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AP_Calculus_BC_CH11_Exam
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1
Series
Error Rate 0%
Correct
Find the limit of the sequence \(a_n = \dfrac{5n^2 - 3n}{2n^2 + n + 4}\).
A
\(0\)
\(\dfrac{5}{2}\)
Correct Answer
C
\(\dfrac{5}{4}\)
D
Does not exist
Explanation
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
Error Rate 100%
Wrong
Which of the following sequences converges?
A
\(a_n = (-1)^n\)
\(a_n = \dfrac{(-1)^n}{n}\)
Correct Answer
C
\(a_n = \sin(n)\)
My Answer
D
\(a_n = n \sin(n)\)
Explanation
\(\dfrac{(-1)^n}{n} arrow 0\) as \(n arrow \infty\) by the Squeeze Theorem.
3
Series
Error Rate 100%
Wrong
Find the sum of the series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{3}{4^n}\).
A
\(\dfrac{3}{4}\)
\(1\)
Correct Answer
C
\(\dfrac{4}{3}\)
D
\(3\)
My Answer
Explanation
Geometric series with \(a = \dfrac{3}{4}, r = \dfrac{1}{4}\). Sum \(= \dfrac{\dfrac{3}{4}}{1 - \dfrac{1}{4}} = 1\)
4
Series
Error Rate 100%
Wrong
The series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^p}\) converges if and only if:
A
\(p > 0\)
B
\(p \geq 1\)
\(p > 1\)
Correct Answer
D
\(p \geq 2\)
My Answer
Explanation
p-series converges iff \(p > 1\). This is a fundamental theorem.
5
Series
Error Rate 100%
Wrong
Using the Integral Test, determine the convergence of \(\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n \ln n}\).
A
Converges
Diverges
Correct Answer
C
Inconclusive
D
Cannot apply
Explanation
\(\displaystyle\int_{2}^{\infty} \dfrac{1}{x \ln x} d x = [\ln(\ln x)]_2^\infty = \infty\). Diverges.
6
Series
Error Rate 100%
Wrong
Using the Comparison Test, \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2 + 1}\):
Converges by comparison with \(\sum \dfrac{1}{n^2}\)
Correct Answer
B
Diverges by comparison with \(\sum \dfrac{1}{n}\)
My Answer
C
Converges by comparison with \(\sum \dfrac{1}{n}\)
D
Cannot be determined by comparison
Explanation
\(\dfrac{1}{n^2 + 1} < \dfrac{1}{n^2}\) and \(\sum \dfrac{1}{n}^2\) converges (p=2>1).
7
Series
Error Rate 100%
Wrong
The alternating series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n^2}\) is:
Absolutely convergent
Correct Answer
B
Conditionally convergent
My Answer
C
Divergent
D
Cannot be determined
Explanation
\(\sum |a_n| = \sum \dfrac{1}{n}^2\) converges (p-series, p=2>1). Thus absolutely convergent.
8
Series
Error Rate 100%
Wrong
Use the Ratio Test on \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n^2}{2^n}\). The series:
Converges, \(L = \dfrac{1}{2}\)
Correct Answer
B
Diverges, \(L = 2\)
C
Inconclusive, \(L = 1\)
D
Converges, \(L = 0\)
Explanation
\(L = lim |a_{n+1}/a_n| = lim \dfrac{(n+1)^2}{2^{n+1}} \cdot \dfrac{2^n}{n^2} = \dfrac{1}{2} < 1\). Converges.
9
Series
Error Rate 100%
Wrong
The radius of convergence for \(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\) is:
A
\(R = 0\)
B
\(R = 1\)
C
\(R = e\)
\(R = \infty\)
Correct Answer
Explanation
This is the series for \(e^x\). By Ratio Test, \(L = 0\) for all \(x\), so \(R = \infty\).
10
Series
Error Rate 100%
Wrong
Find the interval of convergence for \(\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n}\).
A
\((-1, 1)\)
\([-1, 1)\)
Correct Answer
C
\((-1, 1]\)
D
\([-1, 1]\)
Explanation
\(R=1\). At \(x=-1\): alternating harmonic (converges). At \(x=1\): harmonic (diverges). So \([-1, 1)\).
11
Series
Error Rate 100%
Wrong
The Maclaurin series for \(\dfrac{1}{1-x}\) is:
\(\displaystyle\sum_{n=0}^{\infty} x^n\)
Correct Answer
B
\(\displaystyle\sum_{n=0}^{\infty} (-1)^n x^n\)
C
\(\displaystyle\sum_{n=1}^{\infty} x^n\)
D
\(\displaystyle\sum_{n=0}^{\infty} n x^n\)
Explanation
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
Error Rate 100%
Wrong
The Maclaurin series for \(e^x\) is:
A
\(\displaystyle\sum_{n=0}^{\infty} x^n\)
\(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n!}\)
Correct Answer
C
\(\displaystyle\sum_{n=1}^{\infty} \dfrac{x^n}{n}\)
D
\(\displaystyle\sum_{n=0}^{\infty} \dfrac{x^n}{n}\)
Explanation
\(e^x = 1 + x + x^\dfrac{2}{2}! + x^\dfrac{3}{3}! + ... = \displaystyle\sum_{n=0}^\infty \dfrac{x^n}{n!}\)
13
Series
Error Rate 100%
Wrong
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
\(\dfrac{1}{6}\)
\(-\dfrac{1}{6}\)
Correct Answer
Explanation
\(\sin x = x - x^\dfrac{3}{3}! + x^\dfrac{5}{5}! - ...\). Coefficient of \(x^3\) is \(-\dfrac{1}{6}\).
14
Series
Error Rate 100%
Wrong
Using \(T_2(x)\) for \(\cos x\) at \(a = 0\), approximate \(\cos(0.1)\):
A
\(0.990\)
\(0.995\)
Correct Answer
C
\(1.000\)
D
\(0.985\)
Explanation
\(T_2(x) = 1 - x^\dfrac{2}{2}\). \(T_2(0.1) = 1 - 0.\dfrac{01}{2} = 1 - 0.005 = 0.995\).
15
Series
Error Rate 100%
Wrong
Which test is BEST for \(\displaystyle\sum_{n=1}^{\infty} \dfrac{3^n}{n!}\)?
A
Integral Test
B
Comparison Test
Ratio Test
Correct Answer
D
Root Test
Explanation
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
Error Rate 100%
Wrong
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 \(\displaystyle\sum_{n=1}^{\infty} n x^{n-1}\).
(No answer submitted)
Answer
(a) $L = frac(1,3) < 1$, converges. (b) All terms positive, so absolute = conditional. (c) $frac(d,d x)[frac(1,1-x)] = frac(1,(1-x)^2)$
Compare with the answer above and grade yourself:
Explanation
Full solution requires showing Ratio Test limit calculation, understanding of positive series, and differentiation of power series.
17
Series
Error Rate 100%
Wrong
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.
(No answer submitted)
Answer
(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 $< frac(1,5) times 0.5^5$
Compare with the answer above and grade yourself:
Explanation
Integration of geometric series, endpoint convergence testing (harmonic vs alternating harmonic), and AST error bound.