We have, $$f'(x) = {{192{x^3}} \over {2 + {{\sin }^4}\pi x}}$$

Given, $$f\left( {{1 \over 2}} \right) = 0$$ and $$m \le \int\limits_{1/2}^1 {f(x)dx \le m} $$

$$\therefore$$ f(x) is increasing in $$\left( {{1 \over 2},1} \right)$$.

$$\therefore$$ $$f'{(x)_{\max }} = {{192} \over {24}} = 96$$

$$ \Rightarrow 96 = {{f(1) - f(1/2)} \over {1/2}} \Rightarrow f(1) = 96 \times {1 \over 2} = 48$$

$$M = {1 \over 2} \times {1 \over 2} \times 48 = 12$$

$$f'{(x)_{\min .}} = {{\left( {{{192} \over 8}} \right)} \over 3} = {{192} \over {24}}$$

$$ \Rightarrow {{192} \over {24}} = {{f(1) - 0} \over {(1/2)}} \Rightarrow f(1) = {{192} \over {24}} \times {1 \over 2}$$

$$\therefore$$ $$m = {1 \over 2} \times {1 \over 2} \times {{192} \over {24}} \times {1 \over 2} = 1$$