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 induction, Pn is true for all n  . 4ii As ,n 1 1 12 12 12 123( 2)( 3) 3 1 ( 2) ( 2) n an a n n n n un n n n u         

5 The complex number z satisfies the equation 3 3 1 3 i1 z z   . Without the use of a graphing calculator, express z3 in the form rei where r  0 and  <   . Hence find the roots of the equation. [6] 5 3 3 1 3i1 z z   33 3 1 3i 3 i (1 3i) 3i 1 zz z       3 3i 1 1 3i z   2πi π3 i 3 πi 3 2e e 2e  Alternative 3 i3 3i 1 3i+1 3i 1 3i 1 3i+1 3i 1 13 i22 = e z      πi2 π3 3e , n zn   π 2 πi 93e , 0, 1 n zn     π 5π 7πi i i9 9 9e ,e ,ez  

6 The figure below shows a rectangle OACB where 2OA OB . Point D is on AC produced such that : :1AD AC  where  is a constant. The lines OD and AB intersect at point E. It is given that OA a , OB b and OEA  . Find OD in terms of a and b, and show that   2 4OD AB  b . [4] In the case when E is the foot of perpendicular from A to OD, deduce the value of . [2] Using this value of  and given that 4 4 2 a   and 2 1 2 b   , find OE . [2] 6 OD OA AD OA AC ab OD AB       a b b a   22 k     a b a b b a 22  ba (since 0  a b = b a ) 22 4 bb (since | | 2 | |a b )   2 4 b (shown) E is foot of perpendicular from A to OD (i.e. AB OD )  0OD AB    2 40 b 4 (since 0b ) D O A C B E

Method 1 OD 4 2 12 2 4 4 1 0 6 0 2 2 6 1                                          OA ODOE OD  (length of projection of OA on OD ) 42 1 6 4 0 65 21                    65 5 Method 2 Area of OAB = 11( )( ) ( )( )22OB OA OE AB  OA OB OE AB   16 16 4 4 1 4 4 25 16       = (6)(3) 6 5 545  Method 3 OD 4 2 12 2 4 4 1 0 6 0 2 2 6 1                                          65OD  . Since OBE is similar to DAE 1 4 OE OB DE DA 16 5 5 OE OD .

7 The function f is defined by 14f: 1 xx x   , x ¡ , x  k. (i) With the aid of a graph, find the least value of k such that f has an inverse. [2] (ii) Using the least value of k found in (i), (a) find f 1(x) and state its domain, [3] (b) find the exact solution(s) of the equation f(x) = f1(x). [2] Describe a sequence of two transformations which would transform the graph of y = f(x) onto the graph of 24 2 xy x   . [2] 7i Using GC, The least value of k is 1 4 . 7iia Let 1 4 4 1 11 xxy xx    1 4 yx y   f 1(x) = 1 4 x x   1fD [0,4)  7iib Since the graphs of y = f(x) and y = f1(x) intersect on the line y = x f(x) = x 2 14 1 3 1 0 x xx xx        - 1f f 35 both values D and D2x  1 1 4 1 4 1 2 2 4 2 1 1 1 2 1 2 xxxxx x x xy y y x x x x             The transformations (in either order) are - A reflection about the y-axis. [B1] - A scaling of factor 2 parallel to the x-axis. y y = 4 x = 1 x 0

8 (i) Use the substitution 2sinx  , where π0 2 and 01 x , to show that 1d sin (1 ) where is an arbitrary cons tant.1 x x x x x c cx     [5] (ii) The region R is bounded by the curve 1 4 1 xy x   and the lines y = 4x  1 and 1 4x . Find the volume of revolution formed when R is rotated completely about the x-axis, giving your answer in exact form. [5] 8i 2sinx  d 2sin cosd x    2 2 sind 2sin cos d1 1 sin x xx     22sin d (1 cos 2 ) d 1 sin 22 c   sin cos c     For π 2 0  : 2sinx   sin x  2cos 1 sin 1 x    1d sin (1 )1 x x x x x cx      (shown) 8ii From GC, 1 4 1 xy x   and y = 4x – 1 intersect at 1 2x  1 Alternative ¼ ½ x y = 4x – 1 y

Required volume 1 2 1 4 21 4 211π d π(1)1 3 4 x xx    1 2 1 4 ππd 1 12 x xx  1 1 2 1 4 ππ sin (1 ) 12x x x    11 1 1 1 1 1 1 ππ sin (1 ) sin (1 )2 2 2 4 4 4 12                    π 1 π 3 ππ 4 2 6 4 12          π 3 3 7π 12 12   