Question

If $f\left(x\right)=\left|\begin{array}{ccc}1& -1& 0\\ ax& a& -1\\ a{x}^2& ax& a\end{array}\right|$ then $f\left(2\right)-f\left(1\right)=$ $a(k+a),$ then the value of $k$ is

Answer 1

Given $f\left(x\right)=\left|\begin{array}{ccc}1& -1& 0\\ ax& a& -1\\ a{x}^2& ax& a\end{array}\right|$

$\Rightarrow f\left(x\right)=a^2+ax+a^{2}x+a{x}^2$

So, $f\left(2x\right)=a^2+2ax+2a^{2}x+4a{x}^2$
Now, $f\left(2x\right)-f\left(x\right)=a^2+2ax+2a^{2}x+4a{x}^2-(a^2+ax+a^{2}x+a{x}^2)$

$\Rightarrow f\left(2x\right)-f\left(x\right)=ax(1+a+3x)$

$\Rightarrow f\left(2\right)-f\left(1\right)=a(4+a)$

i.e $x=1$

$\therefore k=4$