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Q.
Let $f(x)=\begin{vmatrix}a & -1 & 0 \\ a x & a & -1 \\ a x^{2} & a x & a\end{vmatrix}, a \in R$. Then the sum of which the squares of all the values of a for $2 f^{\prime}(10)-f^{\prime}(5)+100=0$ is :
$f(x)=\begin{vmatrix}a & -1 & 0 \\ a x & a & -1 \\ a^{2} & a x & a\end{vmatrix}$
$f(x)=a \begin{vmatrix}1 & -1 & 0 \\ x & a & -1 \\ x^{2} & a x & a\end{vmatrix}$
$=a\left[1\left(a^{2}+a x\right)+1\left(a x+x^{2}\right)\right]$
$\Rightarrow f(x)=a(x+a)^{2}$
so, $f^{\prime}(x)=2 a(x+a)$
as, $2 f^{\prime}(10)-f^{\prime}(5)+100=0$
$\Rightarrow 2 \times 2 a(10+a)-2 a(5+a)+100=0$
$\Rightarrow 40 a+4 a^{2}-10 a-2 a^{2}+100=0$
$2 a ^{2}+30 a +100=0$
$\Rightarrow a ^{2}+15 a +50=0$
$( a +10)( a +5)=0$
$a =-10$ or $a =-5$
Required $=(-10)^{2}+(-5)^{2}=125$