If $f(x) = \left\{ \begin{array}{r} \frac{1 - \sqrt{2}\sin x}{\pi - 4x},\ if\ x
eq \frac{\pi}{4} \ a,\ \ \ \ \ \ if\ x = \frac{\pi}{4} \end{array} ight.\$ is continuous at $\frac{\pi}{4}$, then $a$ is equal to

Question

If $f(x) = \left\{ \begin{array}{r} \frac{1 - \sqrt{2}\sin x}{\pi - 4x},\ if\ x 
eq \frac{\pi}{4} \ a,\ \ \ \ \ \ if\ x = \frac{\pi}{4} \end{array} ight.\$ is continuous at $\frac{\pi}{4}$, then $a$ is equal to

Dr. Abubacker · Accepted Answer

(d)
$\lim_{x ightarrow \frac{\pi}{4}}{f(x)} = \lim_{x ightarrow \frac{\pi}{4}}\frac{1 - \sqrt{2}\sin x}{\pi - 4x}$
$= \lim_{x ightarrow \frac{\pi}{4}}\frac{- \sqrt{2}\cos x}{4} = \frac{1}{4}$ [by L ‘Hospital’s rule]
Since, $f(x)$ is continuous at $x = \frac{\pi}{4}$
$	herefore\ \lim_{x ightarrow \frac{\pi}{4}}{f(x)} = f\left( \frac{\pi}{4} ight)\ \ \ \  \Rightarrow \ \ \frac{1}{4} = a$

Q. If $f(x) = \left\{ \begin{array}{r} \frac{1 - \sqrt{2}\sin x}{\pi - 4x},\ if\ x \neq \frac{\pi}{4} \\ a,\ \ \ \ \ \ if\ x = \frac{\pi}{4} \end{array} \right.\$ is continuous at $\frac{\pi}{4}$, then $a$ is equal to

Explanation

Reviewed by Dr. Abubacker

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