Question

If $m$ and $M$ are the minimum and the maximum values of $4+\frac{1}{2}{\sin }^{2}2x-2{\cos }^{4}x,x\in R$, then $M-m$ is equal to :

• A. $\frac{15}{4}$
• B. $\frac{9}{4}$
• C. $\frac{7}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$4+\frac{1}{2}si{n}^{2}2x-2co{s}^{4}x$
$4+2\left(1-co{s}^{2}x\right)co{s}^{2}x-2co{s}^{4}x$
$-4\left\{co{s}^{4}x-\frac{co{s}^{2}x}{2}-1+\frac{1}{16}-\frac{1}{16}\right\}$
$-4\left\{{\left(co{s}^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16}\right\}$
$0\le co{s}^{2}x\le 1$
$-\frac{1}{4}\le co{s}^{2}x-\frac{1}{4}\le \frac{3}{4}$
$0\le {\left(co{s}^{2}x-\frac{1}{4}\right)}^2\le \frac{9}{16}$
$-\frac{17}{16}\le {\left(co{s}^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16}\le \frac{9}{16}-\frac{17}{16}$
$\frac{17}{14}\ge -4\left\{{\left(co{s}^{2}x-\frac{1}{4}\right)}^2-\frac{17}{16}\right\}\ge \frac{1}{2}$
$M=\frac{17}{4}$
$m=\frac{1}{2}$
$M-m=\frac{17}{4}-\frac{2}{4}=\frac{15}{4}$