 e 1ln(e e ) ln2 2 2 e 1 1 1 1 1 1ln(e 1) ln(e ) ln(e 1)2 2 2 2 2 2 1 ln(e 1)2 x x x x x x x x F x c x c x c x x c c                              The difference is c d , where c and d are the arbitrary constants for the answers in (ii) and (i) respectively. 5 Suggested solution (i)         2Let 3 2 3 2 3 2 3 y a x a y a x a x a y a x a y a                 1 ff 1 D R \ 2f : , , , 3 a x a x x a x a       