Question

If $\phi(x)=\frac{1}{\sqrt{ x }} \int_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) dt , x>$, then $\emptyset^{\prime}\left(\frac{\pi}{4}\right)$ is equal to :

• A. $\frac{4}{6+\sqrt{\pi}}$
• B. $\frac{8}{6+\sqrt{\pi}}$
• C. $\frac{8}{\sqrt{\pi}}$
• D. $\frac{4}{6-\sqrt{\pi}}$

Answer 1

The Correct Option is B

${\varphi }^{\prime }(x)=\frac{1}{\sqrt{x}}\left[\left(4\sqrt{2}\sin x-3{\varphi }^{\prime }(x)\right)\cdot 1-0\right]-\frac{1}{2}{x}^{-3/2}$
$\underset{\frac{\pi }{4}}{\overset{x}{\int }}\left(4\sqrt{2}\sin t-3{\varphi }^{\prime }(t)\right)dt$
${\varphi }^{\prime }\left(\frac{\pi }{4}\right)=\frac{2}{\sqrt{\pi }}\left[4-3{\varphi }^{\prime }\left(\frac{\pi }{4}\right)\right]+0$
$\left(1+\frac{6}{\sqrt{\pi }}\right){\varphi }^{\prime }\left(\frac{\pi }{4}\right)=\frac{8}{\sqrt{\pi }}$
${\varphi }^{\prime }\left(\frac{\pi }{4}\right)=\frac{8}{\sqrt{\pi }+6}$
So, thr correct option is (B) : $\frac{8}{6+\sqrt{\pi}}$