Question

Let $f$ be a differentiable function satisfying $f^{\prime }\left(x\right)=f\left(x\right)+\underset{0}{\overset{2}{\int }}f\left(x\right)dx$. If $f\left(0\right)=\frac{4-e^2}{3}$, then

• A. $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{3e-e^2-2}{3}$
• B. $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{3e+e^2-2}{3}$
• C. Number of solutions of $f\left(x\right)+x=0$ is $1$
• D. Number of solutions of $f\left(x\right)+x=0$ is $2$

Answer 1

Answer: (A), (C)

The correct options are
A $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\frac{3e-e^2-2}{3}$
C Number of solutions of $f\left(x\right)+x=0$ is $1$

$f^{\prime }\left(x\right)=f\left(x\right)+\underset{0}{\overset{2}{\int }}f\left(x\right)dx$
Let $\underset{0}{\overset{2}{\int }}f\left(x\right)dx=k$
Then, $f^{\prime }\left(x\right)-f\left(x\right)=k$
which is a linear differential equation.
General solution is
$e^{-x}f\left(x\right)=k\int e^{-x}dx$
$\Rightarrow f\left(x\right)=c{e}^x-k$
$\underset{0}{\overset{2}{\int }}f\left(x\right)dx=k\Rightarrow k=\frac{c(e^2-1)}{3}$
So, $f\left(x\right)=c{e}^x-\frac{c(e^2-1)}{3}$

Given, $f\left(0\right)=\frac{4-e^2}{3}$
$\Rightarrow c=1$
$\therefore f\left(x\right)=e^x-\frac{e^2-1}{3}$

Roots of $f\left(x\right)+x=0$
$\Rightarrow f\left(x\right)=-x$
We plot the graphs of $y=e^x-k$, where $\frac{e^2-1}{3}$ and $y=-x$



Clearly, $f\left(x\right)+x=0$ has only one root.

Now, $\underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}(e^x-\frac{e^2-1}{3})dx$
$={[e^x-\frac{(e^2-1)x}{3}]}_0^1=\frac{3e-e^2-2}{3}$