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I had a question in regards to the following question I encountered on a GRE practice exam. The question is as follows:

The following function is defined for all positive $x$: $$f(x) = \int_{x}^{2x} \frac{\sin(t)}{t} \, dt $$ At what value on the interval $(0, 3\pi/2)$ does the function attain a local maximum? A.) $\pi/6$ B.) $\pi/3$ C.) $\pi/2$ D.) $\pi$ E.) $2\pi/3$

I arrived at the correct answer B, but did it slightly differently than the solution manual did and wanted some clarification.

So I calculated $$f'(x) = \frac{\sin(2x) - \sin(x)}{x},$$ set the numerator equal to 0 (since the denominator being equal to zero is outside the domain) and arrived at $\sin(2x) = \sin(x)$. At this point I ruled out A, C, and E since $\sin(2x) \neq \sin(x)$ for those $x$-values and then computed the second derivative and determined that $f''(\pi/3) < 0$ and $f''(\pi)>0$; so the answer is B.

The solution manual used a trig identity I did not know off the top of my head and said that $\sin(2x) - \sin(x) = 0$ implies that $\sin(x)(2 \cos(x) - 1) = 0$ and so $\sin(x) = 0$ or $\cos(x) = 1/2$, which implies that $x = \pi/3$ because $0 < x < 3\pi/2$. Why were they able to rule out the solution $x = \pi$ since after all $\sin(\pi) = 0$ and $0 < \pi < 3\pi/2$?

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  • $\begingroup$ Between $\pi$ and $2 \pi$, the integrand is negative. $\endgroup$ – Patrick Stevens Oct 22 '16 at 19:58
  • $\begingroup$ For $x$ near $\pi$ the factor $2\cos x-1$ is near $-1$, so $\sin x(2\cos x-1)$ goes from negative to positive as $x$ increases through $\pi$, meaning that $f$ has a local minimum at $x=\pi$. $\endgroup$ – Brian M. Scott Oct 22 '16 at 20:00
  • $\begingroup$ @PatrickStevens Why is that relevant? $\endgroup$ – Oiler Oct 22 '16 at 20:03
  • $\begingroup$ @Oiler Because for $x$ small and positive, the integrand is positive and hence the integral is positive; therefore there must be a local maximum of the function before $x=\pi$. (I suppose I'm assuming that exactly one of the given answers is true, in order to conclude that $x=\pi$ is not a local maximum.) $\endgroup$ – Patrick Stevens Oct 22 '16 at 20:05
  • $\begingroup$ @BrianM.Scott I have convinced my self that $x = \pi$ is a local minimum using the second derivative test and that trick is definitely nice to rule that out faster. But why did they not even acknowledge $x = \pi$ as a critical point? $\endgroup$ – Oiler Oct 22 '16 at 20:06

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