If `0 lt x lt pi/2` and `sin^(n) x + cos^(n) x ge 1`, then

Options

A. `[2,oo)`
B. `(-oo,2]`
C. `[-1,1]`
D. `[-1oo,1]`

Answer

B

Explanation

For `0ltxlt(pi)/(2),` we have
`0lt sin lt1,0ltcosxlt1`
`implies(1)/(sinx)gt1and(1)/(cosx)gt1`
`implies((1)/(sinx))^(n)gt1 and ((1)/(cosx))^(n)gt1`
`impliessin^(-n)x+cos^(-n)xgt1,where n inN`
For n=0, we have
`sin^(n)x+cos^(n)x=1+1=2gt1"for all" x in(0,(pi)/(2))`
For n=1, we have
`sin^(n)x+cos^(n)x=sinx+cosx`
`impliessin^(n)x+cos^(n)x=sqrt(1+sin2x)gt1"for all"x in(0,pi//2)`
For n=2, we have `sin^(n)x+cos^(n)x=sin^(2)x+cos^(2)x=1"for all"x in(0,(pi)/(2))`
For `n ge3,` we have
`sin^(n)x+cos^(n)x ltsin^(2)x+cos^(2)x=1"for all"x in(0,(pi)/(2))`
Hence, `n in(-oo2].`