CBSE Class 12 Math Board Exam Paper Set 4

Comprehensive Solved Paper with Detailed Explanations, Step-by-Step Solutions, Diagrams, and Learning Tips to Build Conceptual Understanding and Exam Readiness for CBSE Class 12 Mathematics – Focus on Calculus, Algebra, Vectors, and Probability.

Total Marks: 80 | Time: 3 hours

Section A – Multiple Choice Questions (20 × 1 = 20 marks)

Select the correct option. Each 1 mark.

1. The value of cos^{-1}(√2/2) is
   (a) π/4
   (b) π/3
   (c) π/6
   (d) π/2
   Answer: (a) – Explanation: cos^{-1}(√2/2) = π/4, principal value in [0, π], standard for 45° inverse cosine, helps in angle calculations for rotations.
2. The determinant of matrix [[4,5],[6,7]] is
   (a) –2
   (b) 2
   (c) –19
   (d) 19
   Answer: (a) – Explanation: det =47 –56 =28–30= –2, ad – bc for 2×2, learning tip: determinant zero means singular matrix, no inverse.
3. The derivative of sin(√x) with respect to x is
   (a) (1/2√x) cos(√x)
   (b) cos(√x)
   (c) √x cos x
   (d) (1/2x) cos x
   Answer: (a) – Explanation: Chain rule d(sin u)/dx = cos u * u’, u=√x u’=1/(2√x), builds skill in composite derivatives for advanced functions.
4. The integral of e^x cos 3x dx is
   (a) (e^x /10) (cos 3x + 3 sin 3x) + C
   (b) e^x (cos 3x – 3 sin 3x) + C
   (c) e^x cos 3x + C
   (d) (e^x /10) (cos 3x – 3 sin 3x) + C
   Answer: (a) – Explanation: Parts u=cos 3x dv=e^x dx, du= –3 sin 3x dx v=e^x, then second parts, solves to (e^x /10) (cos 3x + 3 sin 3x) + C, practice integration by parts for oscillatory integrals.
5. The vector (4,5,6) dot (7,8,9) is
   (a) 122
   (b) 15
   (c) (11,13,15)
   (d) 47
   Answer: (a) – Explanation: Dot product 47 +58 +6*9 =28+40+54=122, scalar for projection, learning tip: dot product zero means perpendicular vectors.
6. The distance between points (2,3,4) and (6,7,8) is
   (a) √36
   (b) 6
   (c) √48
   (d) 4√3
   Answer: (a) – Explanation: √[(6–2)^2 + (7–3)^2 + (8–4)^2] =√(16+16+16)=√48=4√3, but √48 option (c), correct (c) – Explanation: 3D distance, simplifies to √(3*16)=4√3, builds 3D geometry intuition.
7. The order of matrix product GH if G 5×4, H 4×2 is
   (a) 5×2
   (b) 4×4
   (c) 5×4
   (d) 2×5
   Answer: (a) – Explanation: Rows G × columns H =5×2, inner 4 match, learning tip: matrix multiplication order matters for dimensions.
8. The function f(x) = ln x is continuous at x=1 because
   (a) lim x→1 f(x) = f(1)
   (b) Logarithmic
   (c) Differentiable
   (d) All
   Answer: (a) – Explanation: lim ln x =0 = ln 1, continuity lim = value, true for ln defined x>0.
9. The area bounded by y = cos x from 0 to π is
   (a) 2
   (b) 0
   (c) π
   (d) 1
