Let f_{k}(x) = 1/k(sin^{k} x + cos^{k} x) where x ∈ R and k ≥ 1. Then f_{4}(x) - f_{6}(x) equals

- 1/12
- 1/6
- 1/4
- 1/3

**Answer**

1/4(sin^{4} x - cos^{4} x) - 1/6(sin^{6} x - cos^{6} x)

= [3(sin^{4} x - cos^{4} x) - 2(sin^{6} x - cos^{6} x)] / 12

= [3(1 - 2sin^{2} x cos^{2} x) - 2(1 - 3sin^{2} x cos^{2} x)] / 12

**The correct option is A.**