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Let f be a function defined by f(x)=(x-1...

Let `f` be a function defined by `f(x)=(x-1)^(2)+1,(xge1)`.
Statement 1: The set `(x:f(x)=f^(-1)(x)}={1,2}`
Statement 2: `f` is a bijection and `f^(-1)(x)=1+sqrt(x-1),xge1`.

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
It is evident from the graph of f(x) that f(x) is a bijection for all `x ge 1` and hence invertible.
Now, `fof^(-1)(x)=x"for all "x ge 1`
`Rightarrow f(f^(-1)(x))=x"for all "x ge 1`
`Rightarrow (f^(-1)(x)-1]^(2)+1=x"for all "x ge 1`
`Rightarrow f^(-1)(x)=1+sqrt(x-1)"for all "x ge 1`
So, statement-2 is correct.
`"Now", [x:f(x)=f^(-1)(x)]={x:f(x)=x}=(x:(x-1)^(2)+1=x)=(1,2)`
Hence, statement-1 is also correct and statement-2 is a correct explantion of statment-1.
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