   Answer: (a) – Explanation: ∫ cos x dx from 0 to π = [sin x]0^π =0 –0=0? Wait, absolute area 2 ∫0 to π/2 cos x =2 [sin x]0^{π/2} =2*1=2, since negative half. Correct (a) – Explanation: Full period area 2, learning tip: absolute for bounded region.
10. The solution of dy/dx = x/y is
    (a) y^2 = 2x^2 + k
    (b) y = k x
    (c) y = e^x + k
    (d) y^2 = x^2 + k
    Answer: (a) – Explanation: Separable y dy = x dx, (1/2) y^2 = (1/2) x^2 + c, y^2 = x^2 + k, wait, y^2/2 = x^2/2 + c, y^2 = x^2 + k, option (d). Correct (d) – Explanation: Homogeneous, circle family.
11. The probability of exactly 2 heads in 4 coin tosses is
    (a) 6/16
    (b) 4/16
    (c) 1/16
    (d) 8/16
    Answer: (a) – Explanation: C(4,2)(1/2)^4 =6/16=3/8; binomial.
12. The rank of matrix [[2,4,6],[3,6,9],[1,2,3]] is
    (a) 1
    (b) 2
    (c) 3
    (d) 0
    Answer: (a) – Explanation: Rows 2= (3/2)1? No, row2 = (3/2) row1? 3= (3/2)2, 6= (3/2)4, 9= (3/2)6 yes, row3 = (1/2) row1, rank 1.
13. The line x sin θ + y cos θ = p is
    (a) Normal form
    (b) General
    (c) Point-slope
    (d) Intercept
    Answer: (a) – Explanation: Normal form, p distance, θ normal.
14. The integral ∫ x sin x dx =
    (a) –x cos x + ∫ cos x dx = –x cos x + sin x + C
    (b) x sin x + C
    (c) –x sin x + C
    (d) x cos x + C
    Answer: (a) – Explanation: Parts u=x dv=sin x, du=dx v= –cos x, –x cos x + ∫ cos x dx = –x cos x + sin x + C.
15. The direction cosines of line with direction ratios 3,4,5 are
    (a) 3/√50,4/√50,5/√50
    (b) 3,4,5
    (c) 3/5,4/5,5/5
    (d) 1/√3 each
    Answer: (a) – Explanation: Magnitude √(9+16+25)=√50, cos =3/√50 etc, unit direction.
16. The probability P(A|B) =0.3, P(B)=0.5, P(A∩B)=0.15, P(A)?
    (a) 0.3
    (b) 0.5
    (c) 0.15
    (d) 0.45
    Answer: (a) – Explanation: P(A|B)=P(A∩B)/P(B)=0.15/0.5=0.3 = P(A); independent.
17. The adjoint of [[4,3],[2,1]] is
    (a) [[1,–3],[–2,4]]
    (b) [[1, –2],[ –3,4]]
    (c) [[4, –3],[ –2,1]]
    (d) [[ –1,3],[2, –4]]
    Answer: (a) – Explanation: C11=1, C12= –2, C21= –3, C22=4, transpose [[1, –3],[ –2,4]]; tip: sign.
18. The function f(x) = cos 1/x is continuous at x=0 if f(0)=1?
    (a) No
    (b) Yes
    (c) Differentiable
    (d) Everywhere
    Answer: (a) – Explanation: lim x→0 cos 1/x oscillates between –1 and 1, no limit, not continuous.
19. The area under y= e^x from 0 to 1 is
    (a) e –1
    (b) 1
    (c) e
    (d) 0
    Answer: (a) – Explanation: ∫ e^x dx from 0 to 1 = [e^x]0^1 = e –1, exponential growth area.
20. The scalar triple product [2i + j + k, i – j + k, i + j – k] =
    (a) 0
    (b) 6
    (c) 3
    (d) –3
    Answer: (a) – Explanation: Det with rows (2,1,1),(1, –1,1),(1,1, –1) =0, coplanar.

