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

Options

A. 1
B. 2
C. 3
D. 4

Answer

A

Explanation

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.