HCI JC 2 H2 Maths 2012 Paper 1 Solution - for Sharing

1 2012 HCI H2 Math Prelim Paper 1 Marking Scheme Qn. Solutions 1 2 2 ( 2)(2 5) ( 5 3)(1)f ( ) ( 2) ' x kx kx xx x       22 2 2 5 4 10 5 3 ( 2) kx x kx kx x x        2 2 47 ( 2) kx kx x   Since f is an increasing function, 2 2 47f ( ) ( 2) ' 0kx kxx x    2 4 7 0kx kx    for all values of x Hence we have 0k  (linear) or for 0k  , 2 ( 4 ) 4 (7) 0kk   (quadratic) 2 16 28 0kk 4 (4 7) 0kk  Hence 70 4 k . 2 (i) and (ii) 2iz OQ a 3 ln lny xy x Differentiate implicitly w.r.t x, M O Re Im ia iaa Locus of z 4  B A a C Q

2 1 d d ln dd 1yy xy x x y y x x x      1 d d ln ln dd yy y x x y x y x x   1d ln ln d yx x y y x yx   22 d (1 ln ) (1 ln ) d 1 ln 1 ln y y x y x x xy x y    For the tangent to be parallel to the y–axis, d 0d x y  2 1 ln 0(1 ln ) y yx   1 ln 0y ey Hence from xyyx , we have ee xx e e0xx   Hence x = 1.32, and since the tangent parallel to the y–axis takes the form xc , hence equation of tangent is 1.32x 4(a)       3i i 1 2i 3i i i i 1 2i 3i 3 i i 2 3i 3 (1 )i (2 ) Comparing Re and Im parts, 3 2 1 131 4 1 i4 zz x y x y x y x y x y x y y y y x x x z                              

3 (b) 12 1 1 4 4 1 1 i2 2 2 4 1 3 1 ppz z z p p            5(i)               1 2 1 2 2 1 2 2 222 22 2 1 ln 1 2 21 2 ... 2 1 2 2 13 1 221 2 2 2 2 ...22 31 4 ... 8 51 ... 2 yx xx xx x x x x x x x xx                                       (ii)     1 1 1 2 1 and ln 1 2 1 1 1 e 1 e 1and2 2 2 2 e 1 1 OR 0.316 0.522 xx xx xx                     (iii) 2 2 2 0 0 5 20 d 1 d 6.67 OR 23y x x x x              Since the integration in  0,2 does not fall into the valid range of 0.316 0.5x   , the approximation is not good. 6(a) Assume the x th day to be the day with maximum amount of goods delivered. For the first x days: ( a = 1000, d = 100) [2(1000) ( 1)(100)]2 (1900 100 )2 x xSx x x     For the remaining (15- x) days: ( a = 1000+ (x - 1)100 – 100 = 800+100x, d = -100)

4 15 (15 ) [2(800 100 ) (15 1)( 100)]2 15 (200 300 )2 x xS x x x x         Since total goods delivered is 21300 tons, (1900 100 )2 x x + 15 (200 300 )2 x x  = 21300 2 31 198 0xx   ( 9)( 22) 0xx   x = 9 or x = 22 (NA, since x  15) Therefore, 9th June was the day with max goods delivered. Goods delivered = 1000 + (9  1)(100) = 1800 tons (b) (i) Series H is a GP with common ratio r. (1 ) 1 narH r   or ( 1) 1 narH r   C is a GP with common ratio 1/r. 1 11(1 ) 11 1 11 nn n rarC ar r r     1 1 1 ( 1) ( 1) 11 nn n n n H a r ar r C r r a ar uu       (b) (ii) 2 3 1 12 1 2 3 ... ( 1) ... ( )( )( )...( ) () n n nn u u u a ar ar ar ar ar          = ( 1) 2 nn nar  Since 21 1 n n H u u a rC  , 12 ... nu u u  ( 1) 2 nn nar  = 21 2() n nar  = 2() n H C 7(a) 1 21 2 32 3 43 ( 1) 1 2(3 ) 2(3 ) 2(3 ) ... 2(3 )n nn u u a u u a u u a u u a                 

