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The difference between the maximum and minimum value of the function `f(x)=3sin^4x-cos^6x` is : |
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Answer» `f(x) = 3sin^4x-cos^6x` `= 3sin^4x-(1-sin^2x)^3` `= 3sin^4x-(1-sin^6x-3sin^2x(1-sin^2x))` `= 3sin^4x-(1-sin^6x-3sin^2x+3sin^4x)` `=sin^ 6x + 3sin^2x -1` `:. f(x) = sin^2x(sin^4x+3) - 1` Now, `f(x)` will be maximum, when `sinx = 1` and will be minimum when `sin x = 0`. `:. f(x)_max = 1(1+3)-1 = 3` `f(x)_min = 0(0+3)-1 = -1` `:. f(x)_max-f_min = 3-(-1) = 4`, which is the required difference. |
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