TJC_H2_MATHS_P1_Solutions

2015 Preliminary Examination H2 Mathematics 9740 Paper 1 (Solutions) 1 The equation of a circle M is given by 22 0x y Ax By C     where A, B and C are real constants. The line y = 2(x + 1) passes through the centre of M and the graph of y = | x | intersects M at the points where x = 2 and x = 8. Find the equation of M. [4] 1 22 22 22 : 0 02 2 4 4 M x y Ax By C A B A Bx y C                       Centre of M : ,22 AB 2( 1)yx  passes through the centre: 2122 2 4 ----- (1 ) BA AB      At intersection between | |yx and M, we have 22 | | | | 0x x Ax B x C     At x = –2, 22 ( 2) (| 2 |) ( 2) (| 2 | ) 0 2 2 8 ----- (2) A B C A B C              At x = –8, 22 ( 8) (| 8 |) ( 8) (| 8 | ) 0 8 8 128 ----- (3) A B C A B C              Solving (2), (3) and (4) using GC: A = 8, B = –12, C = 32 22: 8 12 32 0M x y x y     

2 The diagram below shows the graph of y = g( x). The graph has a minimum point at (0, 2) and a maximum point at 13, 2   . The equations of the asymptotes are x = 1 , y = 0 and y = 2x. On separate diagrams, sketch the graphs of (i) y = g(x), [2] (ii) 1 g( )y x , [2] showing clearly in each case, the equations of the asymptotes and the coordinates of the turning points and axial intercepts, where applicable. 2i (i) y = g(x) 2ii 1 g( )y x y x 1 0 (3, 2) x = 2 y x y = 2 x = 1 3 0 y = g(x) y x y = 2x x = 1 2 (0, 2) 0

3 Without using a calculator, solve the inequality 23 112 x x  . Hence solve 23 112 x x  . [5] 3     2 2 3 112 3 2 1 0 12 3 1 1 1 2 0 111 or 32 x x xx x x x x xx           2 3 112 Replace by , x x xx  111 (no real solution) or 32xx   1 1 1 1 or 2 3 3 2xx     x

4 The sequence of real numbers 1 2 3, , , . . .u u u is defined by 11 2 and , where 1 and .4 nn nu u u a n an      (i) Prove by mathematical induction that for 1.12 ( 2)( 3)n nau nn   [4] (ii) Determine the limit of 1 ( 2) nunn u as .n [2] 4i Let Pn be the statement for 1.12 ( 2)( 3)n nau nn   When n = 1, LHS = 1 (given)ua RHS 12 (1 2)(1 3) aa  = LHS  P1 is true. Assume Pk is true for some k  , i.e. .12 ( 2)( 3)k au kk  When n = k + 1, 1 2LHS 4kk kuu k   2 4 12 ( 2)( 3) k k a kk    12 ( 4)( 3) a kk  12 (( 1) 2)(( 1) 3) a kk     = RHS  Pk is true  Pk+1 is true. Since P1 is true, and Pk is true  Pk+1 is true, by mathematical induc