Humein equation di gayi hai:
Isse solve karne ke liye hum expressions ko simplify karte hain:
Step 1: Term simplify karein Humein pata hai ki 2cos2x=1+cos2x. Aur 2sin4x ko hum aise likh sakte hain: 2sin4x=2(sin2x)2=2(21−cos2x)2 2sin4x=2(41+cos22x−2cos2x)=21+cos22x−2cos2x
Step 2: Values ko original equation mein rakhein λ=cos22x−(21+cos22x−2cos2x)−(1+cos2x)
L.C.M lene par: λ=22cos22x−1−cos22x+2cos2x−2−2cos2x λ=2cos22x−3
Step 3: λ ki range nikaalein Humein pata hai ki cos22x ki value 0 aur 1 ke beech hoti hai:
Ab poori inequality mein se 3 subtract karein:
Ab 2 se divide karein:
Isliye, λ∈[−23,−1].
Sahi Option: (3)
