NCERT Plus Two (Class 12) Mathematics • Chapter 1

Relations and Functions

Q: Give an example of a relation on A = {1, 2, 3} which is symmetric only (neither reflexive nor transitive).

Key (1 Mark): R = {(1, 2), (2, 1)} . • Not reflexive: (1,1) ∉ R . • Symmetric: (1,2) ∈ R ⇔ (2,1) ∈ R . • Not transitive: (1,2), (2,1) ∈ R , yet (1,1) ∉ R .

Q: If A = {1, 2, 3, 4}, how many bijective functions can be defined from A onto itself?

Key (1 Mark): Number of bijections from an n -element set to itself = n! = 4! = 24 .

Q: Check the injectivity and surjectivity of f: ℕ → ℕ given by f(x) = x².

• Injective (1-1): f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = x₂ (since x₁, x₂ ∈ ℕ are positive). Hence One-One . • Surjective (Onto): Take 2 ∈ ℕ (co-domain). There is no x ∈ ℕ with x² = 2 . Hence NOT Onto .

Q: Find g o f and f o g if f(x) = |x| and g(x) = |5x − 2|.

• (g o f)(x) = g(f(x)) = g(|x|) = |5|x| − 2| . • (f o g)(x) = f(g(x)) = f(|5x − 2|) = ||5x − 2|| = |5x − 2| .

Q: Show that the relation R in ℤ given by R = {(a, b) : 2 divides (a − b)} is an equivalence relation. Write the equivalence classes [0] and [1].

• Reflexive: a − a = 0 is divisible by 2 ⇒ (a, a) ∈ R . • Symmetric: 2 | (a − b) ⇒ 2 | −(a − b) = (b − a) ⇒ (b, a) ∈ R . • Transitive: a − b = 2k, b − c = 2m ⇒ a − c = 2(k + m) divisible by 2 ⇒ (a, c) ∈ R . Hence Equivalence Relation ! [0] = all even integers, [1] = all odd integers.

Q: Show that R = {(a, b) : |a − b| is even} on A = {1, 2, 3, 4, 5} is an equivalence relation. Show that all elements of {1, 3, 5} are related to each other.

• Reflexive: |a − a| = 0 (even). • Symmetric: |a − b| = |b − a| . • Transitive: If a − b and b − c are both even, their sum a − c is even. • Since difference between any two elements of {1, 3, 5} is even (2 or 4 or 0), all elements in {1, 3, 5} are related to each other!

Q: Let A = ℝ − {3} and B = ℝ − {1}. Consider f: A → B defined by f(x) = (x − 2)/(x − 3). Show f is one-one and onto, and find f⁻¹.

• One-One: (x₁−2)/(x₁−3) = (x₂−2)/(x₂−3) cross-multiplies to x₁x₂ − 3x₁ − 2x₂ + 6 = x₁x₂ − 2x₁ − 3x₂ + 6 ⇒ x₁ = x₂ . • Onto: For any y ∈ ℝ − {1} , set y = (x−2)/(x−3) ⇒ x(y−1) = 3y−2 ⇒ x = (3y−2)/(y−1) . Note x ≠ 3 (since 3y−2 = 3y−3 is impossible!). • Inverse: f⁻¹(y) = 3y − 2 y − 1 .

Q: Consider f: ℝ₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with f⁻¹(y) = (√(y + 6) − 1) / 3.

Rewrite f(x) = (3x + 1)² − 6 . Define g: [−5, ∞) → ℝ₊ by g(y) = √(y+6) − 1 3 . • (g o f)(x) = √((3x+1)²) − 1 3 = (3x+1) − 1 3 = x (as x ≥ 0 ). • (f o g)(y) = [3( √(y+6) − 1 3 ) + 1]² − 6 = (y + 6) − 6 = y . Thus g o f = I and f o g = I ⇒ f is invertible and f⁻¹ = g !

Reflexive, symmetric, transitive & equivalence relations; one-one, onto, composite & invertible maps.

📄 Official NCERT Textbook PDF

Topic 1.2: Empty, Universal, Reflexive, Symmetric & Transitive Relations

Picture three friends on a Thrissur school bus: A = {1, 2, 3}. Take the relation "lives in the same town as": every student lives in the same town as themselves (Reflexive), if 1 matches 2 then 2 matches 1 (Symmetric), and if 1 matches 2 and 2 matches 3 then 1 matches 3 (Transitive)! Meanwhile, "is 200 years older than" has zero pairs inside class (Empty Relation), and "is at least 0 cm tall" pairs everybody (Universal Relation)!

