We have,
$f(x)+\int_0^x t f(t) d t+x^2=0$
On differentiating, we get
$f^{\prime}(x)+x f(x)+2 x =0$
$\Rightarrow f^{\prime}(x) =-x(f(x)+2)$
$\Rightarrow \frac{f^{\prime}(x)}{f(x)+2} =-x$
On integrating, we get
$\quad \log (f(x)+2)=-\frac{x^2}{2}+C$
$\Rightarrow \quad f(x)=A e^{-x^2 / 2}-2$
$\Rightarrow \quad f(0)=0=A-2 \Rightarrow A=2$
$\therefore \quad \lim _{x \rightarrow \infty} f(x)=-2$
$\qquad \quad \lim ^{-x^2 / 2}-2$
$\Rightarrow(x) \text { is an even function. }$
$f(x) \text { intersect } X \text {-axis at one point }(0,0)$
$\therefore \text { Option (b) is correct. }$
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$f(x)\left\{ \begin{gathered} = 1\,,\,{\text{if}}\,\,\,x > 0 \hfill \\ = - 1\,,\,{\text{if}}\,\,\,x < 0 \hfill \\ = 0\,,\,{\text{if}}\,\,\,x = 0 \hfill \\ \end{gathered} \right.$ then ${\left. {\frac{{dy}}{{dx}}} \right|_{x = \frac{{5\pi }}{4}}}$ is