📐 Class 12 Maths — Important Questions 2026

Calculus (35M) + Algebra (10M) + Vectors & 3D (14M) + Relations (8M) + LP (5M) + Probability (8M)

Complete Chapter-wise Guide with Answer Hints & Step-by-Step Approach — Score 80+!

📅 Board Exam 2026 — Check Date Sheet →

Mathematics is the highest-scoring subject in Class 12 Board Exams — if you know the pattern and practice the right questions, scoring 80+ is absolutely achievable. The key insight is that Calculus alone carries 35 out of 80 marks (44% of the paper), so mastering Integration, Differentiation, and Applications of Derivatives/Integrals is non-negotiable.

This guide covers every chapter with the most expected questions, each with step-by-step answer hints so you know exactly which formula to apply and how to structure your solution. Whether you're from CBSE or RBSE, this is your one-stop revision guide.

📖 Preparation Order: Chapter-wise Notes → Solve these Important Questions → Attempt 200 MCQ Mega Test → Revise formulas using Formula Sheet

⚠️ RBSE Students Note: CBSE Maths = 80 marks Theory. RBSE Maths = 80 marks (Theory 60 + Internal 20). Marks distribution below follows CBSE pattern. RBSE students: focus on Calculus and Algebra — these two sections together = 45/80 marks (56%). Master these and you're already passing!

📊 Class 12 Maths — Marks Distribution 2026

📘 Unit I — Relations & Functions (8 Marks)

Ch 1: Relations & Functions + Ch 2: Inverse Trigonometric Functions

⭐ Ch 1: Relations and Functions [Notes →]

1. 🟡 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even} is an equivalence relation. (3M) ⭐⭐⭐ Most Expected
   💡 Answer Hint: Prove 3 properties: (1) Reflexive: |a−a| = 0, which is even ✓ (2) Symmetric: If |a−b| is even, then |b−a| = |a−b| is also even ✓ (3) Transitive: If |a−b| is even and |b−c| is even, then a−b = 2k and b−c = 2m, so a−c = 2(k+m) which is even ✓. Since all three hold → R is an equivalence relation. Formula: Always prove Reflexive → Symmetric → Transitive in order.
2. 🟡 Check whether the function f : R → R defined by f(x) = x³ is one-one and onto. (3M) ⭐⭐⭐
   💡 Answer Hint: One-one (Injective): If f(x₁) = f(x₂) → x₁³ = x₂³ → x₁ = x₂ ✓. Onto (Surjective): For any y ∈ R, x = y^(1/3) ∈ R and f(x) = y ✓. Since both hold → f is bijective. Tip: For one-one, also check f'(x) = 3x² ≥ 0 (monotonically increasing → one-one).
3. 🟡 Let f : N → N be defined by f(n) = (n+1)/2 if n is odd, n/2 if n is even. Check if f is onto. (3M)
   💡 Answer Hint: For any y ∈ N, we need x ∈ N such that f(x) = y. Take x = 2y (even), then f(2y) = 2y/2 = y ✓. So for every y in codomain, there exists a pre-image → f is onto. But f is NOT one-one: f(1) = 1 and f(2) = 1, so different inputs give same output.

⭐ Ch 2: Inverse Trigonometric Functions [Notes →]

1. 🟡 Prove that: tan⁻¹(1/2) + tan⁻¹(1/3) = π/4 (2M) ⭐⭐⭐ Most Asked
   💡 Answer Hint: Use formula: tan⁻¹x + tan⁻¹y = tan⁻¹[(x+y)/(1−xy)] when xy < 1. Here: tan⁻¹[(1/2 + 1/3)/(1 − 1/6)] = tan⁻¹[(5/6)/(5/6)] = tan⁻¹(1) = π/4. Key: Check xy < 1 before applying. If xy > 1, add π to the result.
2. 🟡 Simplify: tan⁻¹[√(1+x²) − 1]/x (2M) ⭐⭐
   💡 Answer Hint: Put x = tanθ. Then √(1+tan²θ) = secθ. Expression becomes tan⁻¹[(secθ−1)/tanθ] = tan⁻¹[(1−cosθ)/sinθ] = tan⁻¹[tan(θ/2)] = θ/2 = (1/2)tan⁻¹x. Substitution trick: x = tanθ, sinθ, or cosθ depending on the expression inside.

