SRJC H2 MATH P2 Solution

1 [TURN OVER] SERANGOON JUNIOR COLLEGE 2018 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9758/2 17 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 7 printed pages (inclusive of this page) and 1 blank page.

2 Section A: Pure Mathematics [40 marks]. 1 R is the region enclosed by the line 5y =− and the curves 2 25yx x=− + − and () 22 5 141 6 yx ++= as shown in the diagram below. Find the volume generated by region R when it is rotated 2π radians about the x-axis. Leave your answer correct to 2 decimal places. [3] Solution Required volume ( ) () () ( ) 2 21.09 0 22 21 . 0 9 51 6 4 d 25 d 2 5 2xx x x xππ π − −− =− − − + − + − − OR ( ) () () 2 21.09 0 0 222 21 . 0 9 2 5 1 64 d 2 5d 5dxx xx x xππ π − −− − =− − − + − + − − −  = 146.91 Using Shell method Vol () 25 8.36 521 4 4 d 4 yyy yπ − −  +=− − − − + −   = 146.91 2 A curve has parametric equations tanx θ= , 2secy θ= for 02 θπ≤< . The equation of the tangent to the curve at the point P with parameter p is given by ()2si n 2c osyp x p=+ . (i) The tangent at P meets the x and y axes at the points A and B respectively. Find the Cartesian equation of the locus of the mid-point of AB as p varies. [3] (ii) The tangent at P meets the line 2yx= at the point S and the line 2yx=− at the point T. Show that the area of the triangle OST is independent of p, where O is the origin. [4] Solution (i) Coordinates of A: () cot ,0p− Coordinates of B: ()0, 2 cos p x y R

3 [TURN OVER] Midpoint of AB: cot ,cos2 p p− So let cot 2 px =− and cosyp= 1tan 2p x=− and 1sec p y= Since 221t a n s e cpp+= 22 111 4x y += (ii) To find the coordinates of S: ()22 s i n 2 c o sx px p=+ cos 1s i n px p∴= − and 2c o s 1s i n py p= − To find the coordinates of T: ()22 s i n 2 c o sx px p−= + cos 1s i n px p −∴= + and 2c os 1s i n py p= + Method 1 Hence area of triangle OST 22 22 11 cos 2cos cos 2cos 1 sin 2 tan2 1 sin 1 sin 1 sin 1 sin 2 pp p p pppp −    − =+ +     −−++     – is for using formula with the correct angle used. () () 2 111c o s 1 1 5 5 2sin tan cos tan2 1 sin 1 sin 2 2 p pp −−  =   −+    2 2 5c o s 2 1 22 1s i n 55 2 p p  =× × −  = Method 2 Hence area of triangle OST 22 22 11 cos 2 cos cos 2 cos 4 sin tan2 1 sin 1 sin 1 sin 1 sin 3 pp p p pppp −    − =+ +     −−++     – is for using formula with the correct angle used. () () 2 11c o s 4 55 s i nt a n21 s i n 1 s i n 3 p pp −= −+  2 2 5c o s 4 25 1s i n 2 p p  =  −  = 3 Given that 1tanln e xy − = , show that () () 2 2 2 dd1l n 1 2 dd yyxy x xx += + − . [2] (i) Find the Maclaurin’s series for y up to and including the term in 3x . [4]

4 (ii) Deduce the Maclaurin’s series for y, where 1tanln e 1 xy x − =+ − , up to and including the term in 3x . [3] Solution 1tanln e xy − = 1tan 2 1d 1 ed 1 xy yx x − = + () 2 d1l n d yx yyx+= () 2 2 2 ddd d12 l n dd dd yy y yxx y x xxx ++ = + () () 2 2 2 dd1l n 1 2 dd yyxy x xx += + − (i) () () 32 2 2 32 2 dd d d 1 d12 l n 1 2 2 dddd d yy y y yxx y x xyxxx x ++ = + −+ −  When x = 0, ey = , d ed y x = , 2 2 d 2e d y x = and 3 3 d 3e d y x = So the Maclaurin series of y is 23 1ee e e . . . 2yx x x=+ + + + (ii) 11tan tane1 e 1 ••ee e e xx xxy −− +− −== 32 31tane1 2 1 •• ee e e e ... 1 ... e 22 6 x x xx xyx x x − +− −   == + + + + + + + +      32 3 3 22 31. . . 22 2 6 xx x xxx xx x=+ + + + + + + + + + 2351 31 2 ... 26xx x=+ + + + 4 Given that the equation () ( ) ( ) 32f2 2 4 2 8 0zz a z az a=++−+− = has no real solution, explain clearly why a is not a real number. [2] (i) It is known that f has a factor ()2iz + , find a. Hence find all the roots of ()f0 z = , showing your workings clearly. [4] (ii) Deduce the roots of the equation () 23 24 2820 aaa ww w − −−+ − = , where a takes the value obtained in (i). [2] Solution If all coefficients are real, then by co njugate root theorem, roots will occur in conjugate pairs. As f is a polynomial of degree 3, it would mean that it must

5 [TURN OVER] have either 3 real roots or it will be having 1 real root with a pair of conjugate roots. But it is given that f(z) = 0 has no real solution and so it must therefore means that at least one of the coefficients is a complex number. Hence a is not a real number. (i) () 32 ii i if2 2 42 8 022 2 2 aa a  −= − + − + − −+− =     i i2 i2 8044 a aa−−++ − = 978i i44 a−= −  4i=+ So () ( ) ( ) ( ) () 32 2f2 4 i 4 2 i 2 i 2 i 2 2zz z z z zz=+ + + + + =+ + + For ()f0 z = () () 22i 220zz z ++ + = Consider 2 22 0zz ++ = , 24 8 2z −± −= 1iz =− ± Hence the roots are 1i−+ , 1i−− and i 2− (ii) Since () 3222 4 2 8 0za z a z a++−+ − = , 23 24 2820 aa a zz z −− + ++= Replace z by w− 23 24 2820 aa a ww w −− −+ − = So the roots of this equations are 1i− , 1i+ and i 2 5 A hollow metallic ramp, in the shape of a prism, is constructed for the marching contingent to march onto to reach an elevated platform from the ground during the national day parade. The diagram below shows the prism with O as the origin of position vectors and the unit vectors i, j and k are parallel to OA, OC and OE respectively. It is given that OE = CD = 1 m, OA = CB = 2 m and OC = AB = ED = 4 m. j O E A C B D k i 4 m 1 m 2 m

6 A laser beam in the form of a line l has Cartesian equation , 12 ax zy+ == , where a ∈¡ , is emitted onto the plane ABDE. (i) Find, in terms of a, the coordinates of the point of intersection, M, of the laser beam and the plane ABDE. [4] For the following parts of the question assume 0a = . (ii) The laser beam is reflected about the plane ABDE. By finding the foot of perpendicular from ()0,1, 0Q to the plane ABDE, find the equation of the reflected beam. [5] (iii) The path traced out by an ant crawling on the floor OABC is given by 21 02 , 00 ββ −     =+ ∈       r ¡ . Let P be the point on the path, located under the ramp, whereby the a