Question

If m and M are the minimum and the maximum value of $4+\frac{1}{2}si{n}^{2}2x-2co{s}^{4}x,xϵR$ then M-m is equal to

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

Answer 1

Answer: (A)

$\frac{9}{4}$

$4+\frac{1}{2}{\sin }^{2}2x-2{\cos }^{4}x,x\in R$

Let $y=4+\frac{1}{2}{\sin }^{2}2x-2{\cos }^{4}x$

$=4+\frac{1}{2}\left(4{\sin }^{2}x{\cos }^{2}x\right)-2{\cos }^{4}x$

$=4+\frac{1}{2}4(1-{\cos }^{2}x){\cos }^{2}x-2{\cos }^{4}x$

$=4+2{\cos }^{2}x-4{\cos }^{4}x$

$=2(2+{\cos }^{2}x-2{\cos }^{4}x)$

$=2(-2{\cos }^{4}x+{\cos }^{2}x+2)$

$=-4({\cos }^{4}x-\frac{{\cos }^{2}x}{2}-1)$

$=-4({\cos }^{4}x-\frac{{\cos }^{2}x}{2}+\frac{1}{16}-\frac{1}{16}-1)$

$=[4{({\cos }^{2}x-\frac{1}{4})}^2-\frac{17}{16}]$

$=\frac{17}{4}-4{({\cos }^{2}x-\frac{1}{4})}^2$

$y=\frac{17}{4}-4{({\cos }^{2}x-\frac{1}{4})}^2$

Since, $0\le {({\cos }^{2}x-\frac{1}{4})}^2\le \frac{9}{16}$

$y_{max}=M=\frac{17}{4}$

$y_{min}=m=\frac{17}{4}-(4\times \frac{9}{16})=\frac{17}{4}-\frac{9}{4}=\frac{8}{4}$

$\Rightarrow$ $M-m=\frac{17}{4}-\frac{8}{4}=\frac{9}{4}$