MCQ
If a function of defined by
$f(x)=\left\{\begin{array}{cc}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & \text { if } x \neq \frac{\pi}{4} \\k, & \text { if } x=\frac{\pi}{4} \end{array}\right.$
is continuous at $x=\frac{\pi}{4}$, then $k =$
  • $\frac{1}{4}$
  • B
    1
  • C
    $-\frac{1}{4}$
  • D
    2

Answer

Correct option: A.
$\frac{1}{4}$
(A)
Since $f (x)$ is continuous at $x=\frac{\pi}{4}$,
$\therefore \quad f\left(\frac{\pi}{4}\right)=\lim _{x \rightarrow \frac{\pi}{4}} f(x)$
$\Rightarrow k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$
Applying L'Hospital rule on R.H.S., we get
$k =\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} \quad \Rightarrow k =\frac{1}{4}$

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