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 $m\le {\int }_{t/2}^{1}f\left(x\right)dx\le M,$ then the possible value 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)

$m=1,M=12$

Given that, $f^{\prime }\left(x\right)=\frac{192x^3}{2+{\sin }^4\pi x}$ for all $x\in R$

And $x\in (\frac{1}{2},1)$ with x$f\left(\frac{1}{2}\right)=0$

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

$f^{\prime }\left(x\right)=\frac{192x^3}{2+1}=\frac{192x^3}{3}=64x^3$ for $x\in \frac{1}{2}$

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

$f^{\prime }\left(x\right)=\frac{192x^3}{2+0}=\frac{192x^3}{2}=96x^3$ for $x\in 1$

Since,

${\int }_{\frac{1}{2}}^{x}64x^{3}dx\le {\int }_{\frac{1}{2}}^x{f}^{\prime }\left(x\right)dx\le {\int }_{\frac{1}{2}}^{x}96x^{3}dx$

$\Rightarrow {{\left[64\frac{x^4}{4}\right]}_{\frac{1}{2}}}^x\le {{\left[f^{\prime }\left(x\right)\right]}_{\frac{1}{2}}}^x\le {{\left[96\frac{x^4}{4}\right]}_{\frac{1}{2}}}^x$

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

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

If given that, $f\left(\frac{1}{2}\right)=0$ then,

$16x^4-1\le f\left(x\right)\le 24x^4-\frac{3}{2}$, $x\in (\frac{1}{2},0)$

On integrate and we get,

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

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

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

$\Rightarrow \frac{26}{10}\le {\int }_{\frac{1}{2}}^{1}f\left(x\right)dx\le \frac{78}{20}$

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

Comparing that given equation,

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

Then, $m=2.6$ and $M=3.9$

Option (D) is correct.

Because, $m=2.6\le 1$ and $M=3.9\le 12$

Hence, this is the answer.