EMPTY & UNIVERSALEmpty: R = ∅ ⊆ A×A (no pairs)
Universal: R = A×A (all pairs)

REFLEXIVE (R)(a, a) ∈ R for EVERY a ∈ A (every node has a self-loop!)

SYMMETRIC (S)If (a, b) ∈ R, then (b, a) ∈ R (every arrow is two-way!)

TRANSITIVE (T)If (a, b) ∈ R & (b, c) ∈ R, then (a, c) ∈ R (shortcut exists!)

🧠 Memory Trick: R = Self-Mirror (a→a for ALL), S = Two-Way Bridge (a⇄b), T = Shortcut Triangle (a→b→c forces a→c)!

💡 Simple Kid-Check: To test a relation on set A, think like a detective! For Reflexive, check if EVERY element points to itself. For Symmetric, flip every existing arrow — does the reverse arrow exist? For Transitive, look for two-step chains a → b → c and check whether the direct jump a → c is present!

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Geometry Test — Perpendicular Lines (L₁ ⊥ L₂): A line cannot be perpendicular to itself (NOT Reflexive). If L₁ ⊥ L₂, then L₂ ⊥ L₁ (Symmetric!). If L₁ ⊥ L₂ and L₂ ⊥ L₃ in a plane, then L₁ ∥ L₃ (NOT Transitive!).
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Vacuous Truth Shortcut: Suppose R = {(1, 2)} on A = {1, 2, 3}. Is it Transitive? YES! There is NO pair starting with 2, so no two-step chain can break!
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Number Relation on ℝ: R = {(a, b) : a ≤ b²} fails Reflexive (try a = ½: ½ > ¼!), fails Symmetric ((1, 2) ∈ R, (2, 1) ∉ R), and fails Transitive ((3, 2), (2, 1.5) ∈ R, but 3 > 2.25!).

🧒 Super-Easy Example: On A = {1, 2, 3}, test R = {(1,1), (2,2), (3,3), (1,2), (2,3)}.
• Reflexive? YES — (1,1), (2,2), (3,3) are all present.
• Symmetric? NO — (1,2) ∈ R, but (2,1) ∉ R.
• Transitive? NO — (1,2) and (2,3) are in R, yet (1,3) ∉ R!

Watch out for the Missing Self-Loop Trap! On A = {1, 2, 3}, students see {(1,1), (2,2)} and mark it Reflexive — WRONG! Since (3,3) is missing, it is NOT reflexive! Also remember: the Empty Relation ∅ on a non-empty set is Symmetric and Transitive vacuously, NOT Reflexive!

Topic 1.2: Equivalence Relations & Equivalence Classes Partitioning Sets

Imagine dumping red, blue, and green marbles on a table and sorting them under the rule "has the same colour as". Every marble matches its own colour (Reflexive), matching works both ways (Symmetric), and matching chains hold (Transitive)! Because all three pass, it is an Equivalence Relation — automatically slicing the whole table into separate, non-overlapping colour bowls called Equivalence Classes!

EQUIVALENCE RELATIONA relation that is Reflexive + Symmetric + Transitive (R + S + T)!

EQUIVALENCE CLASS [a][a] = {x ∈ A : (x, a) ∈ R} — all elements related to a!

PARTITION RULEFor any a, b: either [a] = [b] OR [a] ∩ [b] = ∅! Union of all classes = A!

🧠 Memory Trick: R + S + T = Equivalence Club! Inside the club, two friends either belong to the EXACT SAME table [a] = [b] or share ZERO overlap [a] ∩ [b] = ∅!

💡 Simple Kid-Check: An equivalence relation sorts a messy set into clean, non-overlapping boxes with zero leftovers. Pick ANY member inside a box and ask for everyone related to them — you get that exact same box back!

Modulo Equivalence on ℤ: R = {(a, b) ∈ ℤ×ℤ : m divides (a − b)} is ALWAYS an equivalence relation!
For m = 2, ℤ splits into just 2 classes: [0] = {..., −2, 0, 2, 4, ...} (Evens) and [1] = {..., −3, −1, 1, 3, ...} (Odds).

🧒 Classic NCERT Example: Let A = {0, 1, 2, ..., 12} and R = {(a, b) : |a − b| is a multiple of 4}.
Find the equivalence class [1]! Collect all x ∈ A where |x − 1| ∈ {0, 4, 8, 12}:
• |1 − 1| = 0 (valid!), |5 − 1| = 4, |9 − 1| = 8.
So [1] = {1, 5, 9}! Notice [1] = [5] = [9], and all 4 classes [0], [1], [2], [3] partition A!

