IJC_H2_MATH_P2_Solution

1    2015 IJC H2 Maths Prelim 2 Paper 2 (Suggested Solution) 1 The parametric equations of a curve are 21e , ettxy  . (i) Find the equation of the normal to the curve at the point   21e, ep pP  . [3] (ii) This normal meets the x-axis at the point Q. Find the cartesian equation of the locus of the mid-point of PQ as p varies. [4] Q1 Solutions (i) d ed tx t  and 2d 2ed ty t  d 2ed ty x  OR 1 e e 1 ...... (1)ttxx   2e .........(2)ty Sub (1) to (2), 2(1 )yx d 2( 1) 2ed ty xx   At point P, tp . Hence of gradient normal at P 1 2ep Equation of the normal:  2 1e( 1 e ) 2e p p pyx    211 1 ee 2e 2 2 p p pyx     OR Let the equation of normal be ym x C 

2    2 2 1e( 1 e ) 2e 11e 2e 2 pp p p p C C       Equation of the normal 211 1 ee 2e 2 2 p p pyx     (ii) At point Q, where 0y , 31e 2 ep px  . Mid-point of PQ: 32 11e e ,e 2 p pp . Let 31e e p px  and 21 e 2 py . Eliminating p: Since 1 2e( 2 )p y , 13 221( 2 ) ( 2 )x yy  .   1 21( 2 )12x yy      22 1( 2 ) 1 2x yy 

3    2 (i) Find the general solution of the differential equation 2 3 2 d 2 d yx x x   . [3] (ii) It is given that 1y when 1x . On a single diagram, sketch three members of the family of solution curves for 0x  . [5] Q2 Solutions (i) 2 23 2 d2 1 d y x xx  2 d1 1 d y Cxx x    1 lnyx C x Dx   (ii) When 1x and 1y  , 110 CD D C     Hence, 1 ln 1yx C xx   y x 0C 0C (1,1) 0C

4    3 (i) Given that sin 2f( ) cos 2 3 xx x   , where x is sufficiently small, find the series expansion of f( )x in ascending powers of x, up to and including the term in 2x . [4] (ii) Use your answer to part (i) to give an approximation for 0 f( ) d n x x in terms of n. Evaluate this approximation in the case where 0.5n , leaving your answer in 6 decimal places. [3] (iii) Use your calculator to fi nd an accurate value for 0.5 0 f( ) dx x , correct to 6 decimal places. Explain why this value is more than the approximation obtained in part (ii). [3] Q3 Solutions (i) Since x is small, sin x x and  2 22cos 2 1 1 2 2 xx x  .  22 sin 2 2 2f( ) cos 2 3 4212 3 xx xx x xx     f( )x    12242xx     121 2142 xx      21 2 1 ...42 xx     211 1 24 4 x x  (ii) 0 f( ) d n x x 2 0 11 1 d24 4 n x xx  23 0 11 1 28 1 2 n xx x    2311 1 28 1 2nn n  When 0.5n ,

5    0.5 23 0 11 1f ( ) d (0.5) (0.5) (0.5)28 1 2 0.29166667 0.291667 (6 dec pl) xx      (iii) For an accurate value, 0.5 0 f ( ) d 0.2932183197 0.293218 (6 dec pl) xx    Method 1: From the above sketch, it can be observed that the graph of f( )yx is higher than the graph of 211 1 24 4y xx  for 00 . 5x  . Hence for 00 . 5x , the area under the curve