Question

If $f\left(x\right)=a{x}^2+bx+c,a,b,c\in R$ and equation $f\left(x\right)-x=0$ has non-real roots $\alpha ,\beta$. Let $\gamma ,\delta$ be the roots of $f\left(f\right(x\left)\right)-x=0$ ($\gamma ,\delta$ are not equal to $\alpha ,\beta$). Then $\left|\begin{array}{ccc}2& \alpha & \delta \\ \beta & 0& \alpha \\ \gamma & \beta & 1\end{array}\right|$ is

• A. 0
• B. purely real
• C. purely imaginary
• D. none of these

Answer 1

The correct option is B purely real

$f\left(x\right)=a{x}^2+bx+c;a,b,c\in R$ and equation $f\left(x\right)-x=0$ has imaginary roots $\alpha ,\beta$ and $\gamma ,\delta$ be the roots of $f\left(f\right(x\left)\right)-x=0$
since $\alpha ,\beta$ are roots of $f\left(x\right)-x=0$
$f\left(\alpha \right)-\alpha =0$ $\Rightarrow f\left(\alpha \right)=\alpha$
$f\left(\beta \right)-\beta =0$ $\Rightarrow f\left(\beta \right)=\beta$
$f\left(f\right(\alpha \left)\right)-\alpha =f\left(\alpha \right)-\alpha =0$
$f\left(f\right(\beta \left)\right)-\beta =f\left(\beta \right)-\beta =0$
$\therefore \alpha ,\beta$ are also roots of $f\left(f\right(x\left)\right)-x=0$ ------(*)
$f\left(x\right)-x=a{x}^2+(b-1)x+c=0$
roots are imaginary.
i.e $\alpha ,\beta$ are conjugate to each other and $D<0$
$\Rightarrow (b-1{)}^2-4ac<0$ -------(1)
$f\left(f\right(x\left)\right)-x=a(a{x}^2+bx+c{)}^2+b(a{x}^2+bx+c)+c-x=0$
$\Rightarrow (a{x}^2+(b-1)x+c)(a^2{x}^2+(ab+a)x+ac+b+1)=0$
$D=(ab+a{)}^2-4a^2(ac+b+1)=a^2\left(\right(b-1{)}^2-4ac)-4a^2<0$ ($\because$ from (1))
$\therefore \gamma ,\delta$ are also imaginary roots and conjugate to each other.
$\left|\begin{array}{ccc}2& \alpha & \delta \\ \beta & 0& \alpha \\ \gamma & \beta & 1\end{array}\right|$ $=-3\alpha \beta +{\alpha }^2\gamma +{\beta }^2\delta$ ------ (2)
$\alpha \beta$ is real
${\alpha }^2\gamma$ is conjugate to ${\beta }^2\delta$
$\Rightarrow {\alpha }^2\gamma +{\beta }^2\delta$ is real
from (2)
$\therefore \left|\begin{array}{ccc}2& \alpha & \delta \\ \beta & 0& \alpha \\ \gamma & \beta & 1\end{array}\right|$ is purely real.
Hence, option B.