Important Questions for Class 12 Maths 2026-27 PDFs with Solutions

CBSE Class 12 Mathematics Important Questions

Class 12 Mathematics Important Questions 2026-27 bring together the most exam-relevant and frequently tested questions from the latest Class 12 CBSE syllabus to help you prepare with focus and score better in board exams. These include different types of questions, such as MCQs, competency-based questions, short-answer and long-answer questions, case-based questions, proof-based questions and HOTS questions.

Students can download Important Questions Class 12 Maths PDFs chapter-wise with detailed step-by-step solutions for free by clicking the links below.

CBSE Class 12 Mathematics 2026-27: A Quick Overview

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Chapter-wise Mathematics Important Questions for CBSE Class 12 PDF

Access our CBSE Class 12 Mathematics Important Questions with Solutions PDF for all chapters in one place. Educart’s team of experts and top academicians carefully curated a wide range of questions, including MCQs, assertion-reason questions, short answers, long answers, HOTS, proof-based questions, and case-based questions - all aligned with the 2026-27 CBSE exam pattern.

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Practice CBSE Class 12 Important Questions after completing the NCERT textbook to understand the chapter-specific concepts effectively. These help students resolve doubts, improve logical thinking, and improve answer-writing skills, which are essential for exam preparation. 

Important Questions for Class 12 Maths 2026-27 with Solutions

This collection of Class 12 Mathematics important questions and answers helps you revise key concepts chapter by chapter. Every solution is written in a clear, exam-ready format so you can understand both the method and the presentation style expected by CBSE examiners.

Chapter 1: Relations and Functions - Important Questions

Q1 (Easy): Define an equivalence relation. Give an example.

Ans: A relation R on a set A is called an equivalence relation if it is reflexive (every element is related to itself), symmetric (if aRb then bRa) and transitive (if aRb and bRc then aRc). Example: The relation "has the same remainder when divided by 3" on the set of integers satisfies all three properties and is therefore an equivalence relation.

Q2 (Medium): Let f: R → R be defined as f(x) = 3x − 5. Show that f is both one-one and onto.

Ans: For one-one: Assume f(x₁) = f(x₂). Then 3x₁ − 5 = 3x₂ − 5, which gives x₁ = x₂. So f is injective. For onto: For any y ∈ R, choose x = (y + 5)/3 ∈ R. Then f(x) = 3 × (y + 5)/3 − 5 = y. Since every element in the codomain has a pre-image, f is surjective. Hence f is bijective.

Q3 (Medium): Let f: A → B and g: B → C be two functions. Prove that if both f and g are one-one, then gof is also one-one.

Ans: Suppose (gof)(x₁) = (gof)(x₂). This means g(f(x₁)) = g(f(x₂)). Since g is one-one, f(x₁) = f(x₂). Since f is one-one, x₁ = x₂. Therefore gof is one-one.

Q4 (HOTS): A function f: N → N is defined by f(n) = n + 1 if n is odd and f(n) = n − 1 if n is even. Determine whether f is bijective and justify your answer carefully.

Ans: For one-one: Consider any two natural numbers n₁ and n₂ such that f(n₁) = f(n₂). If both are odd, n₁ + 1 = n₂ + 1 ⟹ n₁ = n₂. If both are even, n₁ − 1 = n₂ − 1 ⟹ n₁ = n₂. If one is odd and one even, the parities of f(n₁) and f(n₂) would differ, making equality impossible. So f is one-one. For onto: Every odd natural number m = n − 1 for even n and every even number m = n + 1 for odd n. Every element of N is hit. So f is onto. Hence f is bijective.

Chapter 2: Inverse Trigonometric Functions - Important Questions

Q1 (Easy): Write the principal value of sin⁻¹(−1/2).

Ans: We know sin(π/6) = 1/2. Since sin(−π/6) = −1/2 and −π/6 lies in the principal value range [−π/2, π/2], the principal value of sin⁻¹(−1/2) is −π/6.

Q2 (Medium): Prove that tan⁻¹(1/2) + tan⁻¹(1/3) = π/4.

Ans: 

tan⁻¹(a) + tan⁻¹(b) = tan⁻¹((a+b)/(1−ab)) when ab < 1

 tan⁻¹(1/2) + tan⁻¹(1/3) 

= tan⁻¹((1/2 + 1/3)/(1 − 1/6)) 

= tan⁻¹((5/6)/(5/6)) 

= tan⁻¹(1) = π/4

Hence proved.

