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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).
Solution :
Step 1: Establishing Definitions of Equality for Functions
Two functions $f: A \to B$ and $g: A \to B$ are defined as equal if and only if:
- Their domains are identical (given here as $A = \{-1, 0, 1, 2\}$).
- Their codomains are identical (given here as $B = \{-4, -2, 0, 2\}$).
- For every element $x \in A$, the image satisfies $f(x) = g(x)$. [Per the Axiom of Extensionality in Function Theory].
Step 2: Evaluating the Function $f(x) = x^2 - x$ for all $x \in A$
| $x$ | Calculation: $f(x) = x^2 - x$ | Image $f(x)$ |
|---|---|---|
| $-1$ | $(-1)^2 - (-1) = 1 + 1$ | $2$ |
| $0$ | $(0)^2 - 0$ | $0$ |
| $1$ | $(1)^2 - 1$ | $0$ |
| $2$ | $(2)^2 - 2$ | $2$ |
Step 3: Evaluating the Function $g(x) = 2|\frac{x}{2} - 1| - 1$ for all $x \in A$
| $x$ | Calculation: $g(x) = 2|\frac{x}{2} - 1| - 1$ | Image $g(x)$ |
|---|---|---|
| $-1$ | $2|\frac{-1}{2} - 1| - 1 = 2|-1.5| - 1 = 3 - 1$ | $2$ |
| $0$ | $2|0 - 1| - 1 = 2|-1| - 1 = 2 - 1$ | $1$ |
| $1$ | $2|\frac{1}{2} - 1| - 1 = 2|-0.5| - 1 = 1 - 1$ | $0$ |
| $2$ | $2|\frac{2}{2} - 1| - 1 = 2|0| - 1 = 0 - 1$ | $-1$ |
Step 4: Comparative Analysis and Verification
We compare the results generated in Step 2 and Step 3:
- For $x = -1$: $f(-1) = 2$ and $g(-1) = 2$. ($f = g$)
- For $x = 0$: $f(0) = 0$ and $g(0) = 1$. ($f \neq g$)
- For $x = 1$: $f(1) = 0$ and $g(1) = 0$. ($f = g$)
- For $x = 2$: $f(2) = 2$ and $g(2) = -1$. ($f \neq g$)
[By the condition of function equality, $f(x)$ must equal $g(x)$ for all $x \in A$. Since $f(0) \neq g(0)$ and $f(2) \neq g(2)$, the functions fail the equality criterion].
Step 5: Visual Representation
Final Solution: The functions $f$ and $g$ are not equal because $f(x) \neq g(x)$ for at least one $x \in A$ (specifically for $x=0$ and $x=2$).
More Questions from Class 12 Mathematics Relations and Functions Miscellaneous Exercise on Chapter 1
- 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.
- 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.
- 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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