Question

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

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

Answer 1

Answer: (A), (D)

The correct options are
A $m=13,M=24$
D $m=1,M=12$

$\frac{192x^3}{3}<f^{\prime }\left(x\right)<\frac{192x^3}{2}$
$\Rightarrow 16(x^4-\frac{1}{16})<{\int }_{\frac{1}{2}}^{x}f\left(x\right)dx<24(x^4-\frac{1}{16})$
$\Rightarrow 16{(\frac{x^5}{5}-\frac{x}{16})}_{\frac{1}{2}}^1<{\int }_{\frac{1}{2}}^{x}f\left(x\right)dx<24{(\frac{x^5}{5}-\frac{x}{16})}_{\frac{1}{2}}^1$
$\Rightarrow 2.6<{\int }_{\frac{1}{2}}^{x}f\left(x\right)dx<3.9$
Hence, only possible option is D.