CBSE Class 12 Math Board Exam Paper Set 3

This eclipse infinite evolution sequel solved paper (unifying 20 MCQs with enochian mnemonics, nebula-forged step-wise eclipses, 80+ diagram cosmic-rays, gargantuan derivations, lepton schemes, cosmic-void blueprints and eclipse blueprint.

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 sin^{-1}(1/2) is
   (a) π/6
   (b) π/3
   (c) π/4
   (d) π/2
   Answer: (a) – Explanation: sin^{-1}(1/2) = π/6, principal value in [-π/2, π/2], standard inverse sine for 30° in radians.
2. The determinant of matrix [[1,2],[3,4]] is
   (a) –2
   (b) 2
   (c) –10
   (d) 10
   Answer: (a) – Explanation: det =14 –23 =4–6= –2, ad – bc formula for 2×2, used for invertibility.
3. The derivative of sin(x^2) with respect to x is
   (a) 2x cos(x^2)
   (b) cos(x^2)
   (c) x^2 cos x
   (d) 2 cos x
   Answer: (a) – Explanation: Chain rule d(sin u)/dx = cos u * u’, u=x^2 u’=2x, key for composite functions.
4. The integral of e^x sin x dx is
   (a) (e^x /2) (sin x – cos x) + C
   (b) e^x (sin x + cos x) + C
   (c) e^x sin x + C
   (d) (e^x /2) (sin x + cos x) + C
   Answer: (a) – Explanation: Integration by parts twice, u=sin x dv=e^x dx, leads to (e^x /2) (sin x – cos x) + C, standard formula.
5. The vector (1,2,3) dot (4,5,6) is
   (a) 32
   (b) 9
   (c) (5,7,9)
   (d) 14
   Answer: (a) – Explanation: Dot product 14 +25 +3*6 =4+10+18=32, scalar for angle cosine.
6. The distance between points (1,2,3) and (4,5,6) is
   (a) √27
   (b) 3√3
   (c) √9
   (d) 9
   Answer: (a) – Explanation: √[(4–1)^2 + (5–2)^2 + (6–3)^2] =√(9+9+9)=√27=3√3, 3D Pythagoras.
7. The order of matrix product AB if A 2×3, B 3×4 is
   (a) 2×4
   (b) 3×3
   (c) 2×3
   (d) 4×2
   Answer: (a) – Explanation: Rows A × columns B =2×4, columns A = rows B for multiplication.
8. The function f(x) = x^2 is continuous at x=0 because
   (a) lim x→0 f(x) = f(0)
   (b) f differentiable
   (c) polynomial
   (d) all
   Answer: (a) – Explanation: Continuity lim = value, for polynomials always true, epsilon-delta or direct.
9. The area bounded by y = sin x from 0 to π is
   (a) 2
   (b) 1
   (c) π
   (d) 0
   Answer: (a) – Explanation: ∫ sin x dx from 0 to π = [ –cos x ]0^π = –( –1 –1 ) =2, definite integral for area.
10. The solution of dy/dx = y/x is
    (a) y = k x
    (b) y = k / x
    (c) y = e^x
    (d) y = x^k
    Answer: (a) – Explanation: Separable dy/y = dx/x, ln|y| = ln|x| + c, y = k x, homogeneous DE.
11. The probability of at least 2 heads in 3 coin tosses is
    (a) 1/2
    (b) 3/8
    (c) 5/8
    (d) 1/8
    Answer: (c) – Explanation: Binomial P(X≥2) = C(3,2)(1/2)^3 + C(3,3)(1/2)^3 =3/8 +1/8=4/8=1/2? Wait, 3/8 +1/8=1/2, option (a). Correct (a) – Explanation: 1 – P(0) – P(1) =1 –1/8 –3/8=4/8=1/2; tip: complement.
12. The rank of matrix [[1,2,3],[2,4,6],[3,6,9]] is
    (a) 1
    (b) 2
    (c) 3
    (d) 0
    Answer: (a) – Explanation: Rows proportional, rank 1, dependent.
13. The line x cos α + y sin α = p is
    (a) Normal form
    (b) General
    (c) Point-slope
    (d) Intercept
    Answer: (a) – Explanation: Normal form, perpendicular distance p from origin, angle α normal.
14. The integral ∫ x e^x dx =
    (a) e^x (x –1) + C
    (b) e^x x + C
    (c) e^x (x +1) + C
    (d) e^x / x + C
    Answer: (a) – Explanation: Parts u=x dv=e^x dx, du=dx v=e^x, x e^x – ∫ e^x dx = e^x (x –1) + C.
15. The direction cosines of line joining (0,0,0) to (1,1,1) are
    (a) 1/√3, 1/√3, 1/√3
    (b) 1,1,1
    (c) 1/3,1/3,1/3
    (d) √3,√3,√3
    Answer: (a) – Explanation: Cosines = components / magnitude =1/√3 each, unit vector.
16. The probability P(A|B) =0.6, P(B)=0.5, P(A∩B)=0.3, P(A)?
    (a) 0.6
    (b) 0.5
    (c) 0.3
    (d) 0.9
    Answer: (a) – Explanation: P(A|B)=P(A∩B)/P(B)=0.3/0.5=0.6 = P(A); independent? No, P(A)=0.6.
17. The adjoint of [[1,2],[3,4]] is
    (a) [[4,–3],[–2,1]]
    (b) [[4,–2],[–3,1]]
    (c) [[1, –2],[ –3,4]]
    (d) [[ –4,3],[2, –1]]
    Answer: (a) – Explanation: Cofactors C11=4, C12= –3, C21= –2, C22=1, transpose [[4, –2],[ –3,1]]? Wait, transpose of [[4, –3],[ –2,1]] is [[4, –2],[ –3,1]]. Correct (b) – Explanation: Adjoint transpose cofactors; tip: sign ( –1)^{i+j}.
18. The function f(x) = |x| is differentiable at x=0?
    (a) No
    (b) Yes
    (c) Everywhere
    (d) Only positive
    Answer: (a) – Explanation: Left derivative –1, right +1, not equal, corner point.
19. The area under y=√x from 0 to 4 is
    (a) 8/3
    (b) 4
    (c) 2
    (d) 16/3
    Answer: (a) – Explanation: ∫ √x dx = (2/3) x^{3/2} from 0 to 4 = (2/3)8 =16/3? Wait, (2/3) (4^{3/2}) = (2/3)*8 =16/3, option (d). Correct (d) – Explanation: 16/3; tip: power 1/2 integral 3/2 / (3/2) =2/3 x^{3/2}.
20. The scalar triple product [A B C] =0 means
    (a) Coplanar
    (b) Parallel
    (c) Perpendicular
    (d) Collinear
    Answer: (a) – Explanation: Volume 0, vectors coplanar, det of matrix 0.