Q3 (Medium): Simplify: tan⁻¹[√(1 + x²) − x] for x ∈ R.

Ans: 

Substitute x = cot θ

     θ = cot⁻¹(x)

√(1 + cot²θ) = cosec θ

tan⁻¹[cosec θ − cot θ] 

= tan⁻¹[(1 − cos θ)/sin θ] 

= tan⁻¹[tan(θ/2)] 

= θ/2 = (1/2)cot⁻¹(x)

Q4 (HOTS): If sin⁻¹x + sin⁻¹y + sin⁻¹z = π, prove that x√(1−x²) + y√(1−y²) + z√(1−z²) = 2xyz.

Ans: 

Let sin⁻¹x = A

sin⁻¹y = B, sin⁻¹z = C

A + B + C = π

sin A = x, sin B = y, sin C = z 

cos A = √(1−x²), cos B = √(1−y²)

cos C = √(1−z²)

Since A + B + C = π

we know sin 2A + sin 2B + sin 2C = 4 

sin A sin B sin C (a standard identity)

Substituting: 

2 sin A cos A + 2 sin B cos B + 2 sin C cos C = 4xyz

x√(1−x²) + y√(1−y²) + z√(1−z²) = 2xyz

Hence proved.

Chapter 3: Matrices - Important Questions

Q1 (Easy): If A is a 3×3 matrix and B is a 3×2 matrix, what is the order of the product AB? Is BA defined?

Ans: The product AB has order 3×2 because the inner dimensions match (3×3 multiplied by 3×2 gives a matrix of order 3×2). BA would require multiplying a 3×2 matrix by a 3×3 matrix — the inner dimensions 2 and 3 do not match, so BA is not defined.

Q2 (Medium): If A = [[2, 3], [1, 4]], find A² − 6A + 7I and verify whether it equals the zero matrix.

Ans:

A² = [[2,3],[1,4]] × [[2,3],[1,4]] 

     = [[4+3, 6+12],[2+4, 3+16]] 

     = [[7,18],[6,19]]

6A = [[12,18],[6,24]]

7I = [[7,0],[0,7]]

A² − 6A + 7I = [[7−12+7, 18−18+0],[6−6+0, 19−24+7]] 

        = [[2, 0],[0, 2]]

This equals 2I, not the zero matrix

(Note: For the Cayley-Hamilton theorem, A² − 6A + 5I = 0 for this matrix, as the characteristic equation is λ² − 6λ + 5 = 0.)

Q3 (Medium): Prove that for any square matrix A, (A + A') is always symmetric and (A − A') is always skew-symmetric.

Ans: 

Let B = A + A'

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B' = (A + A')' 

= A' + A = B

B' = B

The matrix (A + A') is symmetric

Let C = A − A'

C' = (A − A')' = A' − A = −(A − A') = −C

C' = −C, the matrix (A − A') is skew-symmetric

Q4 (HOTS): If A and B are symmetric matrices of the same order, prove that AB is symmetric if and only if A and B commute (i.e., AB = BA).

Ans: 

Given A and B are symmetric matrices: A' = A and B' = B

(AB)' = B'A' = BA

Now, (AB)' = AB, which means BA = AB. 

AB is symmetric ⟺ AB = BA. 

Hence AB is symmetric if and only if A and B commute

Chapter 4: Determinants (Pages 76-103) - Important Questions

Q1 (Easy): If A is a square matrix of order 3 and |A| = 5, find the value of |2A|.

Ans: For any n×n matrix, |kA| = kⁿ|A|

n = 3 and k = 2, 

|2A| = 2³ × |A| 

= 8 × 5 = 40

Q2 (Medium): Using properties of determinants, prove that: |a+b+2c, a, b; c, b+c+2a, b; c, a, c+a+2b| = 2(a+b+c)³.

Ans: 

Apply C₁ → C₁ + C₂ + C₃ to get a common factor of 2(a+b+c) in the first column.

Factor it out, then apply R₁ → R₁ − R₃ and R₂ → R₂ − R₃

After these operations, the remaining 3×3 determinant evaluates to (a+b+c)², giving the full result 2(a+b+c)³

Q3 (Medium): Find the area of the triangle whose vertices are (1, 0), (6, 0) and (4, 3) using the determinant method.

Ans: 

Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| 

= (1/2)|1(0 − 3) + 6(3 − 0) + 4(0 − 0)|

= (1/2)|−3 + 18| 

= (1/2)(15) = 7.5 square units

Q4 (HOTS): A square matrix A satisfies A³ = I. Using the property of determinants, find the possible values of |A| if A is a real matrix.

