AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems Solutions

AP Calculus BC - Sequences and Series Chapter 11- AP Exam Problems solutions

100 15⎛⎞+ n 1 1 1. A ssnn==⎜⎟,lim ⋅1= 54⎝⎠+ n n→∞ 5 5

2. C I. convergent: p-series with p = 2 >1 II. divergent: Harmonic series which is known to diverge 1 III. convergent: Geometric with r = <1 3

3. A I. Converges by Alternate Series Test n 1⎛⎞3 II Diverges by the nth term test: lim ⎜⎟≠ 0 n→∞ n ⎝⎠2 ∞ 1 L III Diverges by Integral test: dx = lim ln (ln x)=∞ ∫2 xxln L→∞ 2

4. A I. Compare with p-series, p = 2 6 II. Geometric series with r = 7 III. Alternating harmonic series

5. B I. Divergent. The limit of the nth term is not zero. II. Convergent. This is the same as the alternating harmonic series. III. Divergent. This is the harmonic series.

6. A This is the integral test applied to the series in (A). Thus the series in (A) converges. None of the others must be true.

7. D The first series is either the harmonic series or the alternating harmonic series depending on whether k is odd or even. It will converge if k is odd. The second series is geometric and will converge if k < 4 . ∞ 8. A Take the derivative of the general term with respect to x: ∑()−1 n+1 x22n− n=1

23 uu23 ()33xx() 9. E Since euu =+1 + + +" , then ex3x =1++3 + +" 2! 3! 2! 3! 33 9 The coefficient we want is = 3! 2

tt35 10. B The Maclaurin series for sin t is t − +−" . Let tx= 2 . 3! 5! (2xx )35(2 ) (−1)n−1 (2x)2n−1 sin(2x) =2x−+ −"+ +" 3! 5! (2n −1)!

xx3 5 (x23) (x25) x6x 10 11. A sin xx=− + −"" ⇒sin x22=x− + − =x2− + −" 3! 5! 3! 5! 3! 5!

xx35 1315 1 1 12. E sin xx≈− + ; sin1 ≈1− + =−+ 1 3! 5! 3! 5! 6 120

∞∞∞ nn−1n−1 13. D If f(x)==∑axnn , then f′(x) ∑nax=∑ naxn. nnn===001 ∞∞ n−1 f ′(1) ==∑nan1 ∑ nan nn==11

14. A The series is the Maclaurin expansion of e−x. Use the calculator to solve ex−x = 3 .

15. D The center is x =1, so only C, D, or E are possible. Check the endpoints. ∞ (−1)n At x = 0: ∑ converges by alternating series test. n=1 n ∞ 1 At x = 2: ∑ which is the harmonic series and known to diverge. n=1 n ∞ (−1)n 16. C Check x =−1, ∑ which is convergent by alternating series test n=1 n ∞ 1 Check x =1, ∑ which is the harmonic series and known to diverge. n=1 n

x −1 17. C This is a geometric series with r = . Convergence for −11<

n 18. B You may use the ratio test. However, the series will converge if the numerator is (1)− and diverge if the numerator is 1n . Any value of x for which x +21 > in the numerator will make the series diverge. Hence the interval is −31≤

(a) Taylor approach Geometric Approach

f ()2=1 11= − f ′()2=− (2 − 1 )2 =− 1 x−11+−()x 2 23 n n − f ′′ ()2 =−1uu+ −u +"" +−()1 u+ f ′′ ()22=− (212; )3 = = 1 2! where ux=− 2 − f ′′′ ()2 f ′′′ ()26216;=− ( − )4 =− =−1 3! 1 23 nn Therefore =−1 ()xx − 2+()− 2−()x − 2+" +−( 1)()x− 2 +" x −1

(b) Antidifferentiates series in (a): nn+1 1213 14()(−−1x 2) ln x−=1 C+x−() x −+2()x −− 2()x−++ 2 "+" 23 4 n +1 0ln21=−⇒=−C 2

1 4 Note: If C ≠ 0 , “first 4 terms” need not include −()x −2 4

35 1112113 (c) ln=− ln 1 =−+−" 22 22232  111 =−+ −" 2824 1111 since <, −=0.375 is sufficient. 24 20 2 8

Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. n+1 =<1 11 <5 RCn +  , where 2 Alternate Justification: (C −1)n 1 n +12 2

< 11 n +12n+1 1 < when n ≥ 2 20 20. 1992 BC6 Solution

11 (a) 0 << for ln ()nn>1, for ≥3 nnpln () np 1 by p-series test, ∑ converges if p > 1 n p ∞ 1 and by direct comparison, ∑ p converges. n=2 nnln ()

1 ∞ (b) Let fx()= , so series is ∑ fn() xxln n=2 ∞ ⌠ 1 b  dx=lim ln ln x =lim[ln(ln(b))− ln(ln 2)] =∞ b→∞ b→∞ ⌡ 2 xxln 2 Since f (x) monotonically decreases to 0, the integral test shows ∞ 1 ∑ diverges. n=2 nnln

11 (c) >>0 for p <1, nnp ln nnln

∞ 1 ≤< so by direct comparison, ∑ p diverges for 01p n=2 nnln

Let f be a function that has derivatives of all orders for all real numbers. Assumeff (1) 3, cc(1) 2, fc(1) 2 , and f ccc(1) 4 .

(a) Write the second-degree Taylor polynomial for f about x 1 and use it to approximate f (0.7) .

(b) Write the third-degree Taylor polynomial for f about x 1 and use it to approximate f (1.2).

(c) Write the second-degree Taylor polynomial for f c , the derivative of f , about x 1 and use it to approximate f c(1.2) .

1995 BC4 Solution

2 2 (a) Tx2 3 2 x1 x1 2 f 0.7 |3 0.6 0.09 3.69

24 3 (b) Tx3 32 x1 x 1 x1 6 2 f 1.2 |3 0.4 0.04 0.008 2.645 3

c 2 (c) Tx3 2 2 x12 x 1 f c 1.2 |2 0.4 0.08 1.52 22. 1997 BC2

Let Px() 73(4)x 5(4)x 232(4x)6(4)x 4 be the fourth-degree Taylor polynomial for the function f about 4. Assume f has derivatives of all orders for all real numbers.

(a) Find ff(4) and ccc(4).

(b) Write the second-degree Taylor polynomial for f c about 4 and use it to approximate f c(4.3) .

x (c) Write the fourth-degree Taylor polynomial for gx() f(t)dt about 4. ³ 4

(d) Can f (3) be determined from the information given? Justify your answer.

1997 BC2 Solution

(a) fP 4 4 7 f ccc 4 2, f ccc 4 12 3!

23 (b) Px3 37x4 5x 4 2 x 4 2 Px3c 301464 x x 2 f c 4.3| 3 10 0.3 6 0.3 0.54

x (c) Pg43 ,x Ptd()t ³4 x 23 ªº7345544 tt tdt ³4 ¬¼

325314 x474xx 4 x 4 23 2

(d) No. The information given provides values for ff 4, cc 4,ffcc 4, cc 4 andf 4 4 only. 23. 1998 Calculus BC Scoring Guidelines

24.

BC{4 1999

4. The function f has derivatives of all orders for all real numb ers x. Assume f (2) = 3, f (2) = 5,

00 000

f (2) = 3, and f (2) = 8.

