Question

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

• A. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true.

Answer 1

Answer: (A)

$f\left(x\right)={\left(x-1\right)}^2+1,x\ge 1$
$f:[1,\infty )\to [1,\infty )$ is a bijective function
$\Rightarrow y={\left(x-1\right)}^2+1\Rightarrow {\left(x-1\right)}^2=y-1$
$\Rightarrow x=1\pm \sqrt{y-1}\Rightarrow f^{-1}\left(y\right)=1\pm \sqrt{y-1}$
$\Rightarrow f^{-1}\left(x\right)=1+\sqrt{x+1}\left\{\therefore x\ge 1\right\}$
so statement-2 is correct
Now $f\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left(x\right)=x\Rightarrow {\left(x-1\right)}^2+1=x$
$\Rightarrow x^2-3x+2=0\Rightarrow x=1,2$
so statement-1 is correct