$x \in R$ માટે,ધારો કે $\tan^{-1}(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. તો $f: R \rightarrow R$ વિધેય,જે $f(x) = \int_0^{x \tan^{-1} x} \frac{e^{(t-\cos x)}}{1+t^{2023}} dt$ દ્વારા વ્યાખ્યાયિત છે,તેની ન્યૂનતમ કિંમત શોધો.

  • A
    $1$
  • B
    $0$
  • C
    $8$
  • D
    $5$

Explore More

Similar Questions

$\int_0^\pi x \sin^4 x \cos^6 x \, dx =$

જો $F(x) = \int_{x^2}^{x^3} \log t \, dt$ $(x > 0)$ હોય,તો $F'(x) = $

સંકલન $\int_0^{\pi / 2} \sin^5 x \, dx$ નું મૂલ્ય છે

$\int_0^{\pi / 2} \sin^8 x \, dx =$

જો $\varphi (x) = \int_{1/x}^{\sqrt{x}} \sin(t^2) \, dt$ હોય,તો $\varphi'(1) = $

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo