Let `f:(0,1) rarr R` be defined by `f(x)=(b-x)/(1-bx)`, where b is a constant such that `0 lt b lt 1`. Then,

Options

A. f is not invertible on (0,1)
B. `f ne f^(-1) on (0,1) and f' (b)=(1)/(f'(0))`
C. `f = f^(-1) on (0,1) and f' (b)=(1)/(f'(0))`
D. `f^(-1)` is differentiable on (0,1)

Answer

A

Explanation

Let ` x, y in (0,1)` such that
f(x)=f(y)
`Rightarrow (b-x)/(1-bx)=(b-y)/(1-by)`
`Rightarrow (x-y) (b^(2)-1)=0 Rightarrow x=y " "[therefore 0 lt b lt 1 therefore b^(2)-1ne0]`
So, f is an injective map.
Let us now find the range of f.
Let f(x)=y then,
`Rightarrow (b-x)/(1-bx)=y`
`Rightarrow x=(b-y)/(1-by)`
`Rightarrow 0 lt (b-y)/(1-by) lt 1" "[therefore 0 lt x lt 1]`
`Rightarrow (b-y)/(1-by) gt 0 and (b-y)/(1-by) lt 1`
`Rightarrow (y-b)/(by-1) gt 0 and ((b-1)+(y+1))/(by-1) gt 0`
`Rightarrow (y-b)/(by-1) gt 0 and (y+1)/(by-1) lt 0" "[therefore 0 lt b lt 1]`
`Rightarrow -1 lt y lt b`
`Rightarrow "Range(f)"=(-1,b)ne R`
So, f is not subjective. Consequently, f is not invertible on [0,1]