The function $f(x) = x^{1/x}, \: x \neq 0$ has
Answer: B
Solution: Let $y = f(x) = x^{1/x}$ Taking $\log$ on both sides, we get: $\log y = \log(x^{1/x}) \implies \log y = \frac{1}{x}\log x $ Differentiating above w.r.t. x, we get: $\frac{1}{y}\frac{dy}{dx} = \frac{1}{x^2} - \frac{1}{x^2}\log x = 0$ Now, $\frac{dy}{dx} = y \underbrace{ \Bigg(\frac{1}{x^2} - \frac{1}{x^2}\log x \Bigg)}_{\text = 0}= 0 \implies \frac{dy}{dx} = y\frac{1}{x^2}(1-\log x)$ Here, $\frac{1}{x^2}$ can't be zero. So,$1-\log x = 0 \implies \log x = 1 \implies e ^1 = x \implies x = e$ $\therefore$ The function has a maximum at $x = e$ Hence, B is the correct option.
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