Question

A function $f:\left[\frac{1}{2},\infty \right)\to \left[\frac{3}{4},\infty \right)$ defined as, $f\left(x\right)=x^2-x+1$.

Then for what value of $x$ does equation $f\left(x\right)=f^{-1}\left(x\right)$ hold right.

• A. $x=3$
• B. $x=2$
• C. $x=1$
• D. None of these

Answer 1

The correct option is C $x=1$

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

Substitute $f\left(x\right)=y$

$y=x^2-x+1$

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

$\Rightarrow x=\frac{1\pm \sqrt{4y-3}}{2}$

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

$\Rightarrow f^{-1}\left(x\right)=\frac{1\pm \sqrt{4x-3}}{2}$
Now, we will check which value of $x$ from the options gives $f\left(x\right)=f^{-1}\left(x\right)$

Here, $f\left(3\right)=7$

$f^{-1}\left(3\right)=\frac{1\pm 3}{2}=2,-1$

$f\left(x\right)\ne f^{-1}\left(x\right)$ for $x=3$
Similarly for $x=2$, $f\left(x\right)\ne f^{-1}\left(x\right)$

Now, for $x=1$

$f\left(1\right)=1$

$f^{-1}=\frac{1\pm 1}{2}=1,0$

Hence, for $x=1$ , $f\left(x\right)=f^{-1}\left(x\right)$