Question

Let $A=\left\{x \in R ; x \geq \frac{1}{2}\right\}$ and $B=\left\{x \in R ; x \geq \frac{3}{4}\right\}$. If $f: A \rightarrow B$ is defined as $f(x)=x^{2}-x+1$, then the solution set of the equation $f(x)=f^{-1}(x)$ is

• A. $\{1\}$
• B. $\{2\}$
• C. $ {\frac {1}{2}} $
• D. None of these

Answer 1

Answer: (A)

Let $f(x)=y$
Then, $x^{2}-x+1=y$
$\Rightarrow x^{2}-x+(1-y)=0$
$\Rightarrow x=\frac{1 \pm \sqrt{1-4(1-y)}}{2}$
$\Rightarrow x=\frac{1 \pm \sqrt{4 y-3}}{2}$
$\Rightarrow x=\frac{1+\sqrt{4 y-3}}{2} \left[\because x \geq \frac{1}{2}\right]$
$\Rightarrow f^{-1}(y)=\frac{1+\sqrt{4 y-3}}{2}$
$\Rightarrow f^{-1}(x)=\frac{1+\sqrt{4 x-3}}{2}$
Now, $ f(x)=f^{-1}(x)$
$\Rightarrow x^{2}-x+1=\frac{1+\sqrt{4 x-3}}{2}$
Clearly, it is satisfied by $x=1$.
$\therefore x=\{1\}$