PLMGS 4E_5NA_AM P1 Prelim 2017_Worked Solutions

1 Paya Lebar Methodist Girls’ School (Secondary) Department of Mathematics 2017 Preliminary Examination Additional Mathematics Paper 1 (4047/1) Worked Solutions No. Answer 1 State the values between which each of the following must lie (a) the principal value of tan  1 x, Principal value of tan  1 x =  90  tan  1 x  90 Principal value of tan  1 x =  2   tan  1 x  2  (b) the principal value of cos  1 2x. Principal value of cos  1 2x = 0  cos  1 2x  90 Principal value of cos  1 2x = 0  cos  1 2x  2  2 The function f is defined, for all values of x, by f(x) = (x + 3)(1 – 2x)2. Find the range of values of x for which f is a decreasing function. f(x) = (x + 3)(1 – 2x)2 = (x + 3)(1 – 4x + 4x2) = x – 4x2 + 4x3 + 3 – 12x + 12x2 = 4x3 + 8x2 – 11x + 3 f (x) = 12x2 + 16x – 11 For f(x) to be a decreasing function, f (x)  0 12x2 + 16x – 11  0 (6x + 11)(2x – 1)  0

2 No. Answer Range of values of x:  6 11  x  2 1 3 In the expansion of (2x – 1)2 8 1       x p , where p is a positive constant, there is no term in .1 3x Find the possible values of the constant p. (2x – 1)2 8 1       x p = (4x2 – 4x + 1)                                  ...)1(5 8)1(4 8)1(3 8...1 5 3 4 4 3 5 x p x p x p 05 8)4(4 8)4(3 8 543                 ppp 56p3 – 280p4 + 224p5 = 0 56p3 (1 – 5p + 4p2) = 0 56p3 (4p – 1) (p – 1) = 0 p = 0 (N.A. as p  0) OR p = 2 1 OR p = 1

3 No. Answer 4 A curve has the equation y = 4x2 – 24x + 30. (i) Express 4x2 – 24x + 30 in the form [a (x + h)] 2 + k. 4x2 – 24x + 30 = 4     2 1562 xx 4x2 – 24x + 30 = 4      92 15)3( 2x 4x2 – 24x + 30 = 4      2 3)3( 2x 4x2 – 24x + 30 = 4 (x – 3)2 – 6 4x2 – 24x + 30 = [2 (x – 3)]2 – 6 (ii) Show that the minimum point of the curve has coordinates (3, – 6). Since [2 (x – 3)]2 ≥ 0, the lowest value of y = [2 (x – 3)]2 – 6 is – 6 and this occurs when [2 (x – 3)]2 = 0, when x has a value of 3. Hence the minimum point on the curve has coordinates (3, – 6). (iii) Sketch the graph of y =  4x2 – 24x + 30 , indicating clearly the exact x-intercept(s) and y- intercept.  Shape of curve and point (3, – 6)  y-intercept  x-intercepts

4 No. Answer A line of gradient m passes through the point (0, – 10). (iv) Given that 0  m  10, determine the exact value of m, for which the line intersects the graph of y =  4x2 – 24x + 30  at one real and distinct point. For line to intersect curve at 1 real and distinct point only, the point of intersection is .0,2 63        Gradient of line, m = 02 63 )10(0         Gradient of line, m =        2 6610 Gradient of line, m =         66 210 Gradient o