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Let f(x) = [x - 1] + {x}^[x] for x in (1, 3), where [.] denotes the greatest integer function and {.} denotes the fractional part function. Then f^(-1)(x) equals:

1. x + 1 for x in (1, 2), and 2 + sqrt(x - 1) for x in [2, 3)
2. x - 1 for x in (1, 2), and 2 - sqrt(x - 1) for x in [2, 3)
3. x - 1 for x in (0, 1), and 2 - sqrt(x - 1) for x in [1, 2)
4. x + 1 for x in (0, 1), and 2 + sqrt(x - 1) for x in [1, 2)

Correct answer: x + 1 for x in (0, 1), and 2 + sqrt(x - 1) for x in [1, 2)

Solution

On (1,2), [x-1]=0 and {x}^[x] = {x}¹ = x-1, so f(x)=x-1 mapping (1,2)->(0,1); inverse is x+1 on (0,1). On [2,3), [x-1]=1 and {x}^[x]=(x-2)², so f(x)=1+(x-2)² mapping [2,3)->[1,2); inverse is 2+sqrt(x-1) on [1,2).

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