which of the following statements is true by the comparison test?
hint: when ( x e ) we have ( ln(x) 1 ). we know ( -1 leq cos(x) leq 1 ).
( int_{e^{10}}^{infty} \frac{2^{x}}{x} dx ) diverges, because ( int_{e^{10}}^{infty} \frac{2}{x} dx ) diverges.
( int_{e^{10}}^{infty} \frac{ln x}{x^{2}} dx ) diverges, because ( int_{e^{10}}^{infty} \frac{dx}{x^{2}} ) converges.
( int_{e^{10}}^{infty} \frac{ln x}{x^{2}} dx ) converges, because ( int_{e^{10}}^{infty} \frac{1}{x^{2}} dx ) converges.
( int_{1}^{infty} \frac{cos x}{x} dx ) converges, because ( int_{1}^{infty} \frac{1}{x} dx ) converges.
( int_{1}^{infty} \frac{cos x}{x^{2}} dx ) diverges, because ( int_{1}^{infty} \frac{1}{x^{2}} dx ) diverges.

Question

which of the following statements is true by the comparison test?
hint: when ( x > e ) we have ( ln(x) > 1 ). we know ( -1 leq cos(x) leq 1 ).
( int_{e^{10}}^{infty} \frac{2^{x}}{x} dx ) diverges, because ( int_{e^{10}}^{infty} \frac{2}{x} dx ) diverges.
( int_{e^{10}}^{infty} \frac{ln x}{x^{2}} dx ) diverges, because ( int_{e^{10}}^{infty} \frac{dx}{x^{2}} ) converges.
( int_{e^{10}}^{infty} \frac{ln x}{x^{2}} dx ) converges, because ( int_{e^{10}}^{infty} \frac{1}{x^{2}} dx ) converges.
( int_{1}^{infty} \frac{cos x}{x} dx ) converges, because ( int_{1}^{infty} \frac{1}{x} dx ) converges.
( int_{1}^{infty} \frac{cos x}{x^{2}} dx ) diverges, because ( int_{1}^{infty} \frac{1}{x^{2}} dx ) diverges.

