Question

If the functions are defined as $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{1-x}$ then what is the common domain of the following functions: $f+g,f-g,\frac{f}{g},\frac{g}{f}$?

• A. $0<x\le 1$
• B. $0\le x<1$
• C. $0\le x\le 1$
• D. $0<x<1$

Answer 1

The correct option is D

$0<x<1$

Explanation for the correct option

Step 1: Solve for the domain of $f+g$

The given functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=\sqrt{1-x}$.

Thus, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

$\Rightarrow \left(f+g\right)\left(x\right)=\sqrt{x}+\sqrt{1-x}$

As the expression under the square root can not be lesser than zero.

Thus, $x\ge 0$ and $\left(1-x\right)\ge 0$

Hence, $x\ge 0$ and $x\le 1$.

Therefore, the domain of $f+g$ is $0\le x\le 1$.

Step 2: Solve for the domain of $f-g$

Now, $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$

$\Rightarrow \left(f-g\right)\left(x\right)=\sqrt{x}-\sqrt{1-x}$

As the expression under the square root can not be lesser than zero.

Thus, $x\ge 0$ and $\left(1-x\right)\ge 0$

Hence, $x\ge 0$ and $x\le 1$.

Therefore, the domain of $f-g$ is $0\le x\le 1$.

Step 3: Solve for the domain of $\left(\frac{f}{g}\right)$

Now, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$

$\Rightarrow \left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{\sqrt{1-x}}$

As the expression under the square root can not be lesser than zero and also the denominator can not be equal to zero.

Thus, $x\ge 0$ and $\left(1-x\right)>0$

Hence, $x\ge 0$ and $x<1$.

Therefore, the domain of $\left(\frac{f}{g}\right)$ is $0\le x<1$.

Step 4: Solve for the domain of $\left(\frac{g}{f}\right)$

Now, $\left(\frac{g}{f}\right)\left(x\right)=\frac{g\left(x\right)}{f\left(x\right)}$

$\Rightarrow \left(\frac{g}{f}\right)\left(x\right)=\frac{\sqrt{1-x}}{\sqrt{x}}$

As the expression under the square root can not be lesser than zero and also the denominator can not be equal to zero.

Thus, $x>0$ and $\left(1-x\right)\ge 0$

Hence, $x>0$ and $x\le 1$.

Therefore, the domain of $\left(\frac{g}{f}\right)$ is $0<x\le 1$.

Thus, the common domain of the functions $f+g,f-g,\frac{f}{g},\frac{g}{f}$ is $0<x<1$.

Hence, option(D) is the correct option i.e. $0<x<1$