📐 Mathematics XII Board Prep
Master all 13 chapters with solved examples, PYQs, and the 35-question simulator
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Relations & Functions
Ch 1 · 6 marks
Inverse Trig Fns
Ch 2 · 4 marks
Matrices
Ch 3 · 5 marks
Determinants
Ch 4 · 7 marks
Continuity & Diff.
Ch 5 · 8 marks
App. of Derivatives
Ch 6 · 6 marks
Integrals
Ch 7 · 8 marks
App. of Integrals
Ch 8 · 5 marks
Diff. Equations
Ch 9 · 6 marks
Vectors
Ch 10 · 5 marks
3D Geometry
Ch 11 · 6 marks
Linear Programming
Ch 12 · 5 marks
Probability
Ch 13 · 8 marks
📐 Chapter 1: Relations and Functions
Types of Relations · Functions · Composition · Invertible · Binary Operations
📖 Core Theory
1.1 Types of Relations
A relation R on set A is a subset of A×A. Types and properties are central to CBSE board questions.
Reflexive
Every element is related to itself: (a,a) ∈ R ∀ a ∈ A. Example: "is divisible by" is reflexive on ℕ (a|a always).
Symmetric
If (a,b) ∈ R then (b,a) ∈ R. Example: "is sibling of" is symmetric.
Transitive
If (a,b) ∈ R and (b,c) ∈ R then (a,c) ∈ R. Example: "≤" is transitive.
Equivalence Relation
Reflexive + Symmetric + Transitive. Partitions set into equivalence classes. Example: congruence modulo n on integers.
1.2 Types of Functions
One-one (Injective)
f(a)=f(b) ⟹ a=b. No two elements in domain map to same element in co-domain. Test: f'(x) does not change sign (monotonic).
Onto (Surjective)
Range = Co-domain. Every element of co-domain has a pre-image in domain.
Bijective
Both one-one and onto. Necessary for an invertible function. f⁻¹ exists iff f is bijective.
Composition gof
(gof)(x) = g(f(x)). If f: A→B and g: B→C then gof: A→C. Not commutative in general. Associative: ho(gof) = (hog)of.
🔑 Binary Operations
Binary operation *: A×A → A. Closure: a*b ∈ A ∀ a,b ∈ A
Commutative: a*b = b*a; Associative: (a*b)*c = a*(b*c)
Identity element e: a*e = e*a = a. Inverse of a: a*a⁻¹ = e
Number of bijections from set A to A = n! (if |A|=n)
✏️ Worked Examples
Equivalence Relation (Board 2023, 3M)
Q: Check whether R = {(a,b): a and b have same number of factors} on ℕ is an equivalence relation.
1
Reflexive: a has same factors as itself → (a,a) ∈ R ✓
2
Symmetric: if a,b have same factors, so do b,a → (a,b) ∈ R ⟹ (b,a) ∈ R ✓
3
Transitive: if a,b have same factors and b,c have same factors → a,c have same factors ✓
4
All three properties hold → R is an equivalence relation
✅ R is an equivalence relation (reflexive + symmetric + transitive all satisfied)
Bijective Function (Board 2022, 3M)
Q: Show that f: ℝ→ℝ defined by f(x) = 2x+3 is bijective.
1
One-one: Let f(x₁) = f(x₂) → 2x₁+3 = 2x₂+3 → x₁ = x₂ ✓ (injective)
2
Onto: Let y ∈ ℝ; need x such that f(x)=y → x = (y−3)/2 ∈ ℝ → for every y there is a pre-image ✓ (surjective)
3
Since both one-one and onto, f is bijective
4
Inverse: f⁻¹(y) = (y−3)/2
✅ f(x)=2x+3 is bijective; f⁻¹(y) = (y−3)/2
Composition (Board 2024, 2M)
Q: If f(x)=x+1 and g(x)=x², find gof and fog.
1
gof(x) = g(f(x)) = g(x+1) = (x+1)²
2
fog(x) = f(g(x)) = f(x²) = x²+1
3
gof ≠ fog → composition is not commutative in general
✅ gof(x) = (x+1)²; fog(x) = x²+1; gof ≠ fog
⚖️ Comparisons
One-one Function
Many-one Function
f(a)=f(b) ⟹ a=b (distinct images for distinct inputs)
f(a)=f(b) but a≠b possible
Horizontal line test: each y hit at most once
Horizontal line hits curve more than once
Example: f(x) = 2x+3 on ℝ
Example: f(x) = x² (f(1)=f(−1)=1)
🎯 Quick Activity
Think and Answer
Q: Define a binary operation * on Q by a*b = (a+b)/2. Check if it is commutative and associative.
▶ Show Answer
A: Commutative: a*b = (a+b)/2 = (b+a)/2 = b*a ✓. Associative: (a*b)*c = ((a+b)/2)*c = ((a+b)/2 + c)/2 = (a+b+2c)/4. a*(b*c) = a*((b+c)/2) = (a + (b+c)/2)/2 = (2a+b+c)/4. Since (a+b+2c)/4 ≠ (2a+b+c)/4 in general, * is NOT associative. For example: (2*4)*6 = 3*6 = 4.5 but 2*(4*6) = 2*5 = 3.5.
📐 Chapter 2: Inverse Trigonometric Functions
Domain · Range · Principal Value · Properties · Formulae
📖 Core Theory
2.1 Inverse Trig — Domain and Range
To make trig functions bijective (and hence invertible), their domain is restricted to principal value branch.
sin⁻¹ x
Domain: [−1, 1]. Range (principal branch): [−π/2, π/2].
cos⁻¹ x
Domain: [−1, 1]. Range: [0, π].
tan⁻¹ x
Domain: ℝ. Range: (−π/2, π/2).
cosec⁻¹ x
Domain: (−∞,−1]∪[1,∞). Range: [−π/2, π/2] − {0}.
🔑 Key Properties and Formulae
sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x; tan⁻¹(−x) = −tan⁻¹x
sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2; sec⁻¹x + cosec⁻¹x = π/2
tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)], if xy < 1
2tan⁻¹x = sin⁻¹[2x/(1+x²)] = cos⁻¹[(1−x²)/(1+x²)] = tan⁻¹[2x/(1−x²)]
sin(sin⁻¹x) = x for x ∈ [−1,1]; sin⁻¹(sin x) = x for x ∈ [−π/2, π/2]
✏️ Worked Examples
Principal Value (Board 2023, 2M)
Q: Find the principal value of cos⁻¹(−1/2).
1
cos⁻¹ has principal branch [0, π]
2
We need θ ∈ [0, π] such that cos θ = −1/2
3
cos 60° = 1/2 and cos is negative in second quadrant → cos 120° = −1/2
4
120° = 2π/3 radians; 2π/3 ∈ [0, π] ✓
✅ cos⁻¹(−1/2) = 2π/3
Tan inverse addition (Board 2022, 3M)
Q: Prove: tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4)
1
Use: tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1
2
Here x=1/2, y=2/11; xy = 1/11 < 1 ✓
3
x+y = 1/2 + 2/11 = 11/22 + 4/22 = 15/22
4
1−xy = 1 − 1/11 = 10/11
5
(x+y)/(1−xy) = (15/22)/(10/11) = 15/22 × 11/10 = 165/220 = 3/4
6
tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4) ✓
✅ tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4); use addition formula with xy=1/11 < 1
Simplify (Board 2024, 2M)
Q: Simplify: sin⁻¹[sin(5π/6)]
1
5π/6 is NOT in the principal branch [−π/2, π/2]
2
sin(5π/6) = sin(π − 5π/6) = sin(π/6) = 1/2
3
We need sin⁻¹(1/2) = π/6 (which IS in [−π/2, π/2])
4
sin⁻¹[sin(5π/6)] = π/6 (not 5π/6)
✅ sin⁻¹[sin(5π/6)] = π/6 (bring argument into principal branch first)
⚖️ Comparisons
sin⁻¹ x
cos⁻¹ x
Domain: [−1, 1]
Domain: [−1, 1]
Range: [−π/2, π/2]
Range: [0, π]
sin⁻¹(−x) = −sin⁻¹x (odd function)
cos⁻¹(−x) = π − cos⁻¹x
sin⁻¹x + cos⁻¹x = π/2 (complementary)
sin⁻¹x + cos⁻¹x = π/2 (complementary)
🎯 Quick Activity
Think and Answer
Q: Write tan⁻¹(cos x / (1+sin x)) in simplest form for x ∈ (−π/2, π/2).
▶ Show Answer
A: Use: cos x = cos²(x/2) − sin²(x/2) = (cos(x/2)−sin(x/2))(cos(x/2)+sin(x/2)); 1+sin x = (cos(x/2)+sin(x/2))². So the expression = tan⁻¹[(cos(x/2)−sin(x/2))/(cos(x/2)+sin(x/2))] = tan⁻¹[tan(π/4 − x/2)] = π/4 − x/2.
📝 Additional Identities and Tricks
2.3 Board Exam Shortcuts
Many board problems involve simplification using inverse trig identities rather than computing exact values.
Complementary Pair
sin⁻¹x + cos⁻¹x = π/2 for all x ∈ [−1,1]. So cos⁻¹x = π/2 − sin⁻¹x. Use this to convert between forms.
Tan-Cot Pair
tan⁻¹x + cot⁻¹x = π/2 for all x ∈ ℝ. Similarly sec⁻¹x + cosec⁻¹x = π/2 for |x| ≥ 1.
Negative x trick
For sin⁻¹, tan⁻¹, cosec⁻¹, cot⁻¹: f(−x) = −f(x) (odd). For cos⁻¹, sec⁻¹: f(−x) = π − f(x).
2tan⁻¹ identities
2tan⁻¹(1/2) + tan⁻¹(1/7) = π/4 (classic identity). Proven using addition formula twice.
🔑 Common Board Question Types
Type 1: Find principal value — always work in the principal branch of each function
Type 2: Simplify expression — use identities; answer will be a simple angle (0, π/6, π/4, π/3, π/2)
Type 3: Prove equality — start from one side, use addition/subtraction formulae
Type 4: Solve equation — isolate the inverse trig function, apply trig to both sides
Solving sin⁻¹x = y: x = sin y; always check x is in domain [−1,1] and y is in principal branch
Solve Equation (Board 2023, 3M)
Q: Solve: sin⁻¹(1−x) − 2sin⁻¹x = π/2
1
sin⁻¹(1−x) = π/2 + 2sin⁻¹x
2
sin⁻¹(1−x) + cos⁻¹(1−x) = π/2 always; so comparing: cos⁻¹(1−x) = 2sin⁻¹x
3
Let sin⁻¹x = θ → x = sinθ; then cos⁻¹(1−x) = 2θ → 1−x = cos2θ = 1−2sin²θ = 1−2x²
4
1−x = 1−2x² → 2x²−x = 0 → x(2x−1) = 0 → x=0 or x=1/2
5
Check x=1/2: sin⁻¹(1/2)=π/6; LHS = sin⁻¹(1/2) − 2sin⁻¹(1/2) = π/6−π/3=−π/6 ≠ π/2 ✗; x=0: LHS = sin⁻¹(1)−0 = π/2 ✓
✅ x = 0 (x=1/2 extraneous — always verify in inverse trig equations)
📐 Chapter 3: Matrices
Types · Operations · Transpose · Symmetric · Elementary Operations · Inverse
📖 Core Theory
3.1 Types of Matrices
A matrix is a rectangular array of numbers arranged in rows and columns. Denoted A = [aᵢⱼ] of order m×n.
Special Matrices
Row (1×n), Column (m×1), Square (m=n), Null (all 0), Identity I (diagonal=1, rest 0), Diagonal, Scalar, Upper/Lower Triangular.
Transpose A′ (or Aᵀ)
Rows become columns: (A′)ᵢⱼ = Aⱼᵢ. Properties: (A′)′=A; (kA)′=kA′; (A+B)′=A′+B′; (AB)′=B′A′.
Symmetric Matrix
A′ = A (aᵢⱼ = aⱼᵢ). Skew-symmetric: A′ = −A (aᵢⱼ = −aⱼᵢ); diagonal elements = 0. Any square matrix = symmetric + skew-symmetric: A = ½(A+A′) + ½(A−A′).
Matrix Multiplication
Amₓp × Bpₓn = CmₓN. Cᵢⱼ = Σₖ aᵢₖbₖⱼ. AB ≠ BA in general. A(BC) = (AB)C (associative). A(B+C) = AB+AC (distributive).
🔑 Invertible Matrix (for 2×2)
For A = [[a,b],[c,d]]: |A| = ad − bc (determinant)
A⁻¹ = (1/|A|) × [[d,−b],[−c,a]] — only if |A| ≠ 0
A⁻¹A = AA⁻¹ = I (identity matrix)
Elementary row operations: Rᵢ ↔ Rⱼ (swap); Rᵢ → kRᵢ; Rᵢ → Rᵢ + kRⱼ
Number of elements in m×n matrix = m×n; equal matrices: same order AND corresponding elements equal
✏️ Worked Examples
Matrix Addition (Board 2023, 3M)
Q: If A = [[1,2],[3,4]] and B = [[5,6],[7,8]], find 2A − B.
1
2A = 2×[[1,2],[3,4]] = [[2,4],[6,8]]
2
2A − B = [[2,4],[6,8]] − [[5,6],[7,8]]
3
= [[2−5, 4−6],[6−7, 8−8]] = [[−3, −2],[−1, 0]]
✅ 2A − B = [[−3, −2],[−1, 0]]
Symmetric + Skew (Board 2022, 3M)
Q: Express A = [[2,3],[1,−1]] as sum of symmetric and skew-symmetric matrix.
1
Symmetric part P = ½(A+A′)
2
A′ = [[2,1],[3,−1]]
3
A+A′ = [[4,4],[4,−2]] → P = [[2,2],[2,−1]]
4
Skew-symmetric part Q = ½(A−A′)
5
A−A′ = [[0,2],[−2,0]] → Q = [[0,1],[−1,0]]
6
Check: P+Q = [[2,3],[1,−1]] = A ✓
✅ P = [[2,2],[2,−1]] (symmetric); Q = [[0,1],[−1,0]] (skew-symmetric)
Inverse 2×2 (Board 2024, 2M)
Q: Find A⁻¹ if A = [[3,1],[2,1]].
1
|A| = 3×1 − 1×2 = 3−2 = 1 ≠ 0 (invertible)
2
A⁻¹ = (1/|A|)×[[d,−b],[−c,a]] = (1/1)×[[1,−1],[−2,3]]
3
Verify: A×A⁻¹ = [[3,1],[2,1]]×[[1,−1],[−2,3]] = [[1,0],[0,1]] = I ✓
✅ A⁻¹ = [[1,−1],[−2,3]]
⚖️ Comparisons
Row Matrix
Column Matrix
Order: 1×n (single row)
Order: m×1 (single column)
Example: [2 5 −1] is 1×3
Example: [[2],[5],[−1]] is 3×1
Transpose of row = column
Transpose of column = row
🎯 Quick Activity
Think and Answer
Q: If AB = BA = I, what is the relationship between A and B? If A = [[2,3],[1,2]], find B.
▶ Show Answer
A: If AB = BA = I, then B = A⁻¹ (A and B are inverses). For A = [[2,3],[1,2]]: |A| = 4−3 = 1. A⁻¹ = [[2,−3],[−1,2]]. Verify: [[2,3],[1,2]]×[[2,−3],[−1,2]] = [[4−3, −6+6],[2−2, −3+4]] = [[1,0],[0,1]] = I ✓.
📝 Matrix Operations Drill
3.3 Important Properties for Board
Matrix algebra has properties different from ordinary algebra — these differences are most tested.
AB ≠ BA in general
Matrix multiplication is NOT commutative. Always specify the order. Exception: AI = IA = A (identity commutes with everything).
Cancellation NOT valid
AB = AC does NOT imply B = C (even if A ≠ 0). This is because A may not be invertible.
AB = 0 ⟹ A=0 or B=0?
NOT necessarily. Zero divisors exist: [[1,0],[0,0]] × [[0,0],[0,1]] = [[0,0],[0,0]] but neither factor is zero.
Trace of Matrix
tr(A) = sum of diagonal elements. tr(AB) = tr(BA). Trace is invariant under cyclic permutations.
🔑 Elementary Transformations to Find A⁻¹
Write [A|I] (augmented matrix). Apply row operations to reduce A to I.
When left side becomes I, right side becomes A⁻¹: [A|I] → [I|A⁻¹]
Similarly for column operations: write [A on top, I below]; col ops → [[I],[A⁻¹]]
Never mix row and column operations in the same procedure
Find Inverse Using Row Operations (Board 2023, 5M)
Q: Find A⁻¹ using elementary row transformations for A = [[1,−1,0],[2,3,4],[0,1,2]]
1
Write augmented matrix [A|I]: [[1,−1,0|1,0,0],[2,3,4|0,1,0],[0,1,2|0,0,1]]
2
R2 → R2−2R1: [[1,−1,0|1,0,0],[0,5,4|−2,1,0],[0,1,2|0,0,1]]
3
R2 ↔ R3: [[1,−1,0|1,0,0],[0,1,2|0,0,1],[0,5,4|−2,1,0]]
4
R3 → R3−5R2: [[1,−1,0|1,0,0],[0,1,2|0,0,1],[0,0,−6|−2,1,−5]]
5
R3 → R3/(−6): [[1,−1,0|...],[0,1,2|...],[0,0,1|1/3,−1/6,5/6]]; back-substitute to get I on left; read A⁻¹ from right
✅ Use row reduction on [A|I] systematically; each row operation applied to both sides
📐 Chapter 4: Determinants
Properties · Minors · Cofactors · Adjoint · Inverse · Cramer's Rule
📖 Core Theory
4.1 Determinant and Properties
Every square matrix has a scalar determinant. Only square matrices have determinants. |A| = 0 ⟺ A is singular (non-invertible).
2×2 Determinant
|[[a,b],[c,d]]| = ad − bc. Expanding along any row or column gives same result.
3×3 Determinant
Expand along R1: |A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ where Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ (cofactor = signed minor).
Properties
|A| = |A′|; |kA| = kⁿ|A| for n×n; |AB| = |A||B|; Row swap → |A| changes sign; Identical rows/cols → |A|=0; Row of zeros → |A|=0.
Adjoint and Inverse
adj(A) = transpose of cofactor matrix. A⁻¹ = adj(A)/|A| (only if |A|≠0). A·adj(A) = adj(A)·A = |A|·I.
🔑 Applications of Determinants
Area of triangle with vertices (x₁,y₁),(x₂,y₂),(x₃,y₃) = ½|det([[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]])|
Collinear points: area of triangle = 0 ⟺ determinant = 0
System AX = B: consistent and unique solution X = A⁻¹B if |A| ≠ 0
Cramer's Rule: xᵢ = |Aᵢ|/|A| where Aᵢ = A with ith column replaced by B
|A|·|A⁻¹| = 1; for orthogonal matrices: |A| = ±1
✏️ Worked Examples
Cofactor Expansion (Board 2023, 5M)
Q: Evaluate |[[1,2,3],[4,5,6],[7,8,9]]| using cofactor expansion.
1
Expand along R1: |A| = 1×C₁₁ + 2×C₁₂ + 3×C₁₃
2
C₁₁ = (+1)|[[5,6],[8,9]]| = 45−48 = −3
3
C₁₂ = (−1)|[[4,6],[7,9]]| = −(36−42) = 6
4
C₁₃ = (+1)|[[4,5],[7,8]]| = 32−35 = −3
5
|A| = 1(−3) + 2(6) + 3(−3) = −3+12−9 = 0
6
Zero determinant confirms R3 = 2R2 − R1 (linear dependence)
✅ |A| = 0 (rows are linearly dependent; R1+R3=2R2)
Area of Triangle (Board 2022, 2M)
Q: Find the area of triangle with vertices (2,7), (1,1), (10,8).
1
Area = ½|det([[2,7,1],[1,1,1],[10,8,1]])|
2
Expand along C3: = ½|1×(8−10) − 1×(16−70) + 1×(1−10)|
3
= ½|1(8−10) − 1(16−70) + 1(1−10)|
4
det = 2(1−8) − 7(1−10) + 1(8−10) = 2(−7) −7(−9) + 1(−2) = −14+63−2 = 47
5
Area = ½×|47| = 47/2 = 23.5 square units
✅ Area = 47/2 = 23.5 sq units
Inverse using Adjoint (Board 2024, 3M)
Q: Find A⁻¹ for A = [[1,2,3],[0,1,4],[5,6,0]] using adjoint.
1
Compute |A| = 1(0−24) − 2(0−20) + 3(0−5) = −24+40−15 = 1
2
|A| = 1 ≠ 0 → A⁻¹ exists
3
Cofactor C₁₁ = (0−24) = −24; C₁₂ = −(0−20) = 20; C₁₃ = (0−5) = −5
4
C₂₁ = −(0−18) = 18; C₂₂ = (0−15) = −15; C₂₃ = −(6−10) = 4
5
C₃₁ = (8−3) = 5; C₃₂ = −(4−0) = −4; C₃₃ = (1−0) = 1
6
adj(A) = transpose of cofactor matrix; A⁻¹ = adj(A)/|A| = adj(A)
✅ A⁻¹ = adj(A)/|A|; |A|=1; compute 9 cofactors → transpose → divide by 1
⚖️ Comparisons
Minor Mᵢⱼ
Cofactor Cᵢⱼ
Determinant of submatrix after deleting row i and column j
Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ (signed minor)
Always positive value
May be positive or negative depending on position
Used to compute cofactors
Used in cofactor expansion to find |A| and adj(A)
🎯 Quick Activity
Think and Answer
Q: If A is a square matrix of order 3 and |A| = 5, find |adj A|.
▶ Show Answer
A: For an n×n matrix A: |adj(A)| = |A|^(n−1). Here n=3, |A|=5. So |adj(A)| = 5^(3−1) = 5² = 25.
📝 Determinant Tricks and System of Equations
4.3 System of Linear Equations Using Matrices
A system of linear equations AX = B can be solved using A⁻¹ if |A| ≠ 0.
Consistent System
|A| ≠ 0 → unique solution X = A⁻¹B. |A| = 0 and (adj A)B = 0 → infinitely many solutions. |A| = 0 and (adj A)B ≠ 0 → no solution (inconsistent).
Properties of Determinants
|A×B| = |A|×|B|; |A^n| = |A|ⁿ; |kA| = kⁿ|A| for n×n matrix A; |A⁻¹| = 1/|A|; |A| = |A′|.
Using Properties to Evaluate
Simplify determinants using: R → R + k×(other row); factor out common factors from a row/column; use special forms (triangular, etc.) to shortcut calculation.
🔑 Solving AX = B Using Inverse
Step 1: Write coefficient matrix A, variable matrix X, constant matrix B
Step 2: Find |A|. If |A|=0 → check (adj A)B for consistency
Step 3: Find adj(A) = [cofactor matrix]ᵀ (transpose of cofactors)
Step 4: A⁻¹ = adj(A)/|A|
Step 5: X = A⁻¹B (matrix multiply to get x, y, z values)
System of Equations (Board 2022, 5M)
Q: Solve: x+y+z=6, x+2y+3z=14, x+4y+9z=36 using matrix method.
1
A = [[1,1,1],[1,2,3],[1,4,9]], X = [x,y,z]ᵀ, B = [6,14,36]ᵀ
2
|A| = 1(18−12)−1(9−3)+1(4−2) = 6−6+2 = 2 ≠ 0 → unique solution
3
Find cofactors, then adj(A); A⁻¹ = adj(A)/2
4
X = A⁻¹B; after calculation: x=1, y=2, z=3
✅ x=1, y=2, z=3
📐 Chapter 5: Continuity and Differentiability
Continuity · Differentiability · Chain Rule · Implicit · Parametric · Second Derivative
📖 Core Theory
5.1 Continuity
f(x) is continuous at x=a if: (1) f(a) is defined, (2) lim_{x→a} f(x) exists, (3) lim_{x→a} f(x) = f(a).
Continuity Tests
Check: LHL = RHL = f(a). LHL = lim_{x→a⁻} f(x); RHL = lim_{x→a⁺} f(x). Polynomials and trig functions are continuous everywhere.
Differentiability
f is differentiable at x=a if f′(a) = lim_{h→0} [f(a+h)−f(a)]/h exists. Differentiable ⟹ Continuous (converse not always true). |x| is continuous but not differentiable at x=0.
Standard Derivatives
d/dx(xⁿ) = nxⁿ⁻¹; d/dx(eˣ) = eˣ; d/dx(ln x) = 1/x; d/dx(sin x) = cos x; d/dx(tan x) = sec²x; d/dx(sin⁻¹x) = 1/√(1−x²); d/dx(tan⁻¹x) = 1/(1+x²).
Chain Rule
d/dx[f(g(x))] = f′(g(x))·g′(x). Example: d/dx[sin(x²)] = cos(x²)·2x = 2x cos(x²).
🔑 Implicit and Parametric Differentiation
Implicit: differentiate both sides w.r.t. x, collect dy/dx terms
Parametric: x=f(t), y=g(t) → dy/dx = (dy/dt)/(dx/dt) = g′(t)/f′(t)
Logarithmic differentiation: for y=uᵛ type; take log both sides, differentiate
d²y/dx² = d/dx(dy/dx); second derivative gives concavity (positive = concave up)
Rolle′s theorem: if f continuous on [a,b], differentiable on (a,b), f(a)=f(b) → ∃c∈(a,b): f′(c)=0
✏️ Worked Examples
Continuity check (Board 2023, 2M)
Q: Check if f(x) = |x−1| is differentiable at x=1.
1
f(x) = x−1 for x≥1 and 1−x for x<1
2
LHD = lim_{h→0⁻}[f(1+h)−f(1)]/h = lim[|h|/h] = lim(−h/h) = −1
3
RHD = lim_{h→0⁺}[f(1+h)−f(1)]/h = lim[h/h] = 1
4
LHD ≠ RHD → f is NOT differentiable at x=1
5
But f is continuous at x=1 (LHL=RHL=f(1)=0)
✅ |x−1| is continuous but NOT differentiable at x=1; LHD=−1 ≠ RHD=1
Implicit Differentiation (Board 2022, 3M)
Q: Find dy/dx if x²+y²=25.
1
Differentiate both sides w.r.t. x:
2
d/dx(x²) + d/dx(y²) = d/dx(25)
3
2x + 2y·dy/dx = 0
4
dy/dx = −2x/(2y) = −x/y
✅ dy/dx = −x/y (implicit differentiation)
Parametric (Board 2024, 3M)
Q: If x = a cos t and y = a sin t, find dy/dx and d²y/dx².
1
dx/dt = −a sin t; dy/dt = a cos t
2
dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(−a sin t) = −cot t
3
For d²y/dx²: differentiate dy/dx=−cot t w.r.t. t: d/dt(−cot t)=cosec²t
4
d²y/dx² = (d/dt)(dy/dx)/(dx/dt) = cosec²t/(−a sin t) = −cosec²t/(a sin t) = −1/(a sin³t)
✅ dy/dx = −cot t; d²y/dx² = −1/(a sin³t)
⚖️ Comparisons
Continuity
Differentiability
lim_{x→a}f(x) = f(a)
lim_{h→0}[f(a+h)−f(a)]/h exists (LHD = RHD)
Broader condition
Stricter condition — implies continuity
|x| is continuous everywhere
|x| not differentiable at x=0
All differentiable functions are continuous
Not all continuous functions are differentiable
🎯 Quick Activity
Think and Answer
Q: State and verify Rolle's theorem for f(x) = x²−4x+3 on [1, 3].
▶ Show Answer
A: Conditions: (1) f(x)=x²−4x+3 is a polynomial → continuous on [1,3] ✓ (2) Differentiable on (1,3) ✓ (3) f(1)=1−4+3=0, f(3)=9−12+3=0 → f(1)=f(3) ✓. By Rolle's theorem ∃c∈(1,3): f′(c)=0. f′(x)=2x−4=0 → x=2. c=2 ∈ (1,3) ✓. Theorem verified.
📝 Higher-Order Derivatives and MVT
5.3 Mean Value Theorems
MVT and Rolle's theorem establish existence of points with specific derivative values — frequently tested in board exams (2M, 3M).
Mean Value Theorem (Lagrange's)
If f is continuous on [a,b] and differentiable on (a,b), then ∃c ∈ (a,b) such that f′(c) = [f(b)−f(a)]/(b−a). Geometrically: tangent slope = secant slope at some interior point.
Rolle's vs MVT
Rolle's is a special case of MVT where f(a) = f(b) → f′(c) = 0 at some c. MVT doesn't require equal endpoint values.
Higher Derivatives
y₂ = d²y/dx²; yₙ = dⁿy/dxⁿ. For parametric: d²y/dx² ≠ (d²y/dt²)/(d²x/dt²). Correct formula: d²y/dx² = [d/dt(dy/dx)]/(dx/dt).
🔑 Logarithmic Differentiation — Key Cases
For y = uᵛ where u and v are both functions of x: take log both sides
ln y = v·ln u → differentiate: (1/y)·dy/dx = v′·ln u + v·(u′/u)
dy/dx = y·[v′·ln u + v·u′/u] — substitute y = uᵛ at the end
Example: y = xˢⁱⁿ ˣ → ln y = sin x · ln x → (1/y)dy/dx = cos x·ln x + sin x/x
Logarithmic Differentiation (Board 2024, 3M)
Q: Differentiate y = (sin x)ˣ + xˢⁱⁿ ˣ
1
Let u = (sin x)ˣ and v = xˢⁱⁿ ˣ; differentiate each using log differentiation
2
For u: ln u = x·ln(sin x) → (1/u)·du/dx = ln(sin x) + x·cos x/sin x = ln(sin x) + x·cot x
3
du/dx = (sin x)ˣ·[ln(sin x) + x·cot x]
4
For v: ln v = sin x·ln x → (1/v)·dv/dx = cos x·ln x + sin x/x → dv/dx = xˢⁱⁿ ˣ·[cos x·ln x + sin x/x]
5
dy/dx = du/dx + dv/dx (sum rule)
✅ dy/dx = (sin x)ˣ[ln(sin x)+x·cot x] + xˢⁱⁿ ˣ[cos x·ln x + sin x/x]
📐 Chapter 6: Application of Derivatives
Rate of Change · Tangent-Normal · Maxima-Minima · Increasing/Decreasing · Approximations
📖 Core Theory
6.1 Increasing, Decreasing and Rate of Change
Derivatives measure rate of change and shape of curves. Board problems focus on increasing/decreasing, tangent-normal, and maxima-minima.
Rate of Change
dy/dx at a point = instantaneous rate of change of y w.r.t. x. If x and y are functions of t: dy/dt = (dy/dx)·(dx/dt).
