For x in R,let tan^{-1}(x) in left(-frac{pi}{ | vedclass

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

Question

(B) The function is given by $f(x) = \int_0^{x 	an^{-1} x} \frac{e^{t-\cos x}}{1+t^{2023}} dt$.Using Leibniz's rule to find the derivative $f'(x)$:$f

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) The function is given by $f(x) = \int_0^{x 	an^{-1} x} \frac{e^{t-\cos x}}{1+t^{2023}} dt$.Using Leibniz's rule to find the derivative $f'(x)$:$f'(x) = \frac{e^{x 	an^{-1} x - \cos x}}{1 + (x 	an^{-1} x)^{2023}} \cdot \frac{d}{dx}(x 	an^{-1} x) + \int_0^{x 	an^{-1} x} \frac{\partial}{\partial x} \left( \frac{e^{t-\cos x}}{1+t^{2023}} ight) dt$.However,observing the structure of the function,we note that for $x=0$,$f(0) = \int_0^0 \dots dt = 0$.For $x 
eq 0$,$x 	an^{-1} x > 0$ because $	an^{-1} x$ has the same sign as $x$.Since the integrand $\frac{e^{t-\cos x}}{1+t^{2023}}$ is positive for $t \geq 0$,and the upper limit $x 	an^{-1} x$ is always non-negative,the integral $f(x)$ is non-negative for all $x \in R$.Specifically,$f(x) \geq 0$ for all $x$,and $f(0) = 0$.Thus,the minimum value of the function is $0$.

For $x \in R$,let $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $f: R \rightarrow R$ defined by $f(x) = \int_0^{x \tan^{-1} x} \frac{e^{(t-\cos x)}}{1+t^{2023}} dt$ is

Similar Questions

Let the slope of the line $45 x+5 y+3=0$ be $27 r_1+\frac{9 r_2}{2}$ for some $r_1, r_2 \in R$. Then $\lim_{x \rightarrow 3} \left( \int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} dt \right)$ is equal to ...................

If $f(x) = \int_a^x {t^3 e^t \, dt}$,then $\frac{d}{dx} f(x) = $

The derivative of $F(x) = \int_{x^2}^{x^3} \frac{1}{\log t} \, dt$,$(x > 0)$ is

$\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to -

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{\int_0^{x^2} \sec^2 t \, dt}{x \sin x} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module