NYJC H2 Maths P1 Solution

1 NYJC 2014 Preliminary Examination 9740/01 NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION SOLUTIONS Higher 2 MATHEMATICS 9740/01 Paper 1 16th September 2014 3 Hours Additional Materials: Cover Sheet Answer Paper List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. 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 a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

1 1k  Let 22( 1 )yx  12 1 2 (NA) or 1 2 ( 1) xy xy xy x            1f : 1 2 , 2xx x      1ff x x  f x x   2 21 x x   2 31 0xx  Using GC, 3.303 (since domain of f is 1)xx  2 Area of A = 2 0 es i n dx x x   22 0 0 es i n ec o s dxx x xx    22 2 0 0 ee c o s e s i n dxx x xx      22 0 e1 e s i n d x x x       22 0 2e s i n d e 1x xx   22 0 1es i n d e 1 2 x xx     Volume  2 22322 00 3 1 es i n d3 3.19 xx xx x x units          3 23 33xx y y  diff w.r.t x 2d23 33 0d d d yxx yy x y x   2 d2 3 d3 3 yx y x xy   Tangent // x-axis, 23 0xy 2 3 xy  Sub 2 3 xy  into 23 33xx y y 

0 2  ` 3 2 2233 33 xxxx      328 3027 xx  4.00x , 2.67y coordinates ( 4.00, 2.67 ) At 1x ,   2 313 1 3 yy  3 32 0yy  2, 1y At  1, 1 , d d y x is undefined,  equation of normal is 1y At  1, 2 ,       2 213 2d d 31 3 2 y x   d4 d9 y x  Equation of normal 29 14 y x   91 7 44yx 4 (i) ii)

11 22 3Largest possible value of sin tan 113       1.88 rad (3 s.f.) (iii)     arg * 2 arg 2 *ww      arg 2w   1 33Largest value of tan 2    1.20 rad For the half-lines  arg 3 3iw   and xxxxxxxxxxxxxxxxxxxxxxx  arg * 2w   to intersect, 01 . 2 0  xxxxxxxxxxxxxxxxxxxxxxx(3 s.f.). 5 Height of an isosceles =  2 2 2 2 1 4 xax x a Area of one = 2 22 11 44 1 22 xxxa a   Area of base = 2 22 2 11 463 24 x ax a   22 2 2300 3 3 11 44 (1)xa y a x y            Surface Area, 22 1 432 2 4Sx a x y a x y   From (1), 22 100 1 4 y ax     Therefore  2212 3 1 3 1 2 0

22 22 22 100 10032 2 4 11 4 1 4 4Sx a x a x ax ax       22 22 22 22 100 10062 4 11 44 20 1 4 1 4 04 0 0 16 11 44 ax a ax ax aax xaa                 2 2 22 d2 0 0 4 0 0 126d 11 44 1 4 Sa axx xaa        Let d 0d S x  2 2 22 200 400 126 0 11 44 1 4 aax xaa              2 2 22 3 2 3 3 200 400 126 11 44 200 12112 4 200 121112 22 200 200 1232 1 2 1 32 1 1 4 aax xaa xa a xa aa xa aa a                     1 3200 32 1x a  

2 2 3 2 2 2 d2 0 0 4 0 0 126 2 0d 11 4 1 4 4 Sa axx aa         as 300xx  and 211 024aa Recall that 22 100 1 4 y ax     therefore  2 23 100 100 1 1 200 4 43 21 y x ax a a                 2 2 32 133 2 1 2 1 3 212112 41 42 1 a aa aaaa a            Method 1(Graphical): Since 1 2a , therefore 0< 21 21 a a   <1 or 0< 21 21 a a   <1 , [Draw graph to show], Therefore 0 < 213 21 a a   <3 Method 2(Algebraic) 32 1 2 1 2 33 121 2121 aa aaa     Since 1 2a ,  21 2 0a  110 21 2a                                                201 21a                                                  211 0 21a  

 201 1 21a      203 1 3 21a   6 ddi) Let , dd d d3 0 d1 d3 0 tt t t tt wywy e e y e tt ye ytt yey ett         d1 d3 0 11d d d since 2030 30 ln 30 C w tt wt t t tt wt            ln 30 C 0, 0 0 ln 30 C So, C= ln 30 ln 30 ln 30 30ln 30 t t t ye t ty ye t tye                2 2 dii) ed d e A d e A B d0, 2 2 1 Ad So, A = 3 0, 1 B = 0 t t t x t x t xt xt t tx               Hence, e +3t tx 

7 iii) 7 (i) Alterna y Let zx  x   22x y  Comparing 22 4xy From (2), x Sub (3) in ( 2 12 y 4 24yy  22 2y  2 24y  2 2y  2y When When y Hence z ative solution 2 44z  i 38e   y  1 iy  2 i4y  2i4xy  real and im 4 ___(1) 23x y _ (1): 2 4y 2 12 0 16 0 or 2 2y   2 or y , 2x 2 , x 62 i o n: 3i 2 43 i 43 i maginary par 24 3xy ___(3) 4 (rejecte 2 3 6 2  6 or 6z (60. rts, ___(2) ed since y 2i x 0, – e-60 ln3  ) 0)

i2 38e n   , 1, 0n 16i 68e n z   , 1, 0n 5ii 668e , 8e    558c o s i s i n , 8c o s i s i n66 6 6                  31 3 122 i s i n , 22c o s i s i n22 2 2               Hence 62 iz  or 62 iz (ii)  3 *4 4 3 iw  3 44 3 iw  i2 38e n  6i 38e n   6i 92e n w    , 1, 0,1n 75ii i99 92e , 2e , 2e    Method 2: i 38e   i2 38e n   6i 38e n  6i 9*2 e n w   , 75ii i99 92e , 2e , 2ew    (iii) Since v is obtained by a counter clockwise rotation of the point representing through one right angle about the point on the Argand diagram, iv is obtained by a counter clockwise rotation of the point representing 2 iz  through one right angle about the origin.  3 *4 4 3 iw  1, 0,1n 2z  0,1

Hence  2ii ivz  i4 4 3 i i  14 3 4 i  14 3 5 iv  8 (i) Let Pn denotes the proposition: cos n nxu n for all n  . For n = 1, LHS = 1u = cos x = RHS.