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Q1:
Show that the function $f : R \to \{x \in R : – 1 < x < 1\}$ defined by $f(x) = \frac{x}{1 + |x|}$, $x \in R$ is one one and onto function.
Solution :
1. Analysis of the Function
Given the function $f: \mathbb{R} \to (-1, 1)$ defined by $f(x) = \frac{x}{1+|x|}$. To prove $f$ is a bijection, we must demonstrate both injectivity (one-one) and surjectivity (onto).
2. Proof of Injectivity (One-One)
Assume $f(x_1) = f(x_2)$ for $x_1, x_2 \in \mathbb{R}$. We analyze three cases for the signs of $x_1$ and $x_2$:
- Case I: $x_1, x_2 \geq 0$
$f(x) = \frac{x}{1+x}$. Since $f'(x) = \frac{(1+x)-x}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$, the function is strictly increasing. Thus, $f(x_1) = f(x_2) \implies x_1 = x_2$. - Case II: $x_1, x_2 < 0$
$f(x) = \frac{x}{1-x}$. Since $f'(x) = \frac{(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$, the function is strictly increasing. Thus, $f(x_1) = f(x_2) \implies x_1 = x_2$. - Case III: $x_1 \geq 0$ and $x_2 < 0$
If $x_1 \geq 0$, then $f(x_1) = \frac{x_1}{1+x_1} \geq 0$. If $x_2 < 0$, then $f(x_2) = \frac{x_2}{1-x_2} < 0$. Since the outputs have different signs, $f(x_1) \neq f(x_2)$.
[By the definition of an injective function, since $f(x_1) = f(x_2) \implies x_1 = x_2$ in all valid domains, $f$ is one-one].
3. Proof of Surjectivity (Onto)
Let $y \in (-1, 1)$. We seek $x \in \mathbb{R}$ such that $f(x) = y$.
| Case | $y \geq 0$ | $y < 0$ |
|---|---|---|
| Equation | $\frac{x}{1+x} = y$ | $\frac{x}{1-x} = y$ |
| Result | $x = \frac{y}{1-y}$ | $x = \frac{y}{1+y}$ |
For any $y \in (-1, 1)$, the calculated $x$ is a real number. Specifically:
- If $y \in [0, 1)$, $x = \frac{y}{1-y} \geq 0$.
- If $y \in (-1, 0)$, $x = \frac{y}{1+y} < 0$.
[Since every $y$ in the codomain has a pre-image $x$ in the domain, $f$ is onto].
Conclusion
Because the function satisfies the criteria for both injectivity and surjectivity, it is a bijection.
Final Solution: The function $f$ is both one-one and onto, hence it is a bijection.
More Questions from Class 12 Mathematics Relations and Functions Miscellaneous Exercise on Chapter 1
- Q2: Show that the function $f : R \to R$ given by $f(x) = x^3$ is injective.
- Q3: Given a non empty set $X$, consider $P(X)$ which is the set of all subsets of $X$. Define the relation $R$ in $P(X)$ as follows: For subsets $A, B$ in $P(X)$, $ARB$ if and only if $A \subset B$. Is $R$ an equivalence relation on $P(X)$? Justify your answer.
- Q4: Find the number of all onto functions from the set $\{1, 2, 3,......, n\}$ to itself.
- Q5: Let $A = \{– 1, 0, 1, 2\}$, $B = \{– 4, – 2, 0, 2\}$ and $f, g : A \to B$ be functions defined by $f(x) = x^2 – x$, $x \in A$ and $g(x) = 2|\frac{x}{2} - 1| - 1$, $x \in A$. Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f : A \to B$ and $g : A \to B$ such that $f(a) = g(a) \forall a \in A$, are called equal functions).
- Q6: Let $A = \{1, 2, 3\}$. Then number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is
- Q7: Let $A = \{1, 2, 3\}$. Then number of equivalence relations containing $(1, 2)$ is
CBSE Solutions for Class 12 Mathematics Relations and Functions
Chapters in CBSE - Class 12 Mathematics
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