PJC H2 MATH P1 SOLUTIONS

PJC2011 [Turn over] 1 PJC 2011 JC2 H2 Mathematics End of Year Paper 1 Solution 1(a)       12 2 12 2 12 4 3 21 2 2 44 21 2 4 11 4 dtan 2 d tan 2 d tan 222 1 4 d 11 1 6 d 4 1tan 2 d28 1 4 d 1 4 2 11 tan 2 ln 1 428 xvxx x x xx x u x x xx xu xx xx v x xx x xx x C                   1(b) To find point of intersection: 2 4( 1 ) 29 ( 2 ) Solving (1)&(2) by G.C. 3 or 3(NA)2 yx x yx xx        Volume of R about x-axis =  233 22322 00 9 d 4 d 50.89 units22 x xx x x    2(i) 1 2 2 2 222 24 24 11 1 444 13 11 221 ...22 4 2 ! 4 131 ...2 8 128 11 3 2 16 256 x x xx xx xx                        

PJC2011 [Turn over] 2 2(ii)  22 11 1 44 x x x x      24 23 11 31 ...2 16 256 11 1 1 2 2 16 16 xx x xx x        2(iii) 2 14 x  2 4 22 x x   3 ( i ) 23 33 3 34 2 4 2 4 2 4 ...... 2 444 4 4 n n S                23 33 3 34 2 4 ...... 44 4 4 n           31 4348 34 1 4 n       342 4 1 4 n     328 24 4 n   3(ii) 328 24 244 n nS    13 64 n  1ln 6 3ln 4 n   6.23n 7n The ball must bounce at least 7 times for it to travel more than 24 m. 3(iii) 3rd (n+1) 1st 2nd H 3 4 H 2 3 4 H  3 4 n H 

PJC2011 [Turn over] 3 3,0 , 2 84 n nS     Since sum to infinity is 28, the ball will not travel more than 28m. 4(i)   1 32 3 2 AB rr r r         1( 2 ) 3 2:1 1 3: 1 1 11 1 32 3 2 Ar B r rB B rA A rr r r                   44 11 1 32 3 2 NN rr rr r r       11 12 11 23 11 34 11 45 ..... 11 54 11 43 11 32 11 2 NN NN NN N                             4(ii)                 4 1 32 111 1 1 1 1 ...12 23 34 6 5 5 4 4 3 3 2 N r rr NN NN NN NN             

PJC2011 [Turn over] 4                6 0 2 4 11 1 1 1 1 ...2 1 21 32 43 5 6 4 5 1 32 11 22 11 4 N r N r rr NN NN rr N N                    Alternatively,   6 0 1 21 N r rr      4 1 32 N r rr      11 43 32NN NN    11 2N      11 43 32NN NN     11 4N  4(iii) 1,0 2N N    4 1 132r rr     which is a finite value. Hence, the series converges.    44 11 1 lim lim 1 132 32 2 N NNrr rr rr N             5 2ec o sxy x 2d 2e cos sin e cosd xxy x xxx   d es i n 2d xy x yx   2 2 dd e2 c o s 2 es i n 2dd xxy yxxx x   2 2 dd d e2 c o s 2dd d xyy y xyx xx      2 2 dd 22 e c o s 2dd xyy y xxx    (shown) 32 32 dd d 24 e s i n 2 2 e c o s 2dd d xxyy y x xxx x   23 23 dd d0, 1, 1, 1, 5 dd d yy yxy xx x     23 2 5e cos 1 ...... 2! 3! x xxyx x    

PJC2011 [Turn over] 5 23 2 5ec o s 1 26 x x xyx x    ec o s s i n 2x x x ec o s 2 s i n c o sx x xx  22sin e cosxx x 32 3 521 3! 2 6 x xxxx          3 232 2 ...... 3 xxxx  3 2 422 3 xxx  6(i)  3f2 2x x   6(ii) 12let 2 xy x    1 21 2 21 2 12 2 12f: , 2 2 xy y x xy y yx y xxx x          6(iii)  ,gR    ,2fD   Since    ,, 2gfRD    , fg does not exist  2,fR   3,gD   Since    2, 3,fgRD  , gf exists  12gf g 2 12ln 3 2 12 3 6 5ln ln , 2 22 xx x x x xx x xxx                 fyx x y 2x 2y  R2 ,f  

