Question

From given functions, which of the following(s) is a point function

• A. $\sqrt{2x-4}+\sqrt{4-2x}$
• B. $\sqrt{3x-6}+\sqrt{3-x}$
• C. $\sqrt{2x-4}+\ln (2-x)$
• D. $\sqrt{\text{sgn}(-x^2)}$

Answer 1

Answer: (A), (D)

$\sqrt{\text{sgn}(-x^2)}$

We know, A function having a single point in its domain is a point function.
Option $\left(a\right)$ : Let $f_1\left(x\right)=\sqrt{2x-4}+\sqrt{4-2x}$
For domain, $2x-4\ge 0\text{and}4-2x\ge 0$
$\Rightarrow x\ge 2\text{and}x\le 2$
$\Rightarrow$ Domain of $f_1\in \left\{2\right\}$
$\therefore f_1$ is a point function.
Option $\left(b\right)$ :
Let $f_2\left(x\right)=\sqrt{3x-6}+\sqrt{3-x}$
For domain, $3x-6\ge 0\text{and}3-x\ge 0$
$\Rightarrow x\ge 2\text{and}x\le 3$
$\Rightarrow$ Domain of $f_2\in [2,3]$
$\therefore f_2$ is not a point function.
Option $\left(c\right)$ :
$f_3\left(x\right)=\sqrt{2x-4}+\ln (2-x)$
For domain, $2x-4\ge 0\text{and}2-x>0$
$\Rightarrow x\ge 2\text{and}x<2$
$\Rightarrow$ Domain of $f_3\in \left\{\varphi \right\}$
$\therefore f_3$ is not a point function.
Option $\left(d\right)$ :
$f_4\left(x\right)=\sqrt{\text{sgn}(-x^2)}$
For domain, $\text{sgn}(-x^2)\ge 0$
$\Rightarrow \text{sgn}(-x^2)=0\text{or}\text{sgn}(-x^2)=1$
$\Rightarrow -x^2=0\text{or}-x^2>0$
$\Rightarrow x=0\text{or}x\in \left\{\varphi \right\}(\because -x^2\le 0\forall x\in R)$
$\Rightarrow$ Domain of $f_4\in \left\{0\right\}$
$\therefore f_4$ is a point function.