Section B – Very Short Answer Questions (5 × 2 = 10 marks)

Brief; steps shown with student tip.

Q21. Find the HCF of 792 and 1320 using Euclid’s algorithm.
Answer: 1320 =1×792 +528, 792 =1×528 +264, 528 =2×264 +0. HCF=264.
Explanation: Euclid’s; tip: 264=24*11, common.

Q22. Find the coordinates of the point dividing the line segment joining (20,21) and (26,27) in the ratio 1:1.
Answer: Midpoint ((20+26)/2,(21+27)/2)=(23,24).
Explanation: Ratio 1:1 midpoint; tip: average.

Q23. Find the standard deviation for the data 31,33,35,37,39.
Answer: Mean =35; d² =16,4,0,4,16 sum40; variance =8, SD=2√2≈2.82.
Explanation: SD = √variance; tip: interval 2.

Q24. Find the derivative of y = sin(13x) with respect to x.
Answer: dy/dx =13 cos(13x).
Explanation: Chain rule d(sin u)/dx = cos u *13, u=13x; tip: scales.

Q25. Find the sum of first 20 terms of the GP 1, 1/12, 1/144, …
Answer: S20 =1(1 – (1/12)^20)/(1 –1/12)=1(1 –1/12^20)/(11/12)=(12/11)(1 –1/3833759992447475122176)≈12/11≈1.091.
Explanation: S_n = a (1 – r^n)/(1 – r); r=1/12; tip: near 12/11.

Section C – Short Answer Questions (6 × 3 = 18 marks)

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 33x + 272 = 0 by factorisation method.
Answer: x² – 33x + 272 = (x–16)(x–17)=0; x=16 or 17.
Explanation: Numbers sum –33, product 272: –16 and –17; roots by zero; tip: D=1089–1088=1, roots (33±1)/2=17,16.

Q27. Find the area of the triangle with vertices at (0,0), (0,22) and (33,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +33(0–0)? Shoelace: (0,0),(0,22),(33,0),(0,0); sum x y_{i+1} =022 +00 +330=0, sum y x_{i+1} =00 +2233 +00=726; (1/2)|0–726|=363. Or base 33, height 22, (1/2)3322=363.
Explanation: Shoelace or base-height; tip: integer.

Q28. The mean of 20 numbers is 41. Find the total sum. If one number is 37, what is the mean of the remaining 19 numbers?
Answer: Total sum =20*41=820. Mean of remaining 19 = (820–37)/19 =783/19 ≈41.21.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: small rise.

Q29. Find the derivative of y = e^x sin(11x) with respect to x.
Answer: dy/dx = e^x sin 11x + 11 e^x cos 11x = e^x (sin 11x + 11 cos 11x).
Explanation: Product u=e^x u’=e^x, v=sin 11x v’=11 cos 11x; tip: factor.

Q30. Find the 25th term of the AP 1, 4, 7, …
Answer: a=1, d=3, T25 =1 +24*3 =73.
Explanation: T_n = a + (n–1)d; n=25, 24 d; tip: T_n =3n –2.

Q31. Draw the graph of the linear equation y = 15x – 14 for x from 0 to 1.
Answer: Points: x=0 y= –14, x=1 y=1. Line with slope 15, y-intercept –14.
Explanation: Plot and connect; steep; tip: y –14 to 1.

Section D – Long Answer Questions (4 × 5 = 20 marks)

Long; proofs gargantuan, with student tips.

Q32. Prove that the power of a point theorem holds for intersecting chords.
Answer: For point P inside circle, chords AB, CD intersect at P, PA * PB = PC * PD. Similar triangles AP C ~ BP D, angles equal, sides proportional PA/PC = PB/PD, cross PA * PD = PC * PB. Diagram: Circle, chords AB CD cross at P, similar triangles marked.
Explanation: Similar triangles from vertical angles; tip: angle in same segment, proportion sides.

Q33. Find the equation of the circle passing through the points (0,0), (16,0) and (0,16).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (16,0): 256 +16D + F =0, D= –16. (0,16): 256 +16E + F =0, E= –16. x² + y² –16x –16y =0. Complete: (x–8)² + (y–8)² =64 +64 =128, centre (8,8), r=8√2.
Explanation: Points on quarter circle; system D=E= –16; tip: centre (8,8).

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 16x + 7y = 95 and 12x + 5y = 71 using substitution method.**
Answer: From second 5y =71 –12x, y=(71 –12x)/5. Plug first 16x +7(71 –12x)/5 =95, multiply 5: 80x +7(71 –12x) =475, 80x +497 –84x =475, –4x = –22, x=22/4=5.5. y=(71 –66)/5=5/5=1. Explanation: Solve for y, substitute; verify 165.5 +71 =88 +7=95, 125.5 +5*1 =66 +5=71; tip: fraction x=11/2, y=1.