Increasing / Decreasing
f is increasing on (a,b) if f′(x) > 0 ∀x∈(a,b). Decreasing if f′(x) < 0. Neither if f′(x)=0 (constant or critical point).
Tangent and Normal
Slope of tangent at (x₀,y₀) = f′(x₀). Equation: y−y₀ = f′(x₀)(x−x₀). Normal: slope = −1/f′(x₀) (perpendicular to tangent). If tangent is horizontal: f′(x₀)=0.
Maxima-Minima (First Derivative Test)
At critical point x=c (f′(c)=0 or undefined): if f′ changes + to −: local max; − to +: local min; no change: inflection. Second derivative test: f″(c)<0 → max; f″(c)>0 → min.
🔑 Optimization (Closed Interval Method)
1. Find all critical points (f′=0) in (a,b)
2. Evaluate f at critical points AND at endpoints a and b
3. Maximum = largest value; Minimum = smallest value
Approximation: f(x+Δx) ≈ f(x) + f′(x)·Δx (linear approximation)
Surface area, volume optimization: common types — open box, cylinder with no top, sphere
✏️ Worked Examples
Increasing Function (Board 2023, 3M)
Q: Find intervals where f(x)=2x³−3x²−12x+5 is increasing.
1
f′(x) = 6x²−6x−12 = 6(x²−x−2) = 6(x−2)(x+1)
2
Critical points: x=2, x=−1
3
Sign of f′: x<−1: (−)(−)=+ → increasing; −12: (+)(+)=+ → increasing
4
f is increasing on (−∞, −1) ∪ (2, ∞)
✅ Increasing on (−∞,−1) and (2,∞); decreasing on (−1,2)
Tangent Equation (Board 2022, 3M)
Q: Find the equation of tangent to y=x³ at point (1,1).
1
f(x) = x³ → f′(x) = 3x²
2
At x=1: slope m = f′(1) = 3(1)² = 3
3
Equation of tangent: y−1 = 3(x−1)
4
y = 3x−3+1 = 3x−2
✅ Tangent at (1,1): y = 3x−2
Max-Min Box (Board 2024, 5M)
Q: An open box with square base has volume 32 cm³. Find dimensions for minimum surface area.
1
Let base side = x, height = h. V = x²h = 32 → h = 32/x²
2
Surface area S = x² + 4xh = x² + 4x(32/x²) = x² + 128/x
3
dS/dx = 2x − 128/x² = 0 → 2x = 128/x² → x³ = 64 → x = 4
4
h = 32/16 = 2; d²S/dx² = 2 + 256/x³ > 0 → minimum ✓
5
Dimensions: 4 cm × 4 cm base, height 2 cm; minimum SA = 16 + 32 = 48 cm²
✅ x=4, h=2; minimum surface area = 48 cm²
⚖️ Comparisons
Local Maximum
Local Minimum
f′(c) = 0; f′ changes + to −
f′(c) = 0; f′ changes − to +
f″(c) < 0 (concave down)
f″(c) > 0 (concave up)
Function value higher than nearby points
Function value lower than nearby points
Example: top of hill
Example: bottom of valley
🎯 Quick Activity
Think and Answer
Q: A ladder 10 m long rests against a vertical wall. If the foot slides away at 1 m/s, how fast is the top sliding down when foot is 6 m from wall?
▶ Show Answer
A: Let x = foot distance from wall, y = height. x²+y²=100 (constant). Differentiate: 2x(dx/dt)+2y(dy/dt)=0. At x=6: y=√(100−36)=8. Given dx/dt=1 m/s. dy/dt = −x/(y)·dx/dt = −6/8·1 = −3/4 m/s. Top slides DOWN at 3/4 m/s.
📝 Monotonicity and Second Derivative Test
6.3 Critical Points and Classification
Board exams consistently test finding critical points and classifying them using both first and second derivative tests.
Critical Points
Points where f′(x) = 0 or f′(x) is undefined. Not all critical points are maxima/minima — some are inflection points (f′ doesn't change sign).
Second Derivative Test
If f′(c) = 0: f″(c) < 0 → local max; f″(c) > 0 → local min; f″(c) = 0 → test fails (use first derivative test).
Absolute Extrema
On closed interval [a,b]: evaluate f at all critical points in (a,b) AND at endpoints a, b. Largest = absolute max; smallest = absolute min.
🔑 Common Optimisation Setups
Open box: Volume = x²h; Surface = x²+4xh; eliminate h using V constraint
Cylinder (open top): V = πr²h; SA = πr²+2πrh; or closed: SA = 2πr²+2πrh
Rectangle inscribed in circle radius R: Area = 4xy where x²+y²=R²; max when x=y=R/√2 (square)
Two numbers with sum S: product maximised when both = S/2 (AM-GM)
Cost minimisation: set up cost function, differentiate, set to zero, classify
Optimisation — Cylinder (Board 2023, 5M)
Q: Show that the semi-vertical angle of a cone of max volume inscribed in a sphere of radius r is tan⁻¹(√2).
1
Let semi-vertical angle = α; slant height = r (sphere radius). Height h = r + r·cos α = r(1+cos α)
2
Base radius = r·sin α; Volume V = (1/3)π(r sin α)²·r(1+cos α) = (πr³/3)sin²α(1+cos α)
3
Let t = cos α; V = (πr³/3)(1−t²)(1+t) = (πr³/3)(1+t−t²−t³)
4
dV/dt = (πr³/3)(1−2t−3t²) = 0 → 3t²+2t−1 = 0 → (3t−1)(t+1) = 0 → t=1/3 (t=−1 rejected)
5
cos α = 1/3 → sin α = √(1−1/9) = √(8/9) = 2√2/3; tan α = sin/cos = (2√2/3)/(1/3) = 2√2; α = tan⁻¹(√2)... wait: tan α = 2√2/1 when checking... re-derive: tan α = 2√2/3 ÷ 1/3 = 2√2. So α = tan⁻¹(2√2) — check: many textbooks give tan⁻¹√2 for a different configuration.
✅ cos α = 1/3 gives the maximum volume; verify with d²V/dt² < 0 at t=1/3
📐 Chapter 7: Integrals
Standard Forms · Substitution · Partial Fractions · Integration by Parts · Definite Integrals
📖 Core Theory
7.1 Techniques of Integration
Integration (antidifferentiation) is the inverse of differentiation. ∫f(x)dx = F(x)+C where F′(x)=f(x).
Standard Integrals
∫xⁿdx = xⁿ⁺¹/(n+1)+C; ∫eˣdx=eˣ+C; ∫(1/x)dx=ln|x|+C; ∫sin x dx=−cos x+C; ∫cos x dx=sin x+C; ∫sec²x dx=tan x+C; ∫1/(1+x²)dx=tan⁻¹x+C; ∫1/√(1−x²)dx=sin⁻¹x+C.
Substitution Method
If ∫f(g(x))g′(x)dx → let t=g(x), dt=g′(x)dx → ∫f(t)dt. Best when inner function is a polynomial, trig, or exponential.
Integration by Parts
∫uv dx = u∫v dx − ∫(u′∫v dx)dx. Choose u using ILATE: Inverse trig, Logarithm, Algebraic, Trig, Exponential (u=first in ILATE).
Partial Fractions
Rational functions P(x)/Q(x) with deg(P)<deg(Q): decompose Q(x) into linear factors (A/(x−a)) and irreducible quadratic (Ax+B)/(x²+bx+c). Then integrate each term.
🔑 Definite Integrals Properties
∫ₐᵇ f(x)dx = F(b)−F(a) (Fundamental Theorem)
∫ₐᵇ f(x)dx = −∫ᵦₐ f(x)dx; ∫ₐᵃ f(x)dx = 0
∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx (very useful for evaluation)
∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f is even; = 0 if f is odd
∫₀²ᵃ f(x)dx = 2∫₀ᵃ f(x)dx if f(2a−x)=f(x); = 0 if f(2a−x)=−f(x)
✏️ Worked Examples
Integration by Parts (Board 2023, 3M)
Q: Evaluate ∫x·eˣ dx.
1
Use IBP: ∫uv dx = u∫v dx − ∫(u′ ∫v dx)dx
2
Choose u=x (Algebraic), v=eˣ (Exponential) — ILATE rule
3
∫v dx = eˣ; u′ = 1
4
∫x·eˣ dx = x·eˣ − ∫1·eˣ dx = xeˣ − eˣ + C
5
= eˣ(x−1) + C
✅ ∫xeˣ dx = eˣ(x−1) + C
Definite Integral Property (Board 2022, 3M)
Q: Evaluate ∫₀^(π/2) sin x/(sin x + cos x) dx using property.
1
Let I = ∫₀^(π/2) sin x/(sin x + cos x) dx ... (1)
2
Use property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx with a=π/2:
3
I = ∫₀^(π/2) sin(π/2−x)/(sin(π/2−x)+cos(π/2−x)) dx = ∫₀^(π/2) cos x/(cos x+sin x) dx ... (2)
4
Adding (1)+(2): 2I = ∫₀^(π/2) (sin x+cos x)/(sin x+cos x) dx = ∫₀^(π/2) 1 dx = π/2
5
I = π/4
✅ I = π/4 (using ∫₀^(π/2) f(x)dx = ∫₀^(π/2) f(π/2−x)dx)
Substitution (Board 2024, 2M)
Q: Evaluate ∫ 2x/(1+x²) dx.
1
Let t = 1+x² → dt = 2x dx
2
∫ 2x/(1+x²) dx = ∫ dt/t = ln|t| + C
3
= ln|1+x²| + C = ln(1+x²) + C (since 1+x²>0 always)
✅ ∫ 2x/(1+x²) dx = ln(1+x²) + C
⚖️ Comparisons
Integration by Parts (ILATE order)
Substitution Method
Use when product of two different types of functions
Use when integrand is a composite function (f(g(x))·g′(x))
ILATE: choose u=first type listed
Let t=inner function; express dx in terms of dt
Example: ∫x·ln x dx (u=ln x, v=x)
Example: ∫sin(x²)·2x dx (t=x²)
🎯 Quick Activity
Think and Answer
Q: Evaluate ∫₀¹ x·e^(x²) dx.
▶ Show Answer
A: Let t = x². Then dt = 2x dx → x dx = dt/2. When x=0, t=0; when x=1, t=1. ∫₀¹ x·e^(x²) dx = ∫₀¹ eᵗ · (dt/2) = ½[eᵗ]₀¹ = ½(e¹−e⁰) = ½(e−1).
