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Question
Let \[f\left( x \right) = \frac{\tan\left( \frac{\pi}{4} - x \right)}{\cot 2x}, x \neq \frac{\pi}{4} .\] The value which should be assigned to f (x) at \[x = \frac{\pi}{4},\]so that it is continuous everywhere is
Options
1
1/2
2
none of these
Solution
\[\frac{1}{2}\]
If \[f\left( x \right)\] is continuous at \[x = \frac{\pi}{4}\]
If \[\frac{\pi}{4} - x = y\], then
\[x \to \frac{\pi}{4} \text{ and } y \to 0\]
\[\therefore \lim_{y \to 0} \left( \frac{\tan y}{\cot 2\left( \frac{\pi}{4} - y \right)} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \lim_{y \to 0} \left( \frac{\tan y}{\cot\left( \frac{\pi}{2} - 2y \right)} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \lim_{y \to 0} \left( \frac{\tan y}{\tan 2y} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \lim_{y \to 0} \left( \frac{\frac{\tan y}{y}}{\frac{\tan 2y}{y}} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \lim_{y \to 0} \left( \frac{\frac{\tan y}{y}}{\frac{2 \tan 2y}{2y}} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \frac{1}{2} \lim_{y \to 0} \left( \frac{\frac{\tan y}{y}}{\frac{\tan 2y}{2y}} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \frac{1}{2}\left( \frac{\lim_{y \to 0} \frac{\tan y}{y}}{\lim_{y \to 0} \frac{\tan 2y}{2y}} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow \frac{1}{2}\left( \frac{1}{1} \right) = f\left( \frac{\pi}{4} \right)\]
\[ \Rightarrow f\left( \frac{\pi}{4} \right) = \frac{1}{2}\]
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