Question

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

The Correct Option is A

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