1. **State the problem:** We have the function $g(x) = \sqrt{1 - \sin^2 x}$ on the interval $0 \leq x \leq \pi$. We want to determine which of the given statements (A, B, C, D) could be false. 2. **Simplify the function:** Note that $1 - \sin^2 x = \cos^2 x$, so $$g(x) = \sqrt{\cos^2 x} = |\cos x|.$$ On $[0, \pi]$, $\cos x$ is positive on $[0, \frac{\pi}{2}]$ and negative on $[\frac{\pi}{2}, \pi]$, so $$g(x) = \begin{cases} \cos x & 0 \leq x \leq \frac{\pi}{2} \\ -\cos x & \frac{\pi}{2} < x \leq \pi \end{cases}.$$ 3. **Check statements:** - **A and B (Extreme Value Theorem):** Since $g$ is continuous on the closed interval $[0, \pi]$ (absolute value of cosine is continuous), EVT guarantees both a minimum and maximum exist. So A and B are true. - **C (Intermediate Value Theorem):** $g(0) = |\cos 0| = 1$, $g(\pi) = |\cos \pi| = 1$. So $$\frac{g(0) + g(\pi)}{2} = \frac{1 + 1}{2} = 1.$$ Since $g(x) \leq 1$ for all $x$, the value 1 is attained at the endpoints, so IVT holds trivially. C is true. - **D (Mean Value Theorem):** MVT requires $g$ to be differentiable on $(0, \pi)$. But $g(x) = |\cos x|$ is not differentiable at $x = \frac{\pi}{2}$ because $\cos x$ changes sign there, causing a sharp corner in $g$. Therefore, MVT does not apply, and the statement could be false. **Final answer:** Statement D could be false. $$\boxed{\text{D}}$$