AJC_H2_MATH_P2_SOLUTIONS

Anderson Junior College Preliminary Examination 2011 H2 Mathematics Paper 2 (Solutions) Section A: Pure Mathematics [40 marks] 1 (i) d 100 3d 97 , 0 u kut ku k      1 du 1 d97 1 ln 97 ln 97 97 e where 1 97 e (shown) kt kc kt tku ku t ck ku kt kc ku A A e uA k                  (ii) For the processing tank, when t = 0, u = 0. Hence, 1 97 0 97AAk    . Therefore,   1 97 97e ktu k  . Since u = 70 when t = 1, we have   170 97 97e 0.69220 (5 s.f. from GC)k kk    Amount of water in the processing container when t = 2 =   2(0.6922)1 97 97e0.6922  = 105.03 litres Amount of water pumped in when t =2 = 200 litres Hence, amount of water processed = 200 – 105.03 - 6 litres = 89.0 litres (3 s.f.)

(iii) As   0.692219 7, 97 97e 140.13 litres0.6922 0.6922 ttu      The minimum volume of the tank is 141 litres. 2 (a) (i) Method 1 (using differentiation) 1 2 1f' ( ) ( 1 ) s i n 1 x xx x     For 10 2x , f ' () 0 f ()x x is a strictly increasing function and therefore f is a 1 to 1 function. Since f is 1-1, the inverse function f -1 exist. Method 2 (using graphical test): Any horizontal line y = k intersects the graph of f at at most one point  f is a one-one function  the inverse function f -1 exist. ( Or : any horizontal line y =k, 01 4k   , intersects the graph of f at exactly one point  f is a one-one function  the inverse function f -1 exist) To find the value of f -1(1), solve f(x) = 1 1(1 ) s i n 1 1 1 (rejected since 0) or 0 xx xx x      1f( 1 ) 0  (ii) At 0, f '(0) 1 and 1xy  Equation of tangent to the curve y = f(x) at x = 0: y = x +1 (iii) At x = 1, we will use the symmetry properties of the curve of f and f -1 along y = x  Equation of tangent to the curve y = f-1(x) at x = 1: x = y +1  y = x – 1

2  (b)  (b) (i) Range of g = [-11,5] (ii) Different values of a will translate the curve 1= 1+y x along the x-axis by a units to give 1 = 1+y x a Domain of g g [-11, 5] h Range of hg Since domain of h = [-11, 5] and we need range of hg to be subset of (1, ), therefore [-11, 5] must fall in the interval (a, ) From the diagram, it can be observed that the greatest value of a is -12. 3  (i) 22 2 2 2 1 2 330 2( 3 )1 2 c o s c o s 7012                             22 2 22 22 2 2 322 1 4 2 70 52 2 9 2 20 2 1 9 2 11 40 16 0 (11 4)( 4) 0 44 as = (rejected as <0) 11                      y y x =a x 5-11 y=1 x -11 5 y=g(x)

(ii) Using ratio theorem, 42 1 0 31 1 30 3 344 4 21 7 OB OAOM  