Question

Given that $\frac{d}{dx} \int_{0}^{\phi(x)} f(t) dt = f(\phi(x)) \phi'(x)$. For all $x \in (0, \frac{\pi}{2})$, if $\int_{1}^{\cos x} t^2 f(t) dt = \cos 2x$, then $f\left(\frac{1}{\sqrt{2}}\right) =$

• A. $2\sqrt{2}$
• B. $4\sqrt{2}$
• C. $\frac{\pi}{4}$
• D. $-\frac{\pi}{4}$

Hint:

Remember the Leibniz rule for differentiation under the integral sign: $\frac{d}{dx} \int_{a(x)}^{b(x)} f(t, x) dt = f(b(x), x) b'(x) - f(a(x), x) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(t, x) dt$. In this problem, $f(t, x) = t^2 f(t)$ which is independent of $x$, so the last term is zero.

Answer 1

The Correct Option is A

Step 1: Differentiate the given integral equation.
We are given $\int_{1}^{\cos x} t^2 f(t) dt = \cos 2x$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus and the chain rule: $$\frac{d}{dx} \left( \int_{1}^{\cos x} t^2 f(t) dt \right) = \frac{d}{dx} (\cos 2x)$$ $$(\cos x)^2 f(\cos x) \cdot \frac{d}{dx}(\cos x) - (1)^2 f(1) \cdot \frac{d}{dx}(1) = -2 \sin 2x$$
Step 2: Simplify the differentiated equation.
$$(\cos^2 x) f(\cos x) (-\sin x) - 0 = -2 \sin x \cos x$$ $$-\cos^2 x \sin x f(\cos x) = -2 \sin x \cos x$$
Step 3: Solve for $f(\cos x)$.
For $x \in (0, \frac{\pi}{2})$, $\sin x \neq 0$ and $\cos x \neq 0$. Dividing both sides by $-\sin x \cos x$: $$\cos x f(\cos x) = 2$$ $$f(\cos x) = \frac{2}{\cos x}$$
Step 4: Find $f\left(\frac{1}{\sqrt{2}}\right)$.
We need to find the value of $f$ when its argument is $\frac{1}{\sqrt{2}}$. We set $\cos x = \frac{1}{\sqrt{2}}$. This occurs at $x = \frac{\pi}{4}$, which is within the given interval $(0, \frac{\pi}{2})$. Substituting $\cos x = \frac{1}{\sqrt{2}}$ into the expression for $f(\cos x)$: $$f\left(\frac{1}{\sqrt{2}}\right) = \frac{2}{\frac{1}{\sqrt{2}}} = 2 \cdot \sqrt{2} = 2\sqrt{2}$$ Thus, $f\left(\frac{1}{\sqrt{2}}\right) = \boxed{2\sqrt{2}}$.