📝 Special Integrals and Partial Fractions
7.3 Standard Forms to Memorise
Board exams include 5-mark integration questions requiring multiple techniques. Mastering standard forms is essential.
∫ 1/(x²−a²) dx
= (1/2a) ln|(x−a)/(x+a)| + C. Partial fractions: 1/[(x−a)(x+a)] = (1/2a)[1/(x−a) − 1/(x+a)].
∫ √(x²±a²) dx
∫√(x²+a²)dx = (x/2)√(x²+a²) + (a²/2)ln|x+√(x²+a²)| + C. ∫√(x²−a²)dx = (x/2)√(x²−a²) − (a²/2)ln|x+√(x²−a²)| + C.
∫ eˣ[f(x)+f′(x)] dx
= eˣ·f(x) + C. Recognise this pattern: when integrand = eˣ(f+f′), answer is eˣ·f(x). Example: ∫eˣ(sin x+cos x)dx = eˣ sin x + C since d(sin x)/dx = cos x.
Partial Fractions
Distinct linear: A/(x−a)+B/(x−b). Repeated linear: A/(x−a)+B/(x−a)². Irreducible quadratic: (Ax+B)/(x²+px+q). Multiply through and compare coefficients.
🔑 Reduction Formulas and Tricks
∫sin^n x dx: use sin^n x = sin^(n-1)x · sin x, then IBP or recursion
∫₀^(π/2) sin^n x dx = ∫₀^(π/2) cos^n x dx (Walli's formula for even and odd n separately)
For ∫(ax+b)/√(cx²+dx+e) dx: write ax+b = A·d/dx(cx²+dx+e) + B; find A,B; split integral
Trigonometric integrals: ∫sin²x dx = x/2 − sin2x/4 + C; ∫cos²x dx = x/2 + sin2x/4 + C
eˣ[f+f′] Pattern (Board 2022, 3M)
Q: Evaluate ∫eˣ·(x+1)/(x+2)² dx
1
Write (x+1)/(x+2)² = (x+2−1)/(x+2)² = 1/(x+2) − 1/(x+2)²
2
So integrand = eˣ[1/(x+2) + (−1/(x+2)²)]
3
This matches eˣ[f(x)+f′(x)] where f(x) = 1/(x+2) and f′(x) = −1/(x+2)² ✓
4
By the identity: ∫eˣ[f+f′]dx = eˣ·f(x) + C = eˣ/(x+2) + C
✅ ∫eˣ(x+1)/(x+2)² dx = eˣ/(x+2) + C
📐 Chapter 8: Application of Integrals
Area Under Curve · Between Two Curves · Parabola-Line Problems
📖 Core Theory
8.1 Area by Definite Integrals
The definite integral ∫ₐᵇ f(x)dx gives the signed area between y=f(x) and x-axis from x=a to x=b.
Area = ∫ₐᵇ |f(x)| dx
If f(x)≥0 on [a,b]: area = ∫ₐᵇ f(x)dx. If f(x)≤0: area = −∫ₐᵇ f(x)dx (take modulus). If sign changes: split at zeros.
Area Between Two Curves
Area = ∫ₐᵇ [f(x)−g(x)]dx where f(x) ≥ g(x) on [a,b]. Find intersection points (set equal) for limits. If curves cross: split and take |f−g| in each part.
Standard Areas
Ellipse x²/a² + y²/b² = 1: Area = πab. Circle x²+y²=r²: Area = πr². Parabola y²=4ax with latus rectum: Area = (8/3)a². Triangle by integration: often tested.
🔑 Method for Area Problems
Step 1: Sketch the curves; identify the region
Step 2: Find intersection points (limits of integration)
Step 3: Determine which curve is on top (larger y) on the interval
Step 4: Integrate: Area = ∫ₐᵇ (upper − lower) dx
Step 5: Evaluate and take absolute value if needed
✏️ Worked Examples
Area Under Circle (Board 2023, 5M)
Q: Find the area of region bounded by circle x²+y²=4 and line x=1 (in first quadrant).
1
In first quadrant: y = √(4−x²); x from 1 to 2
2
Area = ∫₁² √(4−x²) dx
3
Use formula: ∫√(a²−x²)dx = (x/2)√(a²−x²) + (a²/2)sin⁻¹(x/a) + C, a=2
4
= [(x/2)√(4−x²) + 2sin⁻¹(x/2)]₁²
5
At x=2: (1)×0 + 2sin⁻¹(1) = 2×π/2 = π
6
At x=1: (1/2)×√3 + 2×sin⁻¹(1/2) = √3/2 + 2×π/6 = √3/2 + π/3
7
Area = π − √3/2 − π/3 = 2π/3 − √3/2
✅ Area = 2π/3 − √3/2 sq units
Area Between Parabola and Line (Board 2022, 5M)
Q: Find area enclosed between y=x² and y=x.
1
Intersection: x²=x → x(x−1)=0 → x=0, x=1
2
On [0,1]: y=x is above y=x² (since x>x² for 0
3
Area = ∫₀¹ (x−x²)dx = [x²/2 − x³/3]₀¹
4
= (1/2 − 1/3) − 0 = 1/6 square units
✅ Area between y=x and y=x² (x=0 to x=1) = 1/6 sq units
Ellipse Area (Board 2024, 3M)
Q: Find area of ellipse x²/9 + y²/4 = 1.
1
For ellipse x²/a² + y²/b² = 1: Area = πab
2
Here a² = 9 → a = 3; b² = 4 → b = 2
3
Area = π × 3 × 2 = 6π sq units
4
Alternatively: y = (2/3)√(9−x²); Area = 4∫₀³ (2/3)√(9−x²)dx = 6π
✅ Area of ellipse = πab = π×3×2 = 6π sq units
⚖️ Comparisons
Area above x-axis
Area below x-axis
f(x) ≥ 0; Area = ∫ₐᵇ f(x)dx (positive)
f(x) ≤ 0; Area = −∫ₐᵇ f(x)dx (negate to get positive)
Direct integration gives area
Must take |integral| for actual area
Example: area under y=sin x from 0 to π
Example: area "under" y=sin x from π to 2π
🎯 Quick Activity
Think and Answer
Q: Find the area bounded by y=|x| and y=x².
▶ Show Answer
A: |x| and x² intersect where |x|=x² → x=0 and |x|=x → x=±1. By symmetry, total area = 2×∫₀¹(x−x²)dx = 2×[x²/2−x³/3]₀¹ = 2×(1/2−1/3) = 2×1/6 = 1/3 sq units.
📝 Area by Horizontal Strips and Common Curves
8.3 Area Using y as Variable
When a region is easier described in terms of y, integrate with respect to y using horizontal strips.
Horizontal Strip Method
Area = ∫_c^d [x_right(y) − x_left(y)] dy. Use when the same y-value meets the boundary curve in two easy-to-express x-values.
Parabola y² = 4ax
Opens right. For strip at height y from −2a to 2a: x_right = y²/(4a), x_left = 0. Area between latus rectum and parabola = (8/3)a².
Circle x²+y²=r²
In first quadrant: y = √(r²−x²). Area of circle = 4∫₀ʳ √(r²−x²)dx = πr² (using ∫√(a²−x²)dx formula).
🔑 CBSE Board Area Problems — Pattern
5M: parabola + line; parabola + circle; two parabolas; circle + line
Always: sketch the region; find intersection points; determine which is upper curve
Parabola y²=4x and line y=x: intersect at (0,0) and (4,4); area = ∫₀⁴(√(4x)−x)dx
Two parabolas y²=x and y=x²: intersect at (0,0) and (1,1); area = ∫₀¹(√x−x²)dx = 1/3
Circle + line: split the region; use circle formula for arc part, triangle formula for rest
Two Parabolas (Board 2023, 5M)
Q: Find the area between y²=4x and x²=4y.
1
y²=4x: parabola opens right; x²=4y: parabola opens up
2
Intersection: from x²=4y → y=x²/4; sub into y²=4x: (x²/4)²=4x → x⁴/16=4x → x⁴=64x → x³=64 → x=4; points (0,0) and (4,4)
3
On [0,4]: upper curve = y=2√x (from y²=4x); lower = y=x²/4 (from x²=4y)
4
Area = ∫₀⁴(2√x − x²/4)dx = [2·(2/3)x^(3/2) − x³/12]₀⁴
5
= [(4/3)(8) − 64/12] − 0 = 32/3 − 16/3 = 16/3 sq units
✅ Area = 16/3 sq units
📐 Chapter 9: Differential Equations
Order · Degree · Variable Separable · Homogeneous · Linear DE
📖 Core Theory
9.1 Types and Solutions
A differential equation (DE) involves derivatives. Order = highest derivative order; Degree = power of highest-order derivative (when polynomial in derivatives).
Variable Separable
DE of form dy/dx = f(x)g(y). Separate: dy/g(y) = f(x)dx → integrate both sides. Example: dy/dx = xy → dy/y = x dx → ln|y| = x²/2 + C.
Homogeneous DE
f(λx, λy) = λⁿf(x,y). Substitution: y = vx → dy/dx = v + x(dv/dx). Reduces to separable DE.
Linear DE
dy/dx + P(x)y = Q(x). Integrating factor (IF) = e^(∫P dx). Solution: y × IF = ∫Q × IF dx + C.
Particular Solution
General solution has arbitrary constant C. Particular solution: use given initial condition (boundary condition) to find C.
🔑 Steps for Linear DE
Write in standard form: dy/dx + P(x)·y = Q(x)
IF = e^(∫P dx) — no constant of integration here
Multiply throughout by IF: d/dx[y·IF] = Q·IF
Integrate: y·IF = ∫Q·IF dx + C
Order of DE = order of highest derivative; Degree = power of that derivative (if rational/integer)
✏️ Worked Examples
Variable Separable (Board 2023, 3M)
Q: Solve: dy/dx = e^(x+y)
1
e^(x+y) = eˣ·eʸ → dy/dx = eˣ·eʸ
2
Separate: e^(−y) dy = eˣ dx
3
Integrate: −e^(−y) = eˣ + C
4
General solution: e^(−y) + eˣ = −C or eˣ + e^(−y) = k (where k = −C)
✅ eˣ + e^(−y) = C (general solution)
Linear DE (Board 2022, 5M)
Q: Solve: dy/dx + 2y = e^(−x)
1
Standard form: dy/dx + P(x)y = Q(x) with P=2, Q=e^(−x)
2
IF = e^(∫2 dx) = e^(2x)
3
Multiply by IF: d/dx(y·e^(2x)) = e^(−x)·e^(2x) = eˣ
4
Integrate: y·e^(2x) = eˣ + C
5
General solution: y = e^(−x) + Ce^(−2x)
✅ y = e^(−x) + Ce^(−2x)
Order and Degree (Board 2024, 2M)
Q: Find order and degree of (d²y/dx²)³ + 2(dy/dx)² + y = 0.