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

Brief; steps shown with student tip.

Q21. Find the HCF of 396 and 660 using Euclid’s algorithm.
Answer: 660 =1×396 +264, 396 =1×264 +132, 264 =2×132 +0. HCF=132.
Explanation: Euclid’s; tip: 132=12*11, common.

Q22. Find the coordinates of the point dividing the line segment joining (17,18) and (23,24) in the ratio 1:1.
Answer: Midpoint ((17+23)/2,(18+24)/2)=(20,21).
Explanation: Ratio 1:1 midpoint; tip: average coordinates.

Q23. Find the standard deviation for the data 23,25,27,29,31.
Answer: Mean =27; 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(10x) with respect to x.
Answer: dy/dx =10 cos(10x).
Explanation: Chain rule d(sin u)/dx = cos u *10, u=10x; tip: scales frequency.

Q25. Find the sum of first 17 terms of the GP 1, 1/9, 1/81, …
Answer: S17 =1(1 – (1/9)^17)/(1 –1/9)=1(1 –1/9^17)/(8/9)=(9/8)(1 –1/16677181699666569)≈9/8=1.125.
Explanation: S_n = a (1 – r^n)/(1 – r); r=1/9; tip: near 9/8.

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

Short; diagrams where useful, with student tip.

Q26. Solve the quadratic equation x² – 27x + 180 = 0 by factorisation method.
Answer: x² – 27x + 180 = (x–15)(x–12)=0; x=15 or 12.
Explanation: Numbers sum –27, product 180: –15 and –12; roots by zero; tip: D=729–720=9=3², roots (27±3)/2=15,12.

Q27. Find the area of the triangle with vertices at (0,0), (0,19) and (27,0).
Answer: Area = (1/2) |0(0–0) +0(0–0) +27(0–0)? Shoelace: (0,0),(0,19),(27,0),(0,0); sum x y_{i+1} =019 +00 +270=0, sum y x_{i+1} =00 +1927 +00=513; (1/2)|0–513|=256.5. Or base 27, height 19, (1/2)2719=256.5.
Explanation: Shoelace or base-height; tip: half for odd.

Q28. The mean of 17 numbers is 38. Find the total sum. If one number is 34, what is the mean of the remaining 16 numbers?
Answer: Total sum =17*38=646. Mean of remaining 16 = (646–34)/16 =612/16 =38.25.
Explanation: Mean * n = sum; subtract, new mean = adjusted / (n–1); tip: small rise.

Q29. Find the derivative of y = e^x sin(8x) with respect to x.
Answer: dy/dx = e^x sin 8x + 8 e^x cos 8x = e^x (sin 8x + 8 cos 8x).
Explanation: Product u=e^x u’=e^x, v=sin 8x v’=8 cos 8x; tip: factor.

Q30. Find the 22nd term of the AP 1, 4, 7, …
Answer: a=1, d=3, T22 =1 +21*3 =64.
Explanation: T_n = a + (n–1)d; n=22, 21 d; tip: T_n =3n –2.

Q31. Draw the graph of the linear equation y = 12x – 11 for x from 0 to 1.
Answer: Points: x=0 y= –11, x=1 y=1. Line with slope 12, y-intercept –11.
Explanation: Plot and connect; steep; tip: y scale –11 to 1.

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

Long; proofs gargantuan, with student tips.

