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  • Provide solution for rd sharma maths class 12 chapter 16 Increasing and decreasing functions exercise multiple choice quection, question 17

Provide solution for rd sharma maths class 12 chapter 16 Increasing and decreasing functions exercise multiple choice quection, question 17

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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