DHS Maclaurin Series (9758) (Revision Solutions)

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 1 6. Maclaurin Series (solutions) 1 3 1 ) 1 (x− + −−+ − + = 23 2 3 1 3 1 ) (! 2 ) () ( 1 xx 211 391 xx= −− +  ------(I) 1) 3 )( 3 (−+ +bx ax 1 3 11 ) 1 ( ) 3 )( 3 (−− ++ = bxax       +− −+ − + + = 2 3 1 3 1 3 1 ) (! 2 ) 2 )( 1 () )( 1 ( 1 ) 1 ( bxbxax )1 ( ) 1 ( 2 2 9 1 3 1 3 1 + + − + =x b bx ax + − + − + =2 9 12 9 1 3 1 3 1 ) ( ) ( 1x ab b x b a + − + − + =22 9 1 3 1 ) ( ) ( 1x ab b x b a ------(II) Comparing (I) & (II), 1− = −b a ----(1) 12 − = −ab b -----(2) From (1), 1ab= − Subst a into (2), 1 ) 1 (2 − = − −b b b Thus b = –1, a = –2 2(i) (ii) 2 e 3 5 (1)xyy+ + = −−−−−−−−− Differentiating (1) with respect to x, dd2 e3 0dd d(2 3) e 0 (2)d x x yyy xx yy x ++ = + + = −−−−−−− Differentiating (2) with respect to x, ( ) 2 2 2 2 2 dd d2 (2 3) e 0dd d dd2 2 3 e 0 (Shown) (3)d d x x yy y yxx x yy yx x    + + +=      + + + = −−−−−  When x = 0, 2 3 40 ( 1)( 4) 0 yy yy + −= − += (from (1)) ⇒ y = 1 or − 4 (NA since y > 0)

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 2 d1When 0, d5 yx x= =− (from (2)) 2 2 d 27 d 125 y x =− (from (3)) 21 271 ...5 250y xx= −− + 3(i) (1 ) ln(1 2 )xy x+=+ … (*) Differentiate (*) w.r.t. x : d2(1 ) d 12 yxy xx+ += + … (1) Differentiate (1) w.r.t. x : 2 22 d dd 2(1 ) 2d d d (1 2 ) yyyx x xx x  −+ ++=  + 2 22 dd 4(1 ) 2 0d d (1 2 ) yyx xx x+ ++ = + (shown)… (2) (ii) Differentiate (2) w.r.t. x : 32 32 3 d d 16(1 ) 3 0d d (1 2 ) yyx xx x+ +− = + … (3) When x = 0 From (*) : ln1 0y= = From (1) : d 2d y x = From (2) : 2 2 d 8d y x =− From (3) : 3 3 d 40d y x = Hence, y = 238 400 2 ...2! 3!xx x−++ + + = 23 202 4 ... 3xx x−+ + (iii) 1(1 ) ln(1 2 )yx x −= ++ = ( 21 ...xx−+ + ) 23(2 ) (2 )2 ...23 xxx −++ = 2x + ( 2 2)−− 2x + 8 223  ++ 3x + …

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 3 = 23 202 4 ... 3xx x−+ + (verified to be same as part(ii)) 4(i) ( ) ( ) 2 1 2 2 1 2 2 222 24 1f 9 9 1 139 11 111 221 ...3 2 9 2! 9 1 1 ...3 18 216 x x x x xx xx − − = − = − = −    − −−       = +− − + − +     = ++ + Range of validity: 2 2199 33 x x x − <⇒ < ∴− < < (ii) When 1 2x= , 24 2 11 11 221 ...3 18 21619 2      = +++ −  ( ) 1 1 3505 3 345635 4 3 3456 2073635 23505 3505 ≈  = ×= 5(i) ( ) ( ) 22 5f( ) 12 112 1 x A Bx Cx xxxx += = +++++ Solving 2, 1, 2A BC= −== , ie, 2 22f( ) 12 1 xx xx += −+++ (ii) ( ) ( ) 1 23 23 2 21 212 2 1 2 4 8 ... 2 4 8 16 ... xx xx x xx x − − = −++ = −−+ − + = −+ − + +

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 4 ( ) ( ) 12 2 2 211 x xxx −+ = +++ ( ) ( ) 21 ...2 xx −+= + 232 2 ...xxx=+− − + 23f( ) 5 10 15 ...xxx x∴= −++ (iii) ( ) ( ) ( ) 11 23 200 12 34 0 d 2 3d 12 1 23 23 4 123 7 2341 2 x x xx xx xx x xx ≈ −+ ++ =−+ =−+= ∫∫ (iv) ( ) ( ) 1 20 d 0.164(3s.f.) 12 1 x x xx = ++∫ Range of validity given by 22 1 and 1xx<< i.e, 11 1 22 2xx< ⇒− < < The answer in part (iii) is not an appropriate approximation as the series expansion for ( ) ( ) 212 1 x xx++ is valid only for 1.2x < Hence it cannot be used for approximation for the interval up to x =1. 6(a)(i) ( ) ( ) ( ) 1 2 2 2 2 2 2 21f( ) 1 12 (1 ) 2 1 ... 1 2 3 ...22 111 1 2 3 ...24 5 11 ...24 xxx xx xx xx xx x x − −= − =− −+ −+   =++ + −− + +     = − ++ +− +     = −+ (ii) Equation of the tangent to curve at origin is 5 2 xy=

