QJEEM22Mac4a9: JEE Main 2022 Mathematics - Class 11 JEEM22Mac4a9 Solution: Let be a quadratic polynomial with leading coefficient 1 such…

Let $f(x) = (x - \alpha )(x - \beta )$

It is given that $f(0) = p \Rightarrow \alpha \beta = p$

and $f(1) = {1 \over 3} \Rightarrow (1 - \alpha )(1 - \beta ) = {1 \over 3}$

Now, let us assume that, $\alpha$ is the common root of $f(x) = 0$ and $fofofof(x) = 0$

$fofofof(x) = 0$

$\Rightarrow fofof(0) = 0$

$\Rightarrow fof(p) = 0$

So, $f(p)$ is either $\alpha$ or $\beta$ .

$(p - \alpha )(p - \beta ) = \alpha$

$(\alpha \beta - \alpha )(\alpha \beta - \beta ) = \alpha \Rightarrow (\beta - 1)(\alpha - 1)\beta = 1$ ( $\because$ $\alpha \ne 0$ )

So, $\beta = 3$

$(1 - \alpha )(1 - 3) = {1 \over 3}$

$\alpha = {7 \over 6}$

$f(x) = \left( {x - {7 \over 6}} \right)(x - 3)$

$f( - 3) = \left( { - 3 - {7 \over 6}} \right)(3 - 3) = 25$