🔢 Unit II — Algebra: Matrices & Determinants (10 Marks)

Guaranteed 5-mark question on solving system of equations using matrices!

⭐ Ch 3: Matrices [Notes →]

1. 🟡 If A = [2 3; 1 −4], B = [1 −2; −1 3], verify that (AB)⁻¹ = B⁻¹A⁻¹. (3M) ⭐⭐⭐
   💡 Answer Hint: Step 1: Find AB by matrix multiplication. Step 2: Find (AB)⁻¹ using formula A⁻¹ = (1/|A|) × adj(A). Step 3: Find A⁻¹ and B⁻¹ separately. Step 4: Compute B⁻¹A⁻¹. Step 5: Show (AB)⁻¹ = B⁻¹A⁻¹. Key: For 2×2 matrix [a b; c d], inverse = (1/(ad−bc)) × [d −b; −c a].
2. 🟡 Express matrix A as sum of symmetric and skew-symmetric matrices where A = [3 5; 1 −1]. (2M) ⭐⭐
   💡 Answer Hint: Formula: A = ½(A + Aᵀ) + ½(A − Aᵀ) where ½(A + Aᵀ) is symmetric and ½(A − Aᵀ) is skew-symmetric. Find Aᵀ = [3 1; 5 −1]. Then: Symmetric part = ½[6 6; 6 −2] = [3 3; 3 −1]. Skew part = ½[0 4; −4 0] = [0 2; −2 0]. Verify: sum = A ✓

⭐ Ch 4: Determinants — System of Equations [Notes →]

1. 🟣 Solve the system using matrix method: 2x + 3y + 3z = 5, x − 2y + z = −4, 3x − y − 2z = 3 (5M) ⭐⭐⭐ GUARANTEED 5-mark Question!
   💡 Answer Hint: Write as AX = B where A = [2 3 3; 1 −2 1; 3 −1 −2], X = [x; y; z], B = [5; −4; 3]. Steps: (1) Find |A| by expanding along R1: 2(4+1) − 3(−2−3) + 3(−1+6) = 10+15+15 = 40 ≠ 0 → unique solution exists. (2) Find cofactors A₁₁=5, A₁₂=5, A₁₃=5... (3) Form adj(A) = transpose of cofactor matrix. (4) A⁻¹ = adj(A)/|A|. (5) X = A⁻¹B → get x, y, z values. Common mistake: Sign pattern for cofactors: (+−+/−+−/+−+). Always verify by substituting back.
2. 🟡 Using properties of determinants, prove that: |a−b−c 2a 2a; 2b b−c−a 2b; 2c 2c c−a−b| = (a+b+c)³ (3M) ⭐⭐⭐
   💡 Answer Hint: Apply R₁ → R₁ + R₂ + R₃. First row becomes [a+b+c, a+b+c, a+b+c]. Take (a+b+c) common from R₁. Then apply C₂→C₂−C₁ and C₃→C₃−C₁ to simplify. You'll get a triangular determinant. Strategy for determinant proofs: Always try R₁→R₁+R₂+R₃ or C₁→C₁+C₂+C₃ first to get a common factor.
3. 🟡 Find the area of triangle with vertices (1, 0), (6, 0), (4, 3) using determinants. (2M)
   💡 Answer Hint: Formula: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = ½|1(0−3) + 6(3−0) + 4(0−0)| = ½|−3+18+0| = ½×15 = 7.5 sq units. If area = 0, points are collinear.

