JEE Main 2021 — Functions Question with Solution

Given, p(x) = f(x³) + xg(x³)

We know, x² + x + 1 = (x $$-$$ $$\omega$$) (x $$-$$ $$\omega$$²)

Given, p(x) is divisible by x² + x + 1. So, roots of p(x) is $$\omega$$ and $$\omega$$².

As root satisfy the equation,

So, put x = $$\omega$$

p($$\omega$$) = f($$\omega$$³) + $$\omega$$g($$\omega$$³) = 0

= f(1) + $$\omega$$g(1) = 0 [$$\omega$$³ = 1]

= f(1) + $$\left(-\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)$$ g(1) = 0

$$\Rightarrow$$ f(1) $$-$$ $$\frac{g(1)}{2}+i\left(\frac{\sqrt{3}g(1)}{2}\right)$$ = 0 + i0

Comparing both sides, we get

f(1) $$-$$ $$\frac{g(1)}{2}$$ = 0

and $$\frac{\sqrt{3}}{2}g(1)=0$$ $$\Rightarrow$$ g(1) = 0

So, f(1) = 0

Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0