Let g:(0, infty) rightarrow R be a differenti | vedclass

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

Question

(D) Given the integral equation: $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)

Vedclass · Accepted Answer

$g - g'$ is increasing in $(0, \pi/2)$

Vedclass · Answer

(D) Given the integral equation: $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus: $\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} = \frac{d}{dx} \left( \frac{x g(x)}{e^x + 1} \right)$. Applying the quotient rule on the right side: $\frac{d}{dx} \left( \frac{x g(x)}{e^x + 1} \right) = \frac{(e^x + 1)(g(x) + x g'(x)) - x g(x) e^x}{(e^x + 1)^2} = \frac{(e^x + 1)g(x) + x g'(x)(e^x + 1) - x g(x) e^x}{(e^x + 1)^2} = \frac{g(x)(e^x + 1 - x e^x) + x g'(x)(e^x + 1)}{(e^x + 1)^2}$. Equating the terms: $\frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} = \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} + \frac{x g'(x)(e^x + 1)}{(e^x + 1)^2}$. Subtracting the common term $\frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2}$ from both sides: $\frac{x(\cos x - \sin x)}{e^x + 1} = \frac{x g'(x)}{e^x + 1}$. Since $x > 0$,we have $g'(x) = \cos x - \sin x$. Now,let's analyze the options: $1$. $g'(x) = \cos x - \sin x$. For $x \in (0, \pi/4)$,$\cos x > \sin x$,so $g'(x) > 0$,meaning $g$ is increasing. $2$. $g''(x) = -\sin x - \cos x$. For $x \in (0, \pi/4)$,$g''(x) $3$. Let $h(x) = g(x) + g'(x)$. Then $h'(x) = g'(x) + g''(x) = (\cos x - \sin x) + (-\sin x - \cos x) = -2 \sin x$. Since $-2 \sin x $4$. Let $\phi(x) = g(x) - g'(x)$. Then $\phi'(x) = g'(x) - g''(x) = (\cos x - \sin x) - (-\sin x - \cos x) = 2 \cos x$. Since $2 \cos x > 0$ for $x \in (0, \pi/2)$,$\phi$ is increasing. Thus,option $D$ is correct.

Let $g:(0, \infty) \rightarrow R$ be a differentiable function such that $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$ for all $x > 0$,where $c$ is an arbitrary constant. Then:

Similar Questions

$\int (1+\tan^2 x)(1+2x \tan x) dx =$

$\int \sqrt{1 + \csc x} \, dx$ equals

If $\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} \tan^{-1}(3 \tan x) + C$,then the maximum value of $a \sin x + b \cos x$ is:

Let $5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3$,where $x > 0$. Then $18 \int \limits_1^2 f(x) \, dx$ is equal to:

If $\int \frac{\cos (13 x)-\cos (14 x)}{1+2 \cos (9 x)} d x=\frac{\sin (4 x)}{a}-\frac{\sin (5 x)}{b}+c$,then $a^b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module