Question

If $a^2+b^2+c^2=-2$ and $f(x)=\begin{vmatrix}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{vmatrix}$ then $f(x)$ is a polynomial of degree

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$C _1 \rightarrow C _1+ C _2+ C _3$
$f(x)=\begin{vmatrix}1+2 x+x\left(a^2+b^2+c^2\right) & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1+2 x+x\left(a^2+b^2+c^2\right) & 1+b^2 x & \left(1+c^2\right) x \\ 1+2 x+x\left(a^2+b^2+c^2\right) & \left(1+b^2\right) x & 1+c^2 x\end{vmatrix} \quad\left(\right.$ as $\left.a^2+b^2+c^2=-2\right)$
$\begin{vmatrix}1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1 & 1+b^2 x & \left(1+c^2\right) x \\ 1 & \left(1+b^2\right) x & 1+c^2 x\end{vmatrix}| $
$ R_2 \rightarrow R_2-R_1 \& R_3 \rightarrow R_3-R_1$
$ \begin{vmatrix}1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 0 & 1-x & 0 \\ 0 & 1-x & 1-x\end{vmatrix} \\ f(x)=(1-x)^2=1-2 x+x^2 \Rightarrow(C)$