If f'(x)=f(x)+₀¹ f(x)\,dx, f(0)=1, then f(x)=

Correct Answer is : 2ex3−e+1−e3−e{2 e^{x}}{3-e}+1-e/3-e3−e2ex+3−e1−e

Correct Answer is : 2ex3−e+1−e3−e{2 e^{x}}{3-e}+1-e/3-e3−e2ex+3−e1−e

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BITSAT 2019 Slot 2 Solved Paper

Section: Mathematics

Question : 147 of 150

Marks: +1, -0

f'(x)=f(x)+\int\limits_{0}^{1} f(x)\,dx, f(0)=1

(x)=

Solution:

Given,

f'(x)=f(x)+\int\limits_{0}^{1} f(x)\,dx, f(0)=1

.......(i)

\Rightarrow f''(x)=f'(x)+0

\Rightarrow \frac{f''(x)}{f'(x)}=1

\Rightarrow \int \frac{f''(x)}{f'(x)} dx = \int dx

\Rightarrow \log f'(x)=x+C

\Rightarrow f'(x)=A e^{x}

\Rightarrow f(x)=A e^{x}+K

f(0)=A+K=1

.........(ii)

\therefore A e^{x}=A e^{x}+K+\int\limits_{0}^{1}(A e^{x}+k)\,dx

\Rightarrow k+\left[A e^{x}+K x\right]_{0}^{1}=0

\Rightarrow k+A e-A+K=0

\Rightarrow A(e-1)+2k=0

........(iii)

From Eqs. (ii) and (iii), we get

A=\frac{2}{3-e}, k=\frac{1-e}{3-e}

\therefore f(x)=\frac{2 e^{x}}{3-e}+\frac{1-e}{3-e}

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