(a) Write the third{degree Taylor p olynomial for f ab out x =2anduseittoapproximate f (1:5).

(4)

(b) The fourth derivativeoff satis es the inequality jf (x)j3 for all x in the closed interval [1:5; 2].

Use the Lagrange error bound on the approximation to f (1:5) found in part (a) to explain why

f (1:5) 6= 5.

explain why g must have a relativeminimum at x =0.

8 3

2 3

(x 2) (x 2) (a) T (f; 2)(x)=3+5(x 2) +

3: T (f; 2)(x)

> 3

2 6

< 1 > each error

f (1:5) T (f; 2)(1:5)

1: approximation of f (1:5)

4 3

3 2

(0:5) (0:5) = 3+ 5(0:5) +

2 3

= 4:9583= 4:958

(

1: value of Lagrange Error Bound

j1:5 2j =0:0078125 (b) Lagrange Error Bound =

1: explanation

f (1:5) > 4:9583 0:0078125 = 4:966 > 5

Therefore, f (1:5) 6= 5.

2: T (g; 0)(x)

< 1 > each incorrect, missing,

4 2 2

or extra term

x = T (f; 2)(x +2) = 3+5x +

1: explanation

The co ecientof x in P (x)isg (0). This co ecient

is 0, so g (0)=0.

(0) g

. This co ecient The co ecientof x in P (x)is

is 5, so g (0) = 10 which is greater than 0.

Therefore, g has a relative minimum at x =0.

Note:

< 1 > max for improp er use of + ::: or

equality AP® CALCULUS BC 2001 SCORING GUIDELINES 25. Question 6 A function f is defined by 12 3 n 1 fx()x x2 Lxn L 3 3323 3n 1 for all x in the interval of convergence of the given power series. (a) Find the interval of convergence for this power series. Show the work that leads to your answer. 1 fx() (b) Find lim 3 . x l0 x (c) Write the first three nonzero terms and the general term for an infinite series that 1 represents ¨ fx()dx. 0 (d) Find the sum of the series determined in part (c).

(2)n xn 1 n 2 (2)n x x (a) lim 3 lim 1 ¦£ 1 : sets up ratio test ld n ld ¦ n(1)nx n(1)33n ¦ n1 ¦ 1 : computes limit 3 4 : ¤ d ¦ n 1 ¦ 1 : conclusion of ratio test At x 3 , the series is (1) n , which diverges. ¦ ¦ n0 3 ¥¦ 1 : endpoint conclusion d n 1 At x = 3, the series is , which diverges. n0 3 Therefore, the interval of convergence is 33x . 1 fx() 3 23 42 2 (b) lim lim x xL 1 : answer xxll0x 023333 4 9

1 1 12 32 Ln 1n L¦£ (c) ¨ fx()dx ¨ xx x dx ¦ 1 : antidifferentiation 0 0 3 3323 3n 1 ¦ ¦ 11 1 1 x 1 ¦ of series 23Ln 1 L ¦ = xx23 xn1 x ¦ 3 33 3 x 0 3 : ¤ 1 : first three terms for ¦ 11 1L 1 L ¦ definite integral series = 23 n1 ¦ 3 33 3 ¦ ¦¥ 1 : general term

1 (d) The series representing ¨ fx()dx is a geometric series. 0 1 1 1 Therefore, fx()dx3 . 1 : answer ¨ 1 0 1 2 3 AP® CALCULUS BC 2002 SCORING GUIDELINES 26 Question 6

The Maclaurin series for the function B is given by

@ " xxxx nn " &! $x fx x LL n !" n n on its interval of convergence. (a) Find the interval of convergence of the Maclaurin series for B. Justify your answer. (b) Find the first four terms and the general term for the Maclaurin series for BN=( ). 1 (c) Use the Maclaurin series you found in part (b) to find the value of B = . 3

2x n £ 1 : sets up ratio ¦ n 2 (1)n ¦ (a) lim lim 2xx 2 ¦ 1 : computes limit of ratio nnll@2x n @(2)n ¦ ¦ 1 : identifies interior of interval n 1 ¦ ¦ 11 ¦ of convergence N for x ¦ 22 ¦ ¦ 2 : analysis/conclusion at endpoints @ 5 ¤¦ ¦ At N , the series is which diverges since ¦ 1 : right endpoint n ¦ n ¦ ¦ 1 : left endpoint this is the harmonic series. ¦ @ ¦ ¦ 1 At N , the series is n which ¦ 1 if endpoints not N o ¦ 2 n n ¦ ¦ 1 if multiple intervals ¥¦ converges by the Alternating Series Test. ¦

11 Hence, the interval of convergence is bx . 22

(b) BNa " N&N $N!KK Nn ¦£ 1 : first 4 terms 2 ¤¦ ¦ 1 : general term ¥¦

(c) The series in (b) is a geometric series. ¦£ 1 n ¦ 1 : substitutes N into infinite B a " & KK ¸ ¦ 3 !!! ! ¦ ¦ series from (b) or expresses series "& $ n 2 ¤ K K ¦ !' % ! ¦ from (b) in closed form ¦ $ ¦ 1 : answer for student's series ¥¦

# ! OR 2 11 fxa() for x . Therefore, 12 x 22 $ B a ! # ! AP® CALCULUS BC 2002 SCORING GUIDELINES (Form B)

27. Question 6

1 d xn The Maclaurin series for ln is with interval of convergence b N . 1 N n n 1 (a) Find the Maclaurin series for ln and determine the interval of convergence. 13 N d n (b) Find the value of . n n d n d (c) Give a value of F such that converges, but diverges. Give reasons why p p n n n n your value of F is correct. d 1 d (d) Give a value of F such that diverges, but converges. Give reasons why your p p n n n n value of F is correct.

1¬1 (a) ln ln 13NN 1(3)® ¦£ 1 : series 2 ¤¦ dd ¦ 1 : interval of convergence (3 x)n3n ¥¦ = or (1)nxn n nn n We must have b !N , so interval 11 of convergence is x b. 33

d n ¬ 1 : answer (1) 1 1 (b) ln ln n n 1 (1) ® 2

£ (c) Some F such that bF because ¦ 1 : correct p ¦ ¦ n d n ¦ (1) converges by AST, but the 3 ¤ 1 : reason why p converges n p ¦ n n ¦ 1 d ¦ 1 : reason why diverges ¦ p F-series diverges for F b . ¥¦ n p n n

£ (d) Some F such that bF because the ¦ 1 : correct p ¦ d ¦ 1 1 3 ¦ 1 : reason why diverges F-series diverges for p b 1 and the ¤ F np ¦ n n ¦ 1 d ¦ 1 : reason why converges ¦ F F-series converges for F . ¥¦ n p n n

7 AP® CALCULUS BC 28. 2003 SCORING GUIDELINES Question 6

The function f is defined by the power series d nnnn (11)x224xxx6)( x 2 fx()==1 + ++ + (n++) ""(n) n=0 21! 3!5!7! 21! for all real numbers x. (a) Find f a()0 and f aa()0. Determine whether f has a local maximum, a local minimum, or neither at x = 0. Give a reason for your answer. 1 1 (b) Show that 1 approximates f ()1 with error less than . 3! 100 (c) Show that yf= ()x is a solution to the differential equation xya +=y cos x.

