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  • Need solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 8

Need solution for RD Sharma maths class 12 chapter Continuity exercise 8.2 question 8

Answers (1)

Answer:

 \frac{1}{2}

Given:

f(x)= \begin{cases}\frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x} & x \neq \pi / 4 \\ k & x=\pi / 4\end{cases}

Hint:

 Apply L-Hospital Rule when you get 0/0 from.

Explanation:

At x = \pi/4

\begin{aligned} &\text { L.H.L } =\text { R.H.L }=f\left(\frac{\pi}{4}\right) \quad \ldots(A) \\ &\text { L.H.L }=\lim _{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _{h \rightarrow 0}\left(\frac{\pi}{4}-h\right) \end{aligned}

=\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}+h\right)}{\cot \left(\frac{\pi}{4}-h\right)}=\lim _{h \rightarrow 0} \frac{\tanh }{\tan 2 h}

=\lim _{h \rightarrow 0} \frac{\frac{\tanh }{h}}{\frac{\tan 2 h}{2 h} \times 2}=\frac{1}{2}

Hence F (x) will be continious on

[0, \frac{\pi }{2}]

If

f(\frac{\pi }{4})=\frac{1}{2}

Posted by

Gurleen Kaur

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