Question

Let $f(x)$ be a quadratic polynomial with leading coefficient 1 such that $f(0)=p,p\ne 0$ 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

Answer: 25

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