Question

The number of real roots of the equation $\begin{vmatrix} x^{2}+1 & 2x^{3} & x \\ 2x^{3} & x & x^{2}+1 \\ x & x^{2}+1 & 2x^{3} \end{vmatrix}=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

Cyclic determinant $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}=0\Rightarrow a+b+c=0ora=b=c$
$\Rightarrow x^{2}+1+2x^{3}+x=0$ or $x^{2}+1=2x^{3}=x$ (reject)
$\Rightarrow 2x^{3}+x^{2}+x+1=0$
$f\left(x\right)=2x^{3}+x^{2}+x+1$
$f^{'}\left(x\right)=6x^{2}+2x+1>0$ , $\forall x\in ℝ$ , as $D < 0$ so $f\left(x\right)$ is strictly increasing.
$\therefore f\left(x\right)=0$ has exactly one real root as $f\left(x\right)$ is odd degree polynomial.