Generally, a primitive polynomial of degree ‘m’ is an irreducible polynomial in such a way that it is a factor of xn + 1, where ‘n’ = . . . . . . . . A. 2m – 1 B. $$\frac{m}{n} – 1$$ C. $$\frac{{\left( {m + 1} \right)}}{2}$$ D. m – n – 1

[amp_mcq option1=”2m – 1″ option2=”$$\frac{m}{n} – 1$$” option3=”$$\frac{{\left( {m + 1} \right)}}{2}$$” option4=”m – n – 1″ correct=”option1″]

The correct answer is $\boxed{\frac{{\left( {m + 1} \right)}}{2}}$.

A primitive polynomial of degree $m$ is an irreducible polynomial in such a way that it is a factor of $x^n + 1$, where $n = \frac{{\left( {m + 1} \right)}}{2}$.

Here is a brief explanation of each option:

• Option A: $2m – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $2 \cdot 3 – 1 = 3$.
• Option B: $\frac{m}{n} – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $\frac{3}{2} – 1 = \frac{1}{2}$.
• Option C: $\frac{{\left( {m + 1} \right)}}{2}$ is always a factor of $x^n + 1$. This is because $x^n + 1$ can be factored as $(x + 1)(x^{n – 1} – x^{n – 2} + \cdots + 1)$. The term $x + 1$ is always a factor of $x^n + 1$, and the other terms in the product are all divisible by $2$. Therefore, $x^n + 1$ is always divisible by $\frac{{\left( {m + 1} \right)}}{2}$.
• Option D: $m – n – 1$ is not always a factor of $x^n + 1$. For example, $x^3 + 1$ is not divisible by $3 – 2 – 1 = 0$.

Therefore, the correct answer is $\boxed{\frac{{\left( {m + 1} \right)}}{2}}$.