Bihar Board 12th Math 2026 Question and Answer Key Pdf (Subject Code- 121/327) Download Pdf

Section A (MCQ) Bihar Board 12th Math 2026 Question

1. tan⁻¹(−1/√3) =

(A) π/3
(B) π/6
(C) −π/3
(D) −π/6

Answer: (D) −π/6

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2. 2 tan⁻¹(1/3) =

(A) tan⁻¹(3/2)
(B) tan⁻¹(3/4)
(C) tan⁻¹(4/3)
(D) tan⁻¹(2/3)

Answer: (D) tan⁻¹(2/3)

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3. x ∈ R, cot(tan⁻¹x + cot⁻¹x) =

(A) 1
(B) 1/2
(C) 0
(D) 1/3

Answer: (A) 1

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4. sin(cos⁻¹(3/5)) =

(A) 3/4
(B) 4/5
(C) 3/5
(D) 5/4

Answer: (B) 4/5

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5. tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2) =

(A) π
(B) 2π/3
(C) 3π/4
(D) π/2

Answer: (D) π/2

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6. cos⁻¹(cos 7π/6) =

(A) π/6
(B) π/3
(C) 5π/6
(D) 7π/6

Answer: (C) 5π/6

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7. tan⁻¹2 + tan⁻¹3 =

(A) −π/4
(B) π/4
(C) 3π/4
(D) π

Answer: (C) 3π/4

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8. If |x| ≤ 1, then tan(cos⁻¹x) =

(A) √(1 − x²)/x
(B) x/(1 + x²)
(C) √(1 + x²)/x
(D) √(1 − x²)

Answer: (A) √(1 − x²)/x

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9. tan⁻¹(x/y) − tan⁻¹(x/(x + y)) =

(A) −3π/4
(B) π/2
(C) π/4
(D) π/3

Answer: (C) π/4

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10. If |x| ≤ 1, cos⁻¹((1 − x²)/(1 + x²)) =

(A) 2cos⁻¹x
(B) 2sin⁻¹x
(C) 2tan⁻¹x
(D) tan⁻¹2x

Answer: (A) 2cos⁻¹x

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11. | 23 12 11 |
     | 36 10 26 | =
     | 63 26 37 |

(A) 1
(B) −1
(C) 0
(D) 2

Answer: (C) 0

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12. | a−b b−c c−a |
     | b−c c−a a−b | =
     | c−a a−b b−c |

(A) 1
(B) 0
(C) −1
(D) a + b + c

Answer: (B) 0

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13. | x 4 | = 0 ⇒ x =
     | 2 2x |

(A) ±2
(B) ±1
(C) ±3
(D) 0

Answer: (A) ±2

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14. | cos15° sin15° |
     | sin75° cos75° | =

(A) 1
(B) 0
(C) −1
(D) 1/2

Answer: (B) 0

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15. 1 − P(A′ ∩ B′) =

(A) P(A ∩ B)
(B) P(A ∪ B)
(C) P(A)
(D) P(B)

Answer: (B) P(A ∪ B)

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16. If A, B and C are three independent events, then P(A ∩ B ∩ C) =

(A) P(A) + P(B) + P(C)
(B) P(A) − P(B) − P(C)
(C) P(A)·P(B)·P(C)
(D) None of these

Answer: (C) P(A)·P(B)·P(C)

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17. If P(A)=7/13, P(B)=9/13 and P(A ∩ B)=4/13, then P(A|B) =

(A) 4/9
(B) 4/7
(C) 12/13
(D) 6

Answer: (A) 4/9

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18. If the odds against the event A is 3 : 7, then P(A) =

