Definite Integration Question 20
Question 20 - 31 January - Shift 2
If $\phi(x)=\frac{1}{\sqrt{x}} \int _{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0$, then $\phi^{\prime}\left(\frac{\pi}{4}\right)$ is equal to ,
then $\phi^{\prime}(\frac{\pi}{4})$ is equal to :
(1) $\frac{8}{\sqrt{\pi}}$
(2) $\frac{4}{6+\sqrt{\pi}}$
(3) $\frac{8}{6+\sqrt{\pi}}$
(4) $\frac{4}{6-\sqrt{\pi}}$
Show Answer
Answer: (3)
Solution:
Formula: Leibnitz Theorem, Properties of definite integral
$\phi(x)=\frac{1}{\sqrt{x}} \int _{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
$\sqrt{x} \ \phi(x)=\int _{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t , \quad \phi(\frac{\pi}{4})=0$
Applying Leibnitz Theorem, we get
$ \begin{aligned} & \frac{1}{2\sqrt{x}}\phi(x) +\sqrt{x}\phi^{\prime}(x)=[(4 \sqrt{2} \sin x-3 \phi^{\prime}(x)) \cdot 1-0]\\ & at \quad x= \frac{\pi}{4} \\ & \frac{1}{2\sqrt{\frac{\pi}{4}}}\phi(\frac{\pi}{4}) +\sqrt{\frac{\pi}{4}}\phi^{\prime}(\frac{\pi}{4})=[(4 \sqrt{2} \sin \frac{\pi}{4}-3 \phi^{\prime}(\frac{\pi}{4})) \cdot 1-0]\\ & \Rightarrow \phi^{\prime}(\frac{\pi}{4})=\frac{2}{\sqrt{\pi}}[4-3 \phi^{\prime}(\frac{\pi}{4})]+0 \\ & \Rightarrow (1+\frac{6}{\sqrt{\pi}}) \phi^{\prime}(\frac{\pi}{4})=\frac{8}{\sqrt{\pi}} \\ & \Rightarrow \phi^{\prime}(\frac{\pi}{4})=\frac{8}{\sqrt{\pi}+6} \end{aligned} $
