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Download App for Answer ( f^{-1}(x) )
domain (0,1)
rang (2,3)
So, ( f^{-1}(x)=x+2, quad ) So, ( d )

# 55. Let f:(2,3) - (0,1) be defined by f(x)= x-[x] then f'(x) equals (a) x-2 (b) x+1 (C) x-1 (d) x+2

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