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 5 (b) 23 23 1 11 11 31 1 ...2 22 2 22 2 3 ! 11 11 ...2 8 16 xxxx xx x   + = ++− +− − +     = +− + + 23 23 11 11 ...1 2 8 16 11 1 ...1 2 8 16 ee ee xx xx xx x +− + ++ −+ + = = 23 23 2 2 23 11 1 1 11 1 1e 1 ... ... ... ...2 8 16 2! 2 8 3! 2 1 111 1 1 1 1e 1 ...2 8 2 2 16 2 8 48 xx x xx x xx x      =+−+ + + −++ ++               = + +−+ + + − + +        311e 1 ...2 48xx=++ + 7(i) ( ) 1tan ln 1yx− = + ( ) 2 2 1d 1 d 111 d1 d yy xyyx x x= ⇒+ = +++ (shown) (ii) ( ) 2 2 dd d12 dd d yy yxy xx x+ += ( ) 232 2 32 2 dd d d1 22 2dd d d yy yyxy xx x x + += +  ( ) 43 3 2 43 3 2 d d d dd1 32 6d d d dd y y y yyxy x x x xx+ += + When 0x= , 0y= d 1d y x = , 2 2 d 1d y x =− , 3 3 d 4d y x = , 4 4 d 18d y x =− ( )( ) 234123tan ln 1 ...234xxx x x∴ + = −+−+ (iii) intersection at (0,0)

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 6 8(a) (b) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 8 2 82 82 1 2 21 2 232 1 2 ...2 2! 2 32 2 ... 2 x x x x xx xx − − − − − = −  = −  = −  −− = +− − + − +     = ++ + Expansion is valid for 12 22 x x< ⇒− < < cot 2 4yx π= +  Differentiating w.r.t. x: ( ) ( ) 2 2 2 d 2cosec 2d4 d 2 1 cot 2d4 d 2 1 shownd y xx y xx y yx π π = −+   = −+ +   = −+ Differentiating w.r.t. x: 2 2 dd 4dd yy yxx =− Differentiating w.r.t. x: 232 32 d dd 44d dd y yy yx xx = −−  For x = 0, y = 1, d 4,d y x =− 2 2 d d y x =16, 3 3 d 128d y x = − 23 641 4 8 ... 3y xx x= −+ − +

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 7 ( ) ( ) ( )( ) ( ) 23 12 222 22 2 64Since cot 2 1 4 8 ...43 tan 2 4 14 8 121 4 8 4 8 ...2! 1 4 8 16 ... 1 4 8 ... x xx x x xx xx xx xx x xx π π −  + = −+ − +  +  ≈− + −−= − −+ + −+ + = +− + + = ++ + Thus, 1, 4, 8 ab c= = = 9(i) 2 2 2 2 2 d3 d4 3 4 11 4 12ln (no need modulus d 2 2)42 d yx xx xyx x xx xxc xx −= − −= − = −+ − += −+ + −< < − ⌠⌡ ⌠⌡  When 0, 2xy= = : 12 ln(1) 24 cc= +⇒= 12ln 242 xyx x +∴= −+ +  − (ii) When x = 0, y = 2 and d3 d4 y x =− . ( ) 2 2 2 dd4 22 dd yyx xx xx− −= When x = 0, 2 2 d 0d y x = . ( ) 22 2 2 3 3 2 d d dd4 222 2d d dd yyy yx xx x x xx−−−− = When x = 0, 3 3 322d1 4 d 48 y x +−= = . 3 3 312 ...4 8 3! 312 ...4 48 xyx xx ∴=− +⋅ + = −+ +

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 8 10 ( ) 113tan 3tan 22 2 33 11 1 3 (Shown) xx dy yye edx x x dyxy dx −− = ⇒= = ++ ∴+ = ( ) ( ) ( ) 22 22 221 2 3 1 23 0d y dy dy d y dyxx x x dx dx dxdx dx++ = ⇒ ++ − = ( ) ( ) ( ) ( ) 32 2 2 32 2 32 2 32 1 2 2 3 2 0 1 4 3 2 0 dy dy dy d yx xx dxdx dx dx d y d y dyxx dxdx dx + + +− + = ⇒ + +− + = When 0x = , 1y = , 3dy dx = , 2 2 9dy dx = , 3 3 21dy dx = Hence, ( ) ( ) ( ) ( )23 23 '' 0 ''' 00 ' 0 ...2! 3! 971 3 ...22 ffy f xf x x xx x = ++ + + = ++ + + 12 3tanxxe −+ 12 3 tanxxee − = ( ) 2 232 971 2 ... 1 3 ...2! 2 2 xx xx x  =++ + ++ + +   2251 5 ... 2xx= ++ + 11 [ ] ) 1 ln(sin 1 += − xy --- (1) ) 1 ln(sin + =x y Differentiate with respect to x: 1 1 d dcos += x x yy (shown) --- (2) Differentiate with respect to x: 22 2 ) 1 ( 1 d d)sin(d d d dcos +− = − +xx yyx y x yy 2 2 2 2 ) 1 ( 1 d dsind dcos +− =    − xx yyx yy (shown) --- (3) (i) Differentiate with respect to x: x yyx y x yy d d)sin(d d d dcos 2 2 3 3 − +

Y5 Topical Revision Package 1 6. Maclaurin Series (solutions) 9 3 2 2 2 ) 1 ( 2 d dcosd d d d d d) 2 )((sin +=                 +     − xx yyx y x y x yy 3 3 2 2 3 3 ) 1 ( 2 d dcosd d d dsin3d dcos +=    −          − xx yyx y x yyx yy --- (4) When x = 0, (1): 0 0sin 1 = =−y (2): 1d d1d d) 0cos( = ⇒ =x y x y (3): ( ) 1d d1 1 ) 0sin(d d) 0cos( 2 2 2 2 2 − = ⇒ − =− x y x y (4): ( )( ) ( ) 3d d2 1 ) 0cos(1 1 )