🔥 Unit III — CALCULUS (35 Marks) — 44% of Your Paper!

Continuity & Differentiability + Applications of Derivatives + Integrals + Applications of Integrals + Differential Equations

⭐ Ch 5: Continuity and Differentiability [Notes →]

1. 🟡 Differentiate y = (sin x)^(cos x) + (cos x)^(sin x) with respect to x. (3M) ⭐⭐⭐ Most Expected
   💡 Answer Hint: Let u = (sinx)^cosx and v = (cosx)^sinx. Then y = u + v → dy/dx = du/dx + dv/dx. For u: Take log: log u = cosx · log(sinx). Differentiate: (1/u)du/dx = −sinx·log(sinx) + cosx·(cosx/sinx). So du/dx = (sinx)^cosx[cos²x/sinx − sinx·log(sinx)]. For v: Similarly apply logarithmic differentiation. This is the classic logarithmic differentiation question — appears almost every year!
2. 🟡 If y = (tan⁻¹x)², show that (1+x²)²y₂ + 2x(1+x²)y₁ = 2 (3M) ⭐⭐⭐
   💡 Answer Hint: y₁ = dy/dx = 2tan⁻¹x · 1/(1+x²). So (1+x²)y₁ = 2tan⁻¹x. Now differentiate both sides: (1+x²)y₂ + 2xy₁ = 2/(1+x²). Multiply by (1+x²): (1+x²)²y₂ + 2x(1+x²)y₁ = 2. Tip: For "show that" questions involving y₂, always find y₁ first, multiply to clear denominators, then differentiate again.
3. 🟡 Find the value of k so that the function f(x) is continuous at x = 0: f(x) = (1−cos4x)/x² if x≠0, f(x) = k if x=0. (2M) ⭐⭐⭐
   💡 Answer Hint: For continuity: lim(x→0) f(x) = f(0) = k. Use L'Hôpital or identity: 1−cos4x = 2sin²(2x). So lim = 2sin²(2x)/x² = 2·4·(sin2x/2x)² = 8·1 = k = 8. Key identity: lim(x→0) sinx/x = 1, so lim sin(ax)/(ax) = 1 → sin²(2x)/x² = 4.
4. 🟡 Verify Rolle's Theorem for f(x) = x² + 2x − 8, x ∈ [−4, 2]. (3M) ⭐⭐
   💡 Answer Hint: 3 conditions: (1) f continuous on [−4,2] ✓ (polynomial). (2) f differentiable on (−4,2) ✓. (3) f(−4) = 16−8−8 = 0, f(2) = 4+4−8 = 0, so f(−4) = f(2) ✓. All conditions met. Now f'(x) = 2x+2 = 0 → x = −1 ∈ (−4,2). Rolle's theorem verified. Similarly for Mean Value Theorem: find c where f'(c) = [f(b)−f(a)]/(b−a).

⭐ Ch 6: Application of Derivatives — Maxima/Minima + Rate of Change [Notes →]

