Let f ^ ′ (x) = 192 x ^ 3 2 + ^ 4 for all x ∈ ℝ with f((1)/(2)) = 0...

Assuming $f ( x )$ to more increasing or less increasing
$\dfrac{192{x}^{3}}{3} < {f}^{′}\left(x\right) < \dfrac{192{x}^{3}}{2}$
$64{x}^{3} < {f}^{′}\left(x\right) < 96{x}^{3}$
$∫ 64{x}^{3}dx < ∫ {f}^{′}\left(x\right)dx < ∫ 96{x}^{3}dx$


$16{x}^{4} - 1 < f\left(x\right) < 24{x}^{4} - \dfrac{3}{2}$
${∫}_{\dfrac{1}{2}}^{1}\left(16{x}^{4} - 1\right)dx < {∫}_{\dfrac{1}{2}}^{1}f\left(x\right)dx < {∫}_{\dfrac{1}{2}}^{1}\left(24{x}^{4} - \dfrac{3}{2}\right)dx$
$\dfrac{26}{10} < {∫}_{\dfrac{1}{2}}^{1}f\left(x\right)dx < \dfrac{78}{20}$
Hence option $m = 1, M = 12$ is correct.