Question

If $f\left(x\right)=2\sqrt{x-1}+5\sqrt{1-x}+(x^2+x+1{)}^{3/2}$ exists, then domain of $f\left(x\right)$ is

• A. $[-1,1]$
• B. $\{-1,1\}$
• C. $\left\{1\right\}$
• D. $(-1,1)$

Answer 1

The correct option is C $\left\{1\right\}$

$f\left(x\right)=2\sqrt{x-1}+5\sqrt{1-x}+(x^2+x+1{)}^{3/2}$
$f$ is defined for all those values of $x$ where
$x-1\ge 0,1-x\ge 0$
i.e., $x\ge 1,x\le 1$

and $x^2+x+1\ge 0$
and ${(x+\frac{1}{2})}^2+\frac{3}{4}\ge 0$ (Always true)

$\therefore x\ge 1$ and $x\le 1\Rightarrow x=1$

Hence, domain of $f$ is $\left\{1\right\}$.