Question:

Step 1: Simplify the expression for $f(x)$. Let $\alpha = \text{cosec}^{-1}\pi$. By definition, $\sin \alpha = \frac{1}{\pi}$. Since $\alpha$ is an angle in a right triangle, $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{1}{\pi}\right)^2} = \sqrt{1 - \frac{1}{\pi^2}} = \frac{\sqrt{\pi^2-1}}{\pi}$. Now, expand the term $\cos(x - \text{cosec}^{-1}\pi)$: $$ \cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha $$ $$ \cos(x - \alpha) = \cos x \left(\frac{\sqrt{\pi^2-1}}{\pi}\right) + \sin x \left(\frac{1}{\pi}\right) $$ $$ \cos(x - \alpha) = \frac{1}{\pi}(\sqrt{\pi^2-1}\cos x + \sin x) $$ Substitute this back into the expression for $f(x)$: $$ f(x) = (\sqrt{\pi^2 - 1}\cos x + \sin x) \cdot \frac{1}{\pi}(\sqrt{\pi^2-1}\cos x + \sin x) $$ $$ f(x) = \frac{1}{\pi} (\sqrt{\pi^2 - 1}\cos x + \sin x)^2 $$ Step 2: Determine the minimum and maximum values of $f(x)$. Let $g(x) = \sqrt{\pi^2 - 1}\cos x + \sin x$. The range of an expression of the form $a\cos x + b\sin x$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$. For $g(x)$, $a = \sqrt{\pi^2-1}$ and $b = 1$. The amplitude of $g(x)$ is $\sqrt{(\sqrt{\pi^2-1})^2 + 1^2} = \sqrt{\pi^2 - 1 + 1} = \sqrt{\pi^2} = \pi$. Thus, the range of $g(x)$ is $[-\pi, \pi]$. Since $f(x) = \frac{1}{\pi} (g(x))^2$, we need to find the range of $(g(x))^2$. The square of a function with range $[-\pi, \pi]$ will have a range of $[0, \pi^2]$. Therefore, the range of $f(x)$ is $\left[\frac{0}{\pi}, \frac{\pi^2}{\pi}\right] = [0, \pi]$. The minimum value of $f(x)$ is $m=0$. The maximum value of $f(x)$ is $M=\pi$. Step 3: Evaluate the given statements using $m=0$ and $M=\pi$. 1. Consider the expression $m+M$. $$ m+M = 0 + \pi = \pi $$ 2. Consider the statement $m+M = 0$. Since $m+M = \pi$, and $\pi \neq 0$, the statement $m+M = 0$ is false. 3. Consider the statement $\cos(m+M) = \cos m + \cos M$. Left Hand Side (LHS): $$ \cos(m+M) = \cos(\pi) = -1 $$ Right Hand Side (RHS): $$ \cos m + \cos M = \cos(0) + \cos(\pi) = 1 + (-1) = 0 $$ Since LHS $\neq$ RHS (i.e., $-1 \neq 0$), the statement $\cos(m+M) = \cos m + \cos M$ is false. 4. Consider the statement $\sin(m+M) = \sin m + \sin M$. Left Hand Side (LHS): $$ \sin(m+M) = \sin(\pi) = 0 $$ Right Hand Side (RHS): $$ \sin m + \sin M = \sin(0) + \sin(\pi) = 0 + 0 = 0 $$ Since LHS = RHS (i.e., $0 = 0$), the statement $\sin(m+M) = \sin m + \sin M$ is true.