TJC JC2 H2 Maths 2012 Prelim Paper 2 Solutions

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 1 Section A: Pure Mathematics [40 marks] 1 A curve C is defined by the parametric equations 2 3 cos 3x  and  2 3 1 siny  , where 02 π. (i) Find d d y x in terms of  . [2] (ii) Show that the Cartesian equation of the curve C is 2 2 3 1 2 3 2 3 1 xy     . Hence or otherwise, sketch the curve of C, indicating clearly the x-intercepts in exact form. [3] (iii) The point P 32 3, 3 2   lies on curve C. The region R is bounded by the curve C for 3x , the x-axis and the line segment joining the points P and  3,0 . Show that the area of R is     3 2 0 3 2 3 1 2 6 3 sin d4      . [4] [Solution] (i) d 2 3 sind x   and   d 2 3 1 cosd y    2 3 1 cosd d 2 3 sin y x      11 cot 23   (ii) 32 3 cos 3 cos 23 xx       2 3 1 sin sin 2 3 1 yy      So 22sin cos 1  2 2 3 1 2 3 2 3 1 xy     3 3 33 3x y x

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 2 (iii) At P, 23x  2 3 cos 3 2 3  1cos 23     When 3 3,x 2 3 cos 3 3 3  cos 1, 0    Area of R = 33 23 area of d yx          0 3 13 3 2 3 1 2 3 1 sin 2 3 sin d22           Area of R     3 2 0 3 2 3 1 2 6 3 sin d4      

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 3 2 The diagram above shows th e curve of 2 ex xy . Two points A and B on the curve have coordinates (, 1 2 ) and (, 1 2 ) respectively. A sequence of real numbers 1 2 3, , ,...x x x satisfies the recurrence relation 1 1 e4 nx nx   for n  1. (i) Show algebraically that if the sequence converges, then it converges to either  or . [3] (ii) Show that 1nx   if nx <  . [2] (iii) Show that 1nnxx  if nx <  . [2] (iv) Explain briefly how the results in (ii) and (iii) may be used to deduce that the sequence converges to  when 1 0x  . [2] [Solution] (i) If the sequence converges to, say l, then nx  l and 1nx   l as n  . i.e. 1 1 e4 nx nx    1 e4 ll  --- (A)  21 e2l l   l  or  from the diagram. (ii) If nx <  , eenx  since ex is an increasing function.  11ee44 nx   1 1 e4nx     1nx   [using eqn (A) in (i)]

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 4 (iii) Method 1A: Using given graph Step 1: Consider 1 1 e4 nx n n nx x x    211e2 2 e n n x n x x  Step 2: If nx < , 2 1 e2n n x x  [as seen in given graph 2 ex xy ] 211e022 e n n x n x x    [ e0nx  for all nx  ] 1 0nnxx   1nnxx (shown) Method 1B: Using given graph If nx < , 2 1 e2n n x x  [as seen in given graph 2 ex xy ] 1 e4 nx nx 1nnxx  1nnxx (shown) Method 2: Sketching you own graph Step 1: Consider 1 1 e4 nx n n nx x x    , Step 2: Sketch the graph of 1 e4 xyx . (We usually use a GC to sketch this graph) [ Note: The x-intercepts satisfy 1 e4 x x 21 e2x x orx  ]

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 5 Hence, if nx  , 1 e04 nx nx [as seen in the graph 1 e4 xyx ] 1 0nnxx   1nnxx (shown) (iv) When 1 1 20,x x x     using results in (ii) and (iii) When 2 2 3 ,x x x   using results in (ii) and (iii) Hence 1 2 3 1 ... ... nnx x x x x        So the sequence is strictly increasing and converges to  .

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 6 3 The function f is defined by 1f : , , 0x x x x x   (i) Sketch the graph of  fyx , showing clearly the coordinates of the stat ionary points and the equations of asymptotes, if any. [2] (ii) Given that    g f 2x x b   where 2b , state a sequence of transformations which transforms  fyx to  gyx . [2] Sketch the graph of  gyx , showing clearly the coordinates of the stationary points and the equations of asymptotes in terms of b, if any. [2] On a separate diagram, sketch g ( )'yx and solve in terms of b, the inequality   g ( ) 1 1' x x b b x     . [5] [Solution] (i)   1fy x x x   At stationary points, 2 22 11f ( ) 1 0 1' xxx xx       For 1, 2xy and 1, 2xy  [Students are allowed to sketch graph using GC, however they have to indicate clearly the coordinates of stationary points and the equations of asymptotes. The details are necessary for next part of the question involving transformations] Coordinates of turning points Shape and equations of asymptotes (1, 2) (1, 2) x = 0 x y  fyx yx

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 7 (ii) The sequence of transformations is (1) Translate graph +b units along x-axis followed by (2) Translate graph by +2 units along y-axis Asymptotes: 2y x b   ; xb Turning points:  1, 0b and  1, 4b From sketch, solution is 1 1,b x b x b     b1 b b+1 y = g'(x) y = (x b+1) (b+1x) x y (b+1,4) (b1,0) x = b x y  gyx 2y x b  

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 8 4 (i) Find the roots of the equation 3 8i 0z  , giving them in cartesian form a + ib, where a and b are exact real numbers. [3] (ii) The roots of the equation   3 3 2i 8i 0z    are 1 2 3, andz z z such that      1 2 3Re Re Rez z z . Hence find 1 2 3, andz z z in cartesian form a + ib, where a and b are exact real numbers. [2] (iii) Show 1 2 3, andz z z on an Argand diagram. [1] (iv) Explain why the locus of all points z such that 23z z z z   passes through the point representing 1z . Sketch this locus on your Argand diagram and find the minimum value of z . [5] [Solution] (i) 3 8zi  22288 iki ee    0, 1k  2 632e ki z  or 41 62e ki z   35 6 6 62e ,2e ,2e i i i z                   3 , 2 , 3i i i    (ii)   3 3 2 8 0z i i      3 3 2 8z i i   3 2 3 , 2 , 3z i i i i       2 3 , 3 4 ,z i i i    So 1 2 3 , 3 4 , 2 3z i z i z i    

TJC_JC2_H2_Maths_2012_Prelim_Paper_2_Solutions 9 (iii) (iv) Now 12| | | 3 3 | 2 3z z i     A