ASRJC 9758 2024 Prelim P1 Solutions

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Prelim Paper 1 (100 marks) 9758 9 Sept 2024 3 hours Additional Material(s): List of Formulae (MF 26) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 11 Total This document consists of 13 printed pages and 3 blank pages.

2 1 (a) From the graphs, 2.82x− or 0x . 2 2 3e 3 dx xx − −− = 02 20 (3e 3)d (3e 3)dxx x x x x − − − − + − − = 0222 20 3e 3 3e 322 xx xx xx −    − − − + − −       = 22[3 (3e 2 6)] [(3e 2 6) 3]−− − − + + − − − 223e 3e 10−= + − 2 (i) ( ) ( ) 1 11 2 2 sin sin sind 11d 1 eee x xx xxx x − −− − = − − (ii) 1 11 2 sin sin sindd 1 e ee x xxx x x x x − −− =− − From (1), 1 11 2 2 sin sin sin 1 d d 1 eee x xx xx x c x x − −− − + + = − So 11 11 2 22 sin sin sin sind 1 d 11 ee ee xx xx xx x x x x c xx −− −− = − − − − −− ( ) 1 11 2 2 sin sin sin1d1 21 e ee x xxx x x x D x − −−  = − − + − 3 (i) 3 2 d2 2d d1 2d x t t y t t =+ =− ( ) 22 3 3 12 21d 2d 222 tty t x t t − − == ++ When 1,t = x y 0 ( )2.82,0.179− ( )0,3

3 [Turn Over d1 d4 y x = , 1x= and 3.y = Eqn of tangent at P: ( )131 4yx− = − 1 11 44yx=+ Eqn of normal at P: ( )3 4 1yx− =− − 47yx=− + (ii) Area of OAPB = ( ) ( )1 7 1 117 7 1 42 4 2 4    − − =       units2 4 (a) 3 1 1 2 r rru u r − − = +  ( ) 3 1 11 1 2 rnn rr rr u u r − ==  − = +    1 1 10 2 2 1 1 32 3 1 1 2 r nn nn nn rr uu uu uu uu uu r == −− − −  +−  +−  + = +   +− +−     ( ) 0 22 1 1 214 n nu nnu −=   + +− ( ) ( ) 22 22 1 1 1 1 242 1 = 3 42 n n n n n u n n + + +−=   +    + − (b) ( ) 2 3 9 12 2 rn r r + =  ++    x y 47yx=− + 1 11 44yx=+ 7 11 4 (1, 3) 7 4

4 = 2 3 11 1 2 rrn r r =+ =  +    ( Replace r by 2r− ) 2 10nuu+=− ( ) ( ) ( ) ( ) 2 2 2 2 2 10 2 3 10 11 11 334 2 4 2 n nn +   ++    = − + − − +                 ( ) ( ) 22 2 23 11 30254 2 1024 n nn + ++ = − − +  5 (a) + = −a b a b 22  + = −a b a b ( ) ( ) ( ) ( ) +  + = −  −a b a b a b a b 2 2 2 2 22 +  + = −  +a a b b a a b b 40  =ab 4 cos 0 = ab Since a and b are non-zero vectors, then 90 = , thus ⊥ab (b)(i) Since ( ) ( ) ( ), and  +  +  +a b c b c a c a b are all vectors, the addition of these vectors will lead to a resultant vector. (ii) ( ) ( ) ( ) + +  + +  +a b c b c a c a b =  +  +  +  +  + a b a c b c b a c a c b =  +  +  −  −  − a b a c b c a b a c b c ( ) and  =−   =− b a a b c a a c = 0 will not be given if the student wrote it as a scalar quantity 6 (i) 2 2 22 2 d( ) 2 (1)d Differentiating (1) w.r.t. : d d d d( ) 1 0 d d d d d d d( ) (1 ) 0 (2)d d d yx y ky x x y y y yx y k x x x x y y yx y k x x x + + = + + + + = + + + + =  (ii) sin 2 sin 2 cos cos 2 sin cos 22 2 2x x x x   + = + = 22 22 11 cos 2sin 2 2 (2 )1 2 xx x − =  +  − 

