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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 $(a, b) f^{\prime}(x) \neq 0$
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The correct answer is:
(A) is true, (R) is true and (R) is the correct explanation for $A$.
Given function,
$f(x)=x-\log \left(\frac{1+x}{x}\right), x>0$
Differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{d}{d x}(x)-\frac{d}{d x} \log \left(\frac{1+x}{x}\right)$
$=1-\frac{1}{\frac{1+x}{x}} \cdot \frac{d}{d x}\left(\frac{1+x}{x}\right) \quad$ [from chain rule]
$=1-\frac{x}{1+x} \cdot\left\{\frac{x \frac{d}{d x}(1+x)-(1+x) \frac{d}{d x}(x)}{x^2}\right\}$
$=1-\frac{x}{1+x}\left\{\frac{x-(1+x)}{x^2}\right\}=1-\frac{x}{1+x}\left\{-\frac{1}{x^2}\right\}$
$f^{\prime}(x)=1+\frac{1}{x(x+1)}$
Since $x>0$, therefore there is no possibility of $f^{\prime}(x)$ to be zero at any value of $x$.
So, for the given function $f^{\prime}(x) \neq 0$
Also, for $\forall x \in \mathbf{R} \quad f^{\prime}(x)>0$
i.e., function $f(x)$ is strictly increasing its defined domain.
So, reason correctly explains given assertion.
$f(x)=x-\log \left(\frac{1+x}{x}\right), x>0$
Differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{d}{d x}(x)-\frac{d}{d x} \log \left(\frac{1+x}{x}\right)$
$=1-\frac{1}{\frac{1+x}{x}} \cdot \frac{d}{d x}\left(\frac{1+x}{x}\right) \quad$ [from chain rule]
$=1-\frac{x}{1+x} \cdot\left\{\frac{x \frac{d}{d x}(1+x)-(1+x) \frac{d}{d x}(x)}{x^2}\right\}$
$=1-\frac{x}{1+x}\left\{\frac{x-(1+x)}{x^2}\right\}=1-\frac{x}{1+x}\left\{-\frac{1}{x^2}\right\}$
$f^{\prime}(x)=1+\frac{1}{x(x+1)}$
Since $x>0$, therefore there is no possibility of $f^{\prime}(x)$ to be zero at any value of $x$.
So, for the given function $f^{\prime}(x) \neq 0$
Also, for $\forall x \in \mathbf{R} \quad f^{\prime}(x)>0$
i.e., function $f(x)$ is strictly increasing its defined domain.
So, reason correctly explains given assertion.
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