Ans: 

Since A³ = I

Taking determinants on both sides: 

|A³| = |I| = 1

Using the property |Aⁿ| = |A|ⁿ for square matrices: |A|³ = 1

|A| is a real cube root of 1. The only real cube root of 1 is 1 itself (the other two cube roots are complex). 

So, |A| = 1

Chapter 5: Continuity and Differentiability - Important Questions

Q1 (Easy): Differentiate f(x) = x³ sin x with respect to x.

Ans: Using the product rule: 

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f'(x) = (x³)' sin x + x³ (sin x)' 

       = 3x² sin x + x³ cos x

f'(x) = x²(3 sin x + x cos x)

Q2 (Medium): Find dy/dx if y = (log x)^(sin x).

Ans: 

Taking log of both sides: 

log y = sin x · log(log x)

Differentiating both sides with respect to x:

(1/y)(dy/dx) = cos x · log(log x) + sin x · (1/log x) · (1/x)

dy/dx = y[cos x · log(log x) + sin x/(x log x)] 

= (log x)^(sin x) [cos x · log(log x) + sin x/(x log x)]

Q3 (Medium): Verify Rolle's Theorem for f(x) = x² − 4x + 3 on [1, 3].

Ans: f(x) is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3)

f(1) = 1 − 4 + 3 = 0 

f(3) = 9 − 12 + 3 = 0

Since f(1) = f(3)

all conditions of Rolle's Theorem are satisfied

Now f'(x) = 2x − 4 = 0 gives x = 2

Since 2 ∈ (1, 3), there exists a point c = 2 in the open interval where f'(c) = 0

Rolle's Theorem is verified.

Q4 (HOTS): If xʸ = eˣ⁻ʸ, prove that dy/dx = log x / (1 + log x)².

Ans: 

Taking log of both sides:

y log x = (x − y) · 1 (since log e = 1)

y log x = x − y. 

y log x + y = x

y(1 + log x) = x

y = x/(1 + log x)

Differentiating using the quotient rule: 

dy/dx = [(1 + log x)(1) − x(1/x)] / (1 + log x)² 

= [1 + log x − 1] / (1 + log x)² 

= log x / (1 + log x)²

Hence proved.

Chapter 6: Application of Derivatives - Important Questions

Q1 (Easy): Find the intervals in which f(x) = 2x³ − 9x² + 12x + 1 is strictly increasing.

Ans: 

f'(x) = 6x² − 18x + 12 

= 6(x² − 3x + 2) 

= 6(x − 1)(x − 2)

f'(x) > 0 when (x − 1)(x − 2) > 0 i.e.

when x < 1 or x > 2

So f is strictly increasing on (−∞, 1) ∪ (2, ∞).

Q2 (Medium): Find the point on the curve y = x² that is closest to the point (0, 5).

Ans: 

Let a point on the curve be (t, t²)

Distance D from (0, 5) is given by D² = t² + (t² − 5)²

Let S = D² = t² + t⁴ − 10t² + 25 

= t⁴ − 9t² + 25. dS/dt 

= 4t³ − 18t = 2t(2t² − 9)

dS/dt = 0: t = 0 or t² = 9/2

At t² = 9/2: d²S/dt² > 0 (minimum)

The closest points are (±3/√2, 9/2)

Q3 (Medium): A ladder 5 m long is leaning against a wall. The bottom of the ladder is being pulled away from the wall at 1 m/s. How fast is the top of the ladder sliding down when the bottom is 3 m from the wall?

Ans: 

Let x be the distance of the foot from the wall, and y be the height of the top. 

x² + y² = 25

Differentiating with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0

When x = 3: y = √(25 − 9) = 4: dx/dt = 1 m/s

2(3)(1) + 2(4)(dy/dt) = 0

dy/dt = −3/4 m/s. 

The top is sliding down at 3/4 m/s

Q4 (HOTS): Show that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 times the volume of the sphere.

Ans: 

Let h be the height of the cone and r its base radius

From the sphere: r² = R² − (R − h)² = 2Rh − h²

V = (1/3)πr²h = (1/3)π(2Rh − h²)h 

= (π/3)(2Rh² − h³)

dV/dh = (π/3)(4Rh − 3h²) = 0 

h = 4R/3

r² = 2R(4R/3) − (4R/3)² = 8R²/9

V_max = (π/3)(8R²/9)(4R/3) 

= 32πR³/81

Volume of sphere = (4/3)πR³

Ratio = (32πR³/81)/((4/3)πR³) 

= (32/81) × (3/4) = 8/27

Hence proved.

