Q. Let $f\left(x\right)=\left\{\begin{matrix} a & ; \, x=\frac{\pi }{2} \\ \frac{\sqrt{2 x - \pi }}{\sqrt{9 + \sqrt{2 x - \pi }} - b} & ; \, x>\frac{\pi }{2} \end{matrix}\right$ . If $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$ , then the value of $\frac{a^{2}}{5 b}$ is

NTA AbhyasNTA Abhyas 2020Continuity and Differentiability Report Error

A

B

C

D

Solution:

$\underset{x \rightarrow \left(\frac{\pi }{2}\right)^{+}}{l i m}f\left(x\right)=f\left(\frac{\pi }{2}\right)$
$\Rightarrow \underset{x \rightarrow \frac{\pi }{2}^{+}}{l i m}\frac{\sqrt{2 x - \pi }}{\sqrt{9 + \sqrt{2 x - \pi }} - b}=a$
For the denominator to be zero, $b=3$
Now, on rationalization
$\underset{x \rightarrow \left(\frac{\pi }{2}\right)^{+}}{l i m}\frac{\sqrt{2 x - \pi } \left(\sqrt{9 + \sqrt{2 x - \pi }} + 3\right)}{\left(\sqrt{9 + \sqrt{2 x - \pi }} - 9\right)}=a$
$\therefore a=6$

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