Question

Let $f\left(x\right)=x^2-x+1$, $x\ge \frac{1}{2}$, then the solution of the equation $f(x)=f^{-1}(x)$ is

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

Answer 1

Answer: (A)

Let $y=x^2-x+1$
$\Rightarrow x=\frac{1\pm \sqrt{1-4\left(1-y\right)}}{2}\because x\ge \frac{1}{2}$
$\therefore x=\frac{1}{2}+\sqrt{y-\frac{3}{4}}$
$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$
Now, $x^2-x+1=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$
$\Rightarrow x=1$ (By checking the given possibilities)