Question

Let $f:R\to R$ be a continuous function satisfying
$f\left(x\right)+\underset{0}{\overset{x}{\int }}tf\left(t\right)dt+x^2=0,$ for all $x\in R.$ Then

• A. $\underset{x\to \infty }{lim}f\left(x\right)=2$
• B. $\underset{x\to \infty }{lim}f\left(x\right)=-2$
• C. $f\left(x\right)$ has more than one point in common with the $x-$axis
• D. $f\left(x\right)$ is an odd function

Answer 1

The correct option is B $\underset{x\to \infty }{lim}f\left(x\right)=-2$

$f\left(x\right)+\underset{0}{\overset{x}{\int }}tf\left(t\right)dt+x^2=0$
$f^{\prime }\left(x\right)+xf\left(x\right)+2x=0\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)+2}=-x$
Integrating both side
$\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)+2}dx=-\int xdx$
$\ln \left(f\right(x)+2)=-\frac{x^2}{2}+c_1$
$f\left(x\right)=k{e}^{-\frac{x^2}{2}}-2$
At $x=0f\left(x\right)=0\Rightarrow k=2$
$f\left(x\right)=2(e^{-\frac{x^2}{2}}-1)$
Hence,$\underset{x\to \infty }{lim}f\left(x\right)=-2$