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Let `f : [-1, -1/2] rarr [-1, 1]` is defined by `f(x)=4x^(3)-3x`, then `f^(-1) (x)` is |
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Answer» Correct Answer - 2 Let `cos^(-1) x= theta` where `-1 le x le - 1/2` i.e. `(2pi)/3 le theta le pi` Then `y=4x^(3)-3x=cos 3 theta` where `2pi le 3theta le 3pi` i.e. `y=cos (3theta-2pi)` where `0 le 3 theta-2pi le pi` `:. 3 theta -2pi=cos^(-1) y` i.e. `3 cos^(-1) x-2pi=cos^(-1) y` `3 cos^(-1) x=2pi+cos^(-1) y` `cos^(-1)x=(2pi)/3 + 1/3 cos^(-1) y` `x= cos ((2pi)/3+1/3 cos^(-1) y)` `:. f^(-1) (x) = cos ((2pi)/3+1/3 cos^(-1) x)` |
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