PJC2011 [Turn over] 6 7(i) 42 d3 d dd x ay a tt t t  2 ddd dd d 3 y yt t xtx At 1 2t  , Gradient of tangent at 2 1 12 31 2P  Gradient of normal at 12P At  ,8 a n d2 8 , 2Px a y a a a  Equation of tangent: 128 12ya xa  14 12 3y xa Equation of normal:  21 28y ax a   12 98 y xa 7(ii) 3 14 12 3 aa att   23 32 12 1 16 16 12 1 0 11By G.C, (N.A.), 24 tt tt t     When 1 ,6 4 ,44tx a y a   Hence the tangent cuts the curve again at  64 , 4aa 7(iii) At Q: 0y 1401 612 3x ax a    16 ,0Qa  At R: 0y 4901 2 9 8 6x ax a    49(, 0 )6R a Area of triangle PQR =   14 9 16 226 aa a  = 22145 units6 a x y  8, 2Paa  16 ,0Qa 49(, 0 )6Ra

PJC2011 [Turn over] 7 8(i) d 12d x kxt  1 dd12 x ktx  1 ln 1 22 x kt C   ln 1 2 2x kt D  212 e e kt Dx  212 e e D ktx   212 e , e kt DxA A     21 1e2 ktxA  0, 1tx 111 12 AA    d0, 0.05d xt t  0.05 1 2k 0.05k   0.11 1e2 tx  8(ii)  0.11 1e2 tx  0.1d1 ed2 0 tx t  since 0.1e0 t  for all t, 0.1d1 e0d2 0 tx t   for all t, x is a decreasing function. Amount of X is always decreasing. Alternative Since at 0t  , d 0d x t  and 2 2 d 20d x kt   , x is a decreasing function. Amount of X is always decreasing. 8(iii) When 0.1 1,e 0 , 2 ttx    In the long run, X will not be used up and will stabilise at 0.5kg. 8(iv) x t 0.5 1  0.11 1e2 tx 

PJC2011 [Turn over] 8 9(i) iezr  is a root, iezr  is another root. A quadratic factor of P z    iieezr zr    2ii 2 eezz r z r r    2i i 2 eezz r r     22 2c o szr z r   (shown) 9(ii) 21zi z 21 1 ii 2zz z     21 5arg arg i arg 23 6zz      2z is an anti-clockwise rotation of 1z about the origin by 2  . 9(iii)    22 2 2 5P 2 2c o s 2 2 2c o s 2 36zz z z z           22 1344 4 422zz zz              22 24 2 34zz z z    10(a)(i) Note : ii*e e 2c o s 2 2 R e () zz r r rx z       is a standard result that you may apply directly. Re(z) Im(z) O i 3 1 2ez   21 izz 3  2 2 O (3, 1/3)  1 fy x 1 x y 0.5y 2x 0x

PJC2011 [Turn over] 9 10(a)(ii) 10(b) Let  22 44h 44 1 ( 2 1 )x xx x   . Before C,    22 44h 1221 yx xx    Let   2 4p 12 x x   Before B,  22 44p 2 112 2 xy xx       Let   2 4g 1 x x   Before A,     2 2 44f= g 1 11 yxx xx     Note : The above method done without co mpleting the square for the denominator is shown below. Let  2 4h 44 1x x x  . Before C,    2 2 44h 44 144 1 yx x xxx       Let  2 4p 44 1x x x  Before B, 2 2 44p 22 1 44 122 xy x xxx         Let  2 4g 21x x x  Before A,     2 2 44f= g 1 12 1 1 yx x xxx       Use GC for Checking Y4 and Y5 should coincide if your equation is correct. 2y f( 1 )yx O 1 1 (2, 3) y x

PJC2011 [Turn over] 10 11(i) 22 2 42 4 4x xk k kyx k x kx k      vertical asymptote : x k oblique asymptote : 4yx k 11(ii)   2 2 24d 1d kky x xk    At stationary points,   2 2 24d 10d kky x xk       2 2 24 1 kk xk    22 224 2kk xk x k  22 24 0xk x k k   For C to have 2 stationary points,   2 224 4 0kk k   281 6 0kk