JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Quadratic Equations and Inequalities

Write $f(x)=x^2+bx+p$, so that $f(0)=p$, and the condition $f(1)=\frac{1}{3}$ gives $1+b+p=\frac{1}{3}$, i.e. $b=-\frac{2}{3}-p$. If $\alpha$ is a root common to $f(x)=0$ and $f(f(f(f(x))))=0$, then $f(\alpha )=0$, and the successive images are $f(\alpha )=0,f(0)=p,f(p),f(f(p))$, so the four-fold iterate at $\alpha$ equals $f(f(p))$, which must vanish. A short computation gives $f(p)=p^2+p(b+1)=p^2+p\left(\frac{1}{3}-p\right)=\frac{p}{3}$, and then $f\left(\frac{p}{3}\right)=\frac{-2p^2+7p}{9}$. Setting this equal to zero and using $p\ne 0$ gives $p=\frac{7}{2}$, whence $b=-\frac{25}{6}$ and $f(x)=x^2-\frac{25}{6}x+\frac{7}{2}$. Its roots $3$ and $\frac{7}{6}$ are real, so a genuine common real root exists, and $f(-3)=9+\frac{75}{6}+\frac{7}{2}=25$.