JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1
Let $f(x)=\begin{vmatrix}a&-1&0\\ax&a&-1\\ax^{2}&ax&a\end{vmatrix}$, $a\in\mathbf{R}$. Then the sum of the squares of all the values of $a$, for which $2f'(10)-f'(5)+100=0$, is
Determinant
Solution
$f(x)=a^3+2a^2x+ax^2$ (expand the determinant), so $f'(x)=2a^2+2ax$. $2f'(10)-f'(5)+100=2(2a^2+20a)-(2a^2+10a)+100=2a^2+30a+100=2(a+5)(a+10)=0$. $a=-5,-10$; sum of squares $=25+100=125$. Option (C).