be continuous for some $a, b, c \in R$ and $f ^{\prime}(0)+ f ^{\prime}(2)= e ,$ then the value of of $a$ is
For continuity at $\mathrm{x}=1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
$\Rightarrow \quad a e+b e^{-1}=c \Rightarrow \quad b=c e-a e^{2}$
For continuity at $x=3$
$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$
$\Rightarrow 9 c=9 a+6 c$
$\Rightarrow \quad c=3 a$
$f^{\prime}(0)+f^{\prime}(2)=e$
$\left(a e^{x}-b e^{x}\right)_{x}=0+(2 c x)_{x}=2=e$
$\Rightarrow \quad a-b+4 c=e$
From $(1),(2) \;and\;(3)$
$a-3 a e+a e^{2}+12 a=e$
$\Rightarrow \mathrm{a}\left(\mathrm{e}^{2}+13-3 \mathrm{e}\right)=\mathrm{e}$
$\Rightarrow \mathrm{a}=\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$
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$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$
of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^2=$