SRJC H2 MATH P1 Solution

1 [TURN OVER] SERANGOON JUNIOR COLLEGE 2018 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9758/1 11 Sept 2018 3 hours Additional materials: Writing paper List of Formulae (MF 26) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calcul ator are not allowed in a question, you are required to present the mathematical steps us ing mathematical notatio ns and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. Total marks for this paper is 100 marks. This question paper consists of 6 printed pages (inclusive of this page) and 2 blank pages.

2 Answer all questions [100 marks]. 1 Find 2 2 d25 x xxx++ . [3] Solution 2 22 25 d1 d25 25 xx x xx xx x +=−++ ++ () 22 22 11 d d 3 d 25 14 xx xxxx x +=− − ++ ++  () 21 31ln 2 5 tan 22 xx xx c − +=− + + − +  2 The complex numbers z and w satisfy the equations *2 1 5 izw z+= and 23 1 1wz+= . Find the complex numbers z and w. [6] Solution ()*2 1 5 i 1zw z+= LL () ( )1 11 3 22wz=− LL Subst (2) into (1) gives ()1 11 3 * 2 15i2 zz z −+ = 211 3 21 5 i22zz z−+ = 215 3 15i22zz−= Let izxy=+ () () 2215 3 i1 5 i22 xy x y+= + + Comparing the real and Imaginary parts, we have 15 152 y = and () 2215 3 3 322 2xx y=+ LL 2y = Subst into (3) gives 2 54 0xx −+ = () ( )41 0xx−− = 1x∴= or 4 When 12 iz =+ , ()1 11 3 6i 4 3i2w =− − = − When 42 iz =+ , ()1111 12 6i 3i22w =− − = − −

3 [TURN OVER] 3 Without the use of a graphing calculator, find the range of values of x for which 218 492 xxx ≥++− . [3] Hence find the exact range of values of x for which 218 e4 e 92e xx x ≥+ +− . [3] Solution 218 492 xxx ≥++− () () 218 2 4 9 02 xx x x +− + + ≥− 3218 2 18 02 xx x x ++ + − ≥− () 2 1 02 xx x + ≥− 1x =− or 02 x≤< Replace x by ex So e1x =− or 0e 2 x≤< No solution or ()0 e and e 2xx≤< No solution or () and ln 2xx∈<¡ ln 2x∴< (i) The curve with equation 2 2 19 y x−= undergoes a two-step transformations to become curve C with equation () 22 1 194 xy −−= . State the two transformations involved for the curve 2 2 19 y x−= . [2] (ii) Draw a sketch of the curve C, labelling clearly the equation(s) of its asymptote(s), intersection with the axes and the coordinates of any turning points. [3] (iii) Show that the point () 1, 3−− lies on the line 3ym x m=+ − for all real values of m. [1] (iv) Hence using the diagram drawn in (ii), find the range of values of k such that the equation () () 22 31 194 kx k x+− − −= has 2 negative real roots. [2] Solution (i) Method 1 ‒1 0 2 ‒ ‒ ‒ + 4

4 () 2222 2 2 111 199 4 9 4 xyy x yx −−= → − = → − = (1) Scaling parallel to the x axis by a scale factor of 2. (2) Translation of 1 unit in the positive x direction. Method 2 () 2222 2 2 1111 199 2 9 4 xyy y xx −− = →− − = →− = (1) Translation of 0.5 units in the positive x direction (2) Scaling parallel to the x axis by a scale factor of 2. (iii) When 1x =− , () SR3HS 1 3 LHmm−+ − − === Hence () 1, 3−− lies on the line 3ym x m=+ − for all m. (iv) From the sketch, the range of values of k are 35 32k >+ or 3 2k <− x y

