Question

If $f\left(x\right)=a{x}^2+bx+c$ $a,b,cϵR$ and the equation $f\left(x\right)-x=0$ has imaginary roots $\alpha$ and $\beta$ and $\gamma$ and $\delta$ be the roots of $f\left(f\right(x\left)\right)-x=0,then\left|\begin{array}{ccc}2& \alpha & \delta \\ \beta & 0& \alpha \\ \gamma & \beta & 1\end{array}\right|$ is

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

Answer 1

The correct option is A purely real

f(x)-x > 0 or, f(x)-x < 0 $\forall xϵ$ R
f(f(x)) - f(x) > 0 or f(f(x)) - f(x) < 0
Adding, f(f(x)) - x > 0 or, f(f(x)) - x < 0
$\Rightarrow$ roots of f(f(x)) - x = 0 are imaginary.
Let $z=\left|\begin{array}{ccc}2& \alpha & \delta \\ \beta & 0& \alpha \\ \gamma & \beta & 1\end{array}\right|\overline{z}=\left|\begin{array}{ccc}2& \overline{\alpha }& \overline{\delta }\\ \overline{\beta }& 0& \overline{\alpha }\\ \overline{\gamma }& \overline{\beta }& 1\end{array}\right|=\left|\begin{array}{ccc}2& \beta & \gamma \\ \alpha & 0& \beta \\ \delta & \alpha & 1\end{array}\right|=z$