1. 🟣 Show that the semi-vertical angle of a cone of maximum volume and given slant height is tan⁻¹(√2). (5M) ⭐⭐⭐ CLASSIC 5-marker!
   💡 Answer Hint: Let slant height = l, semi-vertical angle = θ. Then r = l·sinθ, h = l·cosθ. Volume V = (1/3)πr²h = (π/3)l³sin²θ·cosθ. Maximize: dV/dθ = (π/3)l³[2sinθcos²θ − sin³θ] = 0 → 2cos²θ = sin²θ → tan²θ = 2 → θ = tan⁻¹(√2). Check d²V/dθ² < 0 at this point for maximum. Strategy for optimization: Express quantity in ONE variable → differentiate → set = 0 → verify max/min with 2nd derivative test.
2. 🟣 Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (5M) ⭐⭐⭐
   💡 Answer Hint: Let numbers be x and (15−x). S = x² + (15−x)². dS/dx = 2x − 2(15−x) = 4x − 30 = 0 → x = 7.5. d²S/dx² = 4 > 0 → minimum confirmed. Both numbers = 7.5 and 7.5. Key principle: For fixed sum, sum of squares is minimum when numbers are equal!
3. 🟡 Find the intervals where f(x) = 2x³ − 3x² − 36x + 7 is increasing and decreasing. (3M) ⭐⭐⭐
   💡 Answer Hint: f'(x) = 6x² − 6x − 36 = 6(x²−x−6) = 6(x−3)(x+2). Sign analysis: f'(x) > 0 when x < −2 or x > 3 → Increasing: (−∞,−2) ∪ (3,∞). f'(x) < 0 when −2 < x < 3 → Decreasing: (−2,3). Method: factorize f'(x), use number line sign test.
4. 🟡 Find the equation of tangent to the curve y = x³ − 3x + 5 at x = 2. (2M)
   💡 Answer Hint: At x=2: y = 8−6+5 = 7, point (2,7). Slope = dy/dx at x=2: 3(4)−3 = 9. Tangent: y−7 = 9(x−2) → y = 9x−11. Normal slope = −1/9. For tangent ∥ x-axis: dy/dx = 0. For tangent ∥ y-axis: dx/dy = 0.

⭐ Ch 7: Integrals — Integration by Parts, Partial Fractions, Definite Integrals [Notes →]

1. 🟣 Evaluate: ∫ x² / (x² + 4)(x² + 9) dx using partial fractions. (5M) ⭐⭐⭐
   💡 Answer Hint: Put x² = t. Then t/[(t+4)(t+9)] = A/(t+4) + B/(t+9). Solving: t = A(t+9) + B(t+4). At t=−4: −4 = 5A → A = −4/5. At t=−9: −9 = −5B → B = 9/5. So integral = (−4/5)∫dx/(x²+4) + (9/5)∫dx/(x²+9) = (−4/5)·(1/2)tan⁻¹(x/2) + (9/5)·(1/3)tan⁻¹(x/3) + C. Formula: ∫dx/(x²+a²) = (1/a)tan⁻¹(x/a) + C.
2. 🟣 Evaluate: ∫ eˣ(sinx + cosx) dx (3M) ⭐⭐⭐ Direct Formula!
   💡 Answer Hint: Use special formula: ∫eˣ[f(x) + f'(x)]dx = eˣf(x) + C. Here f(x) = sinx, f'(x) = cosx. Direct answer: eˣsinx + C. This formula saves 5 minutes! Verify: d/dx[eˣsinx] = eˣsinx + eˣcosx = eˣ(sinx+cosx) ✓
3. 🟡 Evaluate: ∫₀^(π/2) log(sinx) dx (3M) ⭐⭐⭐
   💡 Answer Hint: Use property: ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx. Let I = ∫₀^(π/2) log(sinx)dx. Replace x by π/2−x: I = ∫₀^(π/2) log(cosx)dx. Add: 2I = ∫₀^(π/2) log(sinx·cosx)dx = ∫₀^(π/2) log(sin2x/2)dx = ∫₀^(π/2) log(sin2x)dx − (π/2)log2. The first integral = I (after substitution). So 2I = I − (π/2)log2 → I = −(π/2)log2. Classic result — memorize it!
4. 🟡 Evaluate: ∫ (3x + 5) / √(x² − 8x + 7) dx (3M) ⭐⭐
   💡 Answer Hint: Write 3x+5 = A·d/dx(x²−8x+7) + B = A(2x−8) + B. Comparing: 2A=3→A=3/2, −8A+B=5→B=17. So integral = (3/2)∫(2x−8)/√(x²−8x+7)dx + 17∫dx/√(x²−8x+7). First part: put t=x²−8x+7, get 3√t. Second: complete square x²−8x+7 = (x−4)²−9, use ∫dx/√(x²−a²) formula. Strategy: For (ax+b)/√(quadratic), always split as A·derivative + B.

