F06 - Numerical Methods - Assignment (Solutions)

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 1 of 7 National Junior College 2016 – 2017 H2 Further Mathematics Topic F6: Numerical Methods (Assignment Solutions) Suggested Solutions 1 Let   2f x x and 2 1 1h n n   ,       f 1 2, 1 2f 1 , 1 2 2f 1 ,2 2f 1 , 1 2f 1 ,1 f 2 1. n n n n n n r n n n r n n n n n                              Using Trapezium Rule,           2 1 1 1 2 d 1 2 2 2 1 2 1 1 2 2 2 1 2 1 3 1 2 2 2 1 2 1 3 1 1 1 1 2 1 3 1 2 shown 2 1 22 3 22 2 22 2 n r xx n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n r                                                                        Since the curve 2y x from 1x to 2x concave upwards, applyi ng the trapezium rule will lead to an overestimation of the actual area as the sum of the areas of the n trapezia given by 1 1 3 1 22 n rn n r     , is more than the actual area, 2 1 2 dxx . x 1 2 www.KiasuExamPaper.com 361

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 2 of 7   2 2 11 2 d 2ln 2ln 2 2ln1 2ln 2 x xx      Since the sum of the areas of the n trapezia, 1 1 3 1 22 n rn n r     , is more than the actual area, 1 1 3 1 2 2ln 22 n rn n r      . By considering n rectangles, 1 1 2 1 2 1 2 1 2total area of rectangles 1 2 3 2 2 2 2 1 2 3 12 n r n n n nn n n n n n n n n n n n n n n n r                                                                           Since the sum of the areas of the n rectangles, 1 12 n r n r  , is less than the actual area, 1 1 1 3 1 12 2ln 2 22 n n r rn n r n r          . x 2 1 11 n 11 n n  21 n  www.KiasuExamPaper.com 362

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMaths / Numerical Methods (Assignment Solutions) Page 3 of 7 2 From the graph of 2 2 2 xy x   and 1 e xy  , the graphs intersect exactly once. Hence, the equation 2 2 1 e2 xx x    has exactly one real root. Let   2 2f 1 e2 xxx x    , applying linear interpolation once,               1 1.9 f 1 1 f 1.9= f 1 f 1.9 1.0676 1.1 to 1 decimal place            Since  fy x has a vertical asymptote at 2x ,  f ' x changes rapidly in the interval  1.9, 1  , 1 is further away from  . Therefore, 1 is not a good approximation to  . 1 e xy  2 2 2 xy x   1y 1x 2y www.KiasuExamPaper.com 363

National Junior College Mathematics Department 2016 2016 – 2017 / H2 FMath