JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
Let $f(x)$ be a quadratic polynomial with leading coefficient $1$ such that $f(0)=p$, $p\ne0$, and $f(1)=\frac13$. If the equations $f(x)=0$ and $f\circ f\circ f\circ f(x)=0$ have a common real root, then $f(-3)$ is equal to ________.
Quadratic Equation · Common Roots
Answer$25$
Solution
Concept. If $f(r)=0$, then $f(f(r))=f(0)=p$, so $f^{(4)}(r)=f(f(p))$; this must be $0$. Formulas used. (1) $f(x)=x^2+bx+p$ with $1+b+p=\frac13$, i.e. $b=-\frac23-p$Step 1 - Follow $r$. $f(r)=0$, $f^{(2)}(r)=p$, $f^{(3)}(r)=f(p)$, $f^{(4)}(r)=f(f(p))=0$. So $f(p)$ is a root of $f$. Step 2 - Compute $f(p)$. By (1), $f(p)=p^2+bp+p=p^2-\frac23p-p^2+p=\frac p3$. Step 3 - Solve. $f\left(\frac p3\right)=0$: $$\frac{p^2}{9}+\left(-\frac23-p\right)\frac p3+p=0\ \Longrightarrow\ -2p^2+7p=0.$$ $p\ne0$, so $p=\frac72$ and $b=-\frac{25}{6}$. Then $f(x)=(x-3)\left(x-\frac76\right)$ has real roots. Step 4 - Value. $f(-3)=9+\frac{25}{2}+\frac72=25$. Answer: 25