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Chapter 7: Problem 48

The set of critical points of the function $f(x)=x-\log x+\int_{2}^{\pi}(1 / z-2-2 \cos 4 z) d z$ is (A) \(\left\\{\frac{\pi}{6}, \frac{n \pi}{2}+\frac{\pi}{6}\right\\}, n \in N\) (B) \(\\{n \pi\\}, n \in N\) (C) $\left\\{\frac{\pi}{2}, \mathrm{n} \pi+\frac{\pi}{6}\right\\}, \mathrm{n} \in \mathrm{N}$ (D) None of these

Short Answer

Expert verified
A. \(\{\log x\}\) B. \(\{x\log x\}\) C. \(\{x\}\) D. None of these Answer: D. None of these Explanation: The function has a critical point x = 1, which is not contained in any of the given sets (A), (B), or (C). Therefore, the correct answer is (D) "None of these".

Step by step solution

01

Find the derivative of \(f(x)\)

To find \(f'(x)\), we need to take the derivative of each term in the function. Notice that the integral term is a constant, so it will become zero when we take its derivative. $$ \frac{d}{dx}(x-\log x) = \frac{d}{dx}(x) - \frac{d}{dx}(\log x) = 1-\frac{1}{x} $$ So, \(f'(x) = 1-\frac{1}{x}\).
02

Solve the equation \(f'(x) = 0\) for the critical points

Now we need to find the critical points by solving: $$ 1-\frac{1}{x} = 0 $$ Adding \(\frac{1}{x}\) to both sides, we get: $$ x = 1 $$ Thus, the set of critical points is \(\{1\}\).
03

Determine the correct set of critical points

Since we found that the set of critical points is \(\{1\}\), we can eliminate options (A), (B), and (C) as they all contain other expressions for their sets. Thus, the correct answer is (D) "None of these".

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