Q110 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

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

  • Correct answer
    A
    (A) is true, (R) is true and (R) is the correct explanation for $A$.
  • B
    (A) is true, $(R)$ is true but $(R)$ is the not the correct explanation for $A$.
  • C
    (A) is true but $(R)$ is false.
  • D
    (A) is false but (R) is true.

Solution

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.

Hence, the correct answer is option: 'A'.

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