Aspire's Library
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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:
$\boxed{\text{Option 3}}$
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
