Given that frac{d}{d x} int_0^{phi(x)} f(t) d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation $\int_1^{\cos x} t^2 f(t) d t = \cos 2 x$. Differentiating both sides with respect to $x$ using the Leibniz rule: $\frac{d}{dx}

Vedclass · Accepted Answer

$4 \sqrt{2}$

Vedclass · Answer

(B) Given the equation $\int_1^{\cos x} t^2 f(t) d t = \cos 2 x$. Differentiating both sides with respect to $x$ using the Leibniz rule: $\frac{d}{dx} \int_1^{\cos x} t^2 f(t) d t = \frac{d}{dx} (\cos 2 x)$ $(\cos x)^2 f(\cos x) \cdot \frac{d}{dx}(\cos x) = -2 \sin 2 x$ $\cos^2 x \cdot f(\cos x) \cdot (-\sin x) = -2 \sin 2 x$ Since $\sin 2 x = 2 \sin x \cos x$,we have: $-\cos^2 x \cdot f(\cos x) \cdot \sin x = -2(2 \sin x \cos x)$ For $x \in \left(0, \frac{\pi}{2}ight)$,$\sin x 
eq 0$ and $\cos x 
eq 0$,so we can divide by $-\sin x \cos x$: $f(\cos x) = \frac{4 \sin x \cos x}{\cos^2 x \sin x} = \frac{4}{\cos x}$ We want to find $f\left(\frac{1}{\sqrt{2}}ight)$. Let $\cos x = \frac{1}{\sqrt{2}}$,which implies $x = \frac{\pi}{4}$. Substituting this into our expression for $f(\cos x)$: $f\left(\frac{1}{\sqrt{2}}ight) = \frac{4}{1/\sqrt{2}} = 4 \sqrt{2}$.

Given that $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. For all $x \in \left(0, \frac{\pi}{2}\right)$,if $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$,then $f\left(\frac{1}{\sqrt{2}}\right)=$

Similar Questions

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x}$ is

Let a differentiable function $f$ satisfy the equation $\int_{0}^{36} f(\frac{tx}{36}) dt = 4\alpha f(x)$. If $y = f(x)$ is a standard parabola passing through the points $(2, 1)$ and $(-4, \beta)$, then $\beta^{\alpha}$ is equal to . . . . . . .

$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)+\sin ^5\left(\frac{3 \pi}{6 n}\right)+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\} = $

Let $f$ be a differentiable function satisfying $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f \left(\frac{\lambda^{2} x}{3}\right) d\lambda$ for $x > 0$ and $f(1) = \sqrt{3}$. If $y = f(x)$ passes through the point $(\alpha, 6)$,then $\alpha$ is equal to $.........$

Let $f, g:(0, \infty) \rightarrow \mathbb{R}$ be two functions defined by $f(x)=\int_{-x}^x(|t|-t^2) e^{-t^2} dt$ and $g(x)=\int_0^{x^2} t^{1/2} e^{-t} dt$. Then the value of $(f(\sqrt{\log_{e} 9}) + g(\sqrt{\log_{e} 9}))$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module