Let g(x) = int_{x}^{2x} frac{f(t)}{t} dt wher | vedclass

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

Question

(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.Using the Leibniz integral rule,we differentiate $g(x)$ with respect to $x$:$g'(x) = \frac{f(2x)}{2

Vedclass · Accepted Answer

$g(x)$ is a constant function

Vedclass · Answer

(C) Given $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$.Using the Leibniz integral rule,we differentiate $g(x)$ with respect to $x$:$g'(x) = \frac{f(2x)}{2x} \cdot \frac{d}{dx}(2x) - \frac{f(x)}{x} \cdot \frac{d}{dx}(x)$$g'(x) = \frac{f(2x)}{2x} \cdot 2 - \frac{f(x)}{x} \cdot 1$$g'(x) = \frac{f(2x)}{x} - \frac{f(x)}{x}$Since it is given that $f(2x) = f(x)$,we substitute this into the expression:$g'(x) = \frac{f(x) - f(x)}{x} = 0$Since $g'(x) = 0$ for all $x > 0$,the function $g(x)$ is a constant function.

Let $g(x) = \int_{x}^{2x} \frac{f(t)}{t} dt$ where $x > 0$ and $f$ is a continuous function such that $f(2x) = f(x)$. Then:

Similar Questions

The value of $\int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$ is

$\int_{-5}^5 x^4\left(25-x^2\right)^{5 / 2} d x=$

If $x \cdot \sin(\pi x) = \int_{0}^{x^2} f(t) \, dt$ where $f$ is a continuous function,then the value of $f(4)$ is:

$\mathop {Limit}\limits_{x \to {x_1}} \,\,\frac{x}{{x - {x_1}}}\,\,\int\limits_{{x_1}}^x {f(t)} \, dt$ is equal to :

Let $f$ be a differentiable function in $\left(0, \frac{\pi}{2}\right)$. If $\int\limits_{\cos x}^{1} t^{2} f(t) d t = \sin^{3} x + \cos x - 1$,then $\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module