(A) 3/10
(B) 7/10
(C) 3/7
(D) 7/3

Answer: (B) 7/10

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19. d/dx (log xⁿ) =

(A) 1/xⁿ
(B) n
(C) 1/x
(D) n/x

Answer: (D) n/x

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20. d²/dx² (sin 2x) =

(A) 4 sin 2x
(B) 4 cos² 2x
(C) −4 sin 2x
(D) 2 sin 4x

Answer: (C) −4 sin 2x

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21. d/dx (e^x−a) =

(A) e^x−a
(B) (x−a)e^x−a
(C) e^x
(D) −e^x−a

Answer: (A) e^x−a

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22. d/dx √(x² + ax + 1) =

(A) (x + a) / (2√(x² + ax + 1))
(B) (2x + a) / (2√(x² + ax + 1))
(C) (2x + a) / √(x² + ax + 1)
(D) 1 / (2√(x² + ax + 1))

Answer: (B) (2x + a) / (2√(x² + ax + 1))

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23. d/dx (sin x²) =

(A) 2x cos x²
(B) cos x²
(C) x² cos x²
(D) x cos x²

Answer: (A) 2x cos x²

24. d/dx (√cot x) =

(A) 1 / (2√cot x)
(B) √cosec²x
(C) −cosec²x / (2√cot x)
(D) cosec²x / (2√cot x)

Answer: (C)

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25. d/dx (tan⁻¹√x + cot⁻¹√x) =

(A) π/2
(B) 0
(C) 1
(D) π

Answer: (A)

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26. d/dx (2 tan⁻¹ x) =

(A) 1 / (1 + x²)
(B) 2 / (1 + x²)
(C) 1 / (2(1 + x²))
(D) 1 / (1 − x²)

Answer: (A)

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27. d/dx (cos √x) =

(A) sin √x
(B) − sin √x / (2√x)
(C) sin √x / (2√x)
(D) 1 / (2√x)

Answer: (B)

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28. d/dx { lim_x→0 (x⁵ − a⁵) / (x − a) } =

(A) a
(B) 0
(C) 5a⁴
(D) 5

Answer: (C)

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29. d/dx (cot⁻¹ x) =

(A) 1 / (1 + x²)
(B) −1 / (1 + x²)
(C) 1 / x
(D) −1 / x

Answer: (B)

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30. The equation of the tangent to the curve y = x² + 4x + 1 at x = 3 is

(A) x + 10y = 8
(B) 10x + y = 8
(C) 10x − y = 8
(D) x − 10y = 8

Answer: (B)

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31. If y = √(sin x + √(sin x + √(sin x + …))), then dy/dx =

(A) 1 / (2y − 1)
(B) cos x / (2y − 1)
(C) sin x / (2y − 1)
(D) (2y − 1) / cos x

Answer: (B)

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32. If xⁿ + yⁿ = aⁿ, then dy/dx =

(A) − xⁿ⁻¹ / yⁿ⁻¹
(B) xⁿ⁻¹ / yⁿ⁻¹
(C) − yⁿ⁻¹ / xⁿ⁻¹
(D) nxⁿ⁻¹

Answer: (A)

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33. If x = a(1 − cosθ), y = a(θ + sinθ), then dy/dx =

(A) tan(θ/2)
(B) −tan(θ/2)
(C) cot(θ/2)
(D) −cot(θ/2)

Answer: (C)

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34. If y = x^x, then dy/dx =

(A) x^x(log x + 1)
(B) log x
(C) log x + 1
(D) nxⁿ⁻¹

Answer: (A)

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35. d/dx [sin⁻¹(3x − 4x³)] =

(A) 3 / √(1 − x²)
(B) 1 / √(1 − x²)
(C) −1 / √(1 − x²)
(D) 1 / (1 − x²)

Answer: (A)

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36. d/dx (sin 3x · cos 5x) =

(A) 4 cos 8x
(B) 4 cos 8x − cos 2x
(C) cos 2x
(D) cos 2x − 4 cos 8x

Answer: (A)

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37. d/dx (cos x³) =

(A) −3x² sin x³
(B) sin x³
(C) 3x² sin x³
(D) 3x²

Answer: (A)

