Let g(t) = int_{-pi/2}^{pi/2} cos left(frac{p | vedclass

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

Question

(D) Given $f(x) = \log_e \left(x + \sqrt{x^2 + 1}\right)$. Note that $f(-x) = \log_e \left(-x + \sqrt{x^2 + 1}\right) = \log_e \left(\frac{1}{\sqrt{x^

Vedclass · Accepted Answer

$\sqrt{2} g(1) = g(0)$

Vedclass · Answer

(D) Given $f(x) = \log_e \left(x + \sqrt{x^2 + 1}\right)$. Note that $f(-x) = \log_e \left(-x + \sqrt{x^2 + 1}\right) = \log_e \left(\frac{1}{\sqrt{x^2 + 1} + x}\right) = -f(x)$. Thus,$f(x)$ is an odd function.$g(t) = \int_{-\pi/2}^{\pi/2} \cos \left(\frac{\pi}{4} t + f(x)\right) \, dx = \int_{-\pi/2}^{\pi/2} \left[ \cos \left(\frac{\pi}{4} t\right) \cos(f(x)) - \sin \left(\frac{\pi}{4} t\right) \sin(f(x)) \right] \, dx$.Since $f(x)$ is odd,$\cos(f(x))$ is an even function and $\sin(f(x))$ is an odd function.Therefore,$\int_{-\pi/2}^{\pi/2} \sin \left(\frac{\pi}{4} t\right) \sin(f(x)) \, dx = 0$.So,$g(t) = \cos \left(\frac{\pi}{4} t\right) \int_{-\pi/2}^{\pi/2} \cos(f(x)) \, dx$.Let $C = \int_{-\pi/2}^{\pi/2} \cos(f(x)) \, dx$. Then $g(t) = C \cos \left(\frac{\pi}{4} t\right)$.$g(1) = C \cos \left(\frac{\pi}{4}\right) = \frac{C}{\sqrt{2}}$ and $g(0) = C \cos(0) = C$.Thus,$g(1) = \frac{g(0)}{\sqrt{2}}$,which implies $\sqrt{2} g(1) = g(0)$.

Let $g(t) = \int_{-\pi/2}^{\pi/2} \cos \left(\frac{\pi}{4} t + f(x)\right) \, dx$,where $f(x) = \log_e \left(x + \sqrt{x^2 + 1}\right)$,$x \in R$. Then which one of the following is correct?

Similar Questions

Assertion $(A)$: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} = \frac{\pi}{12}$
Reason $(R)$: $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$

$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

The value of the integral $\int_0^\infty \frac{\log_e(x)}{x^2+4} dx$ is:

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

The value of $\int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module