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Relations and Functions
NCERT Exemplar Class 12
CBSE
Grade 12
Question:
Show that the signum function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 1$ if $x > 0$, $0$ if $x = 0$, $-1$ if $x < 0$ is neither one-one nor onto.
Step-by-Step Solution
Given: Signum function $f: \mathbb{R} \to \mathbb{R}$ defined by:
$$f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$$
Step 1: Test for Injectivity (One-One):
Observe that $f(1) = 1$ and $f(2) = 1$.
Here $1
eq 2$, but $f(1) = f(2) = 1$.
Since different elements in domain have the same image, $f$ is not one-one. [1.0 Mark]
Step 2: Test for Surjectivity (Onto):
The range of $f$ is $\text{Range}(f) = \{-1, 0, 1\}$.
However, the codomain is $\mathbb{R}$.
Since $\text{Range}(f) = \{-1, 0, 1\}
eq \mathbb{R}$ (e.g. $2 \in \mathbb{R}$ has no pre-image), $f$ is not onto. [1.0 Mark]
Conclusion: Signum function is neither one-one nor onto.
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🎯 Official CBSE Marking Scheme:
Proving not one-one: 1.0 Mark
Proving not onto: 1.0 Mark
$$f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$$
Step 1: Test for Injectivity (One-One):
Observe that $f(1) = 1$ and $f(2) = 1$.
Here $1
eq 2$, but $f(1) = f(2) = 1$.
Since different elements in domain have the same image, $f$ is not one-one. [1.0 Mark]
Step 2: Test for Surjectivity (Onto):
The range of $f$ is $\text{Range}(f) = \{-1, 0, 1\}$.
However, the codomain is $\mathbb{R}$.
Since $\text{Range}(f) = \{-1, 0, 1\}
eq \mathbb{R}$ (e.g. $2 \in \mathbb{R}$ has no pre-image), $f$ is not onto. [1.0 Mark]
Conclusion: Signum function is neither one-one nor onto.
---
🎯 Official CBSE Marking Scheme:
Proving not one-one: 1.0 Mark
Proving not onto: 1.0 Mark
Correct Answer:
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