Question

Let $A=\left\{x\in R;x\ge \frac{1}{2}\right\}$ and $B=\left\{x\in R;x\ge \frac{3}{4}\right\}$. If $f:A\to B$ 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, $x^2-x+1=y$
$\Rightarrow x^2-x+(1-y)=0$
$\Rightarrow x=\frac{1\pm \sqrt{1-4(1-y)}}{2}$
$\Rightarrow x=\frac{1\pm \sqrt{4y-3}}{2}$
$\Rightarrow x=\frac{1+\sqrt{4y-3}}{2}\left[\because x\ge \frac{1}{2}\right]$
$\Rightarrow f^{-1}(y)=\frac{1+\sqrt{4y-3}}{2}$
$\Rightarrow f^{-1}(x)=\frac{1+\sqrt{4x-3}}{2}$
Now, $f(x)=f^{-1}(x)$
$\Rightarrow x^2-x+1=\frac{1+\sqrt{4x-3}}{2}$
Clearly, it is satisfied by $x=1$.
$\therefore x=\{1\}$