⭐ Ch 8: Application of Integrals — Area Under Curves [Notes →]

1. 🟣 Find the area of the region bounded by the curve y² = 4x, the y-axis and the line y = 3. (5M) ⭐⭐⭐
   💡 Answer Hint: y² = 4x → x = y²/4. Area = ∫₀³ x dy = ∫₀³ y²/4 dy = (1/4)[y³/3]₀³ = (1/4)(27/3) = 9/4 sq units. Key insight: When bounded by y-axis, integrate w.r.t. y (express x as function of y). When bounded by x-axis, integrate w.r.t. x. Always draw a rough sketch first!
2. 🟣 Find the area enclosed between the circle x² + y² = 4 and the line x + y = 2. (5M) ⭐⭐⭐ CLASSIC!
   💡 Answer Hint: Find intersection: x+y=2 → y=2−x. Substitute: x²+(2−x)²=4 → 2x²−4x=0 → x=0,2. Area = ∫₀² [√(4−x²) − (2−x)]dx = [x√(4−x²)/2 + 2sin⁻¹(x/2)]₀² − [2x−x²/2]₀² = (π−2) sq units. Steps: (1) Find intersection points (2) Identify upper/lower curves (3) Area = ∫(upper − lower)dx.
3. 🟡 Find the area bounded by the ellipse x²/4 + y²/9 = 1. (3M)
   💡 Answer Hint: By symmetry, area = 4 × ∫₀² (3/2)√(4−x²) dx = 6[x√(4−x²)/2 + 2sin⁻¹(x/2)]₀² = 6[0+2·π/2] = 6π sq units. General formula: Area of ellipse x²/a² + y²/b² = 1 is πab. Here a=2, b=3, so area = π·2·3 = 6π ✓.

⭐ Ch 9: Differential Equations [Notes →]

1. 🟣 Solve: (1 + x²)dy/dx + 2xy = 1/(1 + x²), given y(0) = 0. (5M) ⭐⭐⭐ Most Expected!
   💡 Answer Hint: Divide by (1+x²): dy/dx + 2x/(1+x²)·y = 1/(1+x²)². This is Linear DE: dy/dx + Py = Q. P = 2x/(1+x²), Q = 1/(1+x²)². IF = e^∫Pdx = e^∫2x/(1+x²)dx = e^log(1+x²) = (1+x²). Solution: y·(1+x²) = ∫(1+x²)·1/(1+x²)²dx = ∫dx/(1+x²) = tan⁻¹x + C. Apply y(0)=0: 0=0+C→C=0. Answer: y = tan⁻¹x/(1+x²).
2. 🟡 Solve: x dy − y dx = √(x² + y²) dx (3M) ⭐⭐⭐
   💡 Answer Hint: Rewrite: dy/dx = [y + √(x²+y²)]/x. This is homogeneous DE. Put y = vx → dy/dx = v + xdv/dx. Then: v + xdv/dx = v + √(1+v²) → xdv/dx = √(1+v²) → dv/√(1+v²) = dx/x. Integrate: log|v + √(1+v²)| = log|x| + C → y + √(x²+y²) = Cx². Key identification: if dy/dx = f(y/x), it's homogeneous.
3. 🟡 Find the order and degree of: d²y/dx² + (dy/dx)³ + 6y⁵ = 0 (1M)
   💡 Answer Hint: Order = highest derivative = d²y/dx² → Order = 2. Degree = power of highest order derivative = power of d²y/dx² = Degree = 1. Note: (dy/dx)³ has power 3, but dy/dx is NOT the highest order derivative. Degree is defined only when DE is polynomial in derivatives.

📐 Unit IV — Vectors & Three Dimensional Geometry (14 Marks)

Ch 10: Vector Algebra + Ch 11: 3D Geometry — 2nd highest scoring after Calculus!

