1. Tardigrade
  2. Question
  3. Mathematics
  4. Let f be a function defined by f(x) = (x-1)2 + 1, (x ge 1). Statement - 1: The set x: f(x) = f-1(x) = 1, 2 Statement - 2: f is a bijection and f-1(x) = 1+√x-1, x ge 1.

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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.$

AIEEEAIEEE 2011Relations and Functions - Part 2

Solution:

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