f = x = £ 1 : f a(0) (a) a(0) coefficient of term 0 ¦ ¦ 2 11 ¦ 1 : f aa(0) f aa(0) = 2 (coefficient of x term) = 2() = ¦ 3! 3 4 : ¤ ¦ 1 : critical point answer f x = f = ¦ has a local maximum at 0 because a(0) 0 and ¦ ¦ 1 : reason f aa(0) < 0. ¥¦

n 111 (1) 1 (b) f(1) = 1 + ++"+" 1 : error bound < 3! 5! 7! (2n + 1)! 100 This is an alternating series whose terms decrease in absolute value with limit 0. Thus, the error is less than the 1111 first omitted term, so f (1) ()1 b =<. 3! 5! 120 100

nn 24xx3 6 x5 (1)2 nx 12 ¦£ ya = + ++ + ¦ 1 : series for ya (c) ""n + ¦ 3! 5! 7! (2 1)! ¦ nn ¦ 2xxx2464 6 (1)2 nx 2 ¦ xya = + ++"" + ¦ 1 : series for xya 3! 5! 7! (2n + 1)! ¦ 4 : ¦ 21 41 61 ¤ xya +=y 1 (+)x 2++()x 4 (++)x " ¦ 3! 3! 5! 5! 7! 7! ¦ 1 : series for xya + y ¦ 21n ¬¦ + ( 1)nn ++x 2 ¦ n++n " ¦ (2 1)! (2 1)! ® ¦ 1 : identifies series as cos x n ¥¦ 111 (1) n ¦ = 1 x24+ x x6++"x 2 +" 2! 4! 6! (2n)! = cos x OR OR

3 xf x x n1 n+ ¦£ 1 : series for ( ) xy== xf()x x ++"(1) x 21+ "¦ 3! (2n + 1)! ¦ ¦ 1 : identifies series as sin x ¦ = sin x 4 : ¤ ¦ 1 : handles xya + y xy +=y ()xy =()x = x ¦ a a sin a cos ¦ ¦ 1 : makes connection ¥¦ AP® CALCULUS BC 2003 SCORING GUIDELINES (Form B) 29. Question 6

The function f has a Taylor series about x = 2 that converges to fx() for all x in the interval of ()n + 1! ()n convergence. The nth derivative of f at x = 2 is given by f ()2 = for n p 1, and f ()21.= 3n (a) Write the first four terms and the general term of the Taylor series for f about x = 2. (b) Find the radius of convergence for the Taylor series for f about x = 2. Show the work that leads to your answer. (c) Let g be a function satisfying g ()23= and gxa()= f(x) for all x. Write the first four terms and the general term of the Taylor series for g about x = 2. (d) Does the Taylor series for g as defined in part (c) converge at x = 2 ? Give a reason for your answer.

n 2! 3! 4! ¦£ f ()(2) (a) f(2) = 1 ; f a(2) = ; f aa(2) = ; f aaa(2) = ¦ 1 : coefficients in 3 32 33 ¦ n ¦ ! 23!4! ¦ fx( )=+ 1 (x 2)+ (x 2)2+ (x 2)3+ ¦ first four terms 2 3 ¦ 3 2!3 3!3 3 : ¤¦ x ¦ 1 : powers of ( 2) in (1)!n + n ¦ ++"(2)x + "¦ n !3n ¦ first four terms ¦ 23243 ¦ 1 : general term =+1(x 2)+ (x 2)+ (x 2) + ¥¦ 3 3233 ¦ n + 1 n ++"(2)x + " 3n

n + 2 x n+1 ¦£ 1 : sets up ratio n+1 (2) n + ¦ 3 = 12 x ¦ (b) lim n + lim ¸2 ¦ 1 : limit nld1 x n nldn + 31 ¦ n (2) ¦ 3 3 : ¤¦ 1 : applies ratio test to 1 ¦ =x < 12 when x 23< ¦ conclude radius of 3 ¦ ¦ ¦ convergence is 3 The radius of convergence is 3. ¥¦

(c) g(2) = 3 ; gfa(2) = (2) ; gfaa(2) = a(2) ; gfaaa(2) = aa(2) ¦£ 1 : first four terms 2 : ¤¦ 11 ¦ 1 : general term g(x)=+ 3 ( x 2)+ (x 2)23+ (x 2) + ¥¦ 3 32 1 n+ ++"(2)x 1 + " 3n

(d) No, the Taylor series does not converge at x = 2 1 : answer with reason because the geometric series only converges on the interval x 23.< AP® CALCULUS BC 30. 2004 SCORING GUIDELINES

Question 6

π Let f be the function given by fx()=sin(5x+), and let Px() be the third-degree Taylor polynomial 4 for f about x = 0. (a) Find Px().

(b) Find the coefficient of x22 in the Taylor series for f about x = 0. 111 (c) Use the Lagrange error bound to show that f()−P()<. 10 10 100 x (d) Let G be the function given by Gx()= f (t) dt. Write the third-degree Taylor polynomial ∫0 for G about x = 0.

π 2 4 : Px() (a) f ()0=sin()= 42 −1 each error or missing term π 52 f ′()0=5cos()= 42 ()π deduct only once for sin 4 π 25 2 f ′′()0=−25sin()=− evaluation error 42 π π 125 2 deduct only once for cos() f ′′′()0=−125cos()=− 4 42 evaluation error 252 252 1252 P()x =+ x− x2− x3 −1 max for all extra terms, + ", 22 22!() 23!() misuse of equality

−5222  1 : magnitude (b) 2 :  222()!  1 : sign

11 () 114 (c) f()−P()≤max f4 (c)()() 1 : error bound in an appropriate 10 10 0≤≤c 1 4! 10 10 inequality 4 ≤=<625() 1 1 1 4! 10 384 100

(d) The third-degree Taylor polynomial for G about 2 : third-degree Taylor polynomial for G x = 252 2522  x 0about x = 0 is ⌠ +−ttdt 22 4 ⌡0  −1 each incorrect or missing term 252 252 =x+x23 − x 24 12 −1 max for all extra terms, + ", misuse of equality

7 ® 31. AP CALCULUS BC 2004 SCORING GUIDELINES (Form B)

Question 2

Let f be a function having derivatives of all orders for all real numbers. The third-degree Taylor polynomial for f about x = 2 is given by Tx()=−79()x−223−3()x−2. (a) Find f ()2 and f ′′()2. (b) Is there enough information given to determine whether f has a critical point at x = 2? If not, explain why not. If so, determine whether f ()2 is a relative maximum, a relative minimum, or neither, and justify your answer. (c) Use Tx() to find an approximation for f ()0. Is there enough information given to determine whether f has a critical point at x = 0 ? If not, explain why not. If so, determine whether f ()0 is a relative maximum, a relative minimum, or neither, and justify your answer. () (d) The fourth derivative of f satisfies the inequality fx4 ()≤ 6 for all x in the closed interval [0, 2]. Use the Lagrange error bound on the approximation to f ()0 found in part (c) to explain why f ()0 is negative.

(a) fT()22==() 7  1 : f () 2 = 7 2 :  f ′′()2 1 : f ′′(2)=− 18 =−9 so f ′′()21=− 8  2!

(b) Yes, since fT′′()22==() 0, f does have a critical  1 : states f ′()2 = 0  point at x = 2. 2 :  1 : declares f ()2 as a relative ′′()=− < ()  Since f 2180, f 2 is a relative maximum  maximum because f ′′()2 < 0 value.

