Q67 / 110 · VITEEE 2023 · Mathematics
If \(f^{\prime}(x)=f(x)+\int_0^1 f(x) d x\) and given \(f(0)=1\).
Then \(f(x)=\)
- A\(\frac{2 \mathrm{e}^{\mathrm{x}}}{2-\mathrm{e}}+\left(\frac{1+\mathrm{e}}{1-\mathrm{e}}\right)\)
- Correct answerB\(\frac{2 \mathrm{e}^{\mathrm{x}}}{3-\mathrm{e}}+\left(\frac{1-\mathrm{e}}{3-\mathrm{e}}\right)\)
- C\(\frac{2 \mathrm{e}^{\mathrm{x}}}{2-\mathrm{e}}\)
- D\(\frac{2 \mathrm{e}^{\mathrm{x}}}{3-\mathrm{e}}\)
Solution
Given \(f^{\prime}(x)=f(x)+\int_0^1 f(x) d x\) ...(1)
Differentiating we get,
\(f^{\prime \prime}(x)=f^{\prime}(x)+0 \Rightarrow \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=1\)
On integrating \(\int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} d x=\int d x\)
\(\Rightarrow \ln f^{\prime}(x)=x+c \Rightarrow f^{\prime}(x)=e^{x+c}=k e^x\)
where \(\mathrm{k}=\mathrm{e}^{\mathrm{c}}\), a constant
Again integrate
\(\int f^{\prime}(x) d x=\int k e^x+D \Rightarrow f(x)=k e^x+D\) ...(2)
Put \(x=0\) in (1), \(f(0)=\mathrm{ke}^{\circ}+\mathrm{D} \Rightarrow \mathrm{k}+\mathrm{D}=1\)
(given \(f(0)=1)\) ...(3)
Also from (1), \(f^{\prime}(x)=f(x)+\int_0^1 f(x) d x\)
\(\begin{aligned}
& \Rightarrow \mathrm{ke}^{\mathrm{x}}=\mathrm{ke}^{\mathrm{x}}+\mathrm{D}+\int_0^1\left(\mathrm{ke}^{\mathrm{x}}+\mathrm{D}\right) \mathrm{dx} \\
& \Rightarrow\left[\mathrm{ke}^{\mathrm{x}}+\mathrm{Dx}\right]_0^1+\mathrm{D}=0 \Rightarrow[\mathrm{ke}+\mathrm{D}-\mathrm{k}]+\mathrm{D}=0 \\
& \Rightarrow \mathrm{k}(\mathrm{e}-1)+2 \mathrm{D}=0
\end{aligned}\)
Solving (3) and (4), we get
\(\begin{aligned}
& \mathrm{k}=\frac{2}{3-\mathrm{e}} \text { and } \mathrm{D}=\frac{1-\mathrm{e}}{3-\mathrm{e}} \\
& \therefore \mathrm{f}(\mathrm{x})=\frac{2 \mathrm{e}^{\mathrm{x}}}{3-\mathrm{e}}+\frac{1-\mathrm{e}}{3-\mathrm{e}}
\end{aligned}\)
Hence, the correct answer is option: 'B'.