Chapter 7: Integrals - Important Questions

Q1 (Easy): Evaluate ∫ (1 + tan x)/(1 − tan x) dx.

Ans: 

(1 + tan x)/(1 − tan x) 

= tan(π/4 + x). 

∫ tan(π/4 + x) dx 

= −log|cos(π/4 + x)| + C 

= log|sec(π/4 + x)| + C.

Q2 (Medium): Evaluate ∫₀^(π/2) log(sin x) dx.

Ans: 

Let I = ∫₀^(π/2) log(sin x) dx

Using the property ∫₀^a f(x) dx = ∫₀^a f(a−x) dx

I = ∫₀^(π/2) log(cos x) dx. 

Adding: 2I = ∫₀^(π/2) log(sin x cos x) dx 

= ∫₀^(π/2) log(sin 2x/2) dx 

= ∫₀^(π/2) log(sin 2x) dx − (π/2)log 2

Using substitution and the same property, ∫₀^(π/2) log(sin 2x) dx = I

2I = I − (π/2)log 2, giving I 

= −(π/2)log 2

Q3 (Medium): Evaluate ∫ x/(x² + 3x + 2) dx using partial fractions.

Ans: 

x/(x² + 3x + 2) = x/((x+1)(x+2))

Using partial fractions: x/((x+1)(x+2)) = A/(x+1) + B/(x+2)

Solving: A = −1, B = 2

So the integral = ∫[−1/(x+1) + 2/(x+2)] dx 

= −log|x+1| + 2log|x+2| + C 

= log[(x+2)²/|x+1|] + C.

Q4 (HOTS): Evaluate ∫₀^π x sin x/(1 + cos²x) dx.

Ans: 

Let I = ∫₀^π x sin x/(1 + cos²x) dx. 

Use the property f(π − x) = f(x) for the function: 

eplacing x by π − x: I = ∫₀^π (π − x) sin x/(1 + cos²x) dx. 

Adding both expressions: 2I = π ∫₀^π sin x/(1 + cos²x) dx. 

Let cos x = t; when x = 0, t = 1; when x = π, t = −1. 

2I = π ∫₋₁^1 dt/(1 + t²) = π[tan⁻¹t]₋₁^1 

 = π(π/4 − (−π/4)) = π²/2. 

Therefore, I = π²/4.

Chapter 8: Application of Integrals - Important Questions

Q1 (Easy): Find the area bounded by y = x² and the lines y = 0, x = 1 and x = 3.

Ans: Area = ∫₁³ x² dx = [x³/3]₁³ = 27/3 − 1/3 = 26/3 square units.

Q2 (Medium): Find the area of the region bounded by the parabola y = x² and the line y = x.

Ans: 

For intersection: x² = x gives x = 0 and x = 1

On [0,1], y = x lies above y = x²

Area = ∫₀¹ (x − x²) dx 

= [x²/2 − x³/3]₀¹ 

= 1/2 − 1/3 = 1/6 square units

Q3 (Medium): Find the area of the ellipse x²/a² + y²/b² = 1 using integration.

Ans: 

Area = 4 × ∫₀^a (b/a)√(a² − x²) dx

= (4b/a) × (πa²/4) 

= πab square unit

Q4 (HOTS): Find the area of the region {(x, y): x² + y² ≤ 4, x + y ≥ 2} using integration.

Ans: The circle x² + y² = 4 has centre O(0,0) and radius 2. The line x + y = 2 intersects the circle at (2, 0) and (0, 2). Required area = Area of circular sector from (2,0) to (0,2) minus the area of the triangle with vertices O(0,0), (2,0), (0,2). Sector area = (1/4)π(4) = π. Triangle area = (1/2)(2)(2) = 2. Required area = π − 2 square units.

Chapter 9: Differential Equations - Important Questions

Q1 (Easy): Find the order and degree of the differential equation: (d²y/dx²)³ + (dy/dx)⁴ + y = 0.

Ans: The highest-order derivative present is d²y/dx², so the order is 2. The degree is the power of the highest-order derivative when the equation is in polynomial form, here it is 3. Therefore order = 2 and degree = 3.

Q2 (Medium): Solve the differential equation dy/dx = (x + y + 1)² by an appropriate substitution.

