If f : R is a differentiable function such that f ( x ) + f ( x ) = 0 2 f ( t ) ? d t . If f ( 0 ) = e - 2 , then 2 f ( 0 ) - f ( 2 ) is equal to _____ . [2023]

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Solution

Q.
If $f : R \to R$ is a differentiable function such that $f' (x) + f (x) = \int_{0}^{2} f (t) d t .$ If $f (0) = e^{- 2},$ then $2 f (0) - f (2)$ is equal to _____ . [2023]

Ans.
(1)

$f' (x) + f (x) = \int_{0}^{2} f (t) d t and f (0) = e^{- 2}$

$Let \int_{0}^{2} f (t) d t = K$

$\Rightarrow \frac{d y}{d x} + y = K [Where \frac{d y}{d x} = f' (x), y = f (x)]$

$\Rightarrow y e^{x} = K e^{x} + C;$ $Now f (0) = e^{- 2}$

$\Rightarrow e^{- 2} = K + C \Rightarrow C = e^{- 2} - K$ $\Rightarrow y e^{x} = K e^{x} + e^{- 2} - K$

$\Rightarrow y = K + (e^{- 2} - K) e^{- x}$

$Now, K = \int_{0}^{2} f (x) d x = \int_{0}^{2} (K + (e^{- 2} - K) e^{- x}) d x$

$= 2 K + {[(K - e^{- 2}) e^{- x}]}_{0}^{2}$ $= 2 K + (K - e^{- 2}) (e^{- 2} - 1)$

$\Rightarrow K = 2 K + K e^{- 2} - K - e^{- 4} + e^{- 2}$ $\Rightarrow K = \frac{e^{- 4} - e^{- 2}}{e^{- 2}}$

$\Rightarrow K = e^{- 2} - 1$

$Now, 2 f (0) - f (2) = 2 e^{- 2} - 2 e^{- 2} + 1 = 1$

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