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           Method 1 Since 1 2 1f f, f ( ) ff ( )x x x    Method 2         2 2f ( ) f 3 2 2 2 23 3 3 3 x a x a a a a x a x a a x a x a                             (ii) 2 1 2f ( ) ff ( ) f ( )k k x x x   2 1 2f (2 ) f (2 ) 3 k a a a a     (iii) For maximum value of   2 g 1 2 2 a ax x         ,

2020 JC2 H2 Mathematics Preliminary Examination It occurs at 12 ax  [when 2 1 02 ax       ] From the sketch,  g fR , \ D 2 a a        Therefore, fg exists When we restrict fD , 2 a     Range of fg = 4 ,3a a a     y a 4,2 3 a a a     

6 Suggested solution 1 2 3 5 3 14 9 2 47 27 3 S a b c S a b c S a b c             OR 1 1 2 2 1 3 3 2 5 3 9 6 33 18 u S a b c u S S a b u S S a b                Using GC, 2, 3 and 2a b c   Hence  2 3 3 2n nS n                  1 1 1 1 2 3 3 1 2 2 3 3 2 2 3 3 3 2 2 3 3 2 2 3 1 3 3 4 3 3 n n n n n n n n n u S S n n n n                                           1 2 2 2 2 2 1 1 4 3 3 4 3 3 1 3 3 1 4 3 2 13 1 18 3 1 3 1 6 3 3 15 n n r r r r n n r r r n n n u n n n                                   

2020 JC2 H2 Mathematics Preliminary Examination 7 Suggested solution (a)(i) Since 1 1 iz   is a root,       2 1 i 1 i 1 3 i 0a b                   2i 1 i 1 3 i 0 1 3 2 i 0 Comparing Re and Im parts 1 3 0 1 3 2 0 1 3 a b a a b a a a b b                             (ii)       2 1 3 1 3 1 3 i 0z z                 2 21 3 1 3 1 3 i 1 iz z z z z              Method 1: Comparing z  2 21 3 1 i 3 i z z        Method 2: Comparing “constant”                         2 2 1 3 1 3 i = 1 i 1 3 1 3 i 1 i1 3 1 3 i 1 i 2 1 3 1 3 i 1 i 3 i2 z z                             Method 3: Sum of roots Sum of roots =  1 3   2 2 1 i 1 3 3 i z z       

Method 4: General formula                       2 2 2 1 3 1 3 4 1 1 3 1 3 i 2 1 3 1 2 3 3 4 4 3 4i 4 3i 2 1 3 2 3 4 3i 4i 2 1 3 1 3 2i 2 1 3 1 3 2i 2 1 i (rej) or 3 i z                                           (b)(i) Method 1: i 4 1 2 2i 2 2 or 2 2 cos isin 4 4w e                         5 i6 2 5 53+i 2 or 2 cos isin 6 6w e          5 7i i4 6 12 1 2 4 2 4 2w w e e         1 2 4 2w w  and  1 2 7arg 12w w  Method 2:    1 2 2 1 3 2 1 3 iw w         2 2 1 2 4 1 3 4 1 3 32 4 2w w             1 1 2 1 3 7arg tan 123 1 w w         (ii) Method 1: From (ii), 7 i12 1 2 7 74 2 or 4 2 cos isin 12 12w w e                       1 2 2 1 3 2 1 3 iw w    

2020 JC2 H2 Mathematics Preliminary Examination Hence   7 7 1 34 2 cos 2 1 3 cos12 12 2 2       Otherwise Method 2: Student using geometry approach on    1 2 2 1 3 2 1 3 iw w     Method 3: Student using special angles and addition formula 8 (a) Suggested solution (a)(i) To prove   2 d *d yx xy kx   Consider    ln ln (1)k xy xy k x x       Diff (1) wrt x, 2d 1 d d d y yx y k x xy kx x x         [shown] (a)(ii)  lnk xy x  At stationary point, d 0d y xy kx    [from (*)] So   1ln ln 1 ek xk x xx x         When 1 1 1e , e e k kx y k x          . Therefore,   1 1e , e k   is a stationary point of the curve  lnk xy x  . (b)(i) Given 2 2d d yy x x yx   ---(**) 2 2 d d 2 2d d v yv x y x y x x     Sub into (**): 2 2d 1 d d 2d 2 d d y v vy x x y v vx x x       [shown] (b)(ii) d 1 d 12 1 d 1 dd d 2 2 v v v v xx x v v        v x C   Since 0y when 2x , we have 4v .

4 2 4 C C    4v x      2 22 2 4 4 8 16 v x y x x x          Hence  f 8 16x x  . 9 Suggested solution (i)   1 1 1 1 1 2 1 d tan tan 1 tan1 1 k k k k xx k kx             (ii) From the diagram, we can see that Area of rectangle ABCD < Area under curve from x=k to x=k +1 < Area of rectangle ABEF Hence       1 2 2 2 1 1 1 1 d 1 1 11 1 k k xx kk            1 1 2 2 1 1 tan 1 tan 11 1 k k kk         [from (i)] [shown] Alternative (For one side)     1 1 1 2[f f ....f 1 ] f d k k k k k x xn n n                  --- (1) Note that (0,1) O x y (0,1) O x y k k+1