RI Maclaurin Series C7C Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 1 of 24 Additional Practice Questions for Chapter 7C: Maclaurin Series (Solutions) 1 (a) Find the fifth term in the expansion of 6 3 1 2x x   in descending powers of x . [2] (b) Find the term independent of x in the expansion of 18 19 3 x x   . [3] (c) Find the coefficient of 18x in the expansion of 12 2 13x x   . [3] (d) Expand 21 0(2 3 4 )xx , up to the term in 3x . [3] (e) The coefficient of 2x in the expansion of 6(2 )xk is equal to the coefficient of 5x in the expansion of 8(2 ) kx . Find k . [3] (f) The expansion of 1 n kx up to and including the term in 3x are 22 313 6 6 6xk x h x  , where n is a positive integer. Find the values of n , k and h . [4] (a) 666 33 1 8 44 11 1 1122 2xx x xx x     Fifth term in the expansion of 6 3 1 2x x   in descending powers of x is 4 18 2 4 6 11 5 4 21 6xx x    (b) General term 1818 1(9 ) 3 r rxr x    11818 218 19 3 r rrr xr    For term independent of x, 318 0 122 rr  12 618 1Term independent of 9 12 3 18 564 x       (c) General term 12 2 12 1(3 ) r rxr x   

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 2 of 24 12 12 212 3 rr rxr   For term with 18x , 12 3 18r  10r  18 2 12Coefficient of 3 10 594 x    (d)   10223 4xx       2310 9 2 8 2 7 2 20 10 2 2 3 3 20 23 2 0 10 10 102 2 34 2 34 2 3412 3 term in 2 15360 20 480 11520 9 24 15360 27 term in 1024 15360 124160 691200 term in x xx x x x xx x x x xx x                            (e) The 2x term in the expansion of 6(2 )x k is  2 46 24 x k  . The 5x term in the expansion of 8(2 ) kx is  538 25 kx  . So  2 43 568 2245 kk     45 1560 448 112kk k  (f)    23 (1 ) 1 12 3 nn nn nkx kx kx kx kx             So   22 33 22 311 211 3 6 6 6 23 ! nn nn nnkx k x k x x k x hx       3 36 (1) (1 ) 66 (2)2 (1 ) (2 ) (3)6 nk nn nn n kh         From (2),   2 132 0 12 11 0 12nn n n n    since 0n . From (1), 12 36 3kk  From (3), 33 12 35 9 4 033 nhk        

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 3 of 24 2 Express   2 2 25 1 5 21 2 x x x x   in partial fractions. [2] Hence or otherwise, obtain the expansion of   2 2 25 1 5 21 2 x x x x   in ascending powers of x , giving the first five terms in the expansion. [3] Find the exact set of values of x for which the expansion is valid. [2]       2 22 22 25 1 5 2 1221 2 25 1 5 12 2 xx A B x C x xxx x xA x B x C x            When 2x , 72 9 8AA  . Comparing coefficients of 2x , 15 2 1AB B  . Comparing constants, 22 3AC C    .   2 22 25 1 5 8 3 2 1221 2 xx x x xxx           1 12 234 2322 2 23 4 8 13 1 222 41 22 2 2 312 2 2 34 713 7 24 x xx xx x x xx x x xx x x                              Expansion is valid for 12 22 x x   and 221 11 22 x x    The set of values of x for which the expansion to be valid is 11, 22  .

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 4 of 24 3 MI Promo 9758/2020/PU2/P1/Q6 (a) Using standard series from the List of Formulae (MF26), expand   3el n 1x ax as far as the term in 3x , where a is a non-zero constant. Given that there is no term in 2 ,x determine the coefficient of 3.x [5] (b) Find the expansion of 2 13 9 x x   in ascending powers of x, up to and including the term in 3x . State the set of values of x for which the expansion is valid. [4] Qn Solution (a)   3 23 2 2 3 3 22 33 23 3 2 23 2 23 el n 1 92 71 3 ... ... 26 2 3 393 ...23 2 2 393 ... 23 2 2 x ax xx a x a xxa x ax ax ax a xax ax aa a aax a x x                                Since there is no term in 2 ,x 2 30 2 aa 2 60 0 (rejected is non-zero) or 6. aa aa    Therefore, the coefficient of 32 3 63 69 6 =4532 2x    (b)      1 2 2 2 1 2 2 1 1 2 2 2 1 2 2 13 13 9 9 13 9 1 9 13 9 1 9 1 13 139 x xx x xx xx xx                    

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 5 of 24 Qn Solution   2 2 23 23 11 11 3 1 ...32 9 111 3 1 ...31 8 11 11 3 ...31 8 6 111 ...35 4 1 8 xx xx xxx xx x                        For expansion to be valid, 2 2 19 9 33 x x x     4 9740/2015/01/Q6 (i) Write down the first three non-zero terms in the Maclaurin series for ln(1 2 ) x , where 11 22 x , simplifying the coefficients. [2] (ii) It is given that the three terms found in part (i) are equal to the first three terms in the series expansion of (1 ) cax bx for small x. Find the exact values of the constants a, b and c and use these values to find the coefficient of 4x in the expansion of (1 ) cax bx , giving your answer as a simplified rational number. [6] (i) 23 23(2 ) (2 ) 8ln(1 2 ) 2 ... 2 2 ... 23 3 xxxx x x x      (ii) 2 22 3(1 ) (1 )(1 ) 1 ( ) ... ( ) ... 22 c cc a bccax bx ax bcx bx ax abc x x        By comparing coefficients, a = 2 21abc bc   2 (1 )8 8 5 3 (1 ) (1 ) ,23 33 5 ab c c bb c      33 5 33 3(1 ) (2 )55 55 52 (1 ) 2 ... ...33 ! 3xx x x           Coefficient of 4 104 27x 

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ________________________________________________ Additional Practice Questions for Chapter 7C: Maclaurin Series Page 6 of 24 5 Given that xx y   121 1 where 0.5x . Show that, provided x is non-zero, then   1 12 1yx xx   . [1] Hence (i) express y as a series of ascending powers of x up to and including the term in 2x [4] (ii) show that 160000 79407 101102 10   . [2]  11 2 1 12 1 12 1 12 1 1 12 1(1 2 ) (1 ) xxy xx xx xx x xxx x         (i)            12 12 23 23 1 12 ( 1 ) 311111 222221 ( 2) ( 2) ( 2) . . .22 3 !1 311111 222221 ....22 3 ! yx xx xx x x xx x                 23 2 11 3 7 ...28 1 6 13 7 ...28