If the function is defined as f(x) & = 5^ x -1 2 -x , when x 2 & … - Vidyadip

MCQ

If the function is defined as
$\begin{aligned}f(x) & =\frac{5^{\cos x}-1}{\frac{\pi}{2}-x}, \text { when } x \neq \frac{\pi}{2} \\& =2 \log 5, \text { when } x=\frac{\pi}{2}, \text { then }\end{aligned}$

A
$f (x)$ is continuous at $x=\frac{\pi}{2}$
✓
$f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$
C
$f (x)$ has irremovable discontinuity at $x=\frac{\pi}{2}$
D
none of these

✓

Answer

Correct option: B.

$f (x)$ has removable discontinuity at $x=\frac{\pi}{2}$

(B)
Applying L'Hospital rule, we get
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x}-1}{\frac{\pi}{2}-x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{5^{\cos x} \cdot \log 5(-\sin x)}{-1}$
$=5^{\cos \frac{\pi}{2}} \cdot \log 5 \sin \frac{\pi}{2}=\log 5$
and $f\left(\frac{\pi}{2}\right)=2 \log 5$
$\therefore f (x)$ is discontinuous at $x=\frac{\pi}{2}$.
Here, $\lim _{x \rightarrow \frac{\pi}{2}} f (x)$ exists but not equal to $f \left(\frac{\pi}{2}\right)$.
$\therefore $ the discontinuity at $x=\frac{\pi}{2}$ is removable.

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