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Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 29 maths textbook solution

Answers (1)


 The correct option is (b)


 If a function f  is continuous at x = a, then

\lim _{x \rightarrow 4-} f(x)=\lim _{x \rightarrow 0+} f(x)=f(a)

\text { (i) } \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)=1


 The function

f(x)=\frac{\tan \left(\frac{\pi}{4} x\right)}{\cot 2 x}, x \neq \frac{\pi}{4}

Step 1: Understand that, if f(x) is continuous a

x=\frac{\pi }{4}

Understand that, if

\frac{\pi }{4}-x=y


x\rightarrow \frac{\pi }{4} \text { and }y\rightarrow 0

\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cot 2 x}=f\left(\frac{\pi}{4}\right) \\ &\therefore \lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot 2\left(\frac{\pi}{4}-y\right)}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}

\lim _{y \rightarrow 0}\left(\frac{\tan y}{\cot \left(\frac{\pi}{2}-2 y\right)}\right)=f\left(\frac{\pi}{4}\right)

\begin{aligned} &\lim _{y \rightarrow 0}\left(\frac{\tan y}{\tan 2 y}\right)=f\left(\frac{\pi}{4}\right) \\ &\frac{1}{2} \lim _{y \rightarrow 0}\left(\frac{\frac{\tan y}{y}}{\frac{\tan 2 y}{2 y}}\right)=f\left(\frac{\pi}{4}\right) \end{aligned}

Apply formula (i)

\begin{aligned} &\therefore \frac{1}{2}\left(\frac{1}{1}\right)=f\left(\frac{\pi}{4}\right) \\ &f\left(\frac{\pi}{4}\right)=\frac{1}{2} \end{aligned}

Hence, the correct answer is option (b)

Posted by

Gurleen Kaur

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