Question

The maximum number of real roots of the equation $x^{2n}-1=0$ is

• A. $2$
• B. $3$
• C. $n$
• D. $2n$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Given, $f\left(x\right)=x^{2n}-1$

Replacing $x$ by $-x$, we get,

$f(-x)={(-x)}^{2n}-1=x^{2n}-1$

Thus, $f(-x)=f\left(x\right)$

Hence, $f\left(x\right)$ is symmetrical about $y-$axis.

So, it is clear from the above graph that the curve intersect the $x-$axis at only $2$ points, then the maximum number of real roots is $2$.

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is the correct answer.