JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 2)
The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :
2
$$2(e-1)$$
$$e(e-1)$$
$$2e-1$$
Explanation
$$
f(x)=\int_0^2 e^{|x-t|} d t
$$
For $x>2$
$$ f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right) $$
For $x<0$
$$ f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right) $$
For $x \in[0,2]$
$$ \begin{aligned} & f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\ & =e^{2-x}+e^x-2 \end{aligned} $$
For $x>2$
$$ \left.f(x)\right|_{\min =e^2-1} $$
For $\mathrm{x}<0$
$$ \left.f(x)\right|_{\min =e^2-1} $$
For $x \in[0,2]$
$$ \left.f(x)\right|_{\min }=2(e-1) $$
For $x>2$
$$ f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right) $$
For $x<0$
$$ f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right) $$
For $x \in[0,2]$
$$ \begin{aligned} & f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\ & =e^{2-x}+e^x-2 \end{aligned} $$
For $x>2$
$$ \left.f(x)\right|_{\min =e^2-1} $$
For $\mathrm{x}<0$
$$ \left.f(x)\right|_{\min =e^2-1} $$
For $x \in[0,2]$
$$ \left.f(x)\right|_{\min }=2(e-1) $$
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