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 2685  xy , x = –3.25 B=(–3.25, 0) Gradient of normal 8 5 8 5 )2( 20   x 2.1x C = (1.2, 0) Area of ABC =  45.422 1 4.45 sq units A B C O y x

5 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 5 (a) (i) Write down the general term in the binomial expansion of 182 2 4        x x . [1] (ii) Write down the power of x in this general term. [1] (iii) Hence, or otherwise, determine the term independent of x in the binomial expansion of 182 2 4        x x . [2] (b) In the binomial expansion of   n kx2 , where 3n and k is a constant, the coefficients of x and x2 are equal. Express k in terms of n. [4] (a)(i) General term rr r x xC           2 4 182 18 or rr r x xC             182 18 2 4 (a)(ii) Power of x = r336 or 183 r (a)(iii) 36 – 3r = 0 12r term independent of x = 1212182 12 18 2 4           x xC =18564 or r = 6 (b)   n kx2 Term x =     11 1 2 kxC nn   =   xnk n 1 2   Term x2 =     22 2 2 kxC nn   =     222 22 1 xknn n        221 22 12   nn knnnk      3 1 21 2     n n nn nk    1 2 31    nk nn nk  1 4 or 1 4   n , or 1 4   n

6 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 6 Do not use a calculator in this question. (i) Express 133 326  in the form 3ba , where a and b are integers. [3] A toy is modelled in the shape of a right pyramid with a square base as shown in the diagram. The vertical height AB of the pyramid is  43 cm. Given that the length of the slant edge AC is 133 326  cm, (ii) find an expression for BC2 in the form 3dc , where c and d are integers, [3] (iii) express the volume of the pyramid in the form  3393 2 k cm3, where k is an integer. [2] (i) 133 133 133 326      127 3926   39 (ii)     222 4339 BC 163833318812 BC 326652 BC (iii) Volume =  433 1 2 l =    4323 1 2 BC =  433 2   32665  3104782603653 2  =  3391823 2  Base area = ½  BC  BC  4 A B C B C

7 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 7 (i) Differentiate  12ln 2x with respect to x. [2] (ii) Express   121 124 2 2   xx xx in partial fractions. [4] (iii) Hence evaluate     2 1 0 2 2 d121 124 xxx xx . [4] (i)   12lnd d 2xx 12 4 2 x x (ii)   121 124 2 2   xx xx 121 2  x CBx x A     112124 22  xCBxxAxx sub 1x , A33 , A = 1 sub 0x , C 11 , 0C Sub 1x , B237  , 2B      121 124 2 2 xx xx 12 2 1 1 2  x x x (iii)     2 1 0 2 2 d121 124 xxx xx xx x x d12 2 1 12 1 0 2          2 1 0 22 1 0 12ln2 11ln    xx = 1ln2 1 2 3ln2 11ln2 3ln  608.0 Note: Use reverse   12 412lnd d 2 2  x xxx 1)12ln(12 4 2 2 cxdxx x  cxdxx x  )12ln(2 1 12 2 2 2

8 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 8 The diagram shows a cycling circuit formed from four straight roads OA, AB, BC, and CO. OA = 7 km, AB = 2 km, angle OAB = angle BCO = 90°, and angle COA = θ where  900  . A cyclist cycled along the circuit OABCO. (i) Show that cCOBCABOA   cos5sin9 , where c is a constant to be found. [2] (ii) Express COBCABOA  in the form   cR  cos . [4] (iii) Find the values of  for which the cyclist cycled a total distance of 19 km. [3] (iv) State the maximum possible value for the total distance of the circuit. [1] (i) 7sin AY  AY = 7sin, 7cos OY  OY = 7cos 2sin XB  XB = 2sin, 2cos AX  AX = 2cos  sin2cos7  XBOYOC  cos2sin7  AXAYBC Total  sin2cos7cos2sin727  9cos5sin9   Note: use vertical and horizontal lines to divide diagram. The sides 7 cm and 2 cm are usually the hypotenuse of the right- angled triangle. (ii)     sinsincoscos cossin9cos5 RR R   cos5 R sin9 R 22 95 R 106 5 9tan   945.60   99.60cos106  (iii)   1999.60cos106   900    106 10945.60cos   95.609095.6095.60   Ref  763.13  1.299.609.60    763.13,763.13945.60  2.47,7.74 O C A B θ 2 km 7 km X Y θ

9 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 (iv) Max =  9106 19.3 km

10 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/02 9 The equation of a curve is    x