NYJC_H2_MATH_P2_Answer

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 1 of 10 Qn Solution 1 Let nP be the statement     1 sin 2 1 scos 2 i i 2s n nn r nr    for n  When n = 1, LHS   1 1 cos 2 cos 2 r r   RHS sinsin3 2sin    2cos2 cos22sin ins  1P is true. Assume that kP true for some , 1kk    . i.e.     1 sin 2 1 scos 2 i i 2s n nk r kr    For n = k +1, we want to prove    1 1 sinsin 2 3cos 2 2sin k r kr      LHS       1 11 cos 2 cos 2 cos 2 2 kk rr r r k            sin 2 1 cos 2 22sin sink k         sin 2 1 cos 2 sin 2 in in s2 2s kk              sin 2 1 sin 2 3 sin 2 1 2sin sink k k           sin 2 3 2s n in sik    kP is true 1kP is true Since P1 is true and Pk is true  Pk+1 is true, by the principle of Mathematical Induction, Pn is true for all n  +.     11 sinsin 2 1 cos cos 2 2sin ( 1) 122·22 2 nn rr nn rr          

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 2 of 10 Qn Solution 2(a) (b) Let y = 21 22 x x   (2 2) 2 1 (2 2) 2 1 21 22 y x x x y y yx y          1 21f : , 1 22 xxx x    f(x) = f –1(x)  f 2(x) = x f 1965 (0.5) = 0 (i), (ii) (iii) For equation 11g g( ) g ( )x g x  , 26 x . x y O y = x g( )yx 1 )g (y x (0,2) (16,6) x (2,0) (6,16) x

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 3 of 10 Qn Solution 3(i) (ii) Since 1 52iz   and 1 103iz   , thus 1 4 2iz  satisfies the equation 52iz  and i 03 1z  . (iii) Note that triangle OBC is a right angle triangle. Further B lies on the locus of 52iz  . Thus       2 2 21 1 1 5 10 102 4 2 55 522 5( 1) area              (iv) 1 5sin 5  Thus 2   and 2   . Hence the required range is 2.03 arg( 1.12) 14iz  . A(2,-1) O Re Im C(4,-2) D(2,4)    B(3,1)

2015 NYJC JC2 Prelim Exam 9740/2 Solutions Page 4 of 10 Qn Solution 4 (i) Let 1tanyx  . Then tan yx Differentiate with respect to x, 2 dsec 1 d yy x    2 d1 tan 1 d yy x 2 d1 d1 y xx  (shown) (ii) From graph, area of shaded rectangle < area under curve for 1n x n   .  22 1 11 1d11 n n n n xnx        1 2 1 1 tan1 n n xn        11 2 1 tan tan 11 nnn    (show) (iii) Comparing area of rectangles with area under curve for 0 xn