Given the function:
\[ f(x) = \left( \sqrt{8x - x^2 - 16} \right)^2 + (x - 7)^2. \]
Simplifying:
\[ f(x) = 8x - x^2 - 16 + (x - 7)^2. \]
Expanding \((x - 7)^2\):
\[ f(x) = 8x - x^2 - 16 + x^2 - 14x + 49. \]
Combining like terms:
\[ f(x) = -6x + 33. \]
To find the maximum and minimum values of \(f(x)\), we differentiate with respect to \(x\):
\[ f'(x) = -6. \]
Since the derivative is constant and negative, \(f(x)\) is a linear function that decreases as \(x\) increases. Therefore, the maximum value occurs at the lower bound of the domain of \(x\), and the minimum value occurs at the upper bound.
For the square root to be real, we require:
\[ 8x - x^2 - 16 \geq 0 \quad \implies \quad x^2 - 8x + 16 \leq 0. \]
Solving the quadratic inequality:
\[ (x - 4)^2 \leq 0 \quad \implies \quad x = 4. \]
Substitute \(x = 4\) into \(f(x)\):
\[ f(4) = 8 \cdot 4 - 4^2 - 16 + (4 - 7)^2 = 32 - 16 - 16 + 9 = 9. \]
Thus, the minimum value \(m = 9\).
Given that \(M = 49\):
\[ M^2 - m^2 = 49^2 - 9^2 = 1600. \]
Therefore, the correct answer is 1600.
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
