$(A)$ $f^{\prime \prime}(x)$ vanishes at least twice on $[0,1]$
$(B)$ $f^{\prime}\left(\frac{1}{2}\right)=0$
$(C)$ $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
$(D)$ $\int_0^{1 / 2} f(t) e^{\sin \pi t} d t=\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t$
$ \text { Put } x=1 / 2+x $
$ f\left(\frac{1}{2}+x\right)=f\left(\frac{1}{2}-x\right)$
Hence $f(x+1 / 2)$ is an even function or $f(x+1 / 2) \sin x$ is an odd function.
Also, $f^{\prime}(x)=-f^{\prime}(1-x)$ and for $x=1 / 2$, we have $f^{\prime}(1 / 2)=0$.
Also, $\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t=-\int_{1 / 2}^0 f(y) e^{\sin \pi y} d y$ (obtained by putting, $1-t=y$ ).
Since $f^{\prime}(1 / 4)=0, f^{\prime}(3 / 4)=0$. Also $f^{\prime}(1 / 2)=0$
$\Rightarrow f^{\prime \prime}(x)=0$ atleast twice in $[0,1]$ (Rolle's Theorem)
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Then the number of possible functions $g : A \rightarrow A$ such that $gof=f$ is ...... .