MCQ
The minimum value of the function $f(x)=\int \limits_0^2 e^{|x-t|} d t$ is
- ✓$2(e-1)$
- B$2 e -1$
- C$2$
- D$e(e-1)$
$f(x)=\int \limits_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)$
For $0 < x < 2$
$f(x)=\int \limits_0^x e^{x-t} d t+\int \limits_x^2 e^{t-x} d t=e^x+e^{2-x}-2$
For $x \geq 2$
$f(x)=\int \limits_0^2 e^{x-t} d t=e^{x-2}\left(e^2-1\right)$
For $x \leq 0, f ( x )$ is $\downarrow$ and $x \geq 2, f ( x )$ is $\uparrow$
$\therefore$ Minimum value of $f ( x )$ lies in $x \in(0,2)$
Applying A.M $\geq$ $G.M$
minimum value of $f(x)$ is $2(e-1)$
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Let $x _1< x _2< x _3<\ldots< x _{ n }<\ldots$ be all the points of local maximum of $f$ and $y_1$
$(1)$ $\left|x_n-y_n\right|>1$ for every $n$
$(2)$ $x_1 < y _1$
$(3)$ $x_n \in\left(2 n , 2 n +\frac{1}{2}\right)$ for every $n$
$(4)$ $x_{n+1}-x_n>2$ for every $n$