InterviewSolution
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If the function,`f:[1,oo]to [1,oo]` is defined by `f(x)=3^(x(x-1))`, then `f^(-1)(x)` isA. `((1)/(3))^(x^((x-1)))`B. `(1)/(2){1-sqrt(1+4log_(3)x)}`C. `(1)/(2){1+sqrt(1+4log_(3)x)}`D. not defined |
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Answer» Correct Answer - C It can be checked that f is a bijection and hence invertible. Now, `fof^(-1)=x` `Rightarrow f(f^(-1)(x))=x` `Rightarrow 3^(f^(-1)(x)(f^(-1)(x)-1))=x` `Rightarrow f^(-1)(x)(f^(-1)(x)-1)=log_(3)x` `Rightarrow (f^(-1)(x))^(2)-f^(-1)(x)-log_(3)x=0` `Rightarrow f^(-1)(x)=(1)/(2){1+sqrt(1+4 log _(3)x)}" "[therefore f^(-1)(x) ge 1]` |
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