Ans: 

Let v = x + y + 1

dv/dx = 1 + dy/dx

dv/dx − 1 = v² 

dv/dx = 1 + v²

dv/(1 + v²) = dx

tan⁻¹(v) = x + C

tan⁻¹(x + y + 1) = x + C.

Q3 (Medium): Solve: (x + y) dy/dx = 1.

Ans: 

dx/dy = x + y

dx/dy − x = y. 

This is a linear ODE in x with integrating factor e^(−y)

Multiplying through: 

d/dy[xe^(−y)] = ye^(−y)

xe^(−y) = −ye^(−y) − e^(−y) + C

x = −y − 1 + Ceʸ or x + y + 1 = Ceʸ

Q4 (HOTS): A population of bacteria doubles every 3 hours. If the initial population is 1000, find the population after 9 hours. Also, find the time at which the population will be 8000. Set up and solve the differential equation.

Ans: Let P be the population at time t

dP/dt = kP

P = P₀eᵏᵗ

t = 0, P₀ = 1000

Doubling time is 3 hours: 2000 = 1000e^(3k)

k = (log 2)/3

After 9 hours: P = 1000e^(3k × 3) 

= 1000 × 2³ = 8000

For P = 8000

8000 = 1000e^(kt)

e^(kt) = 8 = 2³

kt = 3 log 2

t = 9 hours

Chapter 10: Vector Algebra - Important Questions

Q1 (Easy): Find the unit vector in the direction of a⃗ = 2î − 3ĵ + 6k̂.

Ans: |a⃗| = √(4 + 9 + 36) = √49 = 7

Unit vector â = a⃗/|a⃗| 

= (1/7)(2î − 3ĵ + 6k̂) 

= (2/7)î − (3/7)ĵ + (6/7)k̂

Q2 (Medium): If a⃗ = î + 2ĵ + 3k̂ and b⃗ = 3î − 2ĵ + k̂, find a⃗ × b⃗ and verify that it is perpendicular to both a⃗ and b⃗.

Ans: 

a⃗ × b⃗ = |î ĵ k̂; 1 2 3; 3 −2 1| 

  = î(2−(−6)) − ĵ(1−9) + k̂(−2−6) 

  = 8î + 8ĵ − 8k̂ 

  = 8(î + ĵ − k̂)

(a⃗ × b⃗)·a⃗ = 8(1 + 2 − 3) = 0

(a⃗ × b⃗)·b⃗ = 8(3 − 2 − 1) = 0

Both dot products are zero, confirming perpendicularity.

Q3 (Medium): Find the projection of a⃗ = 2î + 3ĵ + 2k̂ on b⃗ = î + 2ĵ + k̂.

Ans: Projection of a⃗ on b⃗ = (a⃗·b⃗)/|b⃗|

a⃗·b⃗ = 2 + 6 + 2 = 10

|b⃗| = √(1 + 4 + 1) = √6

Projection = 10/√6 = 10√6/6 = 5√6/3

Q4 (HOTS): Prove that for any three vectors a⃗, b⃗, c⃗: a⃗×(b⃗×c⃗) = (a⃗·c⃗)b⃗ − (a⃗·b⃗)c⃗ (the vector triple product identity). Verify it with a⃗ = î, b⃗ = ĵ, c⃗ = k̂.

Ans: This is the standard BAC-CAB rule.

b⃗×c⃗ = ĵ×k̂ = î

a⃗×(b⃗×c⃗) = î×î = 0⃗

RHS: (a⃗·c⃗)b⃗ − (a⃗·b⃗)c⃗ 

= (î·k̂)ĵ − (î·ĵ)k̂ = 0·ĵ − 0·k̂ = 0⃗

LHS = RHS = 0⃗. The identity is verified.

Chapter 11: Three Dimensional Geometry - Important Questions

Q1 (Easy): Find the direction cosines of the line joining A(2, 3, −1) and B(3, −2, 1).

Ans: Direction ratios: (3−2, −2−3, 1−(−1)) = (1, −5, 2). Magnitude = √(1 + 25 + 4) = √30. Direction cosines: l = 1/√30, m = −5/√30, n = 2/√30.

Q2 (Medium): Find the angle between the planes 2x − y + 2z = 5 and 3x + 6y − 2z = 7.

Ans: Normal to plane 1: n⃗₁ = (2, −1, 2), Normal to plane 2: n⃗₂ = (3, 6, −2). cos θ = |n⃗₁·n⃗₂|/(|n⃗₁||n⃗₂|) = |6 − 6 − 4|/(3 × 7) = 4/21. Therefore θ = cos⁻¹(4/21).

