Question

If $f\left(x\right)={\int }_0^4{e}^{|t-x|}dt(0⩽x⩽4),$ the maximum value of $f\left(x\right)$ is

• A. $e^4-1$
• B. $2(e^2-1)$
• C. $e^2-1$
• D. None of these

Answer 1

The correct option is A $e^4-1$

$\begin{array}{c}f\left(x\right)={\int }_0^4{e}^{\left(t-x\right)}dt\left(0\le x\le 4\right)\\ {\int }_0^x{e}^{\left(x-t\right)}dt+{\int }_x^t{e}^{\left(t-x\right)}dt=e^x{\int }_0^x{e}^{\left(-t\right)}dt+e^{-x}{\int }_0^x{e}^{\left(t\right)}dt\\ e^x{\left[e^{-t}\right]}_x^0+e^{-x}\left[e^4-e^x\right]\\ =\left(e^x-1\right)+e^{-x}\left[e^4-e^x\right]\\ =\left(e^x-1\right)+e^{-x}{e}^4-1\\ f\left(x\right)=e^x+\frac{e^4}{e^x}-2\\ f^{\prime }\left(x\right)=e^x-e^4{e}^{-x}=0\\ e^x=e^{4-x}\\ x=2\\ signof{f}^{\prime }\left(x\right):(-)(+)\\ Clearlyf{\left(x\right)}_{max}occursatx=4\\ f{\left(x\right)}_{max}=e^4+\frac{e^4}{e^4}-2\\ =\left(e^4-1\right)\\ OptionAiscorrectanswer.\end{array}$