Aspire Faculty ID #11982 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow \mathbb{R}$, be two functions such that $h(x) = sgn(g(x))$. Then select which of the following is not true?( $\mathbb{R}$ denotes the set of all real numbers, sgn stands for signum function)

Solution

Let $g:\mathbb{R}\to\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ be such that $h(x) = \operatorname{sgn}(g(x))$. 
 Recall: $\operatorname{sgn}(t) = \begin{cases} 1, & t>0\\ 0, & t=0\\ -1, & t<0 \end{cases}$ 
 Check each statement: 

 1) "The domain of $h(x)$ is the same as the domain of $g(x)$." $\Rightarrow$ True, because $\operatorname{sgn}(g(x))$ is defined for every $x$ where $g(x)$ is defined. 

 2) "The domain of continuity of $h(x)$ equals the domain of continuity of $g(x) - \{x\in\mathbb{R} : g(x)=0\}$." 
 At points where $g(x)\neq 0$, $h(x)$ is locally constant ($1$ or $-1$), hence continuous there (provided $g$ itself is continuous). 
 At points where $g(x)=0$, $h(x)$ jumps from $-1$ to $1$, so it is discontinuous. $\Rightarrow$ 
This statement is true. 

 3) "The domain of $h(x)$ is different from the domain of $g(x)$ at the same point." 
 Since for every $x$ in the domain of $g$, $h(x)=\operatorname{sgn}(g(x))$ is defined, the domains are exactly the same; they never differ. 
 $\Rightarrow$ This statement is false. 

 4) " $h(x)$ is discontinuous at $g(x)=0$." 
 At any $x_0$ where $g(x_0)=0$, the left and right limits of $h(x)$ are $-1$ and $1$, not equal to $h(x_0)=0$. 
 $\Rightarrow$ $h$ is discontinuous there, so this statement is true. 
 Therefore, the statement which is **not true** is: $\boxed{\text{Option 3}}$
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