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AP_Calculus_BC_CH11_Exam
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1
Series
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정답
Find the limit of the sequence \(a_n = \dfrac{5n^2 - 3n}{2n^2 + n + 4}\).
A
\(0\)
\(\dfrac{5}{2}\)
정답
C
\(\dfrac{5}{4}\)
D
Does not exist
해설
Divide numerator and denominator by \(n^2\): \(lim \dfrac{5 - \dfrac{3}{n}}{2 + \dfrac{1}{n} + \dfrac{4}{n}^2} = \dfrac{5}{2}\)
2
Series
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오답
Which of the following sequences converges?
A
\(a_n = (-1)^n\)
\(a_n = \dfrac{(-1)^n}{n}\)
정답
C
\(a_n = \sin(n)\)
내 답
D
\(a_n = n \sin(n)\)
해설
\(\dfrac{(-1)^n}{n} arrow 0\) as \(n arrow \infty\) by the Squeeze Theorem.
3
Series
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오답
Find the sum of the series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{3}{4^n}\).
A
\(\dfrac{3}{4}\)
\(1\)
정답
C
\(\dfrac{4}{3}\)
D
\(3\)
내 답
해설
Geometric series with \(a = \dfrac{3}{4}, r = \dfrac{1}{4}\). Sum \(= \dfrac{\dfrac{3}{4}}{1 - \dfrac{1}{4}} = 1\)
4
Series
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오답
The series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^p}\) converges if and only if:
A
\(p > 0\)
B
\(p \geq 1\)
\(p > 1\)
정답
D
\(p \geq 2\)
내 답
해설
p-series converges iff \(p > 1\). This is a fundamental theorem.
5
Series
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오답
Using the Integral Test, determine the convergence of \(\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n \ln n}\).
A
Converges
Diverges
정답
C
Inconclusive
D
Cannot apply
해설
\(\displaystyle\int_{2}^{\infty} \dfrac{1}{x \ln x} d x = [\ln(\ln x)]_2^\infty = \infty\). Diverges.
6
Series
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오답
Using the Comparison Test, \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 1}\):
Converges by comparison with \(\sum \dfrac{1}{n^2}\)
정답
B
Diverges by comparison with \(\sum \dfrac{1}{n}\)
내 답
C
Converges by comparison with \(\sum \dfrac{1}{n}\)
D
Cannot be determined by comparison
해설
\(\dfrac{1}{n^2 + 1} < \dfrac{1}{n^2}\) and \(\sum \dfrac{1}{n}^2\) converges (p=2>1).
7
Series
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오답
The alternating series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}\) is:
Absolutely convergent
정답
B
Conditionally convergent
내 답
C
Divergent
D
Cannot be determined
해설
\(\sum |a_n| = \sum \dfrac{1}{n}^2\) converges (p-series, p=2>1). Thus absolutely convergent.
8
Series
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오답
Use the Ratio Test on \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n^2}{2^n}\). The series:
Converges, \(L = \dfrac{1}{2}\)
정답
B
Diverges, \(L = 2\)
C
Inconclusive, \(L = 1\)
D
Converges, \(L = 0\)
해설
\(L = lim |a_{n+1}/a_n| = lim \dfrac{(n+1)^2}{2^{n+1}} \cdot \dfrac{2^n}{n^2} = \dfrac{1}{2} < 1\). Converges.
9
Series
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오답
The radius of convergence for \(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\) is:
A
\(R = 0\)
B
\(R = 1\)
C
\(R = e\)
\(R = \infty\)
정답
해설
This is the series for \(e^x\). By Ratio Test, \(L = 0\) for all \(x\), so \(R = \infty\).
10
Series
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오답
Find the interval of convergence for \(\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n}\).
A
\((-1, 1)\)
\([-1, 1)\)
정답
C
\((-1, 1]\)
D
\([-1, 1]\)
해설
\(R=1\). At \(x=-1\): alternating harmonic (converges). At \(x=1\): harmonic (diverges). So \([-1, 1)\).
