Let `A={x inR: x >=1/2} and B={x in R: x>=3/4}.` If `f:A->B` is defined as `f(x)=x^2-x=1,` then the solution set of the equation `f(x)=f^-1(x)` isA. {1}B. {2}C. {1//2}D. None of these

Let `A={x inR: x >=1/2} and B={x in R: x>=3/4}.` If `f:A->B` is define

1.	Let `A={x inR: x >=1/2} and B={x in R: x>=3/4}.` If `f:A->B` is defined as `f(x)=x^2-x=1,` then the solution set of the equation `f(x)=f^-1(x)` isA. {1}B. {2}C. {1//2}D. None of these
Answer» Correct Answer - A Clearly, `f:A to B` is a bijection. This fact can also be observed from the graph of f(x) as it represents an arc of the parabola `y=x^(2)-x+1` lying on the right side of the vertex `(1//2, 3//4)` We know that the curves `y=f(x) and y=f^(-1)(x)` are mirror images of each other in the line mirror y=x. This means that the two curves interest at points lying on the line y=x. `therefore f(x)=f^(-1)(x)` `therefore f(x)=x Rightarrow x^(2)-x+1 Rightarrow (x-1)^(2)=0 Rightarrow x=1`