CBSE Class 12 Previous Year Questions
Board: CBSE Class 12 Previous Year Questions | Chapter-wise Previous Year Questions (2016-2025)
Practice with real exam questions solved by Ajay Yadav (Math King). Download the complete PDF or browse questions below.
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Chapter-wise Questions
12 chapters covering Relations & Functions, Inverse Trigonometry, Matrices, Determinants, Continuity & Differentiability, Application of Derivatives, Integration, Differential Equations, Vector Algebra, 3D Geometry, Linear Programming, and Probability.
Relations and Functions
Q1. Check if f:R→R given by f(x)=x³ is injective. [2020 – 2 marks]
Solution: f(x₁)=f(x₂) ⇒ x₁³=x₂³ ⇒ x₁=x₂. Hence injective (one-one).
Q2. Prove f:R→R given by f(x)=2x is one-one and onto. [2023 – 2 marks]
Solution: One-one: f(a)=f(b) ⇒ 2a=2b ⇒ a=b. Onto: For y∈R, x=y/2 gives f(x)=y. Hence bijective.
Q3. If f(x)=2x+3, find f^{-1}(x). [2019 – 1 marks]
Solution: y=2x+3 ⇒ x=(y-3)/2 ⇒ f^{-1}(x)=(x-3)/2.
Q4. Show f:N→N given by f(x)=2x is one-one but not onto. [2024 – 5 marks]
Solution: One-one: f(a)=f(b)⇒a=b. Not onto: odd numbers like 1 have no preimage as 2x=1 ⇒ x=0.5∉N.
Q5. Check if R={(1,1),(2,2),(3,3)} on {1,2,3} is reflexive, symmetric, transitive. [2022 – 1 marks]
Solution: Reflexive: all (a,a) present. Symmetric: if (a,b) then (b,a) – vacuously true. Transitive: vacuously true. It is an equivalence relation.
Inverse Trigonometric Functions
Q1. Find principal value of sin^{-1}(1/2). [2020 – 1 marks]
Solution: sin(π/6)=1/2 and π/6∈[-π/2,π/2]. Principal value = π/6.
Q2. Find principal value of tan^{-1}(√3). [2019 – 1 marks]
Solution: tan(π/3)=√3 and π/3∈(-π/2,π/2). Principal value = π/3.
Q3. Prove tan^{-1}(1/2)+tan^{-1}(1/3)=π/4. [2023 – 2 marks]
Solution: tan^{-1}(1/2)+tan^{-1}(1/3)=tan^{-1}[(1/2+1/3)/(1-1/6)]=tan^{-1}(1)=π/4.
Q4. Solve tan^{-1}(2x)+tan^{-1}(3x)=π/4. [2019 – 3 marks]
Solution: tan^{-1}[(2x+3x)/(1-6x²)]=π/4 ⇒ 5x/(1-6x²)=1 ⇒ 5x=1-6x² ⇒ 6x²+5x-1=0 ⇒ x=1/6 or x=-1. x=-1 rejected as LHS negative. x=1/6.
Q5. Prove 2tan^{-1}(1/5)+tan^{-1}(1/7)+2tan^{-1}(1/8)=π/4. [2024 – 5 marks]
Solution: Using formula, 2tan^{-1}(1/5)=tan^{-1}(5/12), 2tan^{-1}(1/8)=tan^{-1}(16/63). Sum = tan^{-1}(5/12)+tan^{-1}(1/7)+tan^{-1}(16/63)=π/4.
Matrices
Q1. If A=[[3,1],[-1,2]], show A²-5A+7I=0. [2020 – 2 marks]
Solution: A²=[[8,5],[-5,3]]. A²-5A+7I=[[8,5],[-5,3]]-[[15,5],[-5,10]]+[[7,0],[0,7]]=[[0,0],[0,0]].
Q2. If A=[[1,2],[2,1]], show A²-2A-3I=0 and find A^{-1}. [2019 – 2 marks]
Solution: A²=[[5,4],[4,5]], A²-2A-3I=0. A^{-1}=(A-2I)/3=[[-1/3,2/3],[2/3,-1/3]].
Q3. Find x if matrix [[4,4],[x,1]] is singular. [2018 – 1 marks]
Solution: |A|=0 ⇒ 4(1)-4(x)=0 ⇒ 4-4x=0 ⇒ x=1.
Q4. Express [[3,3,-1],[-2,-2,1],[-4,-5,2]] as sum of symmetric and skew-symmetric. [2024 – 5 marks]
Solution: A = (A+A’)/2 + (A-A’)/2. Symmetric part = [[3,1/2,-5/2],[1/2,-2,-2],[-5/2,-2,2]]. Skew-symmetric = [[0,5/2,3/2],[-5/2,0,3],[-3/2,-3,0]].
Determinants
Q1. Evaluate determinant of [[2,3],[4,5]]. [2019 – 1 marks]
Solution: |A| = 2×5 – 3×4 = 10-12 = -2.
Q2. If |A| = 5 for 3×3 matrix, find |adj A|. [2018 – 1 marks]
Solution: |adj A| = |A|^{n-1} = |A|² = 5² = 25.
Q3. Find x: |[[x,2],[3,4]]| = 2. [2022 – 1 marks]
Solution: 4x-6=2 ⇒ 4x=8 ⇒ x=2.
Q4. If A=[[1,2],[3,4]], find A^{-1} using adjoint. [2019 – 3 marks]
Solution: |A|=-2. adj A=[[4,-2],[-3,1]]. A^{-1}=adj A/|A|=[[-2,1],[3/2,-1/2]].
Q5. Solve system: 3x+4y+5z=18, 2x-y+8z=13, 5x-2y+7z=20 using matrix method. [2017 – 5 marks]
Solution: A=[[3,4,5],[2,-1,8],[5,-2,7]], |A|=-94 ≠0. X=A^{-1}B. x=3, y=1, z=1.
Compiled by Monomath.com | Ajay Yadav (Math King) | 15+ Years Teaching Experience