Question

Let $f^{\prime }\left(x\right)=\frac{192x^3}{2+{\sin }^4\pi x},\forall x\in R$ and $f\left(\frac{1}{2}\right)=0$. If $m\le \underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\le M$, then the values of $m$ and $M$ are

• A. $m=1.3,M=2.4$
• B. $m=0.25,M=4.5$
• C. $m=2.6,M=3.9$
• D. $m=1,M=12$

Answer 1

The correct option is C $m=2.6,M=3.9$

Given: $f^{\prime }\left(x\right)=\frac{192x^3}{2+{\sin }^4\pi x}$
We know
$0\le {\sin }^4\pi x\le 1\Rightarrow 2\le 2+{\sin }^4\pi x\le 3\Rightarrow \frac{1}{3}\le \frac{1}{2+{\sin }^4\pi x}\le \frac{1}{2}\Rightarrow \frac{192x^3}{3}\le f^{\prime }\left(x\right)\le \frac{192x^3}{2}\Rightarrow 64x^3\le f^{\prime }\left(x\right)\le 96x^3\Rightarrow \underset{1/2}{\overset{x}{\int }}64x^{3}dx\le \underset{1/2}{\overset{x}{\int }}{f}^{\prime }\left(x\right)dx\le \underset{1/2}{\overset{x}{\int }}96x^{3}dx\Rightarrow 16[x^4-\frac{1}{16}]\le f\left(x\right)-0\le 24[x^4-\frac{1}{16}]\Rightarrow 16x^4-1\le f\left(x\right)\le 24x^4-\frac{3}{2}$

Now, again integrating, we get
$\underset{1/2}{\overset{1}{\int }}(16x^4-1)dx\le \underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\le \underset{1/2}{\overset{1}{\int }}(24x^4-\frac{3}{2})dx\Rightarrow {[\frac{16x^5}{5}-x]}_{1/2}^1\le \underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\le {[\frac{24x^5}{5}-\frac{3x}{2}]}_{1/2}^1\Rightarrow \frac{26}{10}\le \underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\le \frac{39}{10}\therefore m=2.6,M=3.9$