Accepted Answer

# Brief Explanations: - For the first option: $\int_{e^{10}}^\infty \frac{2}{x}dx$ diverges, but $\frac{2^x}{x} > \frac{2}{x}$ for $x > e^{10}$ (since $2^x > 2$ when $x>1$), so comparison test can't use a smaller divergent integral to conclude divergence of a larger one. - For the second option: $\int_{e^{10}}^\infty \frac{1}{x^2}dx$ converges (p-integral with $p=2>1$). Given $\ln x >1$ for $x>e$, so $\frac{\ln x}{x^2} > \frac{1}{x^2}$ for $x>e^{10}$. But a larger function than a convergent one isn't necessarily divergent—this logic is wrong. - For the third option: $\int_{e^{10}}^\infty \frac{1}{x^2}dx$ converges. Since $\ln x >1$ for $x>e$, $\frac{\ln x}{x^2} > \frac{1}{x^2}$? No, wait—wait, no: actually, for $x>e^{10}$, $\ln x >10$, so $\frac{\ln x}{x^2} > \frac{10}{x^2}$, but $\int \frac{10}{x^2}dx$ still converges. Wait, no—wait, the statement says it converges because $\int \frac{1}{x^2}$ converges. Since $\ln x$ grows slower than any polynomial: for large $x$, $\ln x < x^a$ for any $a>0$. Let $a=0.5$, so $\ln x < x^{0.5}$, so $\frac{\ln x}{x^2} < \frac{x^{0.5}}{x^2} = \frac{1}{x^{1.5}}$, and $\int \frac{1}{x^{1.5}}dx$ converges (p=1.5>1). So $\frac{\ln x}{x^2}$ is bounded above by a convergent integral, so it converges. This statement's logic (using the convergent $\int 1/x^2$ as a comparison, even if the function is larger, but actually we can find an upper bound—wait, the statement says "because $\int 1/x^2$ converges"—since $\ln x$ is positive and $\frac{\ln x}{x^2} = \ln x \cdot \frac{1}{x^2}$, but for large $x$, $\ln x < x^c$ so it's still integrable. So this statement is true. - For the fourth option: $\int_1^\infty \frac{1}{x}dx$ diverges, so the reason is wrong. - For the fifth option: $\int_1^\infty \frac{1}{x^2}dx$ converges, so the reason (saying it diverges because $\int 1/x^2$ diverges) is wrong. Wait, correcting: The third option says $\int_{e^{10}}^\infty \frac{\ln x}{x^2}dx$ converges because $\int_{e^{10}}^\infty \frac{1}{x^2}dx$ converges. Since $\ln x$ is positive, but even though $\ln x >1$, $\frac{\ln x}{x^2}$ is still integrable (as $\ln x$ grows slower than any polynomial). The key is that the comparison test can use an upper bound: if $0 < f(x) < g(x)$ and $\int g(x)dx$ converges, then $\int f(x)dx$ converges. Here, for large $x$, $\ln x < x^{0.5}$, so $\frac{\ln x}{x^2} < \frac{x^{0.5}}{x^2} = \frac{1}{x^{1.5}}$, which converges. Alternatively, integration by parts: $\int \frac{\ln x}{x^2}dx = -\frac{\ln x}{x} + \int \frac{1}{x^2}dx = -\frac{\ln x}{x} - \frac{1}{x} + C$, which tends to 0 as $x o\infty$, so it converges. So the third statement is true. The fourth option's reason is wrong because $\int 1/x dx$ diverges. The fifth option's reason is wrong because $\int 1/x^2 dx$ converges. The first option's reason: $\int 2/x dx$ diverges, but $2^x >2$ so $\frac{2^x}{x} > \frac{2}{x}$, so it diverges—but the statement says it diverges because $\int 2/x$ diverges, which is correct? Wait no: $2^x$ grows exponentially, so $\frac{2^x}{x}$ is way larger than $\frac{2}{x}$, so it does diverge, but is the comparison valid? The comparison test for divergence: if $f(x) > g(x) >0$ and $\int g(x)dx$ diverges, then $\int f(x)dx$ diverges. So $\frac{2^x}{x} > \frac{2}{x}$ for $x>e^{10}$ (since $2^x >2$ when $x>1$), so yes, it diverges. But wait, the third option: let's check again. The third option says it converges because $\int 1/x^2$ converges. Is that valid? Since $\frac{\ln x}{x^2} > \frac{1}{x^2}$, but convergence of a smaller integral doesn't imply convergence of a larger one. Wait, I made a mistake earlier: the comparison test for convergence requires the function to be bounded above by a convergent integral. So the third option's logic is flawed, but the first option's logic is correct? Wait no: $2^x$ at $x=e^{10}$ is $2^{e^{10}}$, which is enormous, so $\frac{2^x}{x}$ is way larger than $\frac{2}{x}$, so the first option's statement is "diverges because $\int 2/x$ diverges"—which is correct by comparison test (divergence). But wait, let's check the third option again: the statement says it converges because $\int 1/x^2$ converges. But since $\frac{\ln x}{x^2} > \frac{1}{x^2}$, that doesn't imply convergence. However, if we take an upper bound: $\ln x < x^a$ for any $a>0$, so take $a=0.5$, then $\frac{\ln x}{x^2} < \frac{x^{0.5}}{x^2} = \frac{1}{x^{1.5}}$, which converges. So the function is bounded above by a convergent integral, so it converges. But the statement's reason is "because $\int 1/x^2$ converges"—which is not the right upper bound, but is the statement true? The question is which statement is TRUE by the comparison test. The first option: $\int 2^x/x dx$ diverges because $\int 2/x dx$ diverges—yes, because $2^x/x >2/x >0$, so by comparison test, divergence. The third option: does it converge? Yes, but is the reason given valid? The reason given is "because $\int 1/x^2 dx$ converges"—but since the function is larger than $1/x^2$, that reason is not valid. The first option's reason is valid: larger than a divergent integral, so it diverges. Wait, but let's compute $\int_{e^{10}}^\infty 2/x dx = 2[\ln x]_{e^{10}}^\infty = \infty$, so it diverges. And $\int 2^x/x dx$ is larger, so it diverges. So the first option's statement is true? But wait, the third option: the function $\ln x/x^2$—does it converge? Yes, as integration by parts shows: $\lim_{b o\infty} [-\ln x/x -1/x]_{e^{10}}^b = 0 - (-\ln e^{10}/e^{10} -1/e^{10}) = (10/e^{10} +1/e^{10}) =11/e^{10}$, which is finite. So it converges. But the reason given is "because $\int 1/x^2 dx$ converges"—is that a valid comparison? No, because the function is larger than $1/x^2$, but if we use an upper bound, it's valid. But the statement says "because $\int 1/x^2 dx$ converges"—so is the statement true? The question is which statement is TRUE. Let's check the options again: First option: True (diverges, reason valid by comparison test). Third option: True (converges, but is the reason valid? The reason says because $\int 1/x^2$ converges—even though the function is larger, but the function does converge. But does the comparison test apply here? The comparison test for convergence requires an upper bound. So the third option's reason is not valid, but the statement's conclusion is correct. The first option's conclusion and reason are both correct. Wait, but let's check the first option: $2^x$ at $x=e^{10}$ is $2^{e^{10}}$, which is way bigger than 2, so $2^x/x >2/x$, so yes, divergence by comparison. So the first option's statement is true. But wait, the hint says when $x>e$, $\ln x>1$. So the third option: $\ln x>1$, so $\ln x/x^2>1/x^2$, but the integral of $1/x^2$ converges, but that doesn't imply the integral of the larger function converges. So the third option's reason is wrong, even though the conclusion is correct. The first option's reason is correct, so the first option is true. Wait, but let's confirm: the question asks which statement is TRUE by the comparison test. So the statement must be both conclusion correct and reason correct (using comparison test). So first option: conclusion (diverges) correct, reason (because $\int 2/x$ diverges) correct (by comparison test for divergence). Third option: conclusion correct, but reason (because $\int1/x^2$ converges) is not a valid comparison (since function is larger), so the statement is not true by the comparison test. So the first option is the correct answer. Wait, but I'm confused. Let's check again: First option: $\int_{e^{10}}^\infty 2^x/x dx$ diverges because $\int_{e^{10}}^\infty 2/x dx$ diverges. Since $2^x >2$ for $x>1$, so $2^x/x >2/x >0$. So by comparison test (divergence), yes, it diverges. So this statement is true. Third option: $\int_{e^{10}}^\infty \ln x/x^2 dx$ converges because $\int_{e^{10}}^\infty1/x^2 dx$ converges. But $\ln x/x^2 >1/x^2$, so comparison test for convergence requires upper bound, not lower bound. So this reason is invalid, so the statement is not true by the comparison test. So the correct answer is the first option. # Answer: $\int_{e^{10}}^\infty \frac{2^x}{x} dx$ diverges, because $\int_{e^{10}}^\infty \frac{2}{x} dx$ diverges.

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