SRJC H2 MATHS P2 Soln

1 [Turn Over SERANGOON JUNIOR COLLEGE 2014 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9740/2 Tuesday 26 Aug 2014 Additional materials: Writing paper List of Formulae (MF15) 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 dark 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 calculator are not allowed in a question, you are required to present the mathematical steps u sing mathematical notations 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 and 1 blank page.

2 [Turn Over Section A: Pure Mathematics [40 marks] 1 The line ℓ with Cartesian equation 4 43 xz  , y = 1 contains the point B with position vector j + 3k. A point A, not lying on ℓ, has position vector 2i + ( 15 )j ‒ k. (i) Given that c denotes a unit vector parallel to ℓ, find AB  c and give a geometrical interpretation of this quantity. [3] (ii) Hence find the shortest distance from A to ℓ. [2] The foot of perpendicular from A to ℓ is denoted by F and the foot of perpendicular from F to AB is denoted by G. (iii) Write down the ratio between the area of ∆AGF and area of ∆BGF. [1] (iv) Hence, deduce the ratio AG:GB and find the position vector of G. [2] Solution (i) 4 4 x = 3 z , y = 1 ⇒ 40 43 xz  , y = 1. Vector equation of ℓ is 44 1 0 , 03                     r . A unit vector c parallel to ℓ = 4 1 05 3    (OR 4 1 05 3    ) 204 11 1 5 0 53 1 3 AB                        c 2 4 1 505 43 4        AB c is the length of projection of AB onto ℓ (or onto c). (iii) 2 55 4 AB      A B ℓ

3 [Turn Over By Pythagoras’ theorem, The shortest distance from A to ℓ = 225 4 3 (iv) AGF and ∆BGF are similar triangles. Since AF corresponds to BF, 2 2 Area of AGF 3 9 Area of BGF 4 16   (v) 1 ( )( )Area of 2 1Area of ( )( )2 AG GFAGF AG BGF BG BG GF   From above, 9 16 AG BG  OG = 9 16 25 OB OA = 20 1 9 1 16 1 525 31          = 32 1 25 16 525 7      2 (a) (i) By using a graphic calculator, find the x-coordinates of the points of intersection of the curves xye and 21yx . Hence solve the inequality 21xex . [2] (ii) Hence, find the exact value of 1 2 2 1 dxe x x   . [3] (b) Find 2 21 d . 47 x xxx   [3] (c) Find   sin ln cos d .x x x . [2] Solution (ai) x = 0 or x = 1.26 0 1.26x 21xex < 0 for 01 x and 21xex > 0 for 20 x   (ii) 1 2 2 1dxe x x   =     01 20 2 1 d 2 1 dxxe x x e x x       A B F G 9 16

4 [Turn Over = 0122 20 xxe x x e x x             = 26 ee (b) 2 21 d 47 x xxx   = 2 2 4 5 d 47 x xxx   =   22 2 4 5 d d 47 23 x xxxx x    =   21 52ln 4 7 tan 33 xx x c       (c)       sin sin ln cos d cos ln cos cos d cos xx x x x x x x x      cos ln cos sin dx x x x    cos ln cos cosx x x C   3 A right circular cone-shaped structure of fixed height h m and semi-vertical angle 60 stands on horizontal ground. A cylindrical tank of radius r m, which is fully filled with a type of liquid chemical, is inscribed inside the cone. (i) Find the value of r in terms of h when the cylindrical tank has a maximum volume. [5] An engineer decides to build the above structure with cylindrical tank of radius r as found in (i). To prevent the liquid chemical from contaminating the ground when leakages occur, an inverted cone of semi-vertical angle of 45 will be attached to the cylindrical tank as shown in the diagram below. mh mr

5 [Turn Over If a crack is found at the centre of the base of the cylindrical tank and the liquid chemical is leaking into the inverted cone at a rate of 0.3 m3/min, (ii) find the exact rate of change of surface area of the liquid chemical in the inverted cone half an hour after the leaking starts. [5] Solution (i) Let d be the vertical length from the vertex of the cone to the top of the cylindrical tank.   2V r h d Since tan 60 r d , we have 3 33 rdr . Hence, 2 3 3V r h r   23 3 3V r h r 2d 23d V rh rr    22 3 0 2 3 0 rh r r h r     2 2 30 (rejected since 0) or 33 r r r h h     2 2 d 2 2 3d V hrr  When 23 3rh , mh mr

6 [Turn Over 2 2 d 2 3 2 2 3d3 2 4 2 (< 0 since is positive) V hhr hh hh        Hence, the cylindrical tank has a maximum volume when 23 3rh . (ii) Let x m be the radius of the water surface in the inverted cone at time t mins and L be the volume of the water in the inverted cone at time t mins. Since the angle is 45 , when the radius of the water surface is x m, the height of the water in the inverted cone is x m as well. 2 32 d 2 d 1d 3d AA x x x LL x x x         30 mins after leaking, volume of chemical in the container 0.3 30 9   m3   319 3 x 1 33x    21 3 d d d d d d d0.3 3 d L L x t x t x t      1 3d m / mind 30 x t    1 1 3 3 d d d d d d d 23d 30 A A x t x t A t          1 231 m / min5 Therefore, the area of liquid surface is increasing at a rate of 1 231 m / min5 . 4 (a) The complex number z is such that 2 3z  and arg( i ) 4z  . Find w in the form iab , where ,ab  , if 23wz  and 2 5arg 6 z w    . [4] (b) Solve the equation z4 = – 81 , giving the roots in the form reiθ, where r > 0 and      . [2]

7 [Turn Over (i) Hence, express z4 + 81 as the product of two quadratic factors with real coefficients, giving each factor in exact non-trigonometrical form. [3] The roots of the equation z4 = – 81 are represented by z1, z2, z3, z4 such that        1 2 3 4arg arg arg argz z z z   . (ii) Explain why the locus of all points z such that | z – z3 | = | z – z2 | passes through the origin. [1] (iii) The points A, B, C and D represent the complex numbers z1, z2, v and z4 respectively, with 3 1 2vz . Find the area enclosed by the points A, B, C and D. [2] Solution (a) 2 33zz       arg arg arg( ) 44 arg( ) 42 3 4 iz i z z              23 23 2 wz wz w    2 5arg 6 52arg( ) arg( ) 6 35arg( ) 2 46 2 3 z w zw w           222 cos sin33 13 wi i                (b) z4 = ‒ 81 = 81ei = ( 2 )81ein 1 424 42 81 e , 0, 1, 2