(c) fT()00≈=() −5  1 : fT() 0 ≈=()0 −5 It is not possible to determine if f has a critical point  1 : declares that it is not 3 : at x = 0 because Tx() gives exact information only   possible to determine = at x 2.  1 : reason

6 4  1 : value of Lagrange error (d) Lagrange error bound =02−=4  4! 2 :  bound ()≤+() =− fT0041  1 : explanation Therefore, f ()0 is negative.

3 AP® CALCULUS BC 32. 2005 SCORING GUIDELINES

Question 6

Let f be a function with derivatives of all orders and for which f ()27.= When n is odd, the nth derivative () ()n − 1! of f at x = 2 is 0. When n is even and n ≥ 2, the nth derivative of f at x = 2 is given by f n ()2.= 3n (a) Write the sixth-degree Taylor polynomial for f about x = 2. (b) In the Taylor series for f about x = 2, what is the coefficient of ()x − 2 2n for n ≥ 1 ? (c) Find the interval of convergence of the Taylor series for f about x = 2. Show the work that leads to your answer.

()=+1! ⋅1 ()−2+3! ⋅1 (−)4 +5! ⋅1 (−)6 (a) Px6 72x x 2x 2⎧ 1 : polynomial about x = 2 322! 344! 366! ⎪ 2 : Px() ⎪ 6 3 : ⎨ −1 each incorrect term ⎪ −1 max for all extra terms, ⎪ ⎩ + ", misuse of equality

(2n − 1!)11 (b) ⋅= 1 : coefficient 32n()2n ! 32n()2n

(c) The Taylor series for f about x = 2 is ⎧ 1 : sets up ratio ∞ ⎪ 1 2n 1: computes limit of ratio fx()=+7(x−2). ⎪ ∑ 2n = 23n ⋅ ⎪ 1: identifies interior of n 1 ⎪ 11 ()n+12 5 : ⎨ interval of convergence ⋅()x −2 2(n + 1)()n+12 ⎪ 1 : considers both endpoints L = lim 3 ⎪ n→∞ 11 2n 1 : analysis/conclusion for ⋅()x −2 ⎪ 2n 2n 3 ⎩⎪ both endpoints 2n 2 23n 2 (x − 2) =⋅−lim ()x 2 = n→∞ 2()n + 13322n 9 L < 1 when x −<23. Thus, the series converges when −< 1 x <5. ∞∞ 32n 11 When x = 5, the series is 7+=+7, ∑2n ∑ nn=1n ⋅ 32 2 =1n ∞ 1 which diverges, because , the harmonic series, diverges. ∑ n n=1 ∞∞ (3)− 2n 11 When x =−1, the series is 7+=+7, ∑2n ∑ n==1123n ⋅ 2 nn ∞ 1 which diverges, because , the harmonic series, diverges. ∑ n n=1 The interval of convergence is ()−1, 5 . AP® CALCULUS BC 33. 2005 SCORING GUIDELINES (Form B)

Question 3

The Taylor series about x = 0 for a certain function f converges to f() x for all x in the interval of convergence. The nth derivative of f at x = 0 is given by

n+1 () ()−1(n +1)! f n ()0 = for n ≥ 2. 5n ()n − 12

The graph of f has a horizontal tangent line at x = 0, and f ()06= . (a) Determine whether f has a relative maximum, a relative minimum, or neither at x = 0. Justify your answer. (b) Write the third-degree Taylor polynomial for f about x = 0. (c) Find the radius of convergence of the Taylor series for f about x = 0. Show the work that leads to your answer.

(a) f has a relative maximum at x = 0 because ⎧ 1 : answer ′()= ′′()< 2 : ⎨ f 00 and f 00. ⎩ 1 : reason

(b) ff()06==,′()00 3 : Px() 3! 6 4! −1 each incorrect term f′′()0=− =− ,f′′′()0= 5122 25 523 2 Note: −1 max for use of extra terms 3!xx234! 31 Px()=66− + =− x2+ x3 52!523!232 25 125

() + fn ()011()−+n 1 (n) (c) u==xnxn ⎧ 1 : general term n n! n (− )2 5n 1 ⎪ 1 : sets up ratio ⎪ n+2 ⎪ ()−12()n ++ 4 : ⎨ 1 : computes limit xn 1 n+ 21 ⎪ 1 : applies ratio test to get un+1 = 5 n ⎪ n+1 un ()−+1 (n 1) ⎩⎪ radius of convergence xn 5n (n − 1)2 n+2n−12 1 = ()()x nn+ 15 un+1 =1 < < lim x 1 if x 5. n→∞ un 5 The radius of convergence is 5. AP® CALCULUS BC 34. 2006 SCORING GUIDELINES

Question 6

The function f is defined by the power series xx223 x3 (−1)n nxn fx()=− + − + + + 23 4Ln + 1L for all real numbers x for which the series converges. The function g is defined by the power series xx23x (−1)n xn gx()=−1 + − + + + 2! 4! 6! LL(2n)! for all real numbers x for which the series converges. (a) Find the interval of convergence of the power series for f. Justify your answer. (b) The graph of yf=−()xg()x passes through the point ()0, −1 . Find y′()0 and y′′()0. Determine whether y has a relative minimum, a relative maximum, or neither at x = 0. Give a reason for your answer.

n+1+ 2 ()−+1()n1xn 1 n + 1 ()n+1 (a) ⋅= ⋅x ⎧ 1 : sets up ratio n + 2 ()− n n ()(n+ 2 n) 1 nx ⎪ 1 : computes limit of ratio 2 ⎪ (n + 1) ⎪ 1 : identifies radius of convergence lim ⋅=x x 5 : n→∞ (n+ 2)()n ⎨ ⎪ 1 : considers both endpoints The series converges when −< 1 x <1. ⎪ 1 : analysis/conclusion for 123 ⎪ When x = 1, the series is −+ − + ⎩ both endpoints 234L This series does not converge, because the limit of the individual terms is not zero.

123 When x =−1, the series is +++ 322 L This series does not converge, because the limit of the individual terms is not zero.

Thus, the interval of convergence is −<11x <.

14 9 1 (b) fx′()=− + x− x2 + and f ′()0.=− ⎧ 1 : y′()0 324 L 2 ⎪ 1 : y′′()0 213 1 ⎪ gx′()=− + x− x2 + and g′()0.=− 4 : ⎨ 2! 4! 6! L 2 ⎪ 1 : conclusion ⎩⎪ 1 : reasoning yf′′()00=−() g′()0=0 4 21 f ′′()0 = and g′′()0==. 3 4! 12 41 Thus, y′′()0=− >0. 312 Since y′()00= and y′′()00> , y has a relative minimum at x = 0. AP® CALCULUS BC 35. 2006 SCORING GUIDELINES (Form B) Question 6 1 The function f is defined by fx()= . The Maclaurin series for f is given by 1 + x3

369 n 3n 11−+xx−x++LL()−x+, which converges to f() x for −< 1 x <1.

(a) Find the first three nonzero terms and the general term for the Maclaurin series for f ′()x .

