question_answer

The maximum number of real roots of the equation \[{{x}^{2n}}-1=0\]is:

A)  2                            

B)         3                            

C) \[n\]                     

D)         \[2n\]

Correct Answer: A

Solution :

Let\[f(x)={{x}^{2n}}-1\] Replacing \[x\]by\[-x,\] we get                 \[f(-x)={{(-x)}^{2n}}-1={{x}^{2n}}-1\]                 \[f(-x)=f(x)\] \[\therefore \]It is symmetrical about the y-axis. It is clear from the figure that curve intersect the real axis (i.e., \[x-\]axis) in two points, therefore the maximum number of possible real roots is 2. Note: The total number of real roots in \[nth\] roots of unity are 2 if \[n\]is even and 1, if \[n\]is odd.