Q104 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics

If a function $f$ defined by $f(x)=\left\{\begin{array}{c}\frac{1-\sqrt{2} \sin x}{\pi-4 x}, \text { if } x \neq \frac{\pi}{4} \\ k, \text { if } x=\frac{\pi}{4}\end{array}\right.$ $x=\frac{\pi}{4}$, then $k=$

  • Correct answer
    A
    $\frac{1}{4}$
  • B
    1
  • C
    $\frac{-1}{4}$
  • D
    2

Solution

We have, $$ f(x)=\left\{\begin{array}{cc} \frac{1-\sqrt{2} \sin x}{\pi-4 x}, & x \neq \frac{\pi}{4} \\ k & , \quad x=\frac{\pi}{4} \end{array}\right. $$ Since, $f(x)$ is continuous at $$ \begin{aligned} & x=\frac{\pi}{4} \\ \therefore & f(\pi / 4)=\lim _{\pi \rightarrow \frac{\pi}{4}} f(x) \\ \Rightarrow \quad k & =\lim _{\pi \rightarrow \frac{\pi}{4}} \frac{1-\sqrt{2} \sin x}{\pi-4 x} \\ \Rightarrow \quad k & =\lim _{\pi \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \cos x}{-4} \text { [using } L^{\prime} \text { hospital Rule] } \\ \Rightarrow \quad k & =\frac{\sqrt{2}}{4} \times \frac{1}{\sqrt{2}} \\ \Rightarrow \quad & k=\frac{1}{4} . \end{aligned} $$

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

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