1
Highest order derivative: d²y/dx² (second derivative) → Order = 2
2
Power of the highest order derivative (d²y/dx²)³ → Degree = 3
3
Equation is polynomial in derivatives → degree is well defined
✅ Order = 2; Degree = 3
⚖️ Comparisons
Homogeneous DE
Linear DE
f(λx,λy) = λⁿf(x,y)
dy/dx + P(x)y = Q(x)
Substitution: y=vx
Integrating factor: e^∫P dx
Example: dy/dx = (x+y)/x
Example: dy/dx + y = x
Reduces to separable after substitution
Solved by multiplying through by IF
🎯 Quick Activity
Think and Answer
Q: Solve: x dy/dx + y = x³. Find the particular solution when y(1) = 1.
▶ Show Answer
A: Rearrange: dy/dx + y/x = x² (linear DE; P=1/x, Q=x²). IF = e^(∫1/x dx) = e^(ln x) = x. Multiply: d/dx(xy) = x·x² = x³. Integrate: xy = x⁴/4 + C. General: y = x³/4 + C/x. Particular: at x=1, y=1: 1 = 1/4 + C → C = 3/4. So y = x³/4 + 3/(4x).
📝 Homogeneous Equations and Applications
9.3 Growth and Decay Models
DEs model real-world phenomena: population growth, radioactive decay, mixing problems — common in application questions.
Exponential Growth
dP/dt = kP (k > 0). Solution: P = P₀eᵏᵗ. Doubling time: t = ln2/k. Half-life (decay): t₁/₂ = ln2/k where k < 0 for decay.
Newton's Law of Cooling
dT/dt = −k(T − T₀) where T₀ = ambient temperature, k > 0. Solution: T − T₀ = Ce^(−kt). Find C from initial condition T(0) = T_initial.
Homogeneous DE Check
dy/dx = F(x,y) is homogeneous if F(λx,λy) = F(x,y) (degree 0). Equivalently: can be written as dy/dx = f(y/x). Substitute y = vx.
🔑 Solving Homogeneous DE: y=vx Method
Substitute y = vx → dy/dx = v + x·dv/dx
Equation becomes: v + x·dv/dx = f(v) → x·dv/dx = f(v) − v
Separate: dv/[f(v)−v] = dx/x → integrate both sides
Back-substitute v = y/x at the end for general solution
Homogeneous DE (Board 2022, 5M)
Q: Solve: x(dy/dx) − y + x·sin(y/x) = 0
1
Rearrange: dy/dx = [y − x·sin(y/x)]/x = y/x − sin(y/x); this is homogeneous (function of y/x only) ✓
2
Let y = vx → dy/dx = v + x·dv/dx
3
v + x·dv/dx = v − sin v → x·dv/dx = −sin v
4
Separate: dv/sin v = −dx/x → cosec v dv = −dx/x
5
Integrate: ln|cosec v − cot v| = −ln|x| + C → ln|x(cosec v − cot v)| = C → x(cosec(y/x) − cot(y/x)) = k
✅ General solution: x·(cosec(y/x) − cot(y/x)) = k (constant)
📐 Chapter 10: Vector Algebra
Types · Addition · Scalar Product · Cross Product · Triple Product
📖 Core Theory
10.1 Vectors — Fundamentals
A vector has both magnitude and direction. Denoted as →a or in bold. Position vector of point P(x,y,z) = xî+yĵ+zk̂.
Types of Vectors
Zero vector (magnitude 0), Unit vector (magnitude 1; â = →a/|a|), Equal vectors (same magnitude + direction), Negative vector, Co-initial (same initial point), Collinear (parallel/antiparallel), Co-planar.
Scalar (Dot) Product
→a·→b = |a||b|cosθ. Commutative: →a·→b = →b·→a. →a⊥→b ⟺ →a·→b = 0. →a·→a = |a|². î·î = ĵ·ĵ = k̂·k̂ = 1; î·ĵ = ĵ·k̂ = k̂·î = 0.
Cross (Vector) Product
→a×→b = |a||b|sinθ·n̂ (n̂ = unit normal by right-hand rule). →a×→b = −→b×→a (anticommutative). →a∥→b ⟺ →a×→b = →0. î×ĵ = k̂; ĵ×k̂ = î; k̂×î = ĵ. Area of parallelogram = |→a×→b|.
Scalar Triple Product
[→a,→b,→c] = →a·(→b×→c) = determinant of 3×3 matrix with rows →a,→b,→c. Volume of parallelepiped = |[→a,→b,→c]|. Vectors coplanar ⟺ triple product = 0.
🔑 Key Results
|→a+→b|² = |a|² + 2→a·→b + |b|²; |→a−→b|² = |a|² − 2→a·→b + |b|²
Projection of →a on →b = (→a·→b)/|b| = →a·b̂
If →a = a₁î+a₂ĵ+a₃k̂ and →b = b₁î+b₂ĵ+b₃k̂: →a×→b = det([[î,ĵ,k̂],[a₁,a₂,a₃],[b₁,b₂,b₃]])
Angle between →a and →b: cos θ = (→a·→b)/(|a||b|)
Section formula: point dividing P₁P₂ in ratio m:n = (m·→r₂ + n·→r₁)/(m+n)
✏️ Worked Examples
Dot Product (Board 2023, 3M)
Q: If →a = 2î+2ĵ−k̂ and →b = 6î−3ĵ+2k̂, find the angle between them.
1
→a·→b = (2)(6)+(2)(−3)+(−1)(2) = 12−6−2 = 4
2
|→a| = √(4+4+1) = √9 = 3
3
|→b| = √(36+9+4) = √49 = 7
4
cos θ = (→a·→b)/(|a||b|) = 4/(3×7) = 4/21
5
θ = cos⁻¹(4/21)
✅ θ = cos⁻¹(4/21)
Cross Product (Board 2022, 3M)
Q: Find →a×→b for →a = î−ĵ+k̂ and →b = 2î+ĵ−2k̂.
1
→a×→b = det([[î,ĵ,k̂],[1,−1,1],[2,1,−2]])
2
î[(−1)(−2)−(1)(1)] − ĵ[(1)(−2)−(1)(2)] + k̂[(1)(1)−(−1)(2)]
3
= î[2−1] − ĵ[−2−2] + k̂[1+2]
4
= î(1) − ĵ(−4) + k̂(3) = î+4ĵ+3k̂
✅ →a×→b = î+4ĵ+3k̂
Coplanar Vectors (Board 2024, 2M)
Q: Show that →a=î+2ĵ−k̂, →b=2î−ĵ+k̂, →c=î−ĵ+2k̂ are coplanar (or not).
1
Vectors are coplanar iff scalar triple product [→a,→b,→c] = 0
2
[→a,→b,→c] = det([[1,2,−1],[2,−1,1],[1,−1,2]])
3
= 1[(−1)(2)−(1)(−1)] − 2[(2)(2)−(1)(1)] + (−1)[(2)(−1)−(−1)(1)]
4
= 1[−2+1] − 2[4−1] + (−1)[−2+1] = −1−6+1 = −6 ≠ 0
5
Vectors are NOT coplanar (linearly independent)
✅ [→a,→b,→c] = −6 ≠ 0 → vectors are NOT coplanar
⚖️ Comparisons
Scalar Product
Vector Product
Result is a scalar
Result is a vector
→a·→b = |a||b|cosθ
|→a×→b| = |a||b|sinθ
Commutative: →a·→b = →b·→a
Anticommutative: →a×→b = −→b×→a
→a⊥→b if →a·→b = 0
→a∥→b if →a×→b = →0
🎯 Quick Activity
Think and Answer
Q: If →a+→b+→c = 0 and |→a|=3, |→b|=5, |→c|=7, find the angle between →a and →b.
▶ Show Answer
A: →a+→b = −→c → |→a+→b|² = |→c|² = 49. |→a|²+2→a·→b+|→b|² = 49. 9+2→a·→b+25 = 49. 2→a·→b = 15. →a·→b = 7.5. cos θ = 7.5/(3×5) = 7.5/15 = 0.5. θ = 60°.
📝 Section Formula and Vector Geometry
10.3 Position Vectors and Geometric Applications
Board problems use vectors to prove geometric properties and find distances, midpoints, and areas.
Section Formula
If P divides AB in ratio m:n (internally): →OP = (m·→OB + n·→OA)/(m+n). Midpoint M: →OM = (→OA + →OB)/2.
Area of Triangle
Area = ½|→AB × →AC| where →AB and →AC are adjacent sides. Same as ½|→a × →b| for triangle with two sides →a and →b.
Collinearity Condition
Points A,B,C are collinear iff →AB and →AC are parallel: →AB = k·→AC for some scalar k, i.e., →AB × →AC = →0.
🔑 Common Vector Identities
→a×(→b+→c) = →a×→b + →a×→c (distributive over addition)
→a×→a = →0; î×î = ĵ×ĵ = k̂×k̂ = →0
|→a×→b|² + (→a·→b)² = |a|²|b|² (Lagrange's identity)
Volume of tetrahedron = (1/6)|[→a,→b,→c]| (scalar triple product)
If →a·→b = →a·→c and →a ≠ →0, it does NOT necessarily mean →b = →c (dot product cancellation invalid)
Area of Triangle (Board 2023, 3M)
Q: Find the area of triangle with vertices A(1,1,2), B(2,3,5), C(1,5,5).
1
→AB = B−A = (1,2,3); →AC = C−A = (0,4,3)
2
→AB × →AC = det([[î,ĵ,k̂],[1,2,3],[0,4,3]]) = î(6−12) − ĵ(3−0) + k̂(4−0) = −6î − 3ĵ + 4k̂
3
|→AB × →AC| = √(36+9+16) = √61
4
Area = ½|→AB × →AC| = √61/2 sq units
✅ Area = √61/2 square units
📐 Chapter 11: Three Dimensional Geometry
Direction Cosines · Line · Plane · Angle · Distance
📖 Core Theory
11.1 Direction Cosines and Lines
3D geometry uses vectors and coordinates. Line through point (x₁,y₁,z₁) in direction (l,m,n): (x−x₁)/l = (y−y₁)/m = (z−z₁)/n.
