Let $f$ be a function defined by $f(x)=(x-1)^2+1,(x \geq 1)$.

Question

Statement - 1: The set $\left\{x: f(x)=f^{\prime}(x)\right\}=\{1,2\}$.

Statement - 2 : $f$ is a bijection and $f^{\prime}(x)=1+\sqrt{x-1}, x \geq 1$.

Let $f$ be a function defined by $f(x)=(x-1)^2+1,(x \geq 1)$.

(1) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

(2) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1

(3) Statement- 1 is true, Statement- 2 is false

(4) Statement- 1 is false, Statement- 2 is true .

Answer 1

SOLUTION : $f(x)=(x-1)^2+1, x \geq 1$

$f:[1, \infty) \rightarrow[1, \infty)$ is a bijective function

$\Rightarrow y=(x-1)^2+1 \Rightarrow(x-1)^2=y-1 \Rightarrow x=1 \pm \sqrt{y-1} \Rightarrow f^{-1}(y)=1 \pm \sqrt{y-1} \\$

$\Rightarrow f^{-1}(x)=1+\sqrt{x-1} \quad\{\therefore x \geq 1\}$

so statement-2 is correct

Now $f(x)=f^{-1}(x) \Rightarrow f(x)=x \Rightarrow(x-1)^2+1=x \quad \Rightarrow x^2-3 x+2=0 \Rightarrow x=1,2$

so statement- 1 is correct 

Hence Option 1 is Correct.