11
Series
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오답
The Maclaurin series for \(\dfrac{1}{1-x}\) is:
\(\displaystyle\sum_{n=0}^{\infty} x^n\)
정답
B
\(\displaystyle\sum_{n=0}^{\infty} (-1)^n x^n\)
C
\(\displaystyle\sum_{n=1}^{\infty} x^n\)
D
\(\displaystyle\sum_{n=0}^{\infty} n x^n\)
해설
This is the geometric series: \(\dfrac{1}{1-x} = 1 + x + x^2 + ... = \displaystyle\sum_{n=0}^\infty x^n\) for \(|x| < 1\).
12
Series
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오답
The Maclaurin series for \(e^x\) is:
A
\(\displaystyle\sum_{n=0}^{\infty} x^n\)
\(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\)
정답
C
\(\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n}\)
D
\(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n}\)
해설
\(e^x = 1 + x + x^\dfrac{2}{2}! + x^\dfrac{3}{3}! + ... = \displaystyle\sum_{n=0}^\infty \dfrac{x^n}{n!}\)
13
Series
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오답
The coefficient of \(x^3\) in the Taylor series for \(\sin x\) centered at \(0\) is:
A
\(\dfrac{1}{3}\)
B
\(-\dfrac{1}{3}\)
C
\(\dfrac{1}{6}\)
\(-\dfrac{1}{6}\)
정답
해설
\(\sin x = x - x^\dfrac{3}{3}! + x^\dfrac{5}{5}! - ...\). Coefficient of \(x^3\) is \(-\dfrac{1}{6}\).
14
Series
오답률 100%
오답
Using \(T_2(x)\) for \(\cos x\) at \(a = 0\), approximate \(\cos(0.1)\):
A
\(0.990\)
\(0.995\)
정답
C
\(1.000\)
D
\(0.985\)
해설
\(T_2(x) = 1 - x^\dfrac{2}{2}\). \(T_2(0.1) = 1 - 0.\dfrac{01}{2} = 1 - 0.005 = 0.995\).
15
Series
오답률 100%
오답
Which test is BEST for \(\displaystyle\sum_{n=1}^{\infty} \dfrac{3^n}{n!}\)?
A
Integral Test
B
Comparison Test
Ratio Test
정답
D
Root Test
해설
Ratio Test is best when factorials are involved. \(L = lim \dfrac{3^{n+1}/(n+1)!}{3^\dfrac{n}{n}!} = lim 3/(n+1) = 0 < 1\).
16
Series
오답률 100%
오답
Consider the series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{3^n}\).
(a) Use the Ratio Test to determine whether the series converges or diverges. Show all steps.
(b) If the series converges, explain why absolute convergence and conditional convergence are the same in this case.
(c) Using the geometric series formula \(\displaystyle\sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}\) for \(|x| < 1\), differentiate both sides and find a closed form for \(\displaystyle\sum_{n=1}^{\infty} n x^{n-1}\).
(미작성)
정답
(a) $L = frac(1,3) < 1$, converges. (b) All terms positive, so absolute = conditional. (c) $frac(d,d x)[frac(1,1-x)] = frac(1,(1-x)^2)$
위 정답과 비교하여 채점하세요:
해설
Full solution requires showing Ratio Test limit calculation, understanding of positive series, and differentiation of power series.
17
Series
오답률 100%
오답
Let \(f(x) = \ln(1 + x)\).
(a) Find the Maclaurin series for \(f(x) = \ln(1 + x)\) by integrating the series for \(\dfrac{1}{1+x}\).
(b) Determine the interval of convergence for the series found in part (a). Be sure to check the endpoints.
(c) Use the first four nonzero terms of the series to approximate \(\ln(1.5)\). Then use the Alternating Series Estimation Theorem to find an upper bound for the error in your approximation.
(미작성)
정답
(a) $ln(1+x) = x - frac(x^2,2) + frac(x^3,3) - frac(x^4,4) + ...$ (b) $(-1, 1]$ (c) $ln(1.5) approx 0.4010$, error $< frac(1,5) times 0.5^5$
위 정답과 비교하여 채점하세요:
해설
Integration of geometric series, endpoint convergence testing (harmonic vs alternating harmonic), and AST error bound.