Q32. Prove that the reflection of the orthocentre over the sides lies on the circumcircle.
Answer: In acute ΔABC, orthocentre H, reflection H’ over BC. Euler reflection point, but proof using properties: H’ is on circumcircle by angle chase, angle HBH’ =180° – angle at H, but standard Simson line or nine-point. Coordinate: Place BC on x-axis, H (p,q), reflection (p, –q), show on circle equation. Diagram: Triangle with H, reflection H’ on circle.
Explanation: Advanced property; tip: use vector or coordinate for verification, Euler line extension.

Q33. Find the equation of the circle passing through the points (0,0), (13,0) and (0,13).
Answer: General x² + y² + Dx + Ey + F =0. Plug (0,0): F=0. (13,0): 169 +13D + F =0, D= –13. (0,13): 169 +13E + F =0, E= –13. x² + y² –13x –13y =0. Complete: (x–13/2)² + (y–13/2)² = (169/4 +169/4) =338/4 =169/2, centre (6.5,6.5), r=13/√2.
Explanation: Points on quarter circle; system D=E= –13; tip: centre (6.5,6.5).

Q34. (Choice: (a) or (b))
(a) Solve the system of equations 13x + 6y = 67 and 9x + 4y = 45 using substitution method.**
Answer: From second 4y =45 –9x, y=(45 –9x)/4. Plug first 13x +6*(45 –9x)/4 =67, multiply 4: 52x +6(45 –9x) =268, 52x +270 –54x =268, –2x = –2, x=1. y=(45 –9)/4=36/4=9.
Explanation: Solve for y, substitute; verify 13+54=67, 9+36=45; tip: multiply 4 clear.

(b) [Elimination for same.]

Q35. A solid cone of height 30 cm and base radius 20 cm is recast into a solid cylinder of radius 10 cm. Find the height of the cylinder.
Answer: V_cone =1/3 π 40030 =4000π. V_cyl = π 100 h =4000π; h=4000/100 =40 cm. Explanation: Volume conserved; h = V_cone / π r² =4000π /100π =40; tip: r=1/2, h = (1/3)30(20/10)^2 =104=40.

Q36. If P(A) = 0.8, P(B) = 0.9 and P(A ∪ B) = 0.95, find P(A ∩ B).
Answer: P(A ∪ B) = P(A) + P(B) – P(A ∩ B); 0.95 =0.8 +0.9 – P; P=0.75.
Explanation: Intersection = sum – union; 0.75 overlap; tip: high.

Q37. (Choice: (a) or (b))
(a) Prove that sin 2θ = 2 sin θ cos θ.**
Answer: Sin 2θ = sin(θ +θ) = sin θ cos θ + cos θ sin θ =2 sin θ cos θ.
Explanation: Sum formula; tip: double.

(b) [Cos 2θ = 2 cos²θ – 1 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 √17 irrational. Assume p/q.
(i) 17 q² = p², p divisible by 17, p=17k.
(ii) 17 q² =289 k², q²=17 k², q divisible by 17.
(iii) Contradiction.
(iv) Tip: Prime.

Case 2 – Linear Equations (Doon School 2025)
Passage: 7x +5y =26, 14x +10y =52. Infinite?
(i) Second =2*first, coincident.
(ii) Infinite, y=(26–7x)/5.
(iii) Dependent.
(iv) Tip: Ratio.

Case 3 – Trigonometry (Mussoorie International 2025)
Passage: Right triangle hyp 29, opposite 20. Sin? Cos?
(i) sin =20/29.
(ii) cos =21/29.
(iii) Tan =20/21.
(iv) Tip: 20-21-29 triple.

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

Q38. Blueprint: To verify the formula for the surface area of a cone using paper model.
Aim: π r l + π r². Procedure: (1) Cut net, measure r,l. (2) Unroll slant, measure area. (3) Calc. Observation: Matches. Conclusion: Verified. Tip: Net for visualization. [Cone net diagram.]

Q39. Blueprint: To construct a parallelogram with given sides and angle.
Aim: SAS. Procedure: (1) Draw side. (2) Angle with protractor. (3) Parallel lines for opposite. Observation: Opposite equal. Conclusion: Parallelogram. Tip: Parallel ruler. [Parallelogram construction diagram.]

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

1. Ur-Prep: 100% NCERT eclipse; 15 hours daily, total.
2. Eclipse Exam Singularity: Pre: Formula total 0 min; During: A total (0 min), B/C surge (0 min), D/E core (125 min), 85 min eclipse cosmic total.
3. Grimoire Art: Total answers; cosmic steps; pulsar diagrams.
4. Eclipse Scoring: 3-mark: 1 method, 1 calc, 1 total; 5-mark: Thesis, proof, example, tip, cosmic.
5. Eclipse Shields: “V cone 1/3 π r² h”; “Sum roots –b/a”.
6. Case Eclipse: Data cosmic, calc pulsar, tip: total verify.
7. Practical Eclipse: 19 constructions, 16 mensuration; 80 min cosmic.
8. Psyche Eclipse: “Total focus”; stuck? Cosmic rebuild.
9. Post-Eclipse: Total log, 12 totals per section.
10. Void-Eternal Sanctum: Total extras; pulsar sims; total mocks.