2017 SNGS Sec 4 AM Prelim P2 Solutions

1 The curve )(f xy  is such that 542)('f 2  xxx . (i) Explain why the curve )(f xy  has no stationary points. [2] (ii) Given that the curve passes through the point (–1, –6), find an expression for )(f x . [3] (i)  5.222)('f 2  xxx   5.112 2  x   312 2  x Since   01 2 x for all real x,   031 2 x for all x. 0)('f x , f(x) has no stationary points. Method 2 0542 2  xx 024 )5)(2(4)4(4 22   acb No real roots. Hence, 0)('f x , f(x) has no stationary points. Note: must conclude with 0)('f x (ii)   dxxxx 542)(f 2 cxxxx  523 2)(f 23 Sub (–1, –6), 3 5c 3 5523 2)(f 23  xxxx

2 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 2 The function f is defined by   82f 234  kxxxx , where k is a constant. It is given that 0)(f x has a repeated root 2. (i) Find the value of k, [2] (ii) Determine, showing all necessary working, the number of real solution(s) of the equation   0f x . [4] (i)       0822222f 234  k 84 k 2k Repeated roots means the roots of the equation are x = 2 and x = 2 (ii)   822f 234  xxxx     )(2)2(f 2 cbxaxxxx  By comparing, a = 1, c = 2,   )2)(44(f 22  bxxxxx Compare x2 term, 2222 2442 xxbxx  2b   )22()2(f 22  xxxx   0)22()2(f 22  xxxx   02 2 x or 0)22( 2  xx 2x (repeated) or     12 21422 2x (no real soln)  1 real solution or 2 real and repeated solutions Repeated root means  2)2(  xx or expand )44( 2  xx 22 0 882 882 882 862 44 82244 2 2 2 23 23 234 2342        xx xx xx xxx xx xxx xxxxx

3 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 3 The roots of the equation 042 2  pxx , where p is a constant, are α and β. The roots of the equation 096 2  qxx , where q is a constant, are   2 and   2 . Find the value of p and of q. [6] 042 2  pxx 22 4   …(1) 2 p …(2) 096 2  qxx   2 +   2 = 6 q …(3)   2    2 = 6 9  = 2 3 ...(4) (2) = (4), 2 3 2 p p = 3 From (3),   2 +   2 =   33 = 6 q Sub (2) and p = 3, 42 3 6 33 qq   …(5)    2 222  =      2 324 = 1   2233   =         2 312 = 1 From (5), 14  q 4q

4 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 4 The diagram shows part of the curve   2 21 50 x y   . The tangent 2685  xy at the point A on the curve cuts the x-axis at B. The normal at A cuts the x-axis at C. Find the area of triangle ABC. [9]     3 212250   xdx dy   3 21 200 xdx dy   , Gradient of tangent 5 8   5 8 21 200 3   x   3 21125 x 521  x 2x 2y A = (–2, 2) Sub y=0 into 2