TMJC H2 Chapter 9 Maclaurin Series Assignment Suggested Solutions 2024

Chapter 9 Maclaurin Series TMJC 2024 Page 1 of 10 H2 Mathematics (9758) Chapter 9 Maclaurin Series Assignment Suggested Solutions 1 2015(9740)/I/6 (i) Write down the first three non -zero terms in the Maclaurin series for  ln 1 2 x , where 1 1 2 2 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 c ax 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 c ax bx , giving your answer as a simplified rational number. [6] (i)       2 3 2 3 ln 1 2 3 82 2 3 2 22 2 x x x xx x x           (ii)       2 2 2 3 ( 1)1 1 2 ( 1) 2 c bx bx bxax ax c cax abcx a c cc b x               By comparing part (i) and (ii), Coefficient of x: 2a Coefficient of 2x : 12 ( 2)abc b a c     Coefficient of 3x : 2( 1) 8 2 3 c c ab  As 1b c , 2 ( 1) 1 8 22 3 c c c        2 1 8( 1) 3c c c      1 8( 1) 3c c      Replace by 2 x x in  ln 1 x in MF26 Replace by x bx and by n c in  1 n x in MF26

Chapter 9 Maclaurin Series TMJC 2024 Page 2 of 10 3 3 8 3 5 c c c    5 3b  Coefficient of 3 4 3 3 8 13 ( 1)( 2) 5 5 5 523! 3! 3 c c cx a b                                104 27 Question only wants coefficient, so don’t give the whole term, i.e. you should not have the ݔସ

Chapter 9 Maclaurin Series TMJC 2024 Page 3 of 10 2 2019(9758)/ACJC Promo/Q7a & 2019(9758)/MI Promo/Q8 (a) Given that  is sufficiently small, show that 2tan 3 4 4 33           . [3] (b) It is given that  f ( ) ln 1 sinx x  . (i) Without the use of a calculator, find f (0), f (0) and f (0). Hence write down the first two non-zero terms in the Maclaurin series for f ( )x . [5] (ii) By substituting 6 x  , show that 2 ln 2 72 6    . [3] (iii) John wants to use the Maclaurin series found in part (i) to estimate the value of f ( )x when x a , where a is a real number close to zero. Suggest one recommendation for John to improve the accuracy of his estimation. [1] 2 ACJC JC1 Promo 9758/2019/Q7a , (a)                2 1 2 2 2 2 tan tan3tan 3 1 tan tan3 3 tan 1 3 tan 3 3 1 23 1 1 ... 2! 3 1 3 3 3 3 ... 3 3 3 3 3 ... 3 4 4 3 ... 1 1 3 3                                                                  Note that 3      is not small, so you cannot do tan 3 3          We need to apply result from MF26. Recall that tan  when  is sufficiently small. Replace by 3 x  in  1 n x in MF26 Always remember to rewrite the term in the denominator as negative power before expanding. 2 tan 33 tan 33                 

Chapter 9 Maclaurin Series TMJC 2024 Page 4 of 10 (bi) Method 1 (Implicit differentiation)         f ( ) ln 1 sin Differentiate wrt , cosf ( ) 1 sin 1 sin f ( ) cos Differentiate wrt , 1 sin f ( ) cos f ( ) sin (1) x x x xx x x x x x x x x x x                                f 0 ln 1 sin 0 0 cos 0f 0 1 1 sin 0 Sub 0 into (1): 1 sin 0 f 0 cos 0 f 0 sin 0 f 0 1 0 f 0 1 x                   2 2 f ( ) 0 (1) 1 ... 2! 1 ...2 xx x x x             Method 2 (using quotient rule – not recommended)  f ( ) ln 1 sin cosf ( ) 1 sin x x xx x            2 2 2 2 1 sin ( sin ) (cos )(cos )f ( ) 1 sin sin sin cos 1 sin x x x xx x x x x x              2 2 2 f (0) ln 1 sin 0 0 cos 0f (0) 1 1 sin 0 sin 0 sin 0 cos 0f (0) 1 1 sin 0             Notice the question asks to find f (0), f (0) and f (0). And only the first two non- zero terms are required, implying one of the values are likely to be 0. Cross-multiply to simplify before differentiate Concept: Use Maclaurin Theorem to find the Maclaurin Series required            2 ( ) f f 0 f 0 f 0 f 0 2! ! n nx xx x n          (in MF 26)

