HCI_JC_2_H2_Maths_2012_Paper 2 Solution - for Sharing

1 © Hwa Chong Institution 2012 9740/02/Prelim/12 HCI H2 Math Prelim Paper 2 Solution Qn. Solutions 1 (i) Since y = 0.5 cuts the graph twice, f is not a one-one function. Therefore 1f does not exist. Restrict the domain to 13 22 x so that the range of h is equal to the range of f.  m = 1 2 , n = 3 2 Since h( 5 6 ) = 1 2 , therefore 1 15h ( )26   . (ii) With domain 13 22 x , hR [ 1, 1] , gD [ 1 , )   , Since hgRD , therefore gh exists. Substituting hR as gD into y = g(x) graph,

2 © Hwa Chong Institution 2012 9740/02/Prelim/12 1 ghR [ , ] aa  2 Let w be the width of the paper used and h be the length of the paper used. Area of paper used, A = wh We have ( 4)( 8) 1352wh   1352 4( 8)w h    Therefore 1352 4( 8)A wh h h     1352 4( 8) h hh 

3 © Hwa Chong Institution 2012 9740/02/Prelim/12 2 d ( 8)(1352) (1352 )(1) 4d ( 8) A h h hh   2 10816 4( 8)h   To minimise the paper used, d 0 d A h  2 10816 4( 8)h  2( 8) 2704h   2704 8 60h    First derivative test: h (60)  60 (60)  d d A h -ve 0 +ve OR Second derivative test: 2 23 d 21632 0d ( 8) A hh  Hence at 60h , the amount of paper used is minimised. At 60h , 1352 452w   30w Hence the dimension of the paper is 30 cm by 60 cm. 3 (i) i i i i 11 e e e e cos isin cos isin 2cos (shown) n n n n nnz z n n n n n                   

4 © Hwa Chong Institution 2012 9740/02/Prelim/12 (ii) 3 3 3 3 3 1By taking 1, 2cos 11cos 2 11cos 2 11 8 1 3 1 3 (shown)8 nz z z z z z z z zz zz                        33 3 3 3 1 3 1cos 3 8 1 1 1= 3 8 1= 2cos3 3 2cos8 13= cos3 cos (shown)44 zz zz zz zz                (iii) Method 1 3 13cos 3 d cos9 cos3 d44 11sin 9 sin 336 4 C            Method 2   32 2 2 3 cos 3 d cos3 cos 3 d cos3 1 sin 3 d cos3 cos3 sin 3 d 11 sin 3 sin 339 C                      

5 © Hwa Chong Institution 2012 9740/02/Prelim/12 4 (a)       22 2 2 50 4 5 0 Since we have cos 60 2 2 5 0 2 5 0 2 24 6 1 or 6 1 rej2 61 AB OP                           b a a b a b a b aa b a b a b aa aa a a (b)     3 4 3 2 2 7 7 Let : , then 11 OB OCOC OE AD AB OD OB OA                 a b a ba Since , , are collinear, for some constant O E D OE kOD k  32 17k     ba ba 3 5 : 3: 5AD AB    5 (i) For x > 0,   2 1x > 0 Since a, b and c are positive, for x > 0,   2 1 c x > 0 and ax b > 0   