Trigonometrical Equations
hard

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

A

$[-2,-1]$

B

$\left[-2,-\frac{3}{2}\right]$

C

$\left[-1,-\frac{1}{2}\right]$

D

$\left[-\frac{3}{2},-1\right]$

(JEE MAIN-2023)

Solution

$\lambda=\cos ^2 2 x-2 \sin ^4 x-2 \cos ^2 x$

$\text { convert all in to } \cos x \text {. }$

$\lambda=\left(2 \cos ^2 x-1\right)^2-2\left(1-\cos ^2 x\right)^2-2 \cos ^2 x$

$=4 \cos ^4 x-4 \cos ^2 x+1-2\left(1-2 \cos ^2 x+\cos ^4 x\right)-$

$2 \cos ^2 x$

$=2 \cos ^4 x-2 \cos ^2 x+1-2$

$=2 \cos ^4 x-2 \cos ^2 x-1$

$=2\left[\cos ^4 x-\cos ^2 x-\frac{1}{2}\right]$

$=2\left[\left(\cos ^2 x-\frac{1}{2}\right)^2-\frac{3}{4}\right]$

$\left.\lambda_{\max }=2\left[\frac{1}{4}-\frac{3}{4}\right]=2 \times\left(-\frac{2}{4}\right)=-1 \text { (max Value }\right)$

$\left.\lambda_{\min }=2\left[0-\frac{3}{4}\right]=-\frac{3}{2} \text { (MinimumValue }\right)$

$\text { So, Range }=\left[-\frac{3}{2},-1\right]$

Standard 11
Mathematics

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