Q3 (Medium): Find the shortest distance between the lines: r⃗ = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂) and r⃗ = (4î + 5ĵ + 6k̂) + μ(2î + 3ĵ + k̂).

Ans: 

Using the formula d = |(a₂⃗ − a₁⃗)·(b₁⃗ × b₂⃗)| / |b₁⃗ × b₂⃗|

b₁⃗ × b₂⃗ = (−3−6, 4−1, 3+6) = (−9, 3, 9)

|b₁⃗ × b₂⃗| = √(81+9+81) = √171 = 3√19

(a₂⃗ − a₁⃗) = (3, 3, 3)

Dot product = −27+9+27 = 9

d = 9/(3√19) = 3/√19 = 3√19/19 units

Q4 (HOTS): Find the equation of the plane passing through the points (1, 1, 0), (1, 2, 1) and (−2, 2, −1). Also find the perpendicular distance from the origin to this plane.

Ans: Two vectors in the plane: b⃗ = (0, 1, 1) and c⃗ = (−3, 1, −1)

Normal n⃗ = b⃗×c⃗ = (−1−1, −3−0, 0+3) = (−2, −3, 3)

Plane equation: −2(x−1) −3(y−1) + 3(z−0) = 0

−2x − 3y + 3z + 5 = 0 

2x + 3y − 3z − 5 = 0

Distance from origin = |0 + 0 − 0 − 5|/√(4+9+9) = 5/√22 = 5√22/22 units

Chapter 12: Linear Programming - Important Questions

Q1 (Easy): Define feasible region and feasible solution in the context of a Linear Programming Problem.

Ans: The feasible region is the set of all points that simultaneously satisfy all the constraints (inequalities) of a linear programming problem, including the non-negativity conditions. Any point that lies within or on the boundary of this region is called a feasible solution. The optimal solution, if it exists, is always found at a corner point (vertex) of the feasible region.

Q2 (Medium): Solve the LPP: Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Ans:

Solving 3x + 5y = 15 

5x + 2y = 10

x = 20/19, y = 45/19

O(0,0), A(2,0), B(20/19, 45/19), C(0,3)

Z(O) = 0, Z(A) = 10

Z(B) = 100/19 + 135/19 = 235/19 ≈ 12.37

Z(C) = 9

Z = 235/19 at x = 20/19, y = 45/19

Q3 (Medium): A manufacturer produces two products X and Y. Product X requires 2 hours of machine time and 1 hour of labour. Product Y requires 1 hour of machine time and 2 hours of labour. Available machine time is 10 hours and labour is 14 hours. Profit on X is ₹30 and on Y is ₹40. Formulate and solve the LPP.

Ans: Let x and y be units of X and Y. Maximise Z = 30x + 40y subject to: 2x + y ≤ 10, x + 2y ≤ 14, x ≥ 0, y ≥ 0. Corner points: (0,0), (5,0), (2,6), (0,7). Z values: 0, 150, 300, 280. Maximum profit is ₹300 by producing 2 units of X and 6 units of Y.

Q4 (HOTS): Explain with a diagram why the optimal solution in a bounded LPP always occurs at a corner point of the feasible region. Can there be infinitely many optimal solutions?

Ans: The objective function Z = ax + by represents a family of parallel lines. As we move these lines in the direction of increasing (or decreasing) Z, the last point of contact with the feasible region gives the optimum. If this last contact is a single vertex, that is the unique optimal solution. If the objective function line happens to coincide with an edge of the feasible region, every point on that edge is optimal giving infinitely many optimal solutions. This occurs when the objective function's slope matches the slope of a boundary edge of the feasible region.

Chapter 13: Probability - Important Questions

Q1 (Easy): A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is a king or a red card.

Ans: 

P(King) = 4/52, P(Red) = 26/52

P(King and Red) = 2/52

P(King or Red) = 4/52 + 26/52 − 2/52 

= 28/52 = 7/13

Q2 (Medium): A box contains 10 bulbs, of which 4 are defective. If 3 bulbs are drawn at random, find the probability that at least one is defective.

Ans: P(at least one defective) = 1 − P(none defective). P(none defective) = C(6,3)/C(10,3) = 20/120 = 1/6. Therefore P(at least one defective) = 1 − 1/6 = 5/6.