5 [Turn Over ( ) 22 2 12 1 4 ... x x − =− = + + 2 2 d0, 1: 2 d d 36 d yx y k x y k x = = = − =− 3642 2 10 3 k k −=  = (iii) 3 2 2 2 3 2 2 2 Differentiating (2) w.r.t. : d d d d d d( ) 1 2 2 0d d d d d d x y y y y y yxy x x x x x x    + + + + + =         32 32 d d d( ) 3 3 0d d d y y yxy x x x + + + = 2 2 ddWhen 0, 1, 1, 3 d d yyxy x x = = = =− , 3 3 d 18 d y x = 2331 3 ... 2y x x x = + − + + 7 (a) 229 54 2 79 0y y x x− − − + = ( ) ( ) 229 6 2 79 0y y x x − − + + = ( ) ( ) 22 9 3 9 1 1 79 0yx    − − − + − + =    ( ) ( ) 22 9 3 1 1yx − − + = ( ) ( ) 22 3 9 1 1yx − − + = ( ) 22 22 1 Replace by 1 Translation of 1 u nit in the negative direction 1 1 Replace by 9 Translation of 9 un its in the p yx x x x yx yy −= + − + = − ( ) ( ) ( ) ( ) 22 22 ositive direction 9 1 1 1Replace by 3 Scaling paralle l to the axis by a factor of 3 3 9 1 1 y yx y y y yx − − + = − − − + = (b)(i) Consider the line yk= . To find the range of y where curve C cannot lie, ( ) ( ) 22 3 9 1 1kx − − + = has no real roots 22 2 2 (3 9) 0x x k + + − − = has no real roots.

6 ( ) 2 4 4 2 3 9 0 k − − −  ( ) ( )( ) 2 3 9 1 0 3 9 1 3 9 1 0 8 10 33 k kk k − −  − + − −   Thus y cannot lie between 8 3 and 10 3 . (ii) (iii) By adding the graph of ( ) 2 13yx= + + , since there are 2 intersections between the 2 curves, so the equation ( ) ( ) 222 29 1 3 54 1 3 2 79 0x x x x   + + − + + − − + =    will have 2 real roots. 8 (a) 242y x x= + − , , 3.5xx ( ) 242y x x =− + − ( 24 2 0 for 3.5x x x+ −   ) 2 24y x x= − − 2( 1 ) 1 4yx= − − − 2( 1 ) 5yx= − − 15xy= + + or 15xy= − + ( )rej 3.5x 15xy= + + 1f ( ) 1 5xx− = + + y x ( )1,3− 1 10 33yx=+ 18 33yx=− + 101, 3 − 81,3 − ( ) ( ) 2 2 2 3 11 1 3 y x− − + = 

7 [Turn Over fR [1.25, )= 1fD [1.25, )− =  (b) 1f ( ) f( )xx− = , 1f fDDx − f ( )xx= , [3.5, )x 2 24x x x= − − If equate 1f − found in (a) to f, can also give the 2 3 4 0xx− − = ( )( )4 1 0xx− + = 1 (rej 3.5) or 4 x x x=−  = (c) )gR 4 e , a−= +  fD [3.5, )= Since e0a−  , 4 e 3.5a− +  gfRD , fg exists. )gR 4 e , a−= +  2f(4 + e ) (4 + e ) 2(4 + e ) 4a a a− − −= − − 216 8e e 8 2e 4a a a− − −= + + − − − 24 6e e aa−−= + + ) 2 fgR 4 6e e , aa−−= + +  (d) 2 g( ) 0 22 x xx  −− AND gDx 2 4e 0 ( 1) 3 ax x +  −− AND 1x− Since 4 e 0, ax x+    1 0 ( 1 3)( 1 3)xx  − + − − AND 1x− ( )1 3 or 1 3xx −  + AND 1x− 13− 13+ + + − 1−

8 – for 1 3 or 1 3xx −  + 1 1 3x−   − OR 13x+ 9 (ai) ( ) i 2 1 1 i 1 i 1 1 2i 1 i e1 i 1 i 2z  ++   =  = + − = =   −+     2 11cos isin i44 22 z    = + = +       12 1 i+ i2 2 4 iiii 8824e e e e ezz   −+ = + = +     3 i82cos 8 e=  (ii) 12zz+= 11 1i 22 ++ 113 2tan 18 2 + = 21 2 2 1 (shown)1 2 + = =