Chapter 9 Maclaurin Series TMJC 2024 Page 5 of 10   2 2 f ( ) 0 (1) 1 ... 2! 1 ...2 xx x x x             (ii) 2 2 2 2 1 When , 6 1ln 1 sin ... 6 6 2 6 1ln 1 ... 2 6 72 1ln ...2 6 72 ln 2 ... 6 72 x                                                 2 2 ln 2 ... 6 72 ln 2 (shown)72 6           (iii) Use more terms in the Maclaurin series of  ln 1 siny x  for the estimation. Substitute 6x  into both the LHS and the RHS of the Maclaurin series expansion in part (i)

Chapter 9 Maclaurin Series TMJC 2024 Page 6 of 10 3 2018(9758)/CJC Prelim/II/1 Given that sinf ( ) e xx  , use the standard series to find the series expansion for f ( )x in the form 2 3 ,a bx cx dx   where a, b, c and d are constants to be determined. Hence show that the first three non-zero terms for the expansion of   2sin 1 e x in ascending powers of x is 2 1 2 2x x  . [4] The function g( )y x satisfies   2d4 1d y yx   and 1y  at 0x . (i) By further differentiation, find the series expansion for , up to and including the term in x3. Hence show that when x is small,     31g f 4 x x x  . [5] (ii) By using the result in (i), justify whether f ( )x is a good approximation to g( )x for values of x close to zero. [1] Q3 2018(9758)/CJC Prelim/II/1   3 sin 2 3 3 ! 3 3 2 3 2 3 3 f e e 1 2 3 ! 3! 1 6 ! 3! 3 6 1 2 ! 2 x xx x xx x xx x x x xx xx                                        1 1, 1, and 0 2a b c d     g( )x Standard series for sin x Standard series for e x Replace 3 by 3! xxx  in e x in MF26

Chapter 9 Maclaurin Series TMJC 2024 Page 7 of 10 From previous result, 2 sine 1 2 x xx    ,        2 2 2 2 22sin 2 2 2 2 2 2 1 1 e 2 31 2 2! 1 2 2 1 1 2 x xx xx x xx x x x                                     (i)   2d4 1d y yx   Differentiating with respect to x,   2 2 d d4 2 1d d y y yx x  Differentiating with respect to x,     3 2 3 2 23 2 3 2 d d d d4 2 1 2d d d d d d d4 2 1 2d d d y y y yyx x x x y y y yx x x                            Sub 0x , 1y , d 1d y x  , 2 2 d 1d y x  , 3 3 d 3 d 2 y x  Using Maclaurin’s formula,   2 3 3g 1 2! 2 3! x xx x             2 3 g 1 2 4 x xx x      2 3 21g( ) f ( ) 1 1 2 4 2 x xx x x x x                     3 4 x (shown) Note that: 2 d d d d d d y y y x x x              , NOT 2 2 d d y x Replace 2 by 2 xxx     and by 2 n  Always remember to rewrite the term in the denominator as negative power before expanding. Concept: Use Maclaurin Theorem to find the Maclaurin Series required            2 ( ) f f 0 f 0 f 0 f 0 2! ! n nx xx x n          (in MF 26)

Chapter 9 Maclaurin Series TMJC 2024 Page 8 of 10 (ii) For values of x close to zero, 31g( ) f ( ) 0 4x x x   Therefore, f ( )x is a good approximation to g( )x for values of x close to zero. Note that: if f ( )x is a good approximation to g( )x , that means that the difference, g( ) f ( )x x , would be very small. And no