Direction Cosines (l,m,n)
Cosines of angles made with positive x, y, z axes respectively. l²+m²+n²=1. If direction ratios are (a,b,c): l=a/√(a²+b²+c²) etc.
Equation of Line
Vector form: →r = →a + λ→b (→a = point, →b = direction vector). Cartesian: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c. Two-point form: (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁).
Equation of Plane
Vector: →r·→n = d (→n = normal vector). Cartesian: ax+by+cz=d. Through three points: use determinant. Normal form: lx+my+nz=p (l,m,n = direction cosines of normal; p = distance from origin).
Angle Between Lines
If direction vectors are →b₁ and →b₂: cos θ = |→b₁·→b₂|/(|b₁||b₂|). Perpendicular if →b₁·→b₂ = 0. Parallel if →b₁ = k→b₂.
🔑 Distances and Skew Lines
Distance from point (x₁,y₁,z₁) to plane ax+by+cz+d=0: |ax₁+by₁+cz₁+d|/√(a²+b²+c²)
Angle between line and plane: sin θ = |→b·→n|/(|b||n|)
Skew lines: not parallel and do not intersect. Shortest distance = |(→a₂−→a₁)·(→b₁×→b₂)|/|→b₁×→b₂|
Foot of perpendicular from P to line: find λ using direction ratio condition, then find point
Distance between parallel planes ax+by+cz=d₁ and ax+by+cz=d₂: |d₁−d₂|/√(a²+b²+c²)
✏️ Worked Examples
Angle Between Lines (Board 2023, 3M)
Q: Find the angle between lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−3)/−1 = (y−2)/4 = (z−5)/5.
1
Direction vectors: →b₁ = (2,3,4); →b₂ = (−1,4,5)
2
→b₁·→b₂ = (2)(−1)+(3)(4)+(4)(5) = −2+12+20 = 30
3
|→b₁| = √(4+9+16) = √29; |→b₂| = √(1+16+25) = √42
4
cos θ = |30|/(√29·√42) = 30/√1218
5
θ = cos⁻¹(30/√1218) ≈ cos⁻¹(0.86)
✅ cos θ = 30/√1218; θ = cos⁻¹(30/√1218)
Distance Point to Plane (Board 2022, 2M)
Q: Find distance from point (3,4,5) to plane 2x+2y−z+3=0.
1
Formula: d = |ax₁+by₁+cz₁+d|/√(a²+b²+c²)
2
= |2(3)+2(4)+(−1)(5)+3|/√(4+4+1)
3
= |6+8−5+3|/√9 = |12|/3 = 4 units
✅ Distance = 4 units
Foot of Perpendicular (Board 2024, 3M)
Q: Find foot of perpendicular from (1,2,3) to line (x−1)/3 = (y−3)/1 = (z−4)/−5.
1
Let foot be at parameter λ: point = (1+3λ, 3+λ, 4−5λ)
2
Vector from given point (1,2,3) to foot = (3λ, 1+λ, 1−5λ)
3
This must be ⊥ to direction (3,1,−5): dot product = 0
4
3(3λ)+1(1+λ)+(−5)(1−5λ) = 0 → 9λ+1+λ−5+25λ = 0 → 35λ = 4 → λ = 4/35
5
Foot = (1+12/35, 3+4/35, 4−20/35) = (47/35, 109/35, 120/35)
✅ λ=4/35; Foot = (47/35, 109/35, 120/35)
⚖️ Comparisons
Line in 3D
Plane in 3D
1 equation in vector form (→r=→a+λ→b)
1 equation in cartesian (ax+by+cz=d)
Determined by point + direction vector
Determined by point + normal vector
Symmetric form: (x−x₁)/l = (y−y₁)/m = (z−z₁)/n
Normal form: lx+my+nz=p
🎯 Quick Activity
Think and Answer
Q: Find the equation of plane through (1,2,3) and perpendicular to the vector î−2ĵ+3k̂.
▶ Show Answer
A: Plane through point →a = î+2ĵ+3k̂ with normal →n = î−2ĵ+3k̂. Vector equation: (→r−→a)·→n = 0 → →r·→n = →a·→n. →a·→n = (1)(1)+(2)(−2)+(3)(3) = 1−4+9 = 6. So →r·(î−2ĵ+3k̂) = 6 → x−2y+3z = 6.
📐 Chapter 12: Linear Programming
Formulation · Feasible Region · Optimal Solution · Types of LPP
📖 Core Theory
12.1 Linear Programming Problem (LPP)
LPP: optimise (maximise/minimise) a objective function Z = ax+by subject to linear constraints, non-negativity restrictions (x≥0, y≥0), and implicit constraints.
Feasible Region
The set of all points satisfying all constraints including non-negativity. Always a convex polygon (or unbounded). The feasible region is the intersection of all constraint half-planes.
Optimal Solution
Corner Point Theorem: optimal value of Z occurs at a corner (vertex) of the feasible region. Evaluate Z at all vertices → take maximum (for maximisation) or minimum (for minimisation).
Infeasible / Unbounded
Infeasible: no point satisfies all constraints (empty feasible region). Unbounded: feasible region extends to infinity; maximisation may have no solution (but minimisation might still have one).
Types of Optimal Solutions
Unique: Z is optimal at exactly one corner. Multiple (alternative): Z is optimal along an entire edge (same value at two adjacent corners). Check if optimal value can be further improved in unbounded region.
🔑 Steps to Solve LPP Graphically
Step 1: Identify decision variables x and y; write objective function Z = ax+by
Step 2: Write all constraints as inequalities (≤ or ≥)
Step 3: Plot each constraint line; shade feasible half-plane
Step 4: Identify feasible region (intersection of all shaded regions)
Step 5: Find all corner points (vertices) of feasible region
Step 6: Evaluate Z at each corner; optimal = max or min value
✏️ Worked Examples
Maximisation LPP (Board 2023, 5M)
Q: Maximise Z=5x+3y subject to: 3x+5y≤15, 5x+2y≤10, x≥0, y≥0.
1
Draw 3x+5y=15 (passes (0,3) and (5,0)) and 5x+2y=10 (passes (0,5) and (2,0))
2
Intersection: solve simultaneously: 3x+5y=15 and 5x+2y=10
3
From 2nd: y=(10−5x)/2; sub in 1st: 3x+5(10−5x)/2=15 → 6x+50−25x=30 → −19x=−20 → x=20/19
4
y=(10−100/19)/2=(90/19)/2=45/19. Corner point: (20/19, 45/19)
5
All corners: (0,0)→Z=0; (2,0)→Z=10; (20/19,45/19)→Z=100/19+135/19=235/19≈12.4; (0,3)→Z=9
6
Maximum Z = 235/19 at (20/19, 45/19)
✅ Max Z = 235/19 ≈ 12.4 at (20/19, 45/19)
LPP Formulation (Board 2022, 5M)
Q: A factory makes two products X,Y. X needs 2hr machine + 1hr labour/unit; Y needs 1hr machine + 3hr labour/unit. Available: 200hr machine, 300hr labour. Profit: ₹40/unit X, ₹60/unit Y. Formulate and solve LPP.
1
Let x=units X, y=units Y. Objective: Maximise Z = 40x+60y
2
Constraints: 2x+y≤200 (machine); x+3y≤300 (labour); x≥0, y≥0
3
Corner points: (0,0)→Z=0; (100,0)→Z=4000; (0,100)→Z=6000
4
Intersection: 2x+y=200 and x+3y=300 → from 1st: y=200−2x; sub: x+3(200−2x)=300 → x+600−6x=300 → −5x=−300 → x=60, y=80
5
Corner (60,80): Z=2400+4800=7200
6
Maximum Z = ₹7200 at x=60, y=80
✅ Max profit ₹7200 when producing 60 units X and 80 units Y
Minimisation (Board 2024, 5M)
Q: Minimise Z=3x+5y subject to x+y≥3, x+3y≥6, x≥0, y≥0.
1
Draw x+y=3 and x+3y=6
2
Feasible region is unbounded (above both lines)
3
Corner points: x+y=3 meets x-axis at (3,0); x+3y=6 meets y-axis at (0,2)
4
Intersection: x+y=3 and x+3y=6 → subtract: 2y=3 → y=3/2, x=3/2; point (3/2,3/2)
5
Z at corners: (3,0)→9; (3/2,3/2)→4.5+7.5=12... recalculate: Z=3(3/2)+5(3/2)=4.5+7.5=12; (0,2)→10
6
Check: Z=3(3)+5(0)=9; Z=3(0)+5(2)=10; Z=3(1.5)+5(1.5)=12; Minimum at (3,0) = 9
✅ Min Z = 9 at (3, 0)
⚖️ Comparisons
Maximisation LPP
Minimisation LPP
Find corner giving HIGHEST Z value
Find corner giving LOWEST Z value
Unbounded region: check if Z can increase indefinitely
Unbounded region: minimisation may still have finite minimum
No issue if feasible region is bounded
For unbounded region: check if line Z=min value intersects feasible region or not
🎯 Quick Activity
Think and Answer
Q: A dealer wants to purchase typewriters and calculators. A typewriter costs ₹2000, calculator ₹500. Spend at most ₹70,000. Max 45 items. Profit ₹300 on typewriter, ₹100 on calculator. How many of each to buy for max profit? Formulate as LPP (do not solve graphically).
▶ Show Answer
A: Decision variables: x = typewriters, y = calculators. Objective: Maximise Z = 300x+100y. Constraints: 2000x+500y ≤ 70000 (i.e. 4x+y ≤ 140); x+y ≤ 45; x≥0, y≥0. Corners of feasible region: (0,0), (35,0), (0,45) and intersection 4x+y=140 with x+y=45: 3x=95 → x=95/3 (not integer; nearest integer solution x=31, y=14 or check corners). Optimal at corner — board expects graphical or simplex solution.
📐 Chapter 13: Probability
Conditional Probability · Multiplication Theorem · Bayes' Theorem · Random Variables
📖 Core Theory
13.1 Conditional Probability and Multiplication Theorem
P(A|B) = P(A∩B)/P(B), provided P(B) > 0. Conditional probability changes sample space from Ω to B.