−+36 −9++(−)n 3n + (b) Use your results from part (a) to find the sum of the infinite series L1− L. 222258 231n x (c) Find the first four nonzero terms and the general term for the Maclaurin series representing f ()tdt. ∫0 12 (d) Use the first three nonzero terms of the infinite series found in part (c) to approximate f ()tdt. What are ∫0 12 the properties of the terms of the series representing f ()tdt that guarantee that this approximation is ∫0 1 within of the exact value of the integral? 10,000

− (a) fx′()=−369x258+ x− x+ +3n(−1)n x31n + 1 : first three terms LL 2 : { 1 : general term

1 (b) The given series is the Maclaurin series for f ′()x with x = . 1 : f ′()x 2 ⎪⎧ − 2 : ⎨ 1 322 1 : f ′() f ′(x)=−()1 +x(3x) ⎩⎪ 2 1 3() 14 16 Thus, the sum of the series is f ′()=− =− . 221 2 7 (1 + ) 8

n + x 1 xx47x 10 ()−1 x31n 1 : first four terms (c) ⎮⌠ dt =x − + − + + + 2 : { 3 LL+ ⌡0 1 + t 7410 31n 1 : general term

1147 12 () () 1122 (d) ⌠ dt ≈− + . 1 : approximation ⎮ 3 ⎧ ⌡0 1 + t 24 7 ⎪ 1 : properties of terms 1 3 : ⎨ The series in part (c) with x = has terms that alternate, decrease in 1 : absolute value of 2 ⎪ < absolute value, and have limit 0. Hence the error is bounded by the ⎩⎪ fourth term 0.0001 absolute value of the next term. ⎛1147⎞ 1 10 12 () () () 11⎜22⎟ 2 1 ⌠ dt −− + < = <0.0001 ⎮ 3 ⎜⎟ ⌡0 1 + t ⎜2 4 7⎟ 10 10240 ⎜⎟ ⎝⎠ AP® CALCULUS BC 2007 SCORING GUIDELINES 36. Question 6

− 2 Let f be the function given by f ()xe= x . (a) Write the first four nonzero terms and the general term of the Taylor series for f about x = 0. 1 −−x2 fx() (b) Use your answer to part (a) to find lim . x→0 x4

x − 2 (c) Write the first four nonzero terms of the Taylor series for edtt about x = 0. Use the first two ∫ 0 12 − 2 terms of your answer to estimate edt t. ∫ 0

12 − 2 (d) Explain why the estimate found in part (c) differs from the actual value of edtt by less than ∫ 0 1 . 200

23 n −−22− 2− 2 46 − 2 ()xx()()x (x) ⎧ x x (a) e x =+1 + + +"+ +"1 : two of 1,−x2 , , − 1! 2! 3! n! ⎪ 26 3 : n ⎨ xx46 ()−1 x2n 1 : remaining terms =−1 x2 + − +"" + + ⎪ 62 n! ⎩⎪ 1 : general term

∞ n+1 − 11−−x22fx() 1 x2 (−) xn 4 (b) =− + + 1 : answer 4 ∑ x 62 n=4 n! ⎛1 −−x2 f()x⎞1 Thus, lim =− . → ⎜4 ⎟ x 0⎝⎠x 2

x 46 − n 2n x −t2 ⌠ ⎛2 tt ()1 t ⎞ (c) edt=−⎜1 t+−++"+"⎟dt ⎧ 1 : two terms ∫ 0 ⎮ ⎜26n ! ⎟⎪ ⌡0 ⎝⎠3 : ⎨ 1 : remaining terms xx35x 7 ⎪ =−x + − +" ⎩ 1 : estimate 31042 Using the first two terms of this series, we estimate that 12 − 2 111 11 edtt ≈−()() = . ∫ 0 238 24

5 12 −t2 −<11 1 ⋅=< 1 1 1 (d) edt () , since ⎧ 1 : uses the third term as ∫ 0 24 2 10 320 200 ⎪ n+12 2 : ⎨ the error bound n 1 ∞ ()−1 () ⎪ 1 : explanation 12 − 2 2 ⎩ edtt = , which is an alternating ∫ 0 ∑ (+ ) n=0 nn!2 1 series with individual terms that decrease in absolute value to 0. AP® CALCULUS BC 37 2007 SCORING GUIDELINES (Form B)

Question 6

Let f be the function given by fx()= 6e−x 3 for all x. (a) Find the first four nonzero terms and the general term for the Taylor series for f about x = 0. x (b) Let g be the function given by g()x= f(t)dt. Find the first four nonzero terms and the general term for the Taylor ∫0 series for g about x = 0. (c) The function h satisfies hx( ) = kf′(ax) for all x, where a and k are constants. The Taylor series for h about x = 0 is given by xx23 xn hx()=+1x++ +"" + + . 2! 3!n ! Find the values of a and k.

n ⎡xx23 x ()−1 xn ⎤⎧ x2 x3 (a) fx()=−61 + − ++"+" 1 : two of 6,−− 2x, , ⎢23 n ⎥⎪ 327 ⎣3 2!3 3!3 n!3 ⎦3 : ⎨ 1 : remaining terms n n ⎪ xx23 ()−16 x =−62x + − +"" + + ⎩⎪ 1 : general term 327 n n!3 −1 missing factor of 6

(b) g()0= 0 and g′()xf= ()x, so ⎧ 1 : two terms n ⎪ ⎡x2x3 x 4 ()−1 xn+1 ⎤3 : ⎨ 1 : remaining terms g()x =−6 x + − ++"+" ⎢23 n ⎥⎩⎪ 1 : general term ⎣6 3!3 4!3 n + 1)( !3 ⎦ n −1 missingss factor of x3x4 ()−16 xn+1 =−6xx2 +− +"+ +" 9 4()27 (n + 1!3)n

(c) fx′()=−2,e−x 3 so hx()=−2ke−ax 3 ⎧ 1 : computes kf′(ax) ⎪ x xx23 xn 1 : recognizes h()x = e , hx()=+1 x+ + +"" + + =ex ⎪ 2! 3!n ! 3 : ⎨ or −ax 3 x ⎪ equates 2 series for hx −=2ke e ⎪ () −a ⎩⎪ 1 : values for ak and = 1 and −=21k 3 1 a =−3 and k =− 2 OR 2 fx′()=−2 + x+", so 3 2 h()x == kf′ (ax) −++2k akx " 3 hx()=+1 x+" 2 −=2k 1 and ak = 1 3 1 k =− and a =−3 2 AP® CALCULUS BC 38. 2008 SCORING GUIDELINES

Question 3

() x hx() hx′() hx′′() hx′′′() hx4 ()

1111 30 42 99 18 488 448 584 22 80 128 3 3 9 753 1383 3483 1125 33 317 2 4 16 16

Let h be a function having derivatives of all orders for x > 0. Selected values of h and its first four derivatives are indicated in the table above. The function h and these four derivatives are increasing on the interval 1 ≤≤x 3. (a) Write the first-degree Taylor polynomial for h about x = 2 and use it to approximate h()1.9 . Is this approximation greater than or less than h()1.9 ? Explain your reasoning. (b) Write the third-degree Taylor polynomial for h about x = 2 and use it to approximate h()1.9 . (c) Use the Lagrange error bound to show that the third-degree Taylor polynomial for h about x = 2 − approximates h()1.9 with error less than 310× 4.

()=+ ( −) ()≈=() () (a) Px1 80 128 x 2 , so hP1.9 1 1.9 67.2 ⎧ 2 : Px1 ⎪ 4 : ⎨ 1 : P (1.9) ()< () ′ 1 Ph1 1.9 1.9 since h is increasing on the interval ⎪ ()< () ⎩ 1 : P1 1.9h 1.9 with reason 13≤≤x .

