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. $\frac{4}{6-\sqrt{\pi}}$

Hint:

For solving such integral equations, always substitute specific values into the equation to simplify the terms. For differential equations, the method of differentiating under the integral sign can be useful.

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) : $\frac{8}{6+\sqrt{\pi}}$

Answer 2

Step 1: Given the equation for \( \phi(x) \): \[ \phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{x}{4}}^{x} \left( 4\sqrt{2} \sin t - 3 \phi(t) \right) dt \] We need to evaluate \( \phi \left( \frac{\pi}{4} \right) \). To do this, first, let's differentiate \( \phi(x) \) with respect to \( x \). Using Leibniz's rule for differentiating under the integral sign, we get: \[ \phi'(x) = \frac{1}{\sqrt{x}} \left[ (4\sqrt{2} \sin x - 3 \phi(x)) \cdot 1 \right] - \frac{1}{2} x^{-3/2} \] Thus, we obtain the expression for \( \phi'(x) \). 
Step 2: Now, let's focus on evaluating \( \phi \left( \frac{\pi}{4} \right) \). For \( x = \frac{\pi}{4} \), the integral simplifies as follows: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \left( 4\sqrt{2} \sin t - 3 \phi(t) \right) dt = 0 \] So, we are left with: \[ \phi \left( \frac{\pi}{4} \right) = \frac{2}{\sqrt{\pi}} \left[ 4 - 3 \phi \left( \frac{\pi}{4} \right) \right] \] Expanding this expression: \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi}} - \frac{6}{\sqrt{\pi}} \phi \left( \frac{\pi}{4} \right) \] Now, solve for \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) + \frac{6}{\sqrt{\pi}} \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi}} \] Factor out \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) \left( 1 + \frac{6}{\sqrt{\pi}} \right) = \frac{8}{\sqrt{\pi}} \] Solve for \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi} \left( 6 + \sqrt{\pi} \right)} \] Thus, the final answer is: \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{6 + \sqrt{\pi}} \]