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<<r . Thus the series is 3 convergent for −<24x < . 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≤<x −. 19. 1990 BC5 Solution (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 1112113 (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 Copyright © 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com 21. 1995 BC4 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 0 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. 2 (c) Write the fourth{degree Taylor p olynomial, P (x), for g (x) = f (x +2) ab out x = 0. Use P to explain why g must have a relativeminimum at x =0. 8 8 3 2 3 (x 2) (x 2) (a) T (f; 2)(x)=3+5(x 2) + 3: T (f; 2)(x) > 3 3 < 2 6 < 1 > each error 4 > : f (1:5) T (f; 2)(1:5) 3 1: approximation of f (1:5) 4 3 3 2 (0:5) (0:5) = 3+ 5(0:5) + 2 3 = 4:9583= 4:958 ( 3 4 1: value of Lagrange Error Bound j1:5 2j =0:0078125 (b) Lagrange Error Bound = 4! 2 1: explanation f (1:5) > 4:9583 0:0078125 = 4:966 > 5 Therefore, f (1:5) 6= 5. 8 2: T (g; 0)(x) (c) P (x)=T (g; 0)(x) > 4 4 > > < < 1 > each incorrect, missing, 3 3 4 2 2 or extra term x = T (f; 2)(x +2) = 3+5x + 2 > > 2 > : 1: explanation 0 The co ecientof x in P (x)isg (0). This co ecient 0 is 0, so g (0)=0. 00 (0) g 2 . This co ecient The co ecientof x in P (x)is 2! 00 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).