Watch out for 0! Students forget that 0 is a multiple of EVERY integer (0 = 4 × 0)! So in [1] for "|a − b| multiple of 4", students sometimes drop 1 itself — WRONG! Every element ALWAYS sits inside its own equivalence class a ∈ [a]! Also, "2 divides (a − b)" on infinite ℤ produces ONLY two equivalence classes, not infinitely many!

Topic 1.3: One-One (Injective) & Many-One Functions (Algebraic + Graphical Tests)

Every student in Kerala gets a unique Aadhaar number — two different students never share the same ID! That VIP 1-to-1 rule makes a function One-One (Injective). Now compare "birth month": Anu and Manu were both born in May — two inputs merge into one output! That sharing makes it Many-One!

ONE-ONE (INJECTIVE)f(x₁) = f(x₂) ⇒ x₁ = x₂
(Distinct inputs always produce distinct outputs!)

MANY-ONETwo different inputs hit the same output: x₁ ≠ x₂ yet f(x₁) = f(x₂)

HORIZONTAL LINE TESTEvery horizontal line y = c cuts the graph at AT MOST 1 point!

🧠 Memory Trick: One-One = Zero Collisions! Never let two input arrows crash into the same target! Total 1-1 maps from m items to n items = ⁿPₘ (0 if m > n).

💡 Simple Kid-Check: To prove a function IS One-One, assume two outputs match (f(x₁) = f(x₂)) and algebraically force x₁ = x₂. To prove it is Many-One, you only need ONE counterexample like f(1) = f(−1) = 1!

Function Rule	Domain → Codomain	Status	Quick Reason
`f(x) = 2x + 3`	`ℝ → ℝ`	One-One	`2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂`
`f(x) = x²`	`ℕ → ℕ`	One-One	`x₁² = x₂² ⇒ x₁ = x₂` (only `+` numbers!)
`f(x) = x²`	`ℤ → ℤ`	Many-One	`f(1) = 1` and `f(−1) = 1`!
`f(x) = x³`	`ℝ → ℝ`	One-One	Cube preserves negative signs!

Counting Shortcut: If n(A) = m and n(B) = n: Total functions = nᵐ | Total One-One functions = ⁿPₘ = n! / (n − m)! (when m ≤ n; 0 if m > n).

NEVER judge a formula without checking its DOMAIN first! f(x) = x² is Many-One on ℝ → ℝ (f(2) = f(−2) = 4), yet the EXACT SAME formula f(x) = x² is One-One on ℕ → ℕ because negative numbers don't exist in ℕ! Also watch x₁² = x₂² on ℝ: taking square root gives x₁ = ±x₂, NOT just x₁ = x₂!

Topic 1.3: Onto (Surjective) vs Into & Bijective Functions

Imagine 4 schoolchildren boarding 3 school buses B = {Bus 1, Bus 2, Bus 3}. If EVERY bus carries at least one student — zero empty buses left over — the boarding is Onto (Surjective)! If even one bus sits empty, it's Into! And when every student gets a private seat AND no bus sits empty, you've achieved perfection: a Bijective map!

ONTO (SURJECTIVE)Range = Co-domain! Every y ∈ B has at least one pre-image x ∈ A with f(x) = y.

INTO FUNCTIONRange ⊂ Co-domain (at least one element in B has NO arrow hitting it).

BIJECTIVEBOTH One-One AND Onto! Number of bijections from n items to n items = n!

🧠 Memory Trick: One-One checks inputs (no two hit same target); Onto checks targets (ZERO lonely unused targets in B)!

💡 Simple Kid-Check: How do you prove a function is Onto in a 3-mark board question? Just reverse the machine! Take ANY target y from the Co-domain, solve f(x) = y for x, and check whether that x legally lives inside your Domain!

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3-Step Onto Proof: Take arbitrary y ∈ Co-domain → Set f(x) = y and solve for x → Verify x ∈ Domain and f(x) = y!
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Example — f: ℕ → ℕ, f(x) = 2x: Take y = 1 ∈ ℕ. Solving 2x = 1 gives x = ½ ∉ ℕ! So 1 has no pre-image ⇒ NOT Onto (Into), even though it IS One-One!
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Finite Equal-Size Magic: If A and B are finite sets with the same number of elements, then One-One ⇔ Onto automatically! (Think: 3 kids in 3 seats — no sharing means no empty seats!)

