The minimum value of the function f ( x ) = 0 2 e | x - t | ? d t is [2023]

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Solution

Q.
$The minimum value of the function f (x) = \int_{0}^{2} e^{| x - t |} d t is$ [2023]

1 2

2 2(e - 1)

3 2e - 1

4 e(e - 1)

Ans.
(2)

Case 1: When $x < 0$

$f (x) = \int_{0}^{2} e^{t - x} d t = e^{- x} {(e^{t})}_{0}^{2} = e^{- x} (e^{2} - 1)$

Case 2: When $0 < x < 2$

$f (x) = \int_{0}^{x} e^{x - t} d t + \int_{x}^{2} e^{t - x} d t \Rightarrow f (x) = e^{x} {(- e^{- t})}_{0}^{x} + e^{- x} {(e^{t})}_{x}^{2}$

$= - e^{x} (e^{- x} - 1) + e^{- x} (e^{2} - e^{x}) = - 1 + e^{x} + e^{2} e^{- x} - 1 = e^{x} + e^{- x} e^{2} - 2$

Case 3: When $x \geq 2$

$f (x) = \int_{0}^{2} e^{(x - t)} d t = e^{x} {(- e^{- t})}_{0}^{2} = - e^{x} (e^{- 2} - 1)$

Hence,

$f (x) =$ ${\begin{matrix} e^{- x} (e^{2} - 1), & x < 0 \\ e^{x} + e^{- x} e^{2} - 2, & 0 \leq x < 2 \\ (1 - e^{- 2}) e^{x}, & x \geq 2 \end{matrix}$

$f (x)$ is decreasing for $x < 0$ and increasing for $x \geq 2$ .

$f (x)$ is also continuous.

For $0 \leq x \leq 2$ , $f (x)$ is minimum at $x = 1$ .

(Using A.M.–G.M. inequality)

$Hence, the minimum value f (x) = f (1) = 2 (e - 1)$

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