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Let fk(x) = 1/k(sin^k x + cos^k x) where...

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

A

`1//6`

B

`1//3`

C

`1//4`

D

`1//12`

Text Solution

Verified by Experts

The correct Answer is:
D

Given, `f_(k)(x)=(1)/(k)(sin^(k)x+cos^(k)x)`, where `x in R` and `k ge 1`
`:.f_(4)(x)-f_(6)(x)=(1)/(4)(sin^(4)x+cos^(4)x)-(1)/(6)(sin^(6)x+cos^(6)x)`
`=(1)/(4)(1-2sin^(2)x.cos^(2)x)-(1)/(6)(1-3sin^(2)x.cos^(2)x)`
`=(1)/(4)-(1)/(6)=(1)/(12)`
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Knowledge Check

  • Let f_(k)(x)=(1//k)(sin^(k)x+cos^(k)x) where x in R and k ge 1 . Then f_(4)(x)-f_(6)(x) equals:

    A
    `1//6`
    B
    `1//3`
    C
    `1//4`
    D
    `1//12`
  • Let f_(k)(x)=1/(k)("sin"^(k)x+"cos"^(k)x) where x inRandkge1 then f_(4)(x)-f_(6)(x) equals

    A
    `1/(6)`
    B
    `1/(3)`
    C
    `1/(4)`
    D
    `1/(12)`
  • Let f_(k)(x)=(1)/(k)(sin^(k)x+cos^(k)x) for k = 1, 2, 3……. Then for all x in R, Then values of f_(4)(x)-f_(6)(x) is equal to

    A
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