AJC_H2_MATH_P1_SOLUTIONS

Anderson Junior College Preliminary Examination 2011 H2 Mathematics Paper 1 (Solutions) 1 12cos dx xx  22 12 4 23 12 4 2 12 4 2cos d22 1 14cos d24 1 1cos 122 xx x x x x xx xx x x xx C               2 Let the price of a X-box console be $ x, a Kinect sensor be $y and a Game DVD be $z. 499 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30 xyz xy z xy z      Aug matrix = 11 1 4 9 9 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30     rref = 1002 4 7 0101 9 9 001 5 3    x = 247 , y = 199 , z = 53 Employees of Company P will pay $[0.9(247)+0.8(199)+0.85(53)] = $426.55 > $426.30. No, it will not be more attractive for employees to purchase all the 3 items from own company. 3      2 0xaxb xxc    , x c , 0x Since  2 0xc as x c ,   0xaxb x     0xx a x b  x a or 0 x b and x c Replace x by ln x . ln x a or 0l n x b  and ln x c 0 ax e or 1 bx e  and cx e

4 (i) tantx  22sec 1dt x tdx   = 2 2 2 11 11 1 dtt t t      = 2 1 12 dtt = 11 tan 2 2 tc  = 11 tan 2 tan 2 x c  (ii) volume is same as Exact volume = 2 4 20 1 d1s i n xx      41 0 1 1 tan 2 tan 2 tan 2 2 x         5 (a) Method 1: sin 2 cos dx xx 1 sin 3 sin d2 1c o s 3 cos23 1c o s 3 cos23 xx x x x C x x C         Method 2: sin 2 cos dx xx y =2 2 1 1s i ny x  2 12 1s i ny x 

1 2 3 [f ( )]2sin cos d [use f ( )[f ( )] d ] 1 2 cos3 n n xx x x xxx c n xC        (b) (i) dd cos 1, 2cos 2dd xy tttt  dd d2 c o s 2 dd dc o s 1 yy t t x tx t  When d 0,d y x  33 1 1 3cos 2 0 2 , , , 22 44 4 4 22 tt t x         At point A, 1 ,142 x y  1y  is the equation of the tangent to the curve at point A. Or Since 0 t  , the maximum and minimum values of y (i.e. sin 2y t ) is 1 and -1. The y- coordinate of point A is 1 and since the tangent to this max pt is a horizontal line ( d 0,d y x  ), therefore the equation of the tangent to the curve at point A is y = 1. (ii) y x A

 1 42 4 4 44 Area 1 d 31 sin 2 cos 1 d4 2 31 sin 2 cos sin 2 d4 2 3 1 1 cos3 cos 2 cos42 3 22 31 2 1 1 43 2 23 2 31 2 2 46 3 yx tt t tt t t x tx                                          6  1 2s i n 2y x    2 2 2cos2 2c o s 2 2s i n 2 dy x y xdx x    Differentiating wrt x:  2 2 2 2 2 2 4c o s 2 22 s i n 2 2( 2 cos 2 ) 4 sin 2 dy d y y x y xdx dx dy d y y x y xdx dx      2 2 2 22 2 2 124 s i n 2 2 4s i n 2 ( ) d y dy dy yxdx y dx dx dy d y y x showndx y dx        (i) 332 2 32 2 24 8s i n 2 8 c o s 2d y dy dy d y dy yx y xdx y dx y dx dx dx           