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

To find \(\phi^{\prime}\left(\frac{\pi}{4}\right)\), we need to differentiate the given function \(\phi(x)\), which is:

\(\phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^x \left(4 \sqrt{2} \sin t - 3 \phi^{\prime}(t)\right) \, dt\)

We can use Leibniz's Rule for differentiation under the integral sign:

Leibniz's Rule states:

\(\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t,x) \, dt\right) = f(b(x),x)\cdot b'(x) - f(a(x),x)\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(t,x) \, dt\)

In this problem:

• \(f(t,x) = 4 \sqrt{2} \sin t - 3 \phi^{\prime}(t)\)
• \(a(x) = \frac{\pi}{4}\) and \(b(x) = x\)

Plug these into Leibniz's Rule:

\(\frac{d}{dx}\left(\phi(x)\right) = \frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) \cdot \int_{\frac{\pi}{4}}^x \left(4 \sqrt{2} \sin t - 3 \phi^{\prime}(t)\right) dt + \frac{1}{\sqrt{x}} \cdot \left(4 \sqrt{2} \sin x - 3 \phi^{\prime}(x)\right)\)

Differentiate \(\frac{1}{\sqrt{x}}\):

\(\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right) = -\frac{1}{2x\sqrt{x}}\)

Substituting back, we have:

\(\phi^{\prime}(x) = -\frac{1}{2x\sqrt{x}} \cdot \int_{\frac{\pi}{4}}^x \left(4 \sqrt{2} \sin t - 3 \phi^{\prime}(t)\right) \, dt + \frac{1}{\sqrt{x}} \cdot \left(4 \sqrt{2} \sin x - 3 \phi^{\prime}(x)\right)\)

When \(x = \frac{\pi}{4}\), the integral \(\int_{\frac{\pi}{4}}^{\frac{\pi}{4}}\) contributes nothing. Therefore:

\(\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \cdot \left(4 \sqrt{2} \sin\left(\frac{\pi}{4}\right) - 3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)\)

Since \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), substitute:

\(\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \cdot \left(4 \sqrt{2} \cdot \frac{\sqrt{2}}{2} - 3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)\)

\(\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{\frac{\pi}{4}}} \cdot \left(4 - 3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)\)

Let \(y = \phi^{\prime}\left(\frac{\pi}{4}\right)\), then:

\(y = \frac{1}{\sqrt{\frac{\pi}{4}}} \cdot (4 - 3y)\)

Solving for \(y\) gives:

\(y(1 + \frac{3}{\sqrt{\frac{\pi}{4}}}) = \frac{4}{\sqrt{\frac{\pi}{4}}}\)

Simplifying both sides:

\(1 + \frac{3}{\sqrt{\frac{\pi}{4}}} = 6/\sqrt{\pi}\) and \(\sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}\)

Then, \(\phi^{\prime}\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}\)

Hence, the correct option is \(\frac{8}{6+\sqrt{\pi}}\).