⭐ Ch 10: Vector Algebra [Notes →]

1. 🟡 Find the area of the parallelogram whose adjacent sides are: a⃗ = 3î + ĵ + 4k̂ and b⃗ = î − ĵ + k̂ (3M) ⭐⭐⭐
   💡 Answer Hint: Area = |a⃗ × b⃗|. Cross product: a⃗×b⃗ = |î ĵ k̂; 3 1 4; 1 −1 1| = î(1+4) − ĵ(3−4) + k̂(−3−1) = 5î + ĵ − 4k̂. |a⃗×b⃗| = √(25+1+16) = √42 sq units. For triangle area, use ½|a⃗×b⃗|.
2. 🟡 If a⃗ + b⃗ + c⃗ = 0 and |a⃗| = 3, |b⃗| = 5, |c⃗| = 7, find the angle between a⃗ and b⃗. (3M) ⭐⭐⭐
   💡 Answer Hint: a⃗+b⃗ = −c⃗. Square both sides: |a⃗|²+|b⃗|²+2a⃗·b⃗ = |c⃗|² → 9+25+2a⃗·b⃗ = 49 → a⃗·b⃗ = 15/2. Now cosθ = (a⃗·b⃗)/(|a⃗||b⃗|) = (15/2)/(3×5) = 1/2 → θ = π/3 (60°). This technique works whenever sum of vectors = 0.
3. 🟡 Find the scalar triple product [a⃗ b⃗ c⃗] where a⃗ = î+2ĵ+3k̂, b⃗ = −î+2ĵ+k̂, c⃗ = 3î+ĵ. (2M)
   💡 Answer Hint: [a⃗ b⃗ c⃗] = |1 2 3; −1 2 1; 3 1 0| = 1(0−1) − 2(0−3) + 3(−1−6) = −1+6−21 = −16. Volume of parallelepiped = |[a⃗ b⃗ c⃗]| = 16 cubic units. If [a⃗ b⃗ c⃗] = 0, vectors are coplanar.

⭐ Ch 11: Three Dimensional Geometry — Lines & Planes [Notes →]

1. 🟣 Find the shortest distance between the lines: r⃗ = (î+2ĵ+k̂) + λ(î−ĵ+k̂) and r⃗ = (2î−ĵ−k̂) + μ(2î+ĵ+2k̂) (5M) ⭐⭐⭐ GUARANTEED Question!
   💡 Answer Hint: Formula: d = |(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)| / |b⃗₁×b⃗₂|. Here a⃗₁=(1,2,1), a⃗₂=(2,−1,−1), b⃗₁=(1,−1,1), b⃗₂=(2,1,2). Step 1: a⃗₂−a⃗₁ = (1,−3,−2). Step 2: b⃗₁×b⃗₂ = |î ĵ k̂; 1 −1 1; 2 1 2| = î(−2−1)−ĵ(2−2)+k̂(1+2) = (−3,0,3). Step 3: |b⃗₁×b⃗₂| = √(9+0+9) = 3√2. Step 4: (a⃗₂−a⃗₁)·(b⃗₁×b⃗₂) = −3+0−6 = −9. d = |−9|/3√2 = 3/√2 = 3√2/2.
2. 🟣 Find the equation of the plane passing through (1, 1, −1), (6, 4, −5), and (−4, −2, 3). (5M) ⭐⭐⭐
   💡 Answer Hint: Method 1 (Determinant): |x−1 y−1 z+1; 6−1 4−1 −5+1; −4−1 −2−1 3+1| = 0 → |x−1 y−1 z+1; 5 3 −4; −5 −3 4| = 0. Expanding: (x−1)(12−12)−(y−1)(20−20)+(z+1)(−15+15) = 0 → 0 = 0. Points are collinear! Method 2: Check if points form a valid plane — find two direction vectors and cross product. If cross product = 0, points are collinear and infinite planes pass through them. Alternative: Use general form ax+by+cz = d and substitute 3 points to get 3 equations.
3. 🟡 Find the angle between the planes: 2x+y−z = 3 and 3x−2y+z = 7. (3M) ⭐⭐⭐
   💡 Answer Hint: Normal vectors: n⃗₁=(2,1,−1), n⃗₂=(3,−2,1). cosθ = |n⃗₁·n⃗₂|/(|n⃗₁||n⃗₂|) = |6−2−1|/(√6·√14) = 3/√84 = 3/(2√21). θ = cos⁻¹(3/2√21). For angle between line and plane: sinθ = |b⃗·n⃗|/(|b⃗||n⃗|). If planes ⊥: n⃗₁·n⃗₂=0. If planes ∥: n⃗₁×n⃗₂=0.
4. 🟡 Find the distance of the point (1, 2, 3) from the plane x − y + z = 5. (2M)
   💡 Answer Hint: Formula: d = |ax₁+by₁+cz₁−d|/√(a²+b²+c²) = |1−2+3−5|/√(1+1+1) = |−3|/√3 = 3/√3 = √3 units. This formula is guaranteed 2 marks — memorize it!

