1.f(x)=[x−1]+{x}xf(x) = [x-1] + \{x\}^xf(x)=[x−1]+{x}x, x∈(1,3)x \in (1,3)x∈(1,3), then f−1(x)f^{-1}(x)f−1(x) is(where [⋅][\cdot][⋅] denotes greatest integer function and {⋅}\{\cdot\}{⋅} denotes fractional part function)a.{x+1x∈(1,2)2+x−1x∈[2,3)\begin{cases} x+1 & x \in (1,2) \\ 2 + \sqrt{x-1} & x \in [2,3) \end{cases}{x+12+x−1x∈(1,2)x∈[2,3)b.{x−1x∈(1,2)2−x−1x∈[2,3)\begin{cases} x-1 & x \in (1,2) \\ 2 - \sqrt{x-1} & x \in [2,3) \end{cases}{x−12−x−1x∈(1,2)x∈[2,3)c.{x−1x∈(0,1)2−x−1x∈[1,2)\begin{cases} x-1 & x \in (0,1) \\ 2 - \sqrt{x-1} & x \in [1,2) \end{cases}{x−12−x−1x∈(0,1)x∈[1,2)d.{x+1x∈(0,1)2+x−1x∈[1,2)\begin{cases} x+1 & x \in (0,1) \\ 2 + \sqrt{x-1} & x \in [1,2) \end{cases}{x+12+x−1x∈(0,1)x∈[1,2)Login to continueOnly logged in users canattempt or see the solution.