Let f be a differentiable function such that f′(x)=f(x)+∫20f(x)dx,f(0)=4−e23, then f(x) is
A
ex−(e2−13)
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B
ex−(e2−23)
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C
ex+e−[arg|z−1|]3
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D
2ex−e2−1[arg(−|z|)] (where [.] greatest integer & z is a complex number)
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Solution
The correct option is Aex−(e2−13) f′(x)=f(x)+k∫f′(x)f(x)+k=∫dxlog(f(x)+k)=x+cf(x)=k1ex−kf(0)=k1−k=4−e23 ....(1) k=∫20f(x)dx3k=k1(e2−1) ....(2)
Solving (1) and (2), we get f(x)=ex−(e2−13)