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.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,