YJC P1 MS

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 1 of 10 1 3 541 xx       2 2 3 5 4 01 3 5 9 4 01 4 9 2 01 4 1 2 01 xx xx x xx x xx x            1 or 1 24xx    Let sinxy , 1sin or 1 sin 2 (rejected)4 0.253 or 2.89 yy yy           2 (i) 2 sin 2 d 2 sin 2 4 cos 2 dx x xe x x e x e x x 2 sin 2 4 cos 2 2 sin 2 d 2 sin 2 4 cos 2 4 2 sin 2 d x x x x x x e x e x e x x e x e x e x x            5 2 sin 2 d 2 sin 2 4 cos 2 12 sin 2 d 2 sin 2 4 cos 25 x x x x x x e x x e x e x C e x x e x e x C           (ii) 2 2 2 2 1 2 2 1d d d2 5 2 5 2 5 xx x x xx x x x x x                 2 2 2 21 1ln 2 5 d 12 11ln 2 5 tan 22 x x x x xx x C              2 1 1 4 2 + + – –

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 2 of 10 3 (a)   5 32 1 0zi   311 25 42 3211 20 510 2 2 , where 2, 1, 0,1, 2 ik ki ze z e k         (b) 3zi 2 cos sin 66 i                 2 cos sin 66 2 cos sin66 n n n zi nn i                     Since nz is purely imaginary,     cos 06 21 , where 62 6 3, n kn k n k k         4 (i)   1 2 1 4 4 x x    1 1 2 2 2 2 141 4 1 1 312 8 128 1 1 3 2 16 256 x xx xx             Range of x: 44 x   (ii) 2 1 1 1 4 3 4 2 16 5 256 544 5              1 1 1 3 2 20 40016 5 5 223 4 400 2235 100      

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 3 of 10 5 (i) 40 2 3yx 320 2yx Area, 21 sin(60 )2A xy x  = 213 22xy x   2 22 2 3 1 320 2 2 2 3320 24 3320 (Shown)42 x x x x x x xx              (ii) dA 3 320 2d 4 2 xx     3320 2 0 42 9.37218 x x     2 2 d 3 3 20d 4 2 A x     A has a maximum value when 9.37218x . Max A     23320 9.37218 9.3721842     93.7218 93.72 (to 2 d.p)   6 (i)  fyx  B(– 4, 0) A(– 2, 0) y = 0 x = 0 x y

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 4 of 10 (ii)   1 fy x (iii)  f1yx 7 (i) ln , 1, 0x t t y t t     d 1 d1 , 1, dd xy t t t   d1 1d1 1 yt xt t   Since 0, 1 0, tt   d 0 for all > 0d1 yt txt Hence C does not have a stationary point (ii) When x = 0, ln 0tt 0.5671432904t (by g.c.) (0, 1.567) y=1 A (– 2, 1) y = 0 x y B (– 4, 1 3 ) C: x = –5 y = 0 x = –1 x y

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 5 of 10 1 0.5671432904 1.5671432904y   When 0, , y 0 + 1=1tx   (iii) When t = 1, x = 1 ln1 1 1 1 2,y   d1 d2 y x  Equation of normal : 2 2( 1)yx   24yx   (iv) Volume generated =   1 22 0.5671432904 11( 1) 1 d 2 (1) 3tt t       = 14.10 (2 decimal places) 8   1ln 1 tan 2yx  11 tan 2yex  Diff w.r.t x,   2 2 d1 (2)d 1 4 d1 4 2 (shown)d y y ye xx yxe x     Diff w.r.t x,             2 2 2 22 22 2 22 22 2 22 2 2 d d d1 4 8 2 d d d d d d1 4 8 1 4d d d d d d1 4 8 1 4 0d d d d d d1 4 8 0 (Shown)d d d yy y yx x e x x x y y yx x x x x x y y yx x x x x x y y yxx x x x                     When x = 0, 2 2 3 3 0 d 2d d 4d d 0d y y x y x y x     222y x x        1 11 tan 2ln ln 1 tan 2 ln 11 x xxx        

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 6 of 10   22 2 122 2 33 2 x x x x xx            9 (i) Area R 2 2 1 11d xx        3 2 0 3 2 0 3 0 3 2 0 3 2 0 3 0 2 1 cos sec tan d sin sec tan d 1sin tan dcos tan d sec 1 d tan tan tan 0 033 3 units3                                       (ii) Area Q 2 0 1 d 1 a y y     1 0 1 sin sin a y a       1 1 sin 3 3 3 sin 3 3 2 a a a         

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 7 of 10 10 (i) 0, ,y x a x a   (ii) (iii) Add in the graph 2 2y x a Number of real roots of the equation 2 222 xxa xa    is 3. 11a (i) Total distance       2 2 2.4 1 0.62 2.1 0.3 n n nn     (ii) 22.1 0.3 420nn 34.0799n Least n = 35 11b (i) Amount of outstanding loan at the end of n months           121.002 250000 1.002 1.002 1.002 1 1.002 250000 1.002 1 1.002 250000 500 1.002 1 n n n n n nn k k k k k             y = 0 xa xa y x 2 a

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 8 of 10 (ii) n = 240 months     240 2401.002 250000 500 1.002 1 0k   1312.61186 $1313 (to nearest dollars)k   12 (i) 2.414, 0.414, 2     (ii) If the sequence converges, then as 1, and nnn x l x l      1 3 3 52 0 5 2 ll ll     From part (i) , the roots of the equation 3 5 2 0xx are ,,   , , or l l l       The sequence converges to either  ,  or  . (iii)   1 331 5 2 5 2n n n n n nx x x x x x        From the given graph, if nx , 3 5 2 0nnxx   1 1 00 nn nn RHS LHS x x xx         From the given graph, if ornnxx  , 3 5 2 0nnxx   1 1 00 nn nn RHS LHS x x xx         (iv)   1x

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 9 of 10 13 (i) Sub 23, 5, 2 into , x y z l    3 2 1 2 b b      (ii) 1 2 -------- (1)s   4 5 1 3 2 1 -------- (2) ss s         From (1) and (2), 2 1 1 0 4 1 5 3 2 1 OB                               Coordinates of B = (0, 5, –1) (iii) 1 4 32 s OC s s      3 1 4 5 4 1 2 3 2 5 2 ss CA s s ss                                     1 00 1 41 1 0 0 5 2 1 4 5 2 0 3 CA s s s ss s                               1 3 2 4 3 7 3 2 3 3 OC          (iv) Let 'C be the point on l3 where C is reflected about l2 '2OC OA OC   32 2 5 7 23 4 3 1                   

Yishun Junior College  JC2 Preliminary Examination 2012  H2 Mathematics 9740/1 Pg 10 of 10 41 ' 2 , 2 22 BC BD                     Normal vector of  = 4 1 0 2 2 6 2 2 6                               Cartesian equation of  : 0 1 0 1 3 1 1 1 1 4yz                               r