JEE Main 202325 Jan, Morning shiftMathematicsEasyIntegral Calculus369× since 2002The minimum value of the function f(x)=∫02e∣x−t∣dtf(x) = \int\limits_0^2 {{e^{|x - t|}}dt}f(x)=0∫2e∣x−t∣dt is :AOption A. 2BOption B. 2(e−1)2(e-1)2(e−1)COption C. e(e−1)e(e-1)e(e−1)DOption D. 2e−12e-12e−1CheckSkip · show solution
