JEE Main 2023 — Mathematics Trigonometry

Given:

$$\lambda ={\mathrm{cos}}^{2}2x-2{\mathrm{sin}}^{4}x-2{\mathrm{cos}}^{2}x$$λ=cos22x−2sin4x−2cos2x

$$\Rightarrow \lambda ={(2{\mathrm{cos}}^{2}x-1)}^{2}-2{(1-{\mathrm{cos}}^{2}x)}^{2}-2{\mathrm{cos}}^{2}x$$⇒λ=(2cos2x−1)2−2(1−cos2x)2−2cos2x

$$\Rightarrow \lambda =4{\mathrm{cos}}^{4}x-4{\mathrm{cos}}^{2}x+1-2(1-2{\mathrm{cos}}^{2}x+{\mathrm{cos}}^{4}x)-2{\mathrm{cos}}^{2}x$$⇒λ=4cos4x−4cos2x+1−2(1−2cos2x+cos4x)−2cos2x

$$\Rightarrow \lambda =2{\mathrm{cos}}^{4}x-2{\mathrm{cos}}^{2}x+1-2$$⇒λ=2cos4x−2cos2x+1−2

$$\Rightarrow \lambda =2{\mathrm{cos}}^{4}x-2{\mathrm{cos}}^{2}x-1$$⇒λ=2cos4x−2cos2x−1

$$\Rightarrow \lambda =2[{\mathrm{cos}}^{4}x-{\mathrm{cos}}^{2}x-\frac{1}{2}]$$⇒λ=2[cos4x−cos2x−21​]

$$\Rightarrow \lambda =2[{({\mathrm{cos}}^{2}x-\frac{1}{2})}^{2}-\frac{3}{4}]$$⇒λ=2[(cos2x−21​)2−43​]

So,

${\lambda }_{\mathrm{max}}=2(\frac{1}{4}-\frac{3}{4})=-1$λmax​=2(41​−43​)=−1 (max Value)

${\lambda }_{\mathrm{min}}=2(0-\frac{3}{4})=-\frac{3}{2}$λmin​=2(0−43​)=−23​ (Minimum Value)

So, range of $\lambda$λ is$[-\frac{3}{2},-1]$[−23​,−1]