Mathematics - Definite Integration Question with Solution | TestHub

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


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