QJEEM23Mad3d6: JEE Main 2023 Mathematics - Class 12 JEEM23Mad3d6 Solution: The minimum value of the function is :

Question

Verified Solution • 2025–26 Session moderate

The minimum value of the function $f(x) = \int\limits_0^2 {{e^{|x - t|}}dt}$ is :

Accepted Answer

f(x)=\int_0^2 e^{|x-t|} d t

For

x&gt;2

f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right)

For

x&lt;0

f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)

For

x \in[0,2]

\begin{aligned}&amp; f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\&amp; =e^{2-x}+e^x-2\end{aligned}

For

x&gt;2

\left.f(x)\right|_{\min =e^2-1}

For

\mathrm{x}&lt;0

\left.f(x)\right|_{\min =e^2-1}

For

x \in[0,2]

\left.f(x)\right|_{\min }=2(e-1)

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