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Question: F(x) =ln(cosx) is Even, Odd or periodic nature...

F(x) =ln(cosx) is Even, Odd or periodic nature

A

Even

B

Odd

C

Periodic

Answer

Even, Periodic

Explanation

Solution

To determine the nature of the function F(x)=ln(cosx)F(x) = \ln(\cos x), we need to check its domain and then evaluate if it's even, odd, or periodic.

1. Domain of the function:

For F(x)=ln(cosx)F(x) = \ln(\cos x) to be defined, two conditions must be met:

  • cosx\cos x must be defined, which is true for all real xx.
  • The argument of the logarithm must be positive: cosx>0\cos x > 0.

The condition cosx>0\cos x > 0 holds when xx lies in intervals of the form (2nππ2,2nπ+π2)(2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}) for any integer nn.
For example, one such interval is (π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}).
The domain of F(x)F(x) is D=nZ(2nππ2,2nπ+π2)D = \bigcup_{n \in \mathbb{Z}} (2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}).

2. Check for Even/Odd nature:

A function F(x)F(x) is even if F(x)=F(x)F(-x) = F(x) for all xx in its domain.
A function F(x)F(x) is odd if F(x)=F(x)F(-x) = -F(x) for all xx in its domain.

First, we check if the domain is symmetric about the origin. If xDx \in D, then cosx>0\cos x > 0. Since cos(x)=cosx\cos(-x) = \cos x, it implies cos(x)>0\cos(-x) > 0. Thus, if xDx \in D, then xD-x \in D. The domain is symmetric about the origin, so the function can be even or odd.

Now, let's evaluate F(x)F(-x):
F(x)=ln(cos(x))F(-x) = \ln(\cos(-x))

Since the cosine function is an even function, cos(x)=cosx\cos(-x) = \cos x.
Therefore,
F(x)=ln(cosx)F(-x) = \ln(\cos x)

This is equal to F(x)F(x).
Since F(x)=F(x)F(-x) = F(x), the function F(x)=ln(cosx)F(x) = \ln(\cos x) is an even function.

3. Check for Periodic nature:

A function F(x)F(x) is periodic if there exists a positive constant TT such that F(x+T)=F(x)F(x+T) = F(x) for all xx in its domain. The smallest such TT is called the period.

We know that the cosine function is periodic with a period of 2π2\pi, meaning cos(x+2π)=cosx\cos(x+2\pi) = \cos x.

Let's evaluate F(x+2π)F(x+2\pi):
F(x+2π)=ln(cos(x+2π))F(x+2\pi) = \ln(\cos(x+2\pi))

Since cos(x+2π)=cosx\cos(x+2\pi) = \cos x,
F(x+2π)=ln(cosx)F(x+2\pi) = \ln(\cos x)

This is equal to F(x)F(x).
Since F(x+2π)=F(x)F(x+2\pi) = F(x), the function F(x)=ln(cosx)F(x) = \ln(\cos x) is a periodic function with a period of 2π2\pi.

Conclusion:

The function F(x)=ln(cosx)F(x) = \ln(\cos x) is both an even function and a periodic function.