Let `f_k(x) = 1/k(sin^k x + cos^k x)` where `x in RR` and `k gt= 1.` Then `f_4(x) - f_6(x)` equals

Options

A. `1/4`
B. `1/12`
C. `1/6`
D. `1/3`

Answer

B

Explanation

We have, `f_(k)(x)=1/k(sin^(k)x+cos^(k)x)`
`thereforef_(4)(x)-f_(6)(x)`
`=1/4(sin^(4)x+cos^(4)x)-1/6(sin^(6)x+cos^(6)x)`
`=1/4{(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x}`
`-1/6{(sin^(2)x+cos^(2)x)^(3)-3sin^(2)x cos^(2)(sin^(2)x+cos^(2)x)}`
`=1/4(1-2sin^(2)xcos^(2)x)-1/6(1-3sin^(2)xcos^(2)x)=1/12`