Q3 (Medium): Two cards are drawn successively without replacement from a pack of 52 cards. Find the probability that both are aces using the concept of conditional probability.

Ans: P(first ace) = 4/52. Given the first is an ace, P(second ace) = 3/51. By multiplication rule: P(both aces) = (4/52) × (3/51) = 12/2652 = 1/221.

Q4 (HOTS): In a town, 40% of people read newspaper A, 60% read newspaper B and 30% read both. A person is selected at random and found to read newspaper B. What is the probability that the person also reads newspaper A? Use Bayes' Theorem if needed or conditional probability directly.

Ans: P(A) = 0.4, P(B) = 0.6, P(A∩B) = 0.3. We need P(A|B) = P(A∩B)/P(B) = 0.3/0.6 = 0.5. So there is a 50% probability that the person also reads newspaper A, given that they read newspaper B.

Chapter Weightage Analysis - Class 12 Mathematics 2026-27

All class 12th Maths students should know the actual distribution of marks in the paper. A balanced approach to studying each chapter sounds like a good practice, but not necessarily a smart one. The distribution below will help you invest your preparation time into the most productive areas.

Understand Unit-wise Marks Distribution

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High-Priority Chapters for 2026-27 Exams

The table below lists chapters that have been the most consistent contributors to board exam marks over the last few years. These chapters offer extensive topical coverage with diverse question types and formats that students encounter in board exams. Students need to have a firm grip on these chapters, especially if they have little time to prepare.

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Hidden Scoring Chapters Students Must Cover

The biggest red flag in preparation is that students often focus on major chapters, like Calculus and Integration, while ignoring smaller-looking ones. That habit can cost them easy marks.

• Inverse Trigonometric Functions: Students think it is easy and hence do not do it. However, questions based on MCQ, simplification, and identities are repeated every year. A few days of practice can help score well.
• Application of Integrals: Students are indifferent to this chapter so they ignore it. But area-based questions of curves, circles, and parabolas are frequently asked questions, helping them earn step marks with diagrams.
• Differential Equations: The pattern of the questions is very standardized, so it may seem challenging at first. By practising  15-20 problems, you can score full marks in this topic.
• Linear Programming: The graph-based method looks difficult, but most questions of linear programming are done in the same way. Once you learn the corner point method, it becomes very easy to score.
• Relations and Functions: Students often leave it after Class 11 revision. But MCQs and short questions from functions, composition, and invertibility are regularly asked.
• Probability: This is not a chapter to skip. Topics like Bayes’ theorem and Bernoulli trials need practice, but they follow clear patterns and are high scoring.

How to Score High in Class 12 Maths Board Exam

Consistency is key! You can score 90+ in the Class 12 Maths board exam if you stay consistent, focus more on practice than learning theory, and don’t skip any chapter.

Step 1

Make a Concept Map: Only memorising formulas is not enough. Organise each chapter into a concept map with definitions, conditions, and when to use each formula. This builds real understanding.

Step 2

Practise Mixed Question Sets Regularly: While practising chapter-wise questions is helpful, it’s equally important to attempt mixed sets of MCQs, numericals, and reasoning questions together. This strengthens thinking capability and writing speed. 

Step 3

Enhance Answer Writing Format: When writing answers, first write the data, then the formula, then substitute values and then give your answer. Clearly present your answer to get full step marks.

Step 4

Check Mistakes After Practice: Once a practice set is finished, analyse your mistakes. Note if they are concept, calculation, or reading errors. Keep a record of mistakes and revise them.

Step 5

Take Sectional Mock Tests: Practise one-hour mock tests on chapters such as Calculus, Algebra and Vectors before full papers. This helps to develop good chapter knowledge and confidence.

Subject-specific Tips for Scoring High

• Drawing neat, well-labelled diagrams can earn potential marks even if not asked. It helps the examiner understand your solution and can improve presentation marks.
• For Calculus questions, state the rule you are using (product, chain, or quotient rule) before solving. This demonstrates clear understanding.
• Sketch the feasible region and label each corner point with corresponding coordinates. Missing labels or points can lose marks.
• Always add the constant of integration (C) in indefinite integrals. Failure to do so may result in loss of marks.
• Be clear about events such as P(A) = “probability that…”. This allows for clarity in your solution, and eliminates confusion.
• When answering matrix questions, always provide the order of the answer, e.g. 2×2, 3×1, etc. It is clear and understandable.

