<s>[INST] Function defined by its own integral at a given point. If $y=f(x)$ is a differentiable function such that 
$$f(x)=x+ \int_0^1x^2zf(z)dz + \int_0^1xz^2f(z)dz$$ Then what is the value of $f\left(\frac{-9}{4}\right)$
The options were 
(A)$\frac{-4}{9}$  (B)$\frac{4}{9}$
(C)$-1$ (D)$0$
First I tried differentiating with respect to $x$. I believe we can take out the x terms from definite integral. I could not deduce anything meaningful from it.
Hence I assumed $x=z$ , but solving required the values of $f(0)$ and $f(1)$, which is not given in the question. 
How can this question be solved?
NOTE:$z$ is NOT a complex variable.
 [/INST] Let $\displaystyle a=\int_0^1zf(z)dz$ and $\displaystyle b=\int_0^1z^2f(z)dz$. Then
$$f(x)=ax^2+(b+1)x$$
We have
\begin{align*}
\int_0^1zf(z)dz&=\int_0^1[az^3+(b+1)z^2]dz\\
a&=\left[\frac{az^4}{4}+\frac{(b+1)z^3}{3}\right]_0^1\\
a&=\frac{a}{4}+\frac{b+1}{3}\\
\frac{3a}{4}&=\frac{b+1}{3}\\
b+1&=\frac{9a}{4}
\end{align*}
So
$$f(x)=ax^2+\frac{9a}{4}x$$
$$f\left(\frac{-9}{4}\right)=a\left(\frac{-9}{4}\right)^2+\frac{9a}{4}\left(\frac{-9}{4}\right)=0$$
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