Answer:
Option B
Explanation:
(b)
Here, $f'(x)=\frac{192x^{3}}{2+\sin^{4}\pi x}$
$\therefore$ $\frac{192x^{3}}{3}\leq f'(x)\leq \frac{192x^{3}}{2}$
On integrating between the limits $\frac{1}{2}$ to x, we get
$\int_{1/2}^{x} \frac{192x^{3}}{3}dx\leq\int_{1/2}^{x} f'(x) dx\leq\int_{1/2}^{x}\frac{192x^{3}}{2} dx$
$\Rightarrow \frac{192}{12}\left(x^{4}-\frac{1}{16}\right)\leq f(x)-f(0)$ $\leq24x^{4}-\frac{3}{2}$
$\Rightarrow 16x^{4}-1\leq f(x) \leq24x^{4}-\frac{3}{2}$
Again integrating between the limits $\frac{1}{2}$ to 1, we get
$\int_{1/2}^{1} (16x^{4}-1)dx\leq\int_{1/2}^{1} f(x) dx\leq\int_{1/2}^{1} \left(24x^{4}-\frac{3}{2}\right)dx$
$\Rightarrow$ $\left[\frac{16x^{5}}{5}-x\right]_{1/2}^1\leq \int_{1/2}^{1} f(x) dx$
$\leq\left[\frac{24x^{5}}{5}-\frac{3}{2}x\right]_{1/2}^1$
$\Rightarrow$ $\left(\frac{11}{2}+\frac{2}{5}\right)\leq\int_{1/2}^{1}f(x) dx\leq\left(\frac{33}{10}+\frac{6}{10}\right) $
$\Rightarrow$ $2.6\leq \int_{1/2}^{1} f(x) dx\leq3.9$
