Question

If A = sin²x + cos⁴x, then for all real x

(a) 1 ≤ A ≤ 2 

(b) 3/4 ≤ A ≤13/16

(c) 3/4 ≤ A ≤ 1 

(d) 13/16 ≤ A ≤ 1

Answer 1

(c) : A = sin²x + cos²x
We have cos⁴x ≤ cos²x
sin²x = sin²x
Adding sin²x + cos⁴x ≤ sin²x + cos²x
∴ A ≤ 1.
Again A = t + (1 – t)² = t² – t + 1, t ≥ 0,
where minimum is 3/4
Thus 3/4 ≤ A ≤ 1 .