🧒 Quick Bijection Check: Show f: ℝ → ℝ, f(x) = 3x − 4 is Bijective.
• 1-1: 3x₁ − 4 = 3x₂ − 4 ⇒ x₁ = x₂ ✓
• Onto: For any y ∈ ℝ, x = y + 43 ∈ ℝ satisfies f(x) = 3(y + 43) − 4 = y ✓! Hence Bijective!

Watch the Co-domain Swap Trap! Take f(x) = x²: from ℝ → ℝ it is NOT Onto (because −1 ∈ ℝ has no real square root), BUT from ℝ → [0, ∞) it IS Onto (every non-negative number has a square root)! Always read the Co-domain set AFTER the arrow → carefully!

Topic 1.4: Composition of Functions (f o g vs g o f)

Think of a two-step Kerala kitchen assembly line! Machine g washes raw mangoes, and Machine f slices whatever comes out! Feed a mango x into g first and pass its output straight into f — together they form ONE composite super-machine (f o g)(x) = f(g(x))! Flip the order to g o f (slicing muddy mangoes before washing) and the result completely changes!

DEFINITIONFor g: A→B & f: B→C, (f o g)(x) = f(g(x)) from A → C!

ORDER MATTERS!In general, f o g ≠ g o f! (Not commutative)

ASSOCIATIVITYh o (g o f) = (h o g) o f always holds!

🧠 Memory Trick: Read (f o g)(x) = f( g(x) ) INSIDE-OUT! Whoever sits closest to x (g) runs FIRST, then f runs SECOND!

💡 Simple Kid-Check: Picture two function boxes connected by a pipe: x enters g, pops out as g(x), and drops straight into f to come out as f(g(x))! To exist, every output of g must fit inside the domain of f (Range(g) ⊆ Domain(f)).

🎲 Two Classic NCERT Computations:
1. Let f(x) = cos x and g(x) = 3x².
• (g o f)(x) = g(f(x)) = g(cos x) = 3 cos²x
• (f o g)(x) = f(g(x)) = f(3x²) = cos(3x²) (Clearly f o g ≠ g o f!)

2. Let f(x) = 8x³ and g(x) = x^1/3.
• (g o f)(x) = (8x³)^1/3 = 2x | (f o g)(x) = 8(x^1/3)³ = 8x!

Biggest beginner blunder: computing (f o g)(x) in left-to-right reading order as g(f(x))! Remember the circle o means **"of"**: (f o g)(x) = f(g(x)) — apply g FIRST (closest to x), then wrap f around it! Also check if Range(g) ⊆ Domain(f) before composing!

Topic 1.4: Invertible Functions & Conditions (f o g = I, g o f = I)

Every morning you put on your socks (f) and then your shoes (g). At night, how do you undo it? You MUST pull off your shoes FIRST (g⁻¹) and your socks SECOND (f⁻¹)! In Maths, an Inverse Function f⁻¹ is a perfect UNDO button that reverses f cleanly — possible ONLY if f is Bijective (One-One and Onto)!

INVERTIBLE CONDITIONf: X → Y is invertible IF AND ONLY IF f is Bijective (1-1 & Onto)!

TWO-WAY UNDO RULEg o f = Iₓ AND f o g = Iᵧ ⇒ g = f⁻¹ (and inverse is unique!)

SHOE-SOCKS LAW(g o f)⁻¹ = f⁻¹ o g⁻¹ (Reverse order!) | (f⁻¹)⁻¹ = f

🧠 Memory Trick: UNDO button needs Bijection! If 2 inputs hit the same output (not 1-1), UNDO doesn't know which one to return; if a target is empty (not Onto), UNDO has nowhere to start!

💡 Simple Kid-Check: To find f⁻¹(y), just solve y = f(x) backwards for x! Graphically, y = f(x) and y = f⁻¹(x) are exact mirror reflections across the 45° line y = x!

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Direct Recipe (for 4-Mark Proofs): Prove f is 1-1 & Onto (OR construct candidate g(y)) → Verify (g o f)(x) = x and (f o g)(y) = y!
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Mirror Symmetry: If (a, b) lies on the graph of f, then swapped pair (b, a) lies on f⁻¹!

🧒 Classic 4-Mark NCERT Inverse: Let f: ℝ₊ → [−5, ∞) be f(x) = 9x² + 6x − 5.
Complete the square: y = (3x + 1)² − 6 ⇒ (3x + 1)² = y + 6.
Since x ≥ 0, take positive root: 3x + 1 = √(y + 6) ⇒ f⁻¹(y) = √(y + 6) − 13!
Check: g(f(x)) = √((3x+1)²) − 13 = x and f(g(y)) = y!