5 Sum of all equations: 1 2 3 ( 1) 1 2(3 3 3 ... 3 ) ( 1) n nu u n a             1 1 11(1 ( ) )332( ) ( 1) 11 3 n nu u n a       1 1 11 (1 ) ( 1)3 1( 1) 3 n n n u n a na           (b) Let nP be the statement denoting 3 3 3 5 5 5 3 3 11 2 ... 3(1 2 ... ) ( 1) 2n n n n         for ,n  When n = 1, LHS = 351 3(1 ) 4 RHS = 331 (1 )(2 )2 = 4 =LHS Therefore, 1P is true. Assume kP is true for some values of ,k  i.e. 3 3 3 5 5 5 3 3 11 2 ... 3(1 2 ... ) ( 1) 2k k k k         To prove 1kP  , i. e. 3 3 3 3 5 5 5 5 33 1 2 ... ( 1) 3(1 2 ... ( 1) ) 1 ( 1) ( 2)2 k k k k kk               LHS = 3 3 3 3 5 5 5 5 3 3 3 5 3 3 2 3 3 2 33 1 2 ... ( 1) 3(1 2 ... ( 1) ) 1 ( 1) ( 1) 3( 1)2 1 ( 1) [ 2 6( 1) ]2 1 ( 1) [ 6 12 8]2 1 ( 1) ( 2)2 k k k k k k k k k k k k k k k kk                               =RHS Therefore 1kP  is true.

6 Since 1P is true, kP is true 1kP  is true. By mathematical induction, nP is true for all n  . 2 5 1 n r r   = 3 3 2 21 1 1[ (2 ) (2 1) (2 ) (2 1) ]3 2 4 n n n n   3 3 2 2 22 2 2 2 1[4 (2 1) (2 1) ]3 1 (2 1) [4 (2 1) 1]3 1 (2 1) (8 4 1)3 n n n n n n n n n n n n             8(i) 22 36xy (ii) d d y kyt  Method 1 Implicit Differentiation w.r.t. time dd2 2 0dd xyxy tt 22 dd dd = ( ) (36 ) (shown) x y y t x t y kyx y k xk xx    Method 2 Implicit Differentiation w.r.t x then using chain rule (rate of change equation) d2 2 0 d yxy x 22 d d d d d d or d d d d d d d =d (36 ) (shown) y y x x x y t x t t y t x ky xt y y k xk xx        

7 (iii) 2 2 2 2 2 24 2 4 4 d 2(36 ) (given)d d 2 d36 12 d 2 d2 36 1 ln 36 22 ln 36 4 ' 36 e 36 e 36 e ( 0) t tt xx tx x xtx x xtx x t C x t C xA x A x A x                       4 Using initial conditions, when 0, 4 4 36 20 36 20e t tx A A x        4 4 For to be 3, 36 9 27 27 36 20e 9e 20 19ln 0.2 s4 20 t t OY OX t             (iv) x t 4 6x

8 Jesse’s model is not appropriate. Based on Jesse’s model, the rod would never fall flat on the ground. 9 (i)      4,0,0 , 6,4,6 , 6,2,0P Q R 20 22 06 12 3 12 4 3 41 PR RQ                                         So the equation of the plane is 3 4 3 3 0 3 12 1 0 1 i.e. 3 3 12x y z                                  r (ii) Method 1 the length of projection of 2 2 2 onto the normal of the plane 43 1= 4 3 3 3 101 24 19 cm19 CP PQR                    = 5.51 cm (correct to 3 sig figs) Method 2 Denote the foot of perpendicular of C to the plane PQR as N. Then 03 43 01 ON                     Sub into the equation of plane PQR, 0 3 3 244 3 3 12 190 1 1                                  72 4 24 24, , , 1919 19 19 19N CN  or 5.51 cm

9 (iii) 30 30 11 1cos 30 19 30 11 76.7                                        (iv) Line PQ: 41 02 03 s                  r , Plane OCGD: 0x At point M, 4 0 4ss    0 8 12 OM      The distance from M to the plane OABC is 12 cm (v)   The point of reflection, ', of about the plane is 6, 4, 6 equation of the reflection plane is 4 2 2 0 2 4 0 0 6 Q Q OABC                                  r       *Note: the plane contains three points, ' 6, 4, 6 , 4,0,0 , 6, 2,0 and three vectors parallel to the plane, 2 2 0 2 , ' 4 , ' 2 0 6 6 All possible answers using one Q P R PR PQ RQ                               point and two vectors will be correct. 10 (a) (i)   44cos sin dx x x =    2 2 2 2cos sin cos sin dx x x x x =   cos 2 1 dxx = 1 sin 2 + 2 xC (ii) 3log (3 1) dxx

10 = ln(3 1) dln 3 x x  dLet ln(3 1), 1 d d3 , d 3 1 vux x u vxxx      3log (3 1) dxx =  13 ln(3 1) ( ) ( ) dln 3 3 1 x x x x x      = 11 ln(3 1) 1 dln 3 3 1x x x x        = 11 ln(3 1) ln(3 1)ln3 3 x x x x C      = 11 ln(3 1)ln 3 3 x x x