If x satisfies the equation left( int_{0}^{1} | vedclass

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

Question

(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$)

Vedclass · Accepted Answer

$\pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$

Vedclass · Answer

(D) Let $I_1 = \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt$. Since $f(t) = \frac{t^2 \sin 2t}{t^2 + 1}$ is an odd function (because $f(-t) = -f(t)$) and the interval $[-3, 3]$ is symmetric about $0$,$I_1 = 0$.Let $I_2 = \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1}$. We can write the denominator as $(t + \cos \alpha)^2 + \sin^2 \alpha$.Thus,$I_2 = \int_{0}^{1} \frac{dt}{(t + \cos \alpha)^2 + \sin^2 \alpha} = \left[ \frac{1}{\sin \alpha} 	an^{-1} \left( \frac{t + \cos \alpha}{\sin \alpha} ight) ight]_0^1$.$I_2 = \frac{1}{\sin \alpha} \left( 	an^{-1} \left( \frac{1 + \cos \alpha}{\sin \alpha} ight) - 	an^{-1} \left( \frac{\cos \alpha}{\sin \alpha} ight) ight)$.Using half-angle identities,$\frac{1 + \cos \alpha}{\sin \alpha} = \frac{2 \cos^2(\alpha/2)}{2 \sin(\alpha/2) \cos(\alpha/2)} = \cot(\alpha/2) = 	an(\pi/2 - \alpha/2)$.Also,$\frac{\cos \alpha}{\sin \alpha} = \cot \alpha = 	an(\pi/2 - \alpha)$.So,$I_2 = \frac{1}{\sin \alpha} (\pi/2 - \alpha/2 - (\pi/2 - \alpha)) = \frac{1}{\sin \alpha} (\alpha/2) = \frac{\alpha}{2 \sin \alpha}$.The equation becomes $\left( \frac{\alpha}{2 \sin \alpha} ight) x^2 - 0 \cdot x - 2 = 0$.$x^2 = \frac{4 \sin \alpha}{\alpha} \Rightarrow x = \pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$.

If $x$ satisfies the equation $\left( \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1} \right) x^2 - \left( \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \right) x - 2 = 0$ for $0 < \alpha < \pi$,then the value of $x$ is

Similar Questions

$A$ polynomial function $f(x)$ satisfying the conditions $f(x) = [f'(x)]^2$ and $\int_{0}^{1} f(x) dx = \frac{19}{12}$ can be:

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be a function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_{0}^{1} f(x)g(x) \, dx$ is

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

If $I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ and $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$,then $I_1 : I_2$ is

The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = -1$,$f'(\log 2) = 31$ and $\int_0^{\log 4} [f(x) - Rx] \, dx = \frac{39}{2}$ are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module