()=+ ( −)+488 ()−2+448 (−)3() (b) Px3 80 128x 2 x 2 x 2 ⎧ 2 : P3 x 6183 : ⎨ () ⎩ 1 : P3 1.9 ()≈=() hP1.9 3 1.9 67.988

(c) The fourth derivative of h is increasing on the interval 1 : form of Lagrange error estimate 2 : () 584 { 13≤≤x , so max h4 ()x= . 1 : reasoning 1.9≤≤x 2 9 4 584 1.9 − 2 Therefore, h()1.9 −P()1.9 ≤ 3 94! − =2.7037× 10 4 − <×3104 AP® CALCULUS BC 39. 2008 SCORING GUIDELINES

Question 6 dy y Consider the logistic differential equation =(6−y). Let yf= ()t be the particular solution to the dt 8 differential equation with f ()08= . (a) A slope field for this differential equation is given below. Sketch possible solution curves through the points ()3, 2 and ()0, 8 . (Note: Use the axes provided in the exam booklet.) (b) Use Euler’s method, starting at t = 0 with two steps of equal size, to approximate f ()1. (c) Write the second-degree Taylor polynomial for f about t = 0, and use it to approximate f ()1. (d) What is the range of f for t ≥ 0 ?

(a) ⎧ 1: solution curve through ()0,8 2 : ⎨ ⎩ 1: solution curve through ()3,2

11 (b) f ( ) ≈+82()−( ) =7 ⎧ 1 : Euler’s method with two steps 22 2 : ⎨ 1 : approximation of f ()1 71 105 ⎩ f ()17≈+(−)( ) = 82 16

dy2 1 dy y dy ⎧ dy2 (c) =(6 −y)+()− ⎪ 2 : dt2 88dt dt ⎪ dt2 4 : ⎨ dy 8 1 : second-degree Taylor polynomial ff()08;= ′ ()0==(6−8)=−2; and ⎪ dt t=0 8 ⎩⎪ 1 : approximation of f ()1 dy2 185 f ′′()0==()−2()−2+()2= 2 882 dt t=0 The second-degree Taylor polynomial for f about 5 t = 0 is Pt()=−28 t+t2. 2 4 29 f()1≈P()1= 2 4 (d) The range of f for t ≥ 0 is 6 <≤y 8 . 1 : answer AP® CALCULUS BC 2008 SCORING GUIDELINES (Form B) 41. Question 6

2x Let f be the function given by fx()= . 1 + x2 (a) Write the first four nonzero terms and the general term of the Taylor series for f about x = 0. (b) Does the series found in part (a), when evaluated at x = 1, converge to f ()1? Explain why or why not. 2x (c) The derivative of ln() 1 + x2 is . Write the first four nonzero terms of the Taylor series for 1 + x2 ln() 1 + x2 about x = 0.

51 (d) Use the series found in part (c) to find a rational number A such that A −ln ()<. Justify 4100 your answer.

1 2 n (a) =+1 uu + +"+u+" ⎧ 1 : two of the first four terms 1 − u ⎪ 3 : ⎨ 1 : remaining terms 1 =−246 + − + +− 2n + 1 xxx"(x)" ⎪ 1 : general term 1 + x2 ⎩ 2x + =−2x2x335+ 2 x − 2 x +"+(−1)2nx21n+" 1 + x2

(b) No, the series does not converge when x = 1 because when 1 : answer with reason x = 1, the terms of the series do not converge to 0.

x 2t (c) ln() 1 +x2 =⌠ dt 1 : two of the first four terms ⎮ 2 ⎧ ⌡0 1 + t 2 : ⎨ 1 : remaining terms x ⎩ =(2t−22t357 +2tt−+")dt ∫ 0 111 =−x2468x+ xx − +" 234

5 112411111618 1 (d) ln ( ) =+ln( 1 ) =( ) −( ) +( ) −( ) +" ⎧ 1 : uses x = 4 42223242 ⎪ 2 3 : ⎨ 124117 1 : value of A Let A =()−()() =. ⎪ 22 232 ⎩ 1 : justification Since the series is a converging alternating series and the absolute values of the individual terms decrease to 0, 5116 111 A −ln ()<() =⋅<. 4 3 2 3 64 100 AP® CALCULUS BC 42. 2009 SCORING GUIDELINES

Question 4

dy Consider the differential equation =−6.x22xy Let yf= ()x be a particular solution to this dx differential equation with the initial condition f ()−=12. (a) Use Euler’s method with two steps of equal size, starting at x =−1, to approximate f ()0. Show the work that leads to your answer. dy2 (b) At the point ()−1, 2 , the value of is −12. Find the second-degree Taylor polynomial for dx2 f about x =−1. (c) Find the particular solution yf= ()x to the given differential equation with the initial condition f ()−=12.

1 ⎛dy ⎞ (a) f (−≈) f()−+1 ⎜⎟⋅Δx1 : Euler’s method with two steps 2 dx ()− 2 : { ⎝⎠1, 2 1 : answer 1 =+⋅24= 4 2

⎛⎞ 1 dy f ()0 ≈−f( ) +⎜⎟⋅Δx 1 2 ⎜dx ()− ,4 ⎟ ⎝2 ⎠ 11 17 ≈+⋅4 = 224

()=+( +)−( +)2 (b) Px2 24x 1 6x 1 1 : answer

dy (c) =−x2 ()6 y 1 : separation of variables dx ⎧ ⎪ 2 : antiderivatives 1 ⎮⌠ dy = x2 dx ⎪ ⌡ 6 − y ∫ 6 : ⎨ 1 : constant of integration ⎪ 1 1 : uses initial condition −ln 6 −yx=3 +C ⎪ 3 ⎩ 1 : solves for y 1 −ln 4 =−+C 3 Note: max 3 6 [1-2-0-0-0] if no constant of 1 C =−ln 4 integration 3 Note: 06if no separation of variables 11 ln 6 −=−yx3 −( −ln 4) 33 1 −(x3 +1) 64−=ye3 1 −(x3 +1) y=−64e3 AP® CALCULUS BC 2009 SCORING GUIDELINES

42. Question 6

xx23 xn The Maclaurin series for ex is ex =1+x + + +"" + + . The continuous function f is defined 62 n! ()− 2 e x 1 − 1 by fx()= for x ≠ 1 and f ()11.= The function f has derivatives of all orders at x = 1. ()x − 1 2

()− 2 (a) Write the first four nonzero terms and the general term of the Taylor series for e x 1 about x = 1. (b) Use the Taylor series found in part (a) to write the first four nonzero terms and the general term of the Taylor series for f about x = 1. (c) Use the ratio test to find the interval of convergence for the Taylor series found in part (b). (d) Use the Taylor series for f about x = 1 to determine whether the graph of f has any points of inflection.

(xx−−1)4(1)6()x−12n 1 : first four terms (a) 1+−(x 1)2 + + ++"+" 2 : { 26 n! 1 : general term

((xxx−−−1)2461)() ()x−1 2n 1 : first four terms (b) 1 ++++"+ +" 2 : { 26 24 ()n + 1! 1 : general term

+ ()x − 1 22n 2 (n+2!) ()n+1! 2 (x−1) (c) lim =lim (x −1)==lim 0 ⎧ 1 : sets up ratio n→→∞(x − 1)2n n∞()n + 2! n→∞n + 2 ⎪ 3 : ⎨ 1 : computes limit of ratio ()n + 1! ⎩⎪ 1 : answer Therefore, the interval of convergence is ()−∞∞,.

