Question

The number of value(s) of $x$ satisfying the equation $x^2-2x=1+\sqrt{1+x},x\in (2,\infty )$ is

• A. $0$
• B. $3$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Let $y=f\left(x\right)=x^2-2x$
$\Rightarrow x^2-2x+1=y+1\Rightarrow (x-1{)}^2=y+1\Rightarrow x=1+\sqrt{y+1}(\because x\in (2,\infty ))$
Now,
$f^{-1}\left(x\right)=1+\sqrt{1+x}$
$\therefore f\left(x\right)=f^{-1}\left(x\right)=x$
$\Rightarrow x^2-2x=x\Rightarrow x^2-3x=0\Rightarrow x=3(\because x\in (2,\infty \left)\right)$

Hence, the number of solution is $1.$