Bounds on Definite Integral of Function Derivative

To determine valid choices for $m$ and $M$ such that $m \le \int_{1/2}^1 f(x) \, dx \le M$, we first analyze the given derivative function $f'(x)$:

$f'(x) = \frac{192x^3}{2 + \sin^4(\pi x)}$

For $x \in \left[ \frac{1}{2}, 1 \right]$, we have $\pi x \in \left[ \frac{\pi}{2}, \pi \right]$, which implies: $0 \le \sin(\pi x) \le 1 \implies 0 \le \sin^4(\pi x) \le 1$

Thus, the denominator $2 + \sin^4(\pi x)$ is bounded by: $2 \le 2 + \sin^4(\pi x) \le 3$

Since $192x^3 > 0$ for all $x \in \left[ \frac{1}{2}, 1 \right]$, taking reciprocals yields the bounds on $f'(x)$: $\frac{192 x^3}{3} \le f'(x) \le \frac{192 x^3}{2}$ $64 x^3 \le f'(x) \le 96 x^3$

Given $f\left(\frac{1}{2}\right) = 0$, we integrate these bounds with respect to $t$ over the interval $\left[ \frac{1}{2}, x \right]$: $\int_{1/2}^x 64 t^3 \, dt \le \int_{1/2}^x f'(t) \, dt \le \int_{1/2}^x 96 t^3 \, dt$

Evaluating the integrals: $16 \left[ t^4 \right]_{1/2}^x \le f(x) - f\left(\frac{1}{2}\right) \le 24 \left[ t^4 \right]_{1/2}^x$ $16 \left( x^4 - \frac{1}{16} \right) \le f(x) \le 24 \left( x^4 - \frac{1}{16} \right)$ $16 x^4 - 1 \le f(x) \le 24 x^4 - \frac{3}{2}$

Now, we integrate $f(x)$ over $x \in \left[ \frac{1}{2}, 1 \right]$ to find lower and upper limits for $I = \int_{1/2}^1 f(x) \, dx$:

$\int_{1/2}^1 (16 x^4 - 1) \, dx \le \int_{1/2}^1 f(x) \, dx \le \int_{1/2}^1 \left( 24 x^4 - \frac{3}{2} \right) \, dx$

1. Lower Bound Calculation: $\int_{1/2}^1 (16 x^4 - 1) \, dx = \left[ \frac{16 x^5}{5} - x \right]_{1/2}^1$ $= \left( \frac{16}{5} - 1 \right) - \left( \frac{16}{5 \cdot 32} - \frac{1}{2} \right)$ $= \frac{11}{5} - \left( \frac{1}{10} - \frac{1}{2} \right) = \frac{11}{5} - \left( -\frac{4}{10} \right) = \frac{11}{5} + \frac{2}{5} = \frac{13}{5} = 2.6$

2. Upper Bound Calculation: $\int_{1/2}^1 \left( 24 x^4 - \frac{3}{2} \right) \, dx = \left[ \frac{24 x^5}{5} - \frac{3}{2} x \right]_{1/2}^1$ $= \left( \frac{24}{5} - \frac{3}{2} \right) - \left( \frac{24}{5 \cdot 32} - \frac{3}{4} \right)$ $= \frac{33}{10} - \left( \frac{3}{20} - \frac{15}{20} \right) = \frac{33}{10} - \left( -\frac{12}{20} \right) = \frac{33}{10} + \frac{6}{10} = \frac{39}{10} = 3.9$

Thus, we have: $2.6 \le \int_{1/2}^1 f(x) \, dx \le 3.9$

For $m \le \int_{1/2}^1 f(x) \, dx \le M$ to hold true, any choice of $m$ must satisfy $m \le 2.6$ and any choice of $M$ must satisfy $M \ge 3.9$.

Checking the given options:

• (A) $m = 13, M = 24$: Invalid, as $13 > 2.6$.
• (B) $m = \frac{1}{4}, M = \frac{1}{2}$: Invalid, as $\frac{1}{2} < 3.9$.
• (C) $m = -11, M = 0$: Invalid, as $0 < 3.9$.
• (D) $m = 1, M = 12$: Valid, since $1 \le 2.6 \le \int_{1/2}^1 f(x) \, dx \le 3.9 \le 12$.

Hence, the correct option is D.