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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
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
    $e(e-1)$
✓

Answer

Correct option: A.
$2(e-1)$
a
For $x \leq 0$

$f(x)=\int \limits_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)$

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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