JEE Main 202126 Feb, Evening shiftMathematicsMediumIntegral Calculus369× since 2002Let f(x)=∫0xetf(t)dt+exf(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}}f(x)=0∫xetf(t)dt+ex be a differentiable function for all x∈\in∈R. Then f(x) equals :AOption A. e(ex−1){e^{({e^{x - 1}})}}e(ex−1)BOption B. 2eex−12{e^{{e^x}}} - 12eex−1COption C. 2eex−1−12{e^{{e^x} - 1}} - 12eex−1−1DOption D. eex−1{e^{{e^x}}} - 1eex−1CheckSkip · show solution
