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QJEEM23Mae559: JEE Main 2023 Mathematics - Class 12 JEEM23Mae559 Solution: Let be a differentiable function such that f ′ ( x ) + f ( x )…

Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function such that f′(x)+f(x)=∫02f(t)dt. If f(0)=e−2f(0)=e^{-2} , then 2f(0)−f(2)2 f(0)-f(2) is equal to ____________.

Step-by-Step Verified Solution

f′(x)+f(x)=kf^{\prime}(x)+f(x)=k

amp;⇒exf(x)=kex+camp;f(x)=k+ce−xamp;k=∫02(k+ce−t)dtamp;k=2k+c⋅e−t−1∣02amp;k=2k+c(e−2−1+1)amp;−k=c(1−1e2)amp;f(x)=ce−x−c(1−1e2)amp;f(0)=c−c+ce2=1e2⇒c=1amp;f(2)=e−2−r(1−e−2)amp;=2e−2−1amp;2f(0)−f(2)=1\begin{aligned}& \Rightarrow e^{x} f(x)=k e^{x}+c \\\\& f(x)=k+c e^{-x} \\\\& k=\int_{0}^{2}\left(k+c e^{-t}\right) d t \\\\& k=2 k+\left.c \cdot \frac{e^{-t}}{-1}\right|_{0} ^{2} \\\\& k=2 k+c\left(\frac{e^{-2}}{-1}+1\right) \\\\& -k=c\left(1-\frac{1}{e^{2}}\right) \\\\& f(x)=c e^{-x}-c\left(1-\frac{1}{e^{2}}\right) \\\\& f(0)=c-c+\frac{c}{e^{2}}=\frac{1}{e^{2}} \Rightarrow c=1 \\\\& f(2)=e^{-2}-r\left(1-e^{-2}\right) \\\\& =2 e^{-2}-1 \\\\& 2f(0)-f(2)=1\end{aligned}

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