43⋅2⋅56 4 (d) fx′′()=+1(x−1)+ () x−1+" ⎧ 1 : f ′′()x 624 2 : ⎨ 1 : answer nn(22 − 1) − ⎩ +−+(x 1)n 22 " (1)!n +

Since every term of this series is nonnegative, fx′′()≥ 0 for all x. Therefore, the graph of f has no points of inflection. AP® CALCULUS BC 43 2009 SCORING GUIDELINES (Form B)

Question 6 The function f is defined by the power series ∞ f()x=+1(x+1) +(x+1)2 +"" +()x+1n+ =∑ (x+1)n n=0 for all real numbers x for which the series converges. (a) Find the interval of convergence of the power series for f. Justify your answer. (b) The power series above is the Taylor series for f about x =−1. Find the sum of the series for f. x 1 (c) Let g be the function defined by g(x)= f(t)dt. Find the value of g(− ) , if it exists, or explain ∫ −1 2 1 why g(− ) cannot be determined. 2

(d) Let h be the function defined by hx()=−f() x2 1. Find the first three nonzero terms and the general 1 term of the Taylor series for h about x = 0, and find the value of h( ). 2

(a) The power series is geometric with ratio ()x + 1. ⎧ 1 : identifies as geometric ⎪ The series converges if and only if x +<11. 3 : ⎨ 1 : x +<1 1 Therefore, the interval of convergence is −< 2 x <0. ⎩⎪ 1 : interval of convergence

OR OR

+ (x + 1)n 1 lim =+

(b) Since the series is geometric, 1 : answer ∞ n 11 fx()=+(x 1) = =− for −<20.x < ∑ 1−(x +1)x n=0 1 1 1− 1 x=− 1 : antiderivative (c) g(−=) ⌠ 2 −dxx =−ln2 = ln 2 2 : { 2 ⌡−1 x x=−1 1 : value

2242n (d) hx()=−f() x 11=+x ++x ""+x + ⎧ 1 : first three terms ⎪ 134 ⎪ 1 : general term h=−=f 3 : ⎨ ( ) ( ) 1 243 ⎪ 1 : value of h( ) ⎩⎪ 2 AP® CALCULUS BC 2010 SCORING GUIDELINES

Question 6 44. cos x − 1 ⎧ for x ≠ 0 ⎪ 2 fx()= ⎨ x 1 ⎪−for x =0 ⎩ 2 The function f, defined above, has derivatives of all orders. Let g be the function defined by x g()x=+1ftdt() . ∫ 0 (a) Write the first three nonzero terms and the general term of the Taylor series for cos x about x = 0. Use this series to write the first three nonzero terms and the general term of the Taylor series for f about x = 0. (b) Use the Taylor series for f about x = 0 found in part (a) to determine whether f has a relative maximum, relative minimum, or neither at x = 0. Give a reason for your answer. (c) Write the fifth-degree Taylor polynomial for g about x = 0. (d) The Taylor series for g about x = 0, evaluated at x = 1, is an alternating series with individual terms that decrease in absolute value to 0. Use the third-degree Taylor polynomial for g about x = 0 to estimate the 1 value of g()1. Explain why this estimate differs from the actual value of g()1 by less than . 6!

24 2n xx n x (a) cos()x =1 − + −"+−()1 +" 1 : terms for cos x 24!( 2!n) ⎧ ⎪ 2 : terms for f 24 2n ⎪ 1 xx n+1 x 3 : ⎨ f()x=− + − +" +() −1 +" 1 : first three terms 24!6! (2n + 2!) ⎪ ⎩⎪ 1 : general term

(b) f ′()0 is the coefficient of x in the Taylor series for f about x = 0, ⎧ 1 : determines f ′()0 2 : ⎨ so f ′()00= . ⎩ 1 : answer with reason

f ′′()0 1 = is the coefficient of x2 in the Taylor series for f about 2! 4! 1 x = 0, so f ′′()0= . 12 Therefore, by the Second Derivative Test, f has a relative minimum at x = 0.

x xx35 1 : two correct terms (c) Px()=1 − + − 2 : { 5 23·4!5·6! 1 : remaining terms

1137 (d) g()11≈− + = 1 : estimate 23·4!72 2 : { 1 : explanation Since the Taylor series for g about x = 0 evaluated at x = 1 is alternating and the terms decrease in absolute value to 0, we know 37 1 1 g()1.−< < 72 5·6! 6! AP® CALCULUS BC 45. 2010 SCORING GUIDELINES (Form B)

Question 6 ∞ ()− n()n ()= 21 x The Maclaurin series for the function f is given by fx ∑ − on its interval of convergence. n=2 n 1 (a) Find the interval of convergence for the Maclaurin series of f. Justify your answer. 4x2 (b) Show that yf= ()x is a solution to the differential equation xy′ −y = for x < R, where R is the + 21 x radius of convergence from part (a).

+ ()2x n 1 ()n +1−1 n−1n−1 (a) lim =lim 2x ⋅=lim 2x⋅=2x ⎧ 1 : sets up ratio n→∞()n n→∞nn→∞n 2x ⎪ 1 : limit evaluation n − 1 ⎪ ⎪ 1 : radius of convergence 1 5 : ⎨ 21x < for x < 1 : considers both endpoints 2 ⎪ ⎪ 1 : analysis and interval of 1 Therefore the radius of convergence is . ⎪ 2 ⎩ convergence ∞()−−n()n∞ =−1 11= 1 When x , the series is ∑−∑−. 2 nn=2nn1=21 This is the harmonic series, which diverges. ∞()−nn ∞(−)n = 1 11= 1 When x , the series is ∑−−∑. 2 nn==22nn11 This is the alternating harmonic series, which converges. 11 The interval of convergence for the Maclaurin series of f is (− ,.⎤ 22⎦⎥

()2()3 () 4 ()− n()n =222xx− +x−+ 12x + ′ (b) y "− " ⎧ 1 : series for y 12 3n 1 ⎪ ′ nn ⎪ 1 : series for xy 16 ()− (21 x) 4 : ⎨ =44xx23− + x 4−+" +" 1 : series for xy′ − y 3n − 1 ⎪ nn−1 ⎩⎪ 1 : analysis with geometric series 64 ()(−nx212) ⋅ ′ =−8xyx1223+ x −"+ +" 3n − 1 nn 64 ()−1nx(2) xy′ =−812 x23 x + x 4−+" +" 3n − 1 xy′ −=y 4 x2 −8 x3 + 16 x 4−"" +(− 12)(nx)n+ − =−4x22(1 2x+−+ 4 x""()− 12n()xn2 +) ∞ − The series −+214x x2 −"+(−1)(nn2x)2 +="∑ (−2x)n is a n=0 1 1 geometric series that converges to for x < . Therefore 1+ 2x 2 1 1 xy′ −=y 4x2 ⋅ for x < . + 21 x 2 AP® CALCULUS BC 46. 2011 SCORING GUIDELINES

Question 6

Let f ()xx=+sin()2 cosx . The graph of yf= ()5 ()x is shown above. (a) Write the first four nonzero terms of the Taylor series for sin x about x = 0, and write the first four nonzero terms of the Taylor series for sin ()x2 about x = 0. (b) Write the first four nonzero terms of the Taylor series for cos x about x = 0. Use this series and the series for sin()x2 , found in part (a), to write the first four nonzero terms of the Taylor series for f about x = 0.

