Question

Let the maximum and minimum values of $\left( \sqrt{8x - x^2 - 12 - 4} \right)^2 + (x - 7)^2, \quad x \in \mathbb{R} \text{ be } M \text{ and } m \text{ respectively}.$ Then $M^2 - m^2$ is equal to _____.

Answer 1

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:

$f(x) = -6x + 33.$

Step 1: Finding Maximum and Minimum Values

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.

Step 2: Calculating the Domain of $x$

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:

$(x - 4)^2 \leq 0 \quad \implies \quad x = 4.$

Step 3: Evaluating $f(x)$ at $x = 4$

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 $m = 9$.

Step 4: Calculating $M^2 - m^2$

Given that $M = 49$:

$M^2 - m^2 = 49^2 - 9^2 = 1600.$

Therefore, the correct answer is 1600.