Watch TWO classic traps! (1) Writing (g o f)⁻¹ = g⁻¹ o f⁻¹ — WRONG! Always reverse order: (g o f)⁻¹ = f⁻¹ o g⁻¹! (2) Confusing f⁻¹(x) with reciprocal 1 / f(x) — f⁻¹ means inverse map, NEVER 1 / f(x)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Give an example of a relation on A = {1, 2, 3} which is symmetric only (neither reflexive nor transitive).

Key (1 Mark): R = {(1, 2), (2, 1)}.
• Not reflexive: (1,1) ∉ R.
• Symmetric: (1,2) ∈ R ⇔ (2,1) ∈ R.
• Not transitive: (1,2), (2,1) ∈ R, yet (1,1) ∉ R.

Q2 (Kerala HSE Say • 1 Marks): If A = {1, 2, 3, 4}, how many bijective functions can be defined from A onto itself?

Key (1 Mark): Number of bijections from an n-element set to itself = n! = 4! = 24.

Q3 (NCERT / DHSE Model • 2 Marks): Check the injectivity and surjectivity of f: ℕ → ℕ given by f(x) = x².

• Injective (1-1): f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = x₂ (since x₁, x₂ ∈ ℕ are positive). Hence One-One.
• Surjective (Onto): Take 2 ∈ ℕ (co-domain). There is no x ∈ ℕ with x² = 2. Hence NOT Onto.

Q4 (Kerala HSE March • 2 Marks): Find g o f and f o g if f(x) = |x| and g(x) = |5x − 2|.

• (g o f)(x) = g(f(x)) = g(|x|) = |5|x| − 2|.
• (f o g)(x) = f(g(x)) = f(|5x − 2|) = ||5x − 2|| = |5x − 2|.

Q5 (Kerala HSE March • 3 Marks): Show that the relation R in ℤ given by R = {(a, b) : 2 divides (a − b)} is an equivalence relation. Write the equivalence classes [0] and [1].

• Reflexive: a − a = 0 is divisible by 2 ⇒ (a, a) ∈ R.
• Symmetric: 2 | (a − b) ⇒ 2 | −(a − b) = (b − a) ⇒ (b, a) ∈ R.
• Transitive: a − b = 2k, b − c = 2m ⇒ a − c = 2(k + m) divisible by 2 ⇒ (a, c) ∈ R.
Hence Equivalence Relation! [0] = all even integers, [1] = all odd integers.

Q6 (NCERT / Kerala HSE Say • 3 Marks): Show that R = {(a, b) : |a − b| is even} on A = {1, 2, 3, 4, 5} is an equivalence relation. Show that all elements of {1, 3, 5} are related to each other.

• Reflexive: |a − a| = 0 (even).
• Symmetric: |a − b| = |b − a|.
• Transitive: If a − b and b − c are both even, their sum a − c is even.
• Since difference between any two elements of {1, 3, 5} is even (2 or 4 or 0), all elements in {1, 3, 5} are related to each other!

Q7 (Kerala HSE March • 4 Marks): Let A = ℝ − {3} and B = ℝ − {1}. Consider f: A → B defined by f(x) = (x − 2)/(x − 3). Show f is one-one and onto, and find f⁻¹.

• One-One: (x₁−2)/(x₁−3) = (x₂−2)/(x₂−3) cross-multiplies to x₁x₂ − 3x₁ − 2x₂ + 6 = x₁x₂ − 2x₁ − 3x₂ + 6 ⇒ x₁ = x₂.
• Onto: For any y ∈ ℝ − {1}, set y = (x−2)/(x−3) ⇒ x(y−1) = 3y−2 ⇒ x = (3y−2)/(y−1). Note x ≠ 3 (since 3y−2 = 3y−3 is impossible!).
• Inverse: f⁻¹(y) = 3y − 2y − 1.

Q8 (Kerala HSE Model • 4 Marks): Consider f: ℝ₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with f⁻¹(y) = (√(y + 6) − 1) / 3.

Rewrite f(x) = (3x + 1)² − 6. Define g: [−5, ∞) → ℝ₊ by g(y) = √(y+6) − 13.
• (g o f)(x) = √((3x+1)²) − 13 = (3x+1) − 13 = x (as x ≥ 0).
• (f o g)(y) = [3(√(y+6) − 13) + 1]² − 6 = (y + 6) − 6 = y.
Thus g o f = I and f o g = I ⇒ f is invertible and f⁻¹ = g!