Let $f : R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0 \quad \forall x$. Then:
(a) $f(x)$ has more than one point in common with $x$-axis
(b) $f(x)$ is odd function
(c) $\lim_{x \to \infty} f(x) = 2$
(d) $\lim_{x \to -\infty} f(x) = -2$
Step-by-Step Solution
Key Concept: Differentiate the given integral equation:
\[
f(x)+\int_0^x t f(t)\,dt+x^2=0.
\]
This gives
\[
f'(x)+x f(x)+2x=0,
\]
a first-order linear differential equation. The value at \(x=0\) comes directly from the original equation:
\[
f(0)=0.
\]
We are given
\[
f(x)+\int_0^x t f(t)\,dt+x^2=0.
\]
Putting \(x=0\),
\[
f(0)=0.
\]
Now differentiate both sides with respect to \(x\):
\[
f'(x)+x f(x)+2x=0.
\]
Thus
\[
f'(x)+x f(x)=-2x.
\]
This is a linear differential equation. Its integrating factor is
\[
e^{\int x\,dx}=e^{x^2/2}.
\]
So
\[
\frac{d}{dx}\left(f(x)e^{x^2/2}\right)
=-2x e^{x^2/2}.
\]
Integrating,
\[
f(x)e^{x^2/2}=-2e^{x^2/2}+C.
\]
Hence
\[
f(x)=-2+Ce^{-x^2/2}.
\]
Using \(f(0)=0\),
\[
0=-2+C,
\]
so
\[
C=2.
\]
Therefore
\[
f(x)=2e^{-x^2/2}-2.
\]
Now
\[
\lim_{x\to-\infty}f(x)
=2\cdot 0-2=-2.
\]
\[
\boxed{\lim_{x\to-\infty}f(x)=-2}
\]
Correct Answer: 4