Assertion (A): The function f(x) = x - log le | vedclass

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(A) Given function: $f(x) = x - \log \left(\frac{1+x}{x}\right), x > 0$. Differentiating with respect to $x$: $f^{\prime}(x) = \frac{d}{dx}(x) - \frac

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$(A)$ is true, $(R)$ is true and $(R)$ is the correct explanation for $(A)$.

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(A) Given function: $f(x) = x - \log \left(\frac{1+x}{x}ight), x > 0$. Differentiating with respect to $x$: $f^{\prime}(x) = \frac{d}{dx}(x) - \frac{d}{dx} \log \left(\frac{1+x}{x}ight) = 1 - \frac{x}{1+x} \cdot \frac{d}{dx} \left( \frac{1}{x} + 1 ight) = 1 - \frac{x}{1+x} \cdot \left( -\frac{1}{x^2} ight) = 1 + \frac{1}{x(1+x)}$. Since $x > 0$, $x(1+x) > 0$, therefore $f^{\prime}(x) = 1 + \frac{1}{x(1+x)} > 1 > 0$ for all $x > 0$. Since $f^{\prime}(x) > 0$ for all $x$ in the domain, the function $f(x)$ is strictly increasing and has no local maximum or minimum. The Reason $(R)$ states that if a function is strictly increasing, then $f^{\prime}(x) 
eq 0$. This is a standard property of strictly increasing functions. Thus, both $(A)$ and $(R)$ are true, and $(R)$ correctly explains $(A)$.

Assertion $(A)$: The function $f(x) = x - \log \left(\frac{1+x}{x}\right), x > 0$ has no maximum. Reason $(R)$: If a function $f(x)$ is strictly increasing in an interval $(a, b)$, then at any point in $(a, b)$, $f^{\prime}(x) \neq 0$. The correct option among the following is

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