QJEEM21Maa38e: JEE Main 2021 Mathematics - Class 11 JEEM21Maa38e Solution: If f(x) and g(x) are two polynomials such that the polynomial…

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

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

Comparing both sides, we get

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

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

So, f(1) = 0

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