38. If f : R → R, where f(x) = 3x − 5, then f⁻¹(x) =

(A) (1/3)(x + 5)
(B) (1/3)(x − 5)
(C) 3x − 5
(D) None of these

Answer: (A)

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39. ∫ (x + 2) dx =

(A) (x + 2)³ + k
(B) x²/2 + k
(C) x²/2 + 2x + k
(D) log(x + 2) + k

Answer: (C)

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40. ∫ cos 2x / (sin x + cos x)² dx =

(A) 2 log(sin x + cos x) + k
(B) log(sin x + cos x) + k
(C) log(sin x − cos x) + k
(D) −1/(sin x + cos x) + k

Answer: (D)

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41. ∫ (1 − cos 2x) / (1 + cos 2x) dx =

(A) tan x + x + k
(B) tan x − x + k
(C) x − tan²x + k
(D) tan(x/2) + k

Answer: (A)

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42. ∫ dx / (x − 1) =

(A) log|x + 1| + k
(B) −log|x + 1| + k
(C) log|x − 1| + k
(D) log x + k

Answer: (C)

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43. ∫ log x dx =

(A) 1/x + k
(B) x log x + k
(C) x log x − x + k
(D) x log x + x + k

Answer: (C)

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44. ∫ cos √x dx =

(A) sin√x + cos√x + k
(B) 1/2 (√x sin√x − cos√x) + k
(C) 2(√x sin√x + cos√x) + k
(D) sin√x + k

Answer: (B)

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45. ∫ e^x(tan⁻¹x + 1/(1 + x²)) dx =

(A) e^x tan⁻¹x + k
(B) e^x / (1 + x²) + k
(C) e^x + k
(D) tan⁻¹x + k

Answer: (A)

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46. ∫ dx / (x² − a²) =

(A) (1/a) tan⁻¹(x/a) + k
(B) (1/2a) log|(x − a)/(x + a)| + k
(C) (1/2a) log|(a + x)/(a − x)| + k
(D) (1/a) log|(x − a)/(x + a)| + k

Answer: (B)

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47. ∫ sin²(x/2) dx =

(A) 1/2 x − 1/2 sin x + k
(B) 1/2 x − 1/2 cos x + k
(C) 1/2 sin x + k
(D) −1/2 sin x + k

Answer: (A)

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48. ∫ tan(sin⁻¹x) / √(1 − x²) dx =

(A) log|sec(sin⁻¹x)| + k
(B) log|cos(sin⁻¹x)| + k
(C) tan(sin⁻¹x) + k
(D) log|sin⁻¹x| + k

Answer: (A)

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49. ∫ dx / (e^x + e^−x) =

(A) cot⁻¹(e^x) + k
(B) tan⁻¹(e^x) + k
(C) log|e^x + 1| + k
(D) sin⁻¹(e^x) + k

Answer: (B)

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50. ∫₀¹ dx / (1 + x²) =

(A) π/2
(B) π/3
(C) π/4
(D) π

Answer: (C)

51. ∫₀¹ x³ / (1 + x⁸) dx =

(A) π/2
(B) π/4
(C) π/8
(D) π/16

Answer: (C) π/8

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52. ∫₁^e (log x)² / x dx =

(A) 1/3
(B) (1/3)e³
(C) (1/3)(e³ − 1)
(D) e³

Answer: (B) (1/3)e³

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53. If A = [4 2 3], then A′ =

(A) [4 2 3]
(B) [3 2 4]
(C) [3 2 4]
(D) [4 2 3]^T

Answer: (D)

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54. ∫₀^π/4 e^{tan x} / cos²x dx =