(d) Let Px4( ) be the fourth-degree Taylor polynomial for f about x = 0. Using information from the graph of 111 y= f()5 ()x shown above, show that P−f<. 4( 4 ) ( 4 ) 3000

xx35x 7 (a) sin xx=− + − +" ⎪⎧ 1 : series for sin x 3! 5! 7! 3 : ⎨ 2 : series for sin x2 xx6101x4 ⎩⎪ () sin x2 =−x2 + − +" () 3! 5! 7!

xx246x (b) cos x =−1 + − +" ⎧ 1 : series for cos x 2! 4! 6! 3 : ⎨ 24 6 ⎩ 2 : series for f ()x xx121x f()x=+1 + − +" 24!6!

f ()6 ()0 (c) is the coefficient of x6 in the Taylor series for f about 1 : answer 6! x = 0. Therefore f (6)()012=− 1.

(d) The graph of y= f()5 ()x indicates that max fx(5)()< 40. 1 : form of the error bound 0 ≤≤x 1 2 : 4 { 1 : analysis Therefore max fx(5)() 0 ≤≤x 1 5 11 4 14011 P4 − f≤⋅<=<. ( 44) ( ) 5!( 4 ) 120⋅ 45 30723000 AP® CALCULUS BC 47. 2011 SCORING GUIDELINES (Form B)

Question 6

Let f ()xx=+ln() 1 3 .

234 n xxx n+1 x (a) The Maclaurin series for ln() 1 + x is x −+−++""()−1⋅+. Use the series to write 234 n the first four nonzero terms and the general term of the Maclaurin series for f. (b) The radius of convergence of the Maclaurin series for f is 1. Determine the interval of convergence. Show the work that leads to your answer. x (c) Write the first four nonzero terms of the Maclaurin series for f ′ t2 . If g()xf= ′ t2 dt, use the first ( ) ∫0 () two nonzero terms of the Maclaurin series for g to approximate g()1. (d) The Maclaurin series for g, evaluated at x = 1, is a convergent alternating series with individual terms that decrease in absolute value to 0. Show that your approximation in part (c) must differ from g(1) by 1 less than . 5

69 12 3n xxx n+1 x 1 : first four terms (a) x3 −+− +"+()−1 ⋅ +" 2 : 23 4 n { 1 : general term

(b) The interval of convergence is centered at x = 0. 2 : answer with analysis 111 1 At x =−1, the series is −−1, − − −"" − − which 234 n diverges because the harmonic series diverges. 111 n+1 1 At x = 1, the series is 11−+−+""+()− ⋅+, the 234 n alternating harmonic series, which converges.

Therefore the interval of convergence is −1 <≤x 1.

(c) The Maclaurin series for f ′()x , f ′()t2 , and gx( ) are ⎧ 1 : two terms for ft′()2 ⎪ ∞ ⎪ 2 n+1 3125n− 2 11 1 : other terms for ft′() fx′()(:∑ −1) ⋅3x =−+−3333x x x x+" 4 : ⎨ n=1 ⎪ 1 : first two terms for g(x) ⎪ ∞ ⎩ 1 : approximation ft′(2 ):1∑(−⋅)n+1 3t6n− 2 =−+− 3333t 4 t 10 t 16 t 22 +" n=1 ∞ n−16 5 11 17 23 n+1 33333xxxxx gx():(−1) ⋅ =− + − +" ∑ 61n − 5 111723 n=1 3318 Thus g()1.≈− = 511 55

(d) The Maclaurin series for g evaluated at x = 1 is alternating, and the 1 : analysis terms decrease in absolute value to 0. 18 3 ⋅ 117 3 1 Thus g()1−< =<. 55 17 17 5 AP® CALCULUS BC 48. 2012 SCORING GUIDELINES

Question 6

The function g has derivatives of all orders, and the Maclaurin series for g is ∞ 21n+ 3 5 n xxxx ()−1=−+−.  n + 32357 n=0 (a) Using the ratio test, determine the interval of convergence of the Maclaurin series for g. 1 (b) The Maclaurin series for g evaluated at x = is an alternating series whose terms decrease in absolute 2 1 17 value to 0. The approximation for g( ) using the first two nonzero terms of this series is . Show that 2 120 1 1 this approximation differs from g( ) by less than . 2 200 (c) Write the first three nonzero terms and the general term of the Maclaurin series for g′()x .

23n+ + + x ⋅23n= 23n⋅2 (a) + ( ) x  1 : sets up ratio 25nn+ x n 12 25+  1 : computes limit of ratio   1 : identifies interior of 23n + 22 5 :  lim ( ) ⋅ x = x interval of convergence n→∞ 25n +   1 : considers both endpoints  xx2 < 1  −<11<  1 : analysis and interval of convergence The series converges when −<11x <.

1111 When x =−1, the series is −+ − + − 3579 This series converges by the Alternating Series Test.

1111 When x = 1, the series is −+−+ 3579 This series converges by the Alternating Series Test.

Therefore, the interval of convergence is −≤11x ≤.

( 1 )5 1172 1 1 1 : uses the third term as an error bound (b) g( ) −< =< 2 : { 2 120 7 224 200 1 : error bound

13 5 n 2n + 1 1 : first three terms (c) g′()x =−xx24+++−()1 ( ) x2n +  2 : { 35 7 2n + 3 1 : general term AP® CALCULUS BC 49. 2013 SCORING GUIDELINES

Question 6

A function f has derivatives of all orders at x = 0. Let Pn ( x) denote the nth-degree Taylor polynomial for f about x = 0. 1 (a) It is known that f ()04= − and that P = −3. Show that f ′=(0) 2. 1( 2 ) 2 1 (b) It is known that f ′′(0) = − and f ′′′()0.= Find Px( ). 3 3 3 (c) The function h has first derivative given by hx′( ) = f(2.x) It is known that h(0) = 7. Find the third-degree Taylor polynomial for h about x = 0.

(a) P( x) =f(0) +fx′′()0=−40+fx( )  1 : uses P( x) 1 2 :  1  1 : verifies f ′(0) = 2 11 P=−4+f′(0) · =−3 1( 22) 1 f ′(0) · = 1 2 f ′()02=

2x2 1x3 (b) Px( ) = −42 + x +− · + ·  1 : first two terms 3 ( 3 ) 2! 3 3!  3 :  1 : third term 1213  −=42++x −xx  1 : fourth term 3 18

(c) Let Qx( ) denote the Taylor polynomial of degree n for h about  2 : applies h′()xf= (2x) n  x = 0. 4 :  1 : constant term  1 : remaining terms 1 2 h′( x) = f ()2x⇒ Q′( x) =−4+2(2x) − (2x) 3 3 xx234 Q( x) =−4x+4· −· +CC;=Q() 0=h() 0= 7 32 33 3 234 Q( x) =74−+x 2x−x 3 9

hx′( ) = f()2,xh′′( x) = 24f′(2,x) h′′′( x) = f ′′(2x) 8 hf′()0= (0) = −4, h′′(02) = f′(0) =4, h′′′(04) = f′′(0) = − 3 23 xx8243 Qx( ) =74−+x 4· −· =−472x+x−x 3 2! 3 3! 9