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Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 14

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f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})


f(x) =tan \: x

To prove:

\text { We have to show that } f(x) \text { is increasing function on } (-\frac{\pi }{2},\frac{\pi }{2})


Condition for increasing function f’(x)>0.



f(x) =tan \: x

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}(\tan x)\\ &\Rightarrow f^{\prime}(x)=\sec ^{2} x\\ \end{aligned}

Now, as given

\begin{aligned} &x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{aligned}

i.e, 4th quadrant to 1st quadrant, where

\begin{aligned} \Rightarrow sec^{2}\: x> 0 \\ \Rightarrow f'(x)> 0 \end{aligned}

Hence condition of f(x) to be increasing.

\text { Hence } f(x) \text { is increasing on interval } x \in (-\frac{\pi }{2},\frac{\pi }{2}).

Hence proved.

Posted by

Gurleen Kaur

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