Summary & JEE Exam Tips — Class 12 Math Notes

Relation: A relation R on a set A is a subset of the Cartesian product $A \times A$ .
Equivalence Relation: A relation is an equivalence relation if it is simultaneously Reflexive ( $(a,a) \in R$ ), Symmetric ( $(a,b) \in R \implies (b,a) \in R$ ), and Transitive ( $(a,b) \in R$ and $(b,c) \in R \implies (a,c) \in R$ ).
Types of Functions:
- One-one (Injective): Every distinct input has a distinct output. ( $f(x_1) = f(x_2) \implies x_1=x_2$ ).
- Onto (Surjective): The range of the function is equal to its codomain.
- Bijective: The function is both one-one and onto.
Composition of Functions: The composition of $f: A \to B$ and $g: B \to C$ is $(g \circ f)(x) = g(f(x))$ . It is not commutative in general.
Invertible Function: A function is invertible if and only if it is bijective. The inverse of $f$ , denoted $f^{-1}$ , reverses the mapping. The graph of $f^{-1}$ is the reflection of the graph of $f$ about the line $y=x$ .

Check All Three Properties: When a question asks to check for an equivalence relation, you must methodically check for reflexivity, symmetry, and transitivity. Don't stop if you find one property is true; check all three. Finding a single counter-example is enough to disprove a property.
Domain and Codomain are Crucial: The properties of a function (one-one, onto) heavily depend on its specified domain and codomain. The function $f(x)=x^2$ is neither one-one nor onto from $\mathbb{R} \to \mathbb{R}$ , but it is bijective from $[0, \infty) \to [0, \infty)$ . Pay close attention to how the function is defined.
How to Prove One-one/Onto:
- To prove One-one: Start with $f(x_1) = f(x_2)$ and algebraically prove that $x_1=x_2$ . Alternatively, show that the derivative $f'(x)$ is either always positive or always negative (strictly monotonic).
- To prove Onto: Take an arbitrary element 'y' from the codomain and set $y=f(x)$ . Solve for x in terms of y. If you can show that for every 'y' in the codomain, there exists an 'x' in the domain, the function is onto.
Inverse Function Shortcut: To find the inverse, simply write $y=f(x)$ , swap the variables x and y, and then solve the new equation for y. This is a fast and reliable method.
Composition Order Matters: Remember that $(g \circ f)(x)$ means applying $f$ first, then $g$ . It's a common mistake to apply them in the wrong order.