📊 Unit V — Linear Programming (5 Marks)

Easiest 5 marks in the paper — Corner Point Method = guaranteed marks!

⭐ Ch 12: Linear Programming [Notes →]

1. 🟣 Maximize Z = 3x + 5y subject to: x + 3y ≤ 12, 3x + y ≤ 12, x, y ≥ 0. (5M) ⭐⭐⭐ Standard Pattern
   💡 Answer Hint: Steps: (1) Graph the constraints on x-y plane. Line 1: x+3y=12 passes through (12,0) and (0,4). Line 2: 3x+y=12 passes through (4,0) and (0,12). (2) Find feasible region (shade towards origin for ≤). (3) Find corner points: O(0,0), A(4,0), B(3,3) [intersection of both lines], C(0,4). (4) Evaluate Z at each: Z(O)=0, Z(A)=12, Z(B)=9+15=24, Z(C)=20. Maximum Z = 24 at (3,3). Always show the table of corner points — it carries marks!
2. 🟡 Minimize Z = x + 2y subject to: 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. (5M)
   💡 Answer Hint: For ≥ constraints, shade away from origin. Corner points: A(6,0), B(0,3) [intersection check carefully]. Evaluate Z: Z(6,0)=6, Z(0,3)=6. Minimum Z = 6 at both points → minimum occurs at all points on line joining (6,0) and (0,3). Key concept: If min/max occurs at two corner points, it occurs at ALL points on the line segment joining them (infinite solutions). LP = guaranteed 5 marks if you draw graph correctly!

🎲 Unit VI — Probability (8 Marks)

Bayes' Theorem (3-5M) + Mean & Variance of Random Variable (3-5M)

⭐ Ch 13: Probability — Bayes' Theorem & Distributions [Notes →]

