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  • explain solution RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 12 maths

explain solution RD Sharma class 12 chapter 8 Continuity exercise Fill in the blanks question 12 maths

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Answer: f\left(\frac{\pi}{4}\right)=\frac{-1}{2}

Hint: You must know about L-hospital’s rule

Given:f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right] \text { and } f(x) \text { is continuous in }\left[0, \frac{\pi}{2}\right]

Solution:

            f(x) is continuous in \left [ 0,\frac{\pi }{2} \right ]

            \begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right) \\ &\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=f\left(\frac{\pi}{4}\right) \end{aligned}

            Applying L-Hospital’s rule

            \begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^{2} x}{4}=f\left(\frac{\pi}{4}\right) \\ &\frac{-2}{4}=f\left(\frac{\pi}{4}\right) \end{aligned}                                        \quad\left [\because \sec ^{2} \frac{\pi}{4}=2\right]

 

            

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