Question

Let $f^{\prime }\left(x\right)=\frac{192x^3}{2+si{n}^4\pi x}$ for all $x\in R$ with $f\left(\frac{1}{2}\right)=0$. If $m\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le M$, 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 $m=1,M=12$

Given$f^{\prime }\left(x\right)=\frac{192x^3}{2+{\sin }^4\pi x}$

$\Rightarrow f\left(x\right)=\int \frac{192x^3}{2+{\sin }^4\pi x}dx$

$f\left(\frac{1}{2}\right)=0$(given) and $m\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le M$

As $\frac{1}{2}\le x\le 1$

Hence ${\sin }^4\pi x$ will range from $0$ to $1$ as $\sin \frac{\pi }{2}=1$ and $\sin \pi =0$

Hence $2+{\sin }^4\pi x$ will range from $2+0=2$ to $2+1=3$

Hence we have $\frac{192t^3}{3}\le f^{\prime }\left(x\right)\le \frac{192t^3}{2}$

$\Rightarrow \frac{192}{3}{\int }_{\frac{1}{2}}^x{t}^{3}dt\le {\int }_{\frac{1}{2}}^x{f}^{\prime }\left(x\right)dx\le \frac{192}{2}{\int }_{\frac{1}{2}}^x{t}^{3}dt$

$\Rightarrow \frac{192}{3}{\left[\frac{t^4}{4}\right]}_{\frac{1}{2}}^x\le {\left[f\left(x\right)\right]}_{\frac{1}{2}}^x\le \frac{192}{2}{\left[\frac{t^4}{4}\right]}_{\frac{1}{2}}^x$

$\Rightarrow 16[x^4-\frac{1}{16}]\le f\left(x\right)\le 24[x^4-\frac{1}{16}]$

$\Rightarrow 16x^4-1\le f\left(x\right)\le 24x^4-\frac{3}{2}$

$\Rightarrow {\int }_{\frac{1}{2}}^1(16x^4-1)dx\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le {\int }_{\frac{1}{2}}^1(24x^4-\frac{3}{2})dx$

$\Rightarrow \frac{16}{5}(1-\frac{1}{32})-\frac{1}{2}\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le \frac{24}{5}(1-\frac{1}{32})-\frac{3}{2}\times \frac{1}{2}$

$\Rightarrow \frac{26}{10}\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le \frac{39}{10}$ on simplification

If ${\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\ge 12\Rightarrow {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\ge 1$

If ${\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le \frac{39}{10}\Rightarrow {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le 12$

Hence $m=1$ and $M=12$