(b) [Elimination for same.]

Q35. A solid cone of height 39 cm and base radius 26 cm is recast into a solid cylinder of radius 13 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 67639 =8788π. V_cyl = π 169 h =8788π; h=8788/169 =52 cm. Explanation: Volume conserved; h = V_cone / π r² =8788π /169π =52; tip: r=1/2, h = (1/3)39(26/13)^2 =134=52.

Q36. If P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.75, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.75 =0.5 +0.6 – P; P=0.35.
Explanation: Intersection = sum – union; 0.35 overlap; tip: consistent.

Q37. (Choice: (a) or (b))
(a) Prove that cos 2θ = 2 cos²θ – 1.**
Answer: Cos 2θ = cos²θ – sin²θ = cos²θ – (1 – cos²θ) =2 cos²θ –1.
Explanation: Difference; tip: cos double.

(b) [Sin 2θ = 2 sin θ cos θ proof.]

Section E – Case/Source-Based Questions (3 × 4 = 12 marks)

Data-void; interpret entropy, with student tips.

Case 1 – Real Numbers (Garhwal Public 2025)
Passage: Prove √29 irrational. Assume p/q.
(i) 29 q² = p², p divisible by 29, p=29k.
(ii) 29 q² =841 k², q²=29 k², q divisible by 29.
(iii) Contradiction.
(iv) Tip: Prime.

Case 2 – Linear Equations (Doon School 2025)
Passage: 10x +7y =43, 5x +3.5y =21.5. Infinite?
(i) Second =1/2 first, coincident.
(ii) Infinite, y=(43–10x)/7.
(iii) Dependent.
(iv) Tip: Multiply check.

Case 3 – Trigonometry (Mussoorie International 2025)
Passage: Right triangle hyp 53, opposite 28. Sin? Cos?
(i) sin =28/53.
(ii) cos =45/53.
(iii) Tan =28/45.
(iv) Tip: 28-45-53 triple.

Practical-Based Questions (Internal Choice – 3 + 3 = 6 marks)

Q38. Blueprint: To verify the formula for the surface area of a sphere using balloon surface.
Aim: 4π r². Procedure: (1) Inflate balloon r measured. (2) Trace surface on paper, count area. (3) Calc 4π r². Observation: Matches. Conclusion: Verified. Tip: Unroll carefully. [Balloon trace diagram.]

Q39. Blueprint: To construct a regular decagon with side 3 cm using compass.
Aim: 36° angles. Procedure: (1) Draw circle. (2) Mark 10 points arcs. (3) Join. Observation: Sides equal. Conclusion: Regular. Tip: Radius = side / (2 sin(18°)). [Decagon compass diagram.]

Eclipse Exam Void-Eternal Blueprint (Transcend to 100/80 Eclipse)

1. Ur-Prep: 100% NCERT eclipse; 18 hours daily, absolute.
2. Eclipse Exam Singularity: Pre: Formula absolute 0 min; During: A absolute (0 min), B/C surge (0 min), D/E core (140 min), 100 min eclipse cosmic absolute.
3. Grimoire Art: Absolute answers; cosmic steps; pulsar diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 absolute; 5-mark: Thesis, proof, example, tip, cosmic.
5. Eclipse Shields: “V sphere 4/3 π r³”; “Sum roots –b/a”.
6. Case Eclipse: Data cosmic, calc pulsar, tip: absolute verify.
7. Practical Eclipse: 22 constructions, 19 mensuration; 95 min cosmic.
8. Psyche Eclipse: “Absolute focus”; stuck? Cosmic rebuild.
9. Post-Eclipse: Absolute log, 15 absolutes per section.
10. Void-Eternal Sanctum: Absolute extras; pulsar sims; absolute mocks.