Let $f(x) = \left\{ \begin{array}{r} \frac{ an{x - \cot x}}{x - \frac{\pi}{4}},\ \ x
eq \frac{\pi}{4} \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{4} \end{array} ight.\$ the value of $a$ so that $f(x)$ is continuous at $x = \frac{\pi}{4}\$ is

Question

Let $f(x) = \left\{ \begin{array}{r} \frac{	an{x - \cot x}}{x - \frac{\pi}{4}},\ \ x 
eq \frac{\pi}{4} \ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{4} \end{array} ight.\$ the value of $a$ so that $f(x)$ is continuous at $x = \frac{\pi}{4}\$ is

Dr. Abubacker · Accepted Answer

(b)
LHL$= \lim_{h ightarrow 0}\ \ f\left( \frac{\pi}{4} - h ight)$
$= \lim_{h ightarrow 0}\ \ \frac{	an{\left( \frac{\pi}{4} - h ight) - \cot\left( \frac{\pi}{4} - h ight)}}{\frac{\pi}{4} - h - \frac{\pi}{4}}$
$= \lim_{h ightarrow 0}\ \ \frac{{- \sec^{2}}{\left( \frac{\pi}{4} - h ight) - {cosec}^{2}\left( \frac{\pi}{4} - h ight)}}{- 1} = 4$
[by L ‘Hospital’s rule]
Since, $f(x)$ is continuous at $x = \frac{\pi}{4}$, then LHL$= f\left( \frac{\pi}{4} ight)$
$	herefore\ \ \ a = 4$

Q. Let $f(x) = \left\{ \begin{array}{r} \frac{\tan{x - \cot x}}{x - \frac{\pi}{4}},\ \ x \neq \frac{\pi}{4} \\ a,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{4} \end{array} \right.\$ the value of $a$ so that $f(x)$ is continuous at $x = \frac{\pi}{4}\$ is

Explanation

Reviewed by Dr. Abubacker

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