How to Use Class 12 Mathematics Important Questions for Best Results

Having a good understanding of the basics and using these questions for practice can only make Important Questions for CBSE Maths Class 12 useful. They should be used in conjunction with the NCERT textbook, concepts, and examples, and not as a substitute for learning the chapter.

A step-by-step approach is given to leverage Maths Important Questions most effectively:

Step 1: Solve like a Real Exam

Write complete answers on paper with proper steps and final answer. Avoid casual or your own logical solutions. It helps you develop the habit of writing exams and practice presenting calculations and reasoning clearly within the limited time.

Step 2: Active Recall Practice

Attempt the question without referring to the answer. This helps you to gain a better understanding and exam confidence as you learn to recall the correct concept, formula and method without relying on solutions.

Step 3: Note Difficult Questions

Circle the questions that looked easy, but you made mistakes in. These indicate your weak points and should be revised, particularly if the same type of error occurs more than once during practice.

Step 4: Fix Only the Incorrect Step

Do not rewrite the entire solution if you make an error. Just fix the part where you went wrong and comprehend it. Find out why the error happened - it can be due to an incorrect formula, a calculation error, or missing steps.

Step 5: Prepare a Short Revision List

After practice, note 5 learning points and/or mistakes that you made. This will aid you with revision later and provide you with a short list of ideas or errors to review before tests and examinations.

Why Solving Important Questions for CBSE Class 12 Mathematics is Essential

Important questions are the connection between reading Maths and actually solving Maths in exams. They act like a helpful resource for students to apply concepts, practice different question types, and determine areas that need more practice.

1. Move from chapter thinking to exam thinking

You just stop doing “chapter work” and begin thinking about the pattern and marking scheme of the exam and the structure of the answer. This also gives you an idea of how much time to spend on each question type and how to clearly present your answers.

2. Identifies repeated patterns

Many questions are similar to the board questions. After a bit of practice, you will notice the common methods when the question is different in the exam, for example, when it comes to integration, determinants, probability or 3D geometry, etc.

3. Applies theory to practice

If you learn the definition, formula, concept, etc., you apply it in problems. This will develop a real understanding and help in determining which method or formula is needed to solve a question.

4. Prepare you for mixed questions

In exams, questions come from different chapters sequentially. Solving important questions helps you switch topics without confusion and apply the right concept, rather than relying on the sequence in which you read the chapters.

5. Assists you in comprehending priorities

Of course, when solving, you naturally understand which topics need more practice, which are easier, which are harder and what kinds of questions you find chellenging. Thereby, you can invest more time in revising those areas while still covering the full syllabus.

Frequently Asked Questions

Q1. Which chapters are most important for Class 12 Maths 2026-27?

Ans. One of the most important subjects to concentrate on is Calculus, particularly Continuity and Differentiability, Integrals and Applications of Derivatives. Students should also pay due attention to Matrices, Determinants, Probability and Three-Dimensional Geometry.

Q2. Which questions from Class 12 Maths important questions should be practised?

Ans. Solve a variety of MCQs, assertion-reason, short-answer, long-answer, case-based, proof-based, and numerical questions. This will help you prepare for various types of questions, rather than only one type.

Q3. What are the most important chapters in class 12th maths that require more practice?

Ans. The chapters like Integrals, Continuity and Differentiability, Applications of Derivatives, Probability and Three-Dimensional Geometry require regular practice, as students might have lengthy questions as well as application-based questions.

Q4. Are important questions useful if I have studied NCERT Class 12 Maths?

Ans. Yes. After completing and practising NCERT questions, important questions can help you in solving different variations of these concepts, make you faster and also help you to know where you are lagging in the topics.

Q5. How many important questions should I solve in a day in Class 12 maths?

Ans. There is no definitive number. Solve approximately 10-15 quality questions every day and try to solve more as the exam nears. Work on the process more than on getting the answers right – and get them wrong.

Q6. How to prepare Integrals for 12th Maths?

Ans. First work through the standard integration formulas; then work through problems using a variety of techniques including substitution, integration by parts and partial fractions. Practice questions frequently to identify which strategy to employ without spending time.

Q7. Can I prepare using important questions for class 12th maths board exam?

Ans. Important questions are beneficial for practice but not a substitute for NCERT books, sample papers, and previous year's papers. Learn the concepts, then practice them using important questions, which will help you revise them.

Q8. How can I download the important questions of Class 12 Maths PDF?

Ans. Important Questions for Class 12 Maths PDF with Solutions can be downloaded using the links given above. You can download, print, practice on paper, review the errors and refine your strategy.

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