Question

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

Answer: (D)

$f^{\prime }(x)=\frac{192x^3}{2+{\sin }^4(\pi x)}\forall x\in R;f\left(\frac{1}{2}\right)=0$
Now $64x^3\le f^{\prime }(x)\le 96x^3\forall x\in \left[\frac{1}{2},1\right]$
So $16x^4-1\le f(x)\le 24x^4-\frac{3}{2}\forall x\in \left[\frac{1}{2},1\right]$
$\frac{16}{5}\cdot \frac{31}{32}-\frac{1}{2}\le {\int }_{y^2}^{1}f(x)dx\le \frac{24}{5}\cdot \frac{31}{32}-\frac{3}{4}$
$\Rightarrow \frac{26}{10}\le \underset{1/2}{\overset{1}{\int }}f(x)dx\le \frac{78}{20}$ hence (D)
(1) is incorrect as $\underset{1/2}{\overset{1}{\int }}f(x)dx\le \frac{78}{20}$
(2) is incorrect as $\underset{1/2}{\overset{1}{\int }}f(x)dx>\frac{26}{10}$
(3) is incorrect as $\underset{y^2}{\overset{1}{\int }}f(x)dx>0$