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Answers (1)

Correct option (c)

Hint: If $f(x)$  is monotonic increasing function then ${f}'(x)>0$

Given: $f(x)=2 x-\tan ^{-1}-\log \left(x+\sqrt{x^{2}+1}\right)$

Explanation:It is given that

$f(x)=2 x-\tan ^{-1}-\log \left(x+\sqrt{x^{2}+1}\right)$ …..(i)

Differentiate (i) w.r.t  x

\begin{aligned} &f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right) \\ &f^{\prime}(x)=2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}} \end{aligned}

\begin{aligned} &f^{\prime}(x)=\frac{1+2 x^{2}}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}} \\ &f^{\prime}(x)=\frac{1+2 x^{2}-\sqrt{x^{2}+1}}{1+x^{2}} \end{aligned}

$\because f(x)$ is monotonic increasing then ${f}'(x)>0$

\begin{aligned} &\frac{1+2 x^{2}-\sqrt{x^{2}+1}}{1+x^{2}}>0 \\ &\Rightarrow 1+2 x^{2}-\sqrt{x^{2}+1}>0 \\ &\Rightarrow 1+2 x^{2}>\sqrt{x^{2}+1} \end{aligned}

Squaring on both sides, we get

$\left(1+2 x^{2}\right)^{2}>x^{2}+1$

\begin{aligned} &1+4 x^{2}+4 x^{4}>x^{2}+1 \\ &\therefore 3 x^{2}+4 x^{4}>0 \forall x \in R \end{aligned}

Thus, the function is monotonically increasing when  \begin{aligned} x \in R \end{aligned}

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