Question

The set of all values of $\lambda$ for which the equation ${\cos }^{2}2x-2{\sin }^{4}x-2{\cos }^{2}x=\lambda$

• A. $[-2,-1]$
• B. $\left[-2,-\frac{3}{2}\right]$
• C. $\left[-1,-\frac{1}{2}\right]$
• D. $\left[-\frac{3}{2},-1\right]$

Answer 1

Answer: (D)

$\lambda ={\cos }^{2}2x-2{\sin }^{4}x-2{\cos }^{2}x$
convert all in to $\cos x$.
$\lambda ={\left(2{\cos }^{2}x-1\right)}^2-2{\left(1-{\cos }^{2}x\right)}^2-2{\cos }^{2}x$
$=4{\cos }^{4}x-4{\cos }^{2}x+1-2\left(1-2{\cos }^{2}x+{\cos }^{4}x\right)-2{\cos }^{2}x$
$=2{\cos }^{4}x-2{\cos }^{2}x+1-2$
$=2{\cos }^{4}x-2{\cos }^{2}x-1$
$=2\left[{\cos }^{4}x-{\cos }^{2}x-\frac{1}{2}\right]$
$=2\left[{\left({\cos }^{2}x-\frac{1}{2}\right)}^2-\frac{3}{4}\right]$
${\lambda }_{\max }=2\left[\frac{1}{4}-\frac{3}{4}\right]=2\times \left(-\frac{2}{4}\right)=-1\text{ (max Value })$
${\lambda }_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2}(\text{ MinimumValue })$
$\text{ So, Range }=\left[-\frac{3}{2},-1\right]$