1. 🟣 Bag I contains 3 red, 4 black balls. Bag II contains 5 red, 6 black balls. One bag is chosen at random and a ball is drawn. It turns out to be red. Find the probability that it came from Bag I. (5M) ⭐⭐⭐ CLASSIC Bayes'!
   💡 Answer Hint: Bayes' Formula: P(E₁|A) = P(E₁)·P(A|E₁) / [P(E₁)·P(A|E₁) + P(E₂)·P(A|E₂)]. P(E₁) = P(Bag I) = 1/2, P(E₂) = 1/2. P(Red|Bag I) = 3/7, P(Red|Bag II) = 5/11. P(Bag I|Red) = (1/2)(3/7) / [(1/2)(3/7) + (1/2)(5/11)] = (3/14) / (3/14 + 5/22) = (3/14) / (33+35)/154 = (33/154)/(68/154) = 33/68. Pattern: Always make a tree diagram first — it prevents sign errors!
2. 🟡 A random variable X has the following distribution. Find k, E(X), and Var(X): P(X=0)=3k, P(X=1)=4k, P(X=2)=5k, P(X=3)=6k, P(X=4)=7k (3M) ⭐⭐⭐
   💡 Answer Hint: ΣP = 1: 3k+4k+5k+6k+7k = 25k = 1 → k = 1/25. E(X) = ΣxP(x) = 0(3/25)+1(4/25)+2(5/25)+3(6/25)+4(7/25) = (4+10+18+28)/25 = 60/25 = 12/5. E(X²) = 0+4/25+20/25+54/25+112/25 = 190/25. Var(X) = E(X²)−[E(X)]² = 190/25 − 144/25 = 46/25.
3. 🟡 Find the probability that exactly 2 defective items are drawn when 4 items are drawn from a lot of 10 containing 3 defective items. (3M) ⭐⭐
   💡 Answer Hint: This is Hypergeometric distribution (without replacement). P = C(3,2)×C(7,2)/C(10,4) = 3×21/210 = 63/210 = 3/10. If with replacement → use Binomial: P = C(4,2)(3/10)²(7/10)² = 6·9·49/10000. Know when to use Binomial (with replacement, fixed probability) vs Hypergeometric (without replacement).

📈 Previous Year Analysis — Chapter-wise Frequency (2020-2025)

Based on CBSE Board papers 2020-2025 (12 sets), these topics appear most frequently as 3-mark and 5-mark questions:

For chapter-wise PYQs: Previous Year Questions Bank | Practice papers: Sample Papers 2026

🏆 Score 80+ in Maths — The Complete Strategy

⚠️ Top 15 Mistakes Students Make in Maths Board Exam — Avoid These!

Based on CBSE examiner reports and analysis of student answer sheets, these are the most common reasons for losing marks — even by students who know the concepts:

Calculus Mistakes: (1) Forgetting "+C" in indefinite integrals — this costs 1 mark every time. (2) Wrong sign in cofactor matrix — the sign pattern is (+−+/−+−/+−+), not all positive. (3) Not checking second derivative condition in maxima/minima — just finding critical points is not enough, you must prove it's a maximum or minimum. (4) Using wrong limits in definite integrals — always sketch the region first. (5) Confusing d/dx and ∫dx — some students differentiate when asked to integrate and vice versa. Practice with our Integration notes to avoid this.

Vectors & 3D Mistakes: (6) Mixing up dot product (scalar result) and cross product (vector result) — refer to our Vector Algebra notes for clarity. (7) Wrong formula for shortest distance — there are different formulas for parallel lines vs skew lines. For parallel: d = |b⃗×(a⃗₂−a⃗₁)|/|b⃗|. For skew: d = |(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)|/|b⃗₁×b⃗₂|. (8) Forgetting to take absolute value in distance formula — distance is always positive.

Algebra & Probability Mistakes: (9) Not verifying solution by substituting back in system of equations — this catches calculation errors and earns extra marks. (10) Confusing P(A|B) with P(B|A) in Bayes' theorem — always draw a tree diagram first. (11) Using Binomial formula when sampling is without replacement (should use Hypergeometric). (12) Wrong matrix multiplication order — AB ≠ BA in general. Check dimensions: (m×n) × (n×p) = (m×p). See our Matrices chapter notes for practice.

General Exam Mistakes: (13) Not reading the question carefully — "find" vs "prove" vs "show that" require different approaches. "Prove" means you must show every step. "Find" means you can jump steps if the answer is clear. (14) Skipping steps in 5-mark questions — CBSE gives marks for each step, so even if you know the answer, show all working. An answer without working gets 0-1 marks. (15) Poor time management — spending 20+ minutes on one difficult question while leaving easy 3-mark questions unanswered. Our timed practice tests help you build speed and accuracy together.

🎯 Practice Resources — Class 12 Maths

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