Question

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

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

Answer 1

Answer: (A)

$x=1$

$y=x^2-x+1\Rightarrow x^2-x+(1-y)=0$

Here, $a=1,b=-1$ and $c=1-y$

$x=\frac{1\pm \sqrt{1-4(1-y)}}{2}[\because x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}]$

$\because x>\frac{1}{2},\therefore x=\frac{1}{2}+\sqrt{y-\frac{3}{4}}$

$\Rightarrow f^{-1}\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$

Now, $x^2-x+1=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$

Since the graphs of the original and inverse functions can intersect only on the straight line $y=x,$ therefore

$x=f\left(x\right)\Rightarrow x=x^2-x+1$

$\Rightarrow x^2-2x+1=0$

$\Rightarrow {(x-1)}^2=0$

$\Rightarrow x=1$