5 [TURN OVER] 5 (i) Find ()ln 1 dxx+ for 1x >− . Show your working clearly. [2] (ii) The curve C is defined by the parametric equations () ()22 l n 12 , 22 l n 11xt t y t t=− + + = − − + + where 1t >− . Another curve L is defined by the equation () 2 21 6xy=+ − . The graphs of C and L intersect at the point A()2,1 as shown in the diagram below. Find the area of the shaded region bounded by C, L and the line 1l n 4y =− − , express your answer in the form 3 262 164233 AAA+− − , where A is an exact real constant. [6] Solution (i) () ()ln 1 d ln 1 d 1 xxxx x x x+=+ − + () () () 1ln 1 1 d 1 ln 1 ln 1 xx x x xx x x c =+ − − + =+ − + + +  () () 1l n1xx x c=+ +− + (ii) Area = () 11 2 1l n 4 1l n 4 d2 1 6 dCxy y y −− −− −+ −  () () 130 1 1l n 4 21222 2 l n 1 2 d 6 13 ytt t y t −−  +=+ − + − − − −  +  () () 31 0 2l n 411 64 1 ln 1 1 d 6 6 6 ln 4 13 3tt t t  −=+ − + + − − − + +  +   () () () 311 1 00 0 ln 1 2 ln 4 2042 d 4 l n 1 d 4 d 6 l n 4 13 3 ttt t t t t +=+ − +− − − + − +   L C 0

6 ( ) () () () () () 311 122 00 0 2l n 4202 2 4 1 ln 1 2 ln 1 6 ln 4 33tt t t t  =+ −+ + − − + − − + −       () [] () () 3 2 2l n 42018 8 4 2ln 2 1 2 ln 2 6ln 4 33=− − − − + − + () () 3 2 2l n 4201 4 8l n2 2 l n2 6l n4 33=− − + − + () () 3 2 16 ln 262 4l n2 2 l n233=+ − − 6 (a) (i) The twelfth, eighth and fifth terms of an arithmetic progression are three consecutive terms of a converging geometric progression of positive terms with common ratio r. Find the value of r. [3] (ii) Take the value of r to be 3 4 . If the difference between the sum of the first n terms of the geometric progression and its sum to infinity is less than 0.3% of the sum to infinity, find the least value of n. [3] (b) A convergent geometric sequence of positive terms, G has first non-zero term a and common ratio r. (i) The sum of the first n odd-numbered terms of G is equal to the sum of all terms after the ()21n − th term of G. Show that 22 121 0nnrr −+− = . [2] (ii) In another sequence H, each term is the reciprocal of the corresponding term of G. If the nth term of G and H is denoted by nu and nv respectively, show that a new sequence whose nth term is ln n n u v    , is an arithmetic progression. [2] Solution (ai) Let b and d be the first term and common difference of the arithmetic progression. () 1 11 1nar b d− =+ LL ()72nar b d=+ LL () 1 43nar b d+ =+ LL Equation () ( )12− gives 1 4nnar ar d− −= Equation () ()23− gives 1 3nnar ar d +−= Hence () () 1143 nn n nar ar ar ar+−−= − 1147 30nn nar ar ar+− −+ =

7 [TURN OVER] () 12 47 3 0nar r r− −+= () ( )43 10rr−− = Since a and r are non-zero 3 4r = Since it is a converging GP. (ii) ()1 0.00311 1 nar aa rr r − −< −− −  or ()1 0.00311 1 naraa rr r − −< −− −  Since nSS∞ > as a and r are positive. 3 0.0034 n  < ()ln 0.003 3ln 4 n >  20.19n > Hence least n is 21. (bi) () 124 2 2 12 ... ... n nnaa r a r a r a r a r − −+++ + = + + () 2 21 2 1 11 n nar ar rr −− =−− for sum of GP Since 0a ≠ , 1r ≠ () () 22 11 11 1 nnrr rr r −− =−+ − () 22 111 nnrr r −−= + () 22 1 2 22 1 1 2 1 0 shown nn n nn rr r rr − − −= + +− = (ii) 1 1 ln ln nn nn uu vv − −    −      11 1 1 ln ln ln ln ln ln nnn n nn nn uvu v uv uv −− − − =