JEE Main 2022 — Functions Question with Solution

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

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

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

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, $$\beta =3$$

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

$$\alpha =\frac{7}{6}$$

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

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