JEE 2022 — Mathematics PYQ

1. Analyze the given equation:

The given equation is $x^4+x^2+1=0$.

We can factorize this expression by completing the square:

Using the identity $a^2-b^2=(a-b)(a+b)$:

2. Relate to Cube Roots of Unity:

Since $\alpha$ is a root, it must satisfy either $x^2+x+1=0$ or $x^2-x+1=0$.

• If ${\alpha }^2+\alpha +1=0$, then $\alpha$ is a complex cube root of unity ($\omega$ or ${\omega }^2$). In this case, ${\alpha }^3=1$.

• If ${\alpha }^2-\alpha +1=0$, then $(\alpha +1)({\alpha }^2-\alpha +1)=0$, which means ${\alpha }^3+1=0$, so ${\alpha }^3=-1$. Consequently, ${\alpha }^6=1$.

In both scenarios, we notice that the powers in the question are multiples of $3$. Let's check ${\alpha }^3$:

From $x^4+x^2+1=0$, multiply by $(x^2-1)$:

This also implies ${\alpha }^3$ can be $1$ or $-1$. However, if ${\alpha }^3=1$, then ${\alpha }^2+\alpha +1=0$. If we plug this into the original equation:

${\alpha }^4+{\alpha }^2+1=\alpha ({\alpha }^3)+{\alpha }^2+1=\alpha (1)+{\alpha }^2+1=0$. This works.

3. Evaluate the required expression:

The expression is $E={\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$.

Notice that all exponents are divisible by $3$:

• $1011=3\times 337$

• $2022=3\times 674$

• $3033=3\times 1011$

Let's assume ${\alpha }^2+\alpha +1=0$, so ${\alpha }^3=1$:

If we assume ${\alpha }^2-\alpha +1=0$, so ${\alpha }^3=-1$:

In both cases, the result is the same.

Final Answer:

The value is 1. Therefore, the correct option is (A).