Multiplication Theorem
P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B). For independent events: P(A∩B) = P(A)·P(B). Events A,B independent iff P(A|B) = P(A).
Total Probability
If B₁,B₂,...,Bₙ partition sample space Ω: P(A) = ΣP(Bᵢ)·P(A|Bᵢ). Used when A occurs through one of several mutually exclusive scenarios.
Bayes' Theorem
P(Bᵢ|A) = P(Bᵢ)·P(A|Bᵢ) / Σ P(Bⱼ)·P(A|Bⱼ). "Posterior probability" given new evidence A. Denominatior = P(A) by total probability theorem.
Random Variable (RV)
Function X: Ω → ℝ assigning real value to each outcome. Discrete RV: finite/countable values. p(X=xᵢ) = pᵢ ≥ 0, Σpᵢ = 1. E(X) = Σxᵢpᵢ (mean). Var(X) = E(X²) − [E(X)]².
🔑 Binomial Distribution
X~B(n,p): n independent trials, p=P(success). P(X=r) = ⁿCᵣ·pʳ·(1−p)^(n−r), r=0,1,...,n
Mean = np; Variance = np(1−p) = npq where q=1−p
P(A∩B) = 0 → A,B mutually exclusive; P(A∩B) = P(A)·P(B) → A,B independent (NOT the same)
P(A∪B) = P(A)+P(B)−P(A∩B) (addition theorem)
P(A′) = 1−P(A); P(A∩B′) = P(A)−P(A∩B)
✏️ Worked Examples
Bayes' Theorem (Board 2023, 5M)
Q: Bag I: 3R,4B; Bag II: 5R,6B. Bag selected randomly, one ball drawn = Red. Find P(Bag I was selected).
1
Let B₁=Bag I, B₂=Bag II selected. P(B₁)=P(B₂)=1/2
2
P(R|B₁) = 3/7; P(R|B₂) = 5/11
3
By total probability: P(R) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77
4
By Bayes': P(B₁|R) = P(B₁)·P(R|B₁)/P(R) = (1/2)(3/7)/(34/77)
5
= (3/14)×(77/34) = 231/476 = 33/68
✅ P(Bag I | Red ball drawn) = 33/68
Independent Events (Board 2022, 3M)
Q: A and B are independent with P(A)=1/4 and P(B)=2/3. Find P(A∩B), P(A∪B), P(A|B).
1
P(A∩B) = P(A)×P(B) = (1/4)(2/3) = 1/6 (independent events)
2
P(A∪B) = P(A)+P(B)−P(A∩B) = 1/4+2/3−1/6 = 3/12+8/12−2/12 = 9/12 = 3/4
3
P(A|B) = P(A∩B)/P(B) = (1/6)/(2/3) = 1/4 = P(A) (confirms independence)
✅ P(A∩B)=1/6; P(A∪B)=3/4; P(A|B)=1/4 (equals P(A) — confirms independence)
Binomial Distribution (Board 2024, 3M)
Q: A die is thrown 6 times. Find P(at least 1 six).
1
n=6, p=P(six)=1/6, q=5/6
2
P(at least 1 six) = 1 − P(no sixes) = 1 − P(X=0)
3
P(X=0) = ⁶C₀·(1/6)⁰·(5/6)⁶ = (5/6)⁶ = 15625/46656
4
P(at least 1 six) = 1 − 15625/46656 = 31031/46656 ≈ 0.665
✅ P(at least 1 six) = 1−(5/6)⁶ = 31031/46656 ≈ 0.665
⚖️ Comparisons
Mutually Exclusive Events
Independent Events
A∩B = φ (cannot occur simultaneously)
A∩B ≠ φ (can occur simultaneously, but don't affect each other's probability)
P(A∩B) = 0
P(A∩B) = P(A)·P(B)
P(A∪B) = P(A)+P(B)
P(A∪B) = P(A)+P(B)−P(A)P(B)
Mutually exclusive events are usually NOT independent (unless one has probability 0)
Independent events are NOT mutually exclusive (unless P(A)=0 or P(B)=0)
🎯 Quick Activity
Think and Answer
Q: Two cards are drawn without replacement from a deck of 52 cards. Find the probability that both are aces.
▶ Show Answer
A: P(1st ace) = 4/52 = 1/13. P(2nd ace | 1st was ace) = 3/51 = 1/17 (only 3 aces left in 51 cards). P(both aces) = P(1st ace) × P(2nd ace|1st ace) = (1/13)×(1/17) = 1/221.
📝 Practice Papers
Full-length 80-mark board-pattern papers · Mathematics XII
MATHEMATICS XII — PRACTICE PAPER 1
Time: 3 Hours | Max Marks: 80 | CBSE 2024-25
SECTION A — MCQ (18×1 + 2 Assertion = 20)
Q1. The relation R = {(a,b): |a−b| is divisible by 4} on Z is: (a) Reflexive only (b) Symmetric only (c) Transitive only (d) Equivalence [Ans: d]
Q2. sin⁻¹(sin 3π/4) = ? (a) 3π/4 (b) π/4 (c) −π/4 (d) 5π/4 [Ans: b]
Q3. If A is a square matrix of order 3 and |A|=3, then |2A| = ? (a) 6 (b) 12 (c) 24 (d) 3 [Ans: c] [2³×|A|=8×3=24]
Q4. d/dx[log(sin x)] = ? (a) cot x (b) −cot x (c) cos x (d) −cos x [Ans: a]
Q5. ∫sin²x dx = ? (a) x/2 − sin2x/4 + C (b) x/2 + sin2x/4 + C (c) −x/2 + cos2x/4 + C (d) sin2x/4 [Ans: a]
Q6. Order of differential equation: (y″)² + (y′)³ + y = 0 is: (a) 1 (b) 2 (c) 3 (d) 6 [Ans: b]
Q7. Vector a⃗ = 2î−ĵ+2k̂. |a⃗| = ? (a) 3 (b) 5 (c) 7 (d) 9 [Ans: a] [√(4+1+4)=3]
Q8. Direction cosines of a line making equal angles with all axes: (a) 1,1,1 (b) 1/√3 each (c) 0,0,0 (d) 1/√2 each [Ans: b]
Q9-20. [Further MCQs + 2 Assertion-Reason on LPP, Probability, Bayes theorem, conditional probability] [12M]
SECTION B — VSA (5×2=10)
Q21. Find the domain of f(x) = cos⁻¹(2x−1). [2M]
Q22. If A = [2,3;1,4], find A + Aᵀ and show it is symmetric. [2M]
Q23. Differentiate: y = sin(x²) with respect to x. [2M]
Q24. Evaluate: ∫₀^(π/2) sin x/(sin x + cos x) dx. [2M]
Q25. Find the unit vector in direction of 3î + 4ĵ. [2M]
SECTION C — SA (6×3=18)
Q26. Check if f: R→R, f(x) = 3x−5 is bijective. Find f⁻¹. [3M]
Q27. Solve: tan⁻¹(x−1) + tan⁻¹(x) + tan⁻¹(x+1) = tan⁻¹(3x). [3M]
Q28. Find the value of k for continuity: f(x) = {kx², x≤2; 3, x>2} at x=2. [3M]
Q29. Evaluate: ∫x·eˣ dx using integration by parts. [3M]
Q30. Solve the differential equation: dy/dx + y/x = x². [3M]
Q31. A bag contains 4 red and 3 blue balls. Two are drawn without replacement. Find P(both red). [3M]
SECTION D — LA (4×5=20) + Case (2×4=8) + Source (4)
Q32. Using matrices, solve: x + y + z = 6; 2x + y − z = 1; x − y + 3z = 7. [5M]
Q33. Find local maxima and minima of f(x) = x³ − 3x² + 3x − 5. Hence find point of inflection. [5M]
Q34. Find the area between y = x² and y = x using integration. [5M]
Q35. Find the equation of the plane through (2,3,−1), (3,−2,4) and (1,2,−3). Find angle it makes with x-axis. [5M]
Q36-37. Case studies: (i) LPP — maximise profit given constraints (ii) Probability — Bayes theorem applied to disease testing. [8M]
MATHEMATICS XII — PRACTICE PAPER 2
Alternate emphasis
All Sections
Q1-20. MCQ + AR: Relations & Functions (composite), Matrix (skew-symmetric), Determinants (minors, cofactors), Derivatives (parametric), Integrals (substitution), Vector cross product, 3D distance formula, Conditional probability. [20M]
Q21-25. VSA: Inverse trigonometry (principal value), Adjoint matrix, Rolle's theorem verification, Definite integral property, Scalar triple product. [10M]
Q26-31. SA: Equivalence relation proof, Matrix equations, L'Hospital's rule, Integration by partial fractions, Homogeneous ODE, Probability distributions. [18M]
Q32-37. LA + Case: Determinant method (Cramer's rule), Application of derivatives (rate of change), Volume of solid of revolution, Line in 3D (skew lines), LPP graphical, Binomial distribution case study. [32M]
MATHEMATICS XII — PRACTICE PAPER 3 (Challenge)
Hardest questions — pre-board level
Q1. Prove: tan⁻¹(1/2) + tan⁻¹(2/11) = tan⁻¹(3/4). [3M]
Q2. If A = [1,2;3,4], prove A² − 5A + 2I = 0. Hence find A⁻¹. [4M]
Q3. Differentiate: x^(sin x) + (sin x)^cos x. [4M]
Q4. Evaluate: ∫₀^π x·sin x/(1+cos²x) dx. [5M]
Q5. Solve: d²y/dx² − 5dy/dx + 6y = 0. [4M]
Q6. Using vectors, prove that the diagonals of a rhombus are perpendicular. [4M]
Q7. Find the shortest distance between lines: r⃗ = (î+ĵ) + t(2î−ĵ+k̂) and r⃗ = (2î+ĵ−k̂) + s(3î−5ĵ+2k̂). [5M]
Q8. LPP: A manufacturer makes two products X and Y. X needs 3 hrs machine time + 2 hrs labour. Y needs 1 hr + 3 hrs. Available: 12 hrs machine time, 15 hrs labour. Profit: X = Rs500, Y = Rs700. Find max profit. [6M]
Q9. A random variable X has the distribution P(X=k) = C(n,k)(0.6)^k(0.4)^(n-k). If mean=6, find n and P(X≥1). [5M]
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