Let the function f be given by f(x)=(x-1)²+1 for x≥ 1.

Statement-1: The solution set of f(x)=f⁻¹(x) is {1,2}.
Statement-2: The function f is one-to-one and onto, and its inverse is f⁻¹(x)=1+√(x-1), with x≥ 1.

Question

Let the function f be given by f(x)=(x-1)²+1 for x≥ 1.

Statement-1: The solution set of f(x)=f⁻¹(x) is {1,2}.
Statement-2: The function f is one-to-one and onto, and its inverse is f⁻¹(x)=1+√(x-1), with x≥ 1.

Accepted Answer

Statement-1 is correct, Statement-2 is correct; Statement-2 correctly explains Statement-1.. f is increasing on x>=1, so f(x)=f^-1(x) reduces to f(x)=x: (x-1)^2+1=x -> x^2-3x+2=0 -> x=1,2, giving solution set {1,2}. The inverse is indeed f^-1(x)=1+sqrt(x-1) for x>=1, and that one-to-one/onto property is exactly why the equation collapses to f(x)=x. So Statement-1 and Statement-2 are both correct and Statement-2 correctly explains Statement-1.

Let the function f be given by f(x)=(x-1)²+1 for x≥ 1. Statement-1: The solution set of f(x)=f⁻¹(x) is {1,2}. Statement-2: The function f is one-to-one and onto, and its inverse is f⁻¹(x)=1+√(x-1), with x≥ 1.

Solution

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