1.

Find the tange of `f(x)=cos^4x+sin^2x-1`.

Answer» Correct Answer - [-1/4, 0]
`f(x)=cos^4x+sin^2x-1`
`=cos^4x+1-cos^2x-1`
`=cos^4-cos^2x`
`=(cos^2x-1//2)^2-1//4`
Now, `0lecos^2xle1`
`rArr -1//2cos^2x-1//2le1//2`
`rArr-1//4le(cos^2x-1//2)^2-1//4le0`


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