JEE Main 2022 — Mathematics Algebra

Let $f(x)=(x-\alpha )(x-\beta )$f(x)=(x−α)(x−β)

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

and $f(1)=\frac{1}{3}\Rightarrow (1-\alpha )(1-\beta )=\frac{1}{3}$f(1)=31​⇒(1−α)(1−β)=31​

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

$$fofofof(x)=0$$fofofof(x)=0

$$\Rightarrow fofofof(\alpha )=0$$⇒fofofof(α)=0

$$\Rightarrow fofof(0)=0$$⇒fofof(0)=0

$$\Rightarrow fof(p)=0$$⇒fof(p)=0

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

Now assuming $(p-\alpha )(p-\beta )=\alpha$(p−α)(p−β)=α

$$\Rightarrow (\alpha \beta -\alpha )(\alpha \beta -\beta )=\alpha \Rightarrow (\beta -1)(\alpha -1)\beta =1$$⇒(αβ−α)(αβ−β)=α⇒(β−1)(α−1)β=1

$\Rightarrow \frac{\beta }{3}=1$⇒3β​=1 $(\text{as }(1-\alpha )(1-\beta )=\frac{1}{3})$(as (1−α)(1−β)=31​)

So, $\beta =3$β=3

Now finding $\alpha$α by putting the value of $\beta \text{ in }(1-\alpha )(1-\beta )=\frac{1}{3}$β in (1−α)(1−β)=31​,

$$\Rightarrow (1-\alpha )(1-3)=\frac{1}{3}$$⇒(1−α)(1−3)=31​

$$\Rightarrow \alpha =\frac{7}{6}$$⇒α=67​

So, $f(x)=(x-\frac{7}{6})(x-3)$f(x)=(x−67​)(x−3)

So, $f(-3)=(-3-\frac{7}{6})(-3-3)=25$f(−3)=(−3−67​)(−3−3)=25