Question

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

• A. $m=13,M=24$
• B. $m=\frac{1}{4},M=\frac{1}{2}$
• C. $m=-11,M=0$
• D. $m=1,M=12$

Answer 1

The correct option is D $m=1,M=12$

As $x\in (\frac{1}{2},0)\Rightarrow \frac{1}{2}\le x\le 1$
$\therefore 8\le f^{\prime }\left(x\right)\le 96\underset{\frac{1}{2}}{\overset{x}{\int }}8dx\le \underset{\frac{1}{2}}{\overset{x}{\int }}{f}^{\prime }\left(x\right)dx\le \underset{\frac{1}{2}}{\overset{x}{\int }}96dx$
$(8x-4)\le f\left(x\right)\le (96x-48)\underset{\frac{1}{2}}{\overset{1}{\int }}(8x-4)dx\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx\le \underset{\frac{1}{2}}{\overset{1}{\int }}(96x-48)dx$
$[4x^2-4x{]}_{\frac{1}{2}}^1\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx\le [48x^2-48x{]}_{\frac{1}{2}}^1$
$1\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)dx\le 12$
$m=1,M=12$