Q106 / 160 · TS EAMCET 2006 · Mathematics

If $f(x)= \begin{cases}\frac{1-\sqrt{2} \sin x}{\pi-4 x} & \text { if } x \neq \frac{\pi}{4} \\ a & \text { if } x=\frac{\pi}{4}\end{cases}$ is continuous at $\frac{\pi}{4}$, then $a$ is equal to

  • A
    4
  • B
    2
  • C
    1
  • Correct answer
    D
    $1 / 4$

Solution

$\because \quad f(x)= \begin{cases}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, & \text { if } x \neq \frac{\pi}{4} \\ a, & \text { if } x=\frac{\pi}{4}\end{cases}$ $\lim _{x \rightarrow \frac{\pi}{4}} f(x)=\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x}$ $=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4}$ (By L' Hospital's rule) $=\frac{\sqrt{2} \cdot \frac{1}{\sqrt{2}}}{4}=\frac{1}{4}$ Since $f(x)$ is continuous at $x=\frac{\pi}{4}$. $\therefore \quad \lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right)$ $\Rightarrow \quad \frac{1}{4}=a$

Hence, the correct answer is option: 'D'.

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