11. The function \( f \) and \( g \) are defined by \[ f(x)=\cos x, \quad 0 \leq x \leq \pi \quad \text { and } \quad g(x)=1-x^{2}, \quad x \in \mathbb{R} \] (a) Solve the equation \( f \circ g(x)=\frac{1}{2} \). (b) Determine the values of \( x \) for which \( f^{-1} \circ g(x) \) is not defined.

To solve the equation \( f \circ g(x) = \frac{1}{2} \), we first express it in terms of the definitions of \( f \) and \( g \): 1. **Step 1:** Substitute \( g(x) \) into \( f(x) \): \[ f \circ g(x) = f(g(x)) = f(1 - x^2) = \cos(1 - x^2) \] 2. **Step 2:** Set the equation: \[ \cos(1 - x^2) = \frac{1}{2} \] 3. **Step 3:** We know that \( \cos(\theta) = \frac{1}{2} \) when \( \theta = \frac{\pi}{3} + 2k\pi \) or \( \theta = \frac{5\pi}{3} + 2k\pi \) for any integer \( k \). Therefore, we have: \[ 1 - x^2 = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 1 - x^2 = \frac{5\pi}{3} + 2k\pi \] 4. **Step 4:** For each case, solve for \( x \): - For \( 1 - x^2 = \frac{\pi}{3} + 2k\pi \): \[ x^2 = 1 - \frac{\pi}{3} - 2k\pi \] \[ x = \pm \sqrt{1 - \frac{\pi}{3} - 2k\pi} \] - For \( 1 - x^2 = \frac{5\pi}{3} + 2k\pi \): \[ x^2 = 1 - \frac{5\pi}{3} - 2k\pi \] \[ x = \pm \sqrt{1 - \frac{5\pi}{3} - 2k\pi} \] Now, we need to find permissible values of \( k \) such that \( x^2 \geq 0 \) leads to valid \( x \) values. --- **(b)** To determine where \( f^{-1} \circ g(x) \) is not defined, we need to understand the range of \( g(x) \) and the requirement for \( f^{-1} \). 1. **Step 1:** Analyze \( g(x) = 1 - x^2 \): - The maximum value occurs at \( x = 0 \), yielding \( g(0) = 1 \). - As \( |x| \to \infty \), \( g(x) \to -\infty \). - Therefore, the range of \( g(x) \) is \( (-\infty, 1] \). 2. **Step 2:** Find the range of \( f(x) = \cos(x) \) for \( 0 \leq x \leq \pi \): - The range is \( [-1, 1] \). - The inverse function \( f^{-1}(y) \) is defined for \( y \in [-1, 1] \). 3. **Step 3:** Identify where \( f^{-1} \) becomes undefined: - Since \( g(x) \) produces values in \( (-\infty, 1] \), any \( g(x) < -1 \) will make \( f^{-1} \) undefined. - Solve \( 1 - x^2 < -1 \): \[ 1 < x^2 \] \[ |x| > 1 \] Thus, \( f^{-1} \circ g(x) \) is not defined for \( x < -1 \) or \( x > 1 \).