(A) e − 1
(B) e + 1
(C) 1/e + 1
(D) 1/e − 1

Answer: (A) e − 1

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55. ∫₀^π/2 cos²x dx =

(A) π/2
(B) π
(C) π/4
(D) 1

Answer: (C) π/4

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56. ∫₀^π/3 cos³x dx =

(A) 3√3/8
(B) √3/8
(C) 3/8
(D) 1/8

Answer: (B) √3/8

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57. ∫₀^2π |sin x| dx =

(A) 2
(B) 4
(C) 1
(D) 3

Answer: (B) 4

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58. ∫₀^π/2 sin x / (sin x + cos x) dx =

(A) π
(B) π/2
(C) 0
(D) π/4

Answer: (D) π/4

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59. ∫_−π^π sin⁵x dx =

(A) 3π/4
(B) 2π
(C) 5π/6
(D) 0

Answer: (D) 0

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60. ∫₀^a √x / √(a − x + √x) dx =

(A) a
(B) a/2
(C) 2a
(D) 3a

Answer: (A) a

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61. ∫₀¹ e^√x / √x dx =

(A) 2(e − 1)
(B) e − 1
(C) 2(e + 1)
(D) e + 1

Answer: (A) 2(e − 1)

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62. ∫₀¹ x e^x dx =

(A) 1
(B) 0
(C) 2
(D) −1

Answer: (B) 0

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63. ∫₁⁴ dx / √x =

(A) 1
(B) −2
(C) 2
(D) −1

Answer: (C) 2

64. ∫₀^a √(a² − x²) dx =

(A) π/4
(B) a²/4
(C) πa²/4
(D) π

Answer: (C) πa²/4

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65. ∫₋₂² |x| dx =

(A) 4
(B) 3
(C) 2
(D) 0

Answer: (A) 4

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66. The order and degree of the differential equation
(d²s/dt²)² + (ds/dt)³ + 4 = 0 is

(A) order = 2, degree = 1
(B) order = 2, degree = 2
(C) order = 1, degree = 2
(D) order = 1, degree = 1

Answer: (B) order = 2, degree = 2

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67. The integrating factor of the differential equation
(1 + x²) dy/dx + y = e^tan−1x is

(A) e^tan−1x
(B) e^sin−1x
(C) tan⁻¹x
(D) sin⁻¹x

Answer: (A)

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68. The solution of the differential equation dy/dx = e^x+y is

(A) e^x + e^−y = k
(B) e^x + e^y = k
(C) e^−x + e^y = k
(D) e^−x + e^−y = k

Answer: (B)

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69. The solution of the differential equation
x (dy/dx) = cot y is

(A) x cos y = k
(B) x tan y = k
(C) x sec y = k
(D) x sin y = k

Answer: (C)

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70. If the operation * is defined as a * b = a + 2b, then (2 * 3) * 4 is

(A) 30
(B) 20
(C) 16
(D) 15

Answer: (B) 20

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71. If A = [ 1 −1
−1 1 ], then A³ =

(A) 3A
(B) 4A
(C) 2A
(D) None of these

Answer: (D)

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72. If A = [ cosα sinα
−sinα cosα ] and A + A′ = I, then α =

(A) π
(B) π/3
(C) 3π/2
(D) π/6

Answer: (D) π/6

73. If A = [ 3 −5
−1 2 ], then adj A =

(A) [ 2 5
1 3 ]
(B) [ 2 3
1 5 ]
(C) [ 1 3
2 5 ]
(D) None of these

Answer: (B)

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74. If A = [1 2 3 4] and B = [ 1
2
3
4 ], then AB =

(A) [30]
(B) [10]
(C) [20]
(D) [40]

Answer: (A)

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75. A = [a_ij]_m×n is a square matrix if

(A) m = n
(B) m < n
(C) m > n
(D) None of these

Answer: (A)

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76. If A = [ 3 −4
1 −1 ], then A + A′ =

(A) [ 6 −3
−3 −2 ]
(B) [ 6 3
3 2 ]
(C) [ 6 3
−3 −2 ]
(D) [ 6 −3
−3 2 ]

Answer: (A)

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77. f : A → B will be an onto function if