UPSC NDA Math Model Paper 3

We have to find the maximum value of 1−4sin2xcos2x
Let f(x)=1−4sin2xcos2x
=1−(2‌sin‌x‌cos‌x)2
=1−sin22x (∵sin‌2‌θ=2‌sin‌θ‌cos‌θ)
For maximum value of f(x), sin 2x should be minimum
As we know that,
Minimum value of sin2θ is 0
Therefore, Maximum value of f(x) = 1 – 0 = 1