# Question aadaa

Mar 26, 2017

Range $\left(- \infty , 0\right]$

#### Explanation:

Given: f(x)=ln(cos(sqrt(π^2-x^2)))#

Find the range.

Before we find the range, we need to find the domain.

The argument of the logarithm must be greater than. This means that the cosine must be greater 0:

$\cos \left(\sqrt{{\pi}^{2} - {x}^{2}}\right) > 0$

This implies:

$0 \le \sqrt{{\pi}^{2} - {x}^{2}} < \frac{\pi}{2}$

$0 \le {\pi}^{2} - {x}^{2} < {\pi}^{2} / 4$

$- {\pi}^{2} \le - {x}^{2} < - 3 {\pi}^{2} / 4$

${\pi}^{2} \ge {x}^{2} > 3 {\pi}^{2} / 4$

$\pi \ge x > \sqrt{3} \frac{\pi}{2} \mathmr{and} - \sqrt{3} \frac{\pi}{2} < x \le - \pi$

This makes the cosine function vary from 1 to nearly 0

When the argument of the natural logarithm goes from nearly 0 to 1, its range varies from $- \infty \to 0$