Binomial TheoremHard
Question
Let f : (0, 1) → R be defined by f(x) 
, where b is a constant such that 0 < b < 1. Then
, where b is a constant such that 0 < b < 1. Then Options
A. f is not invertible on (0, 1)
B.f ≠ f-1 on (0, 1) and f′(b) =
C.f = f -1 on (0, 1) and f′(b) =
D.f -1 is differentiable on (0, 1)
Solution
Let y =
0 < x < 1 ⇒ 0 <
< 1 > 0 ⇒ y < b or y > 
- 1 < 0 ⇒ – 1 < y < ∴ - 1 < y < b.
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