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JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1
f(x)+f′(x)=∫02f(t)dtf(x)+f^{\prime}(x)=\int\limits_{0}^2 f(t) d t and f(0)=e−2f(0)=e^{-2}, then the value of f(2)−2f(0)f(2)-2 f(0) is:
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Correct option: B
−1-1
Option B (-1) is correct because after solving the differential equation and using the given condition, the value of f(2)-2f(0) is -1.

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