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.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
