Question

Let $f(x)$ be a quadratic polynomial with leading coefficient 1 such that $f (0)= p , p \neq 0$ and $f (1)=\frac{1}{3}$. If the equation $f ( x )=0$ and fofofof $(x)=0$ have a common real root, then $f(-3)$ is equal to. ...............

Answer 1

Let $f( x )= x ^2+ bx + p$
$f (1)=\frac{1}{3} \Rightarrow 1+ b + p =\frac{1}{3} .....$(1)
Assume common root be $\alpha$
$ f (\alpha)=0 \& f(f(f(f(\alpha))))=0 $
$ \Rightarrow f ( f ( f (0)))=0 $
$ \Rightarrow f ( f ( p ))=0 $
$\Rightarrow f \left( p ^2+ bp + p \right)=0 $
$ \Rightarrow f ( p ( p + b +1))=0 $
$ \Rightarrow f \left(\frac{ p }{3}\right)=0$
$ \Rightarrow \frac{p^2}{9}+b \cdot \frac{p}{3}+p=0$
$ \Rightarrow \frac{p}{9}+\frac{b}{3}+1=0$
$ p+3 b+9=0.....$(2)
From (1) & (2) $\Rightarrow p =\frac{7}{2}$
Now, $f(-3)=9-3 b+p$
$=9-(-p-9)+p $
$ =18+2 p =18+2 \times \frac{7}{2}=25$