Q98 / 225 · JEE Main 2005 · Mathematics

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

  • A
    1
  • B
    0
  • C
    3
  • Correct answer
    D
    2

Solution

$$ \begin{aligned} & f(x)=\left|\begin{array}{ccc} 1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1+\left(a^2+b^2+c^2+2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ 1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & 1+c^2 x \end{array}\right| \text {, Applying } C_1 \rightarrow C_1+C_2+C_3 \\ & =\left|\begin{array}{ccc} 1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\ 1 & 1+b^2 x & \left(1+c^2\right) x \\ 1 & \left(1+b^2\right) x & 1+c^2 x \end{array}\right| \because a^2+b^2+c^2+2=0 \\ & f(x)=\left|\begin{array}{ccc} 0 & x-1 & 0 \\ 0 & 1-x & x-1 \\ 1 & \left(1+b^2\right) x & 1+c x \end{array}\right| ; \text { Applying } R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & f(x)=(x-1)^2 \end{aligned} $$ Hence degree $=2$.

Hence, the correct answer is option: 'D'.

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