Q.Find the domain of $g(x) = \cos^{-1}(x^2 - 1)$. Hence, find the value of $x$ for which $g(x) = \frac{\pi}{3}$. Also, write the range of $\cos^{-1} x$ other than its principal branch.

Concept: The domain of the inverse cosine function, $\cos^{-1}(y)$, is $[-1, 1]$. Its principal range is $[0, \pi]$.

1. Domain of $g(x)$: For $g(x) = \cos^{-1}(x^2 - 1)$ to be defined, the argument $x^2 - 1$ must satisfy $-1 \le x^2 - 1 \le 1$.

   This inequality simplifies to $0 \le x^2 \le 2$.

   The condition $x^2 \le 2$ implies $-\sqrt{2} \le x \le \sqrt{2}$.

2. Value of $x$ for $g(x) = \frac{\pi}{3}$:

   Given $g(x) = \cos^{-1}(x^2 - 1) = \frac{\pi}{3}$.

   Taking the cosine of both sides, we get $x^2 - 1 = \cos\left(\frac{\pi}{3}\right)$.

   Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$, we have $x^2 - 1 = \frac{1}{2}$.

   Solving for $x^2$: $x^2 = 1 + \frac{1}{2} = \frac{3}{2}$.

   Therefore, $x = \pm\sqrt{\frac{3}{2}} = \pm\frac{\sqrt{6}}{2}$. Both values are within the domain $[-\sqrt{2}, \sqrt{2}]$.

3. Range of $\cos^{-1} x$ other than its principal branch: …

The function $g(x) = \cos^{-1}(x^2 - 1)$ is an inverse trigonometric function. To understand its domain and how to solve for specific values, we must first recall the fundamental properties of the inverse cosine function.

The inverse cosine function, denoted as $\cos^{-1} x$ or $\operatorname{arccos} x$, is defined as the inverse of the cosine function. For $\cos^{-1} x$ to be a well-defined function, the domain of the cosine function must be restricted. The principal branch of $\cos x$ is typically restricted to the interval $[0, \pi]$, where it is one-to-one.

This means that the argument inside $\cos^{-1}(\cdot)$ must always be between $-1$ and $1$, inclusive.

Let's apply this understanding to the given problem.

1. Find the domain of $g(x) = \cos^{-1}(x^2 - 1)$

For $g(x)$ to be defined, its argument, $x^2 - 1$, must lie in the interval $[-1, 1]$.

So, we must satisfy the inequality:

This compound inequality can be broken down into two separate inequalities:

a) $x^2 - 1 \ge -1$

b) $x^2 - 1 \le 1$

Let's solve each part:

a) $x^2 - 1 \ge -1$

Adding $1$ to both sides:

$x^2 \ge 0$

This inequality is true for all real numbers $x$, since the square of any real number is always non-negative. So, $x \in (-\infty, \infty)$.

b) $x^2 - 1 \le 1$

Adding $1$ to both sides:

$x^2 \le 2$

Taking the square root of both sides (and remembering to consider both positive and negative roots):

$-\sqrt{2} \le x \le \sqrt{2}$

This means $x \in [-\sqrt{2}, \sqrt{2}]$.

For $g(x)$ to be defined, both conditions must be met. Therefore, we need to find the intersection of the solutions from (a) and (b):

$(-\infty, \infty) \cap [-\sqrt{2}, \sqrt{2}] = [-\sqrt{2}, \sqrt{2}]$

Thus, the domain of $g(x)$ is $[-\sqrt{2}, \sqrt{2}]$.

2. Find the value of $x$ for which $g(x) = \frac{\pi}{3}$

We are given $g(x) = \frac{\pi}{3}$, which means:

To solve for $x$, we take the cosine of both sides:

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.

So, the equation becomes:

Add $1$ to both sides:

Take the square root of both sides:

We can rationalize the denominator: