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 of line, m = 66 66 66 20    Gradient of line, m =   30 6620  Gradient of line, m =   63 24/3 6212/663 2 

5 No. Answer 5 (i) Factorise completely 2x3 + 7x2 + 4x – 4. Let f(x) = 2x3 + 7x2 + 4x – 4 42 142 172 122 1f 23                  = 0 (2x – 1) is a factor of f(x). f(x) = 2x3 + 7x2 + 4x – 4 f(x) = (2x – 1)(x2 + 4x + 4) f(x) = (2x – 1)(x + 2)2 (ii) Express 4472 313212 23 2   xxx xx in partial fractions. 2 2 )2()12( 313212   xx xx = 2)2(212   x C x B x A = 2 2 )2()12( )12()2)(12()2(   xx xCxxBxA 12x2 + 32x + 31 = A(x + 2)2 + B(2x – 1)(x + 2) + C(2x – 1) When x = 2 1 , 22 2 5312 1322 112              A 504 25 A A = 8 When x = – 2, 12(– 2)2 + 32(– 2) + 31 = C(– 5) – 5C = 15 C = – 3

6 No. Answer When x = 0, 12(0)2 + 32(0) + 31 = 8(2)2 + B(2)(– 1) + (– 3)( – 1) 35 – 2B = 31 B = 2 2 2 )2()12( 313212   xx xx = 2)2( 3 2 2 12 8   xxx 6 The equation of a curve is y = ax2 – 3x + 4 – a, where a is a constant. (i) In the case where a = – 2, find the set of values of x for which the curve lies completely below the line y = – 3. When a = – 2, y = – 2x2 – 3x + 6 If the curve lies completely below the line y = – 3, – 2x2 – 3x + 6  – 3 2x2 + 3x – 6  3 2x2 + 3x – 9  0 (2x – 3)(x + 3)  0 Set of values of x: x   3 OR x  2 3

7 No. Answer (ii) In the case where a = 3, show that the line y = 3x – 2 is a tangent to the curve. When a = 3, y = 3x2 – 3x + 1 When the curve and line intersects, 3x2 – 3x + 1 = 3x – 2 3x2 – 6x + 3 = 0 x2 – 2x + 1 = 0 b2 – 4ac = (– 2)2 – 4(1)(1) b2 – 4ac = 0 Since discriminant = 0, the line y = 3x – 2 intersects the curve at only 1 real and distinct point. Hence the line is a tangent to the curve. (iii) Determine if there is any other value of a for which the line y = 3x – 2 intersects the curve at only one point. When the curve and line intersects, ax2 – 3x + 4 – a = 3x – 2 ax2 – 6x + (6 – a) = 0 For line to intersect curve at only 1 point, b2 – 4ac = 0. (– 6)2 – 4(a)(6 – a) = 0 4a2 – 24a + 36 = 0 a2 – 6a + 9 = 0 (a – 3)2 = 0 a = 3 [part (ii)] There is only 1 real and repeated root for a, i.e. a = 3. Hence there is no other value of a for which the line intersects the curve at one point only.

8 No. Answer 7 (i) Prove that .tan)1(sin)tan(sec sin    )1(sin)tan(sec sin    = )1(sincos sin cos 1 sin            = )1(sincos sin1 sin           =   2sin1 cossin  =   2cos cossin =   cos sin = tan  = RHS (proven) (ii) Find all the values of  between 0 and  for which .sec1)1(sin)tan(sec sin 2     2sec1)1(sin)tan(sec sin  tan  = 1 – sec2  tan  = 1 – (1 + tan2 ) tan2  + tan  = 0 tan  (tan  + 1) = 0 tan  = 0 (no solution) OR tan  =  1

9 No. Answer Since tan   0,  lies in the 2nd quadrant. Consider tan  = 1,  = .4   = 4    = 4 3 rad. / 2.36 rad. 8 An auction house claimed that it is worthwhile to invest in their art pieces as the value of one of their art pieces has been increasing exponentially since it was produced. The value, $V, of this art piece is related to t, the number of years since it was produced at the start of the year 1995. The variables V and t can be modelled by the equation V = 10 000 + ae kt, where a and k are constants. The table below gives values of V and t at the start of some of the years 2000 to 2015. Year 2000 2005 2010 2015 t years 5 10 15 20 $V 16 000 20 260 27 545 40 000 (i) Plot a suitable straight line graph to show that the model is valid for the years 2000 to 2015. V = 10 000 + ae kt V – 10 000 = ae kt ln (V – 10 000) = ln a + ln e kt ln (V – 10 000) = kt + ln a Vertical axis: ln (V – 10 000) Horizontal axis: t Gradient: k Vertical axis-intercept: ln a

10 No. Answer t years 5 10 15 20 $V 16 000 20 260 27 545 40 000 ln (v – 10 000) 8.70 9.24 9.77 10.31  ln (V – 10 000) = kt + ln a  Table of values  Straight line graph with suitable scale (ii) Estimate the value of a and of k. ln a = 8.1875 ln a = e 8.1875 ln a  3595.72 ln a = 3600 (3 s.f.) OR 3596 (nearest whole no.) k = 75.075.13 25.8625.9   k = 104 11 / 0.106 (3 s.f.) (iii) A claim was made that in the year 2065, this art piece will increase in value by 500 times