Question

Let $f\left(x\right)=x,g\left(x\right)=\frac{1}{x}\text{and}h\left(x\right)=f\left(x\right)g\left(x\right).$ Then, $h\left(x\right)=1$ if and only if

• A. $x$ is a real number
• B. $x$ is a rational number
• C. $x$ is an irrational number
• D. $x$ is a non-zero real number

Answer 1

The correct option is D

$x$ is a non-zero real number

Explanation for the correct option.

Finding the condition for $h\left(x\right)=1$

Given, $f\left(x\right)=x,g\left(x\right)=\frac{1}{x}\text{and}h\left(x\right)=f\left(x\right)g\left(x\right).$

Putting $h\left(x\right)=1$

$$\begin{gathered} \Rightarrow f\left(x\right)g\left(x\right)=1[\because h(x)=f(x\left)g\right(x\left)\right] \\ =x\left(\frac{1}{x}\right) \\ \because x\ne 0 \end{gathered}$$

We observe that domain of $f$ is $\mathrm{ℝ}$ and domain of $g$ is $\mathrm{ℝ}$

Therefore domain of $h=$domain of $f$ $\cap$ domain of $g$

Domain of $h=\mathrm{ℝ}\cap \mathrm{ℝ}=\mathrm{ℝ}$

But we also saw that $x$ cannot take the value $0$.

This means $h\left(x\right)=1$ only in the domain $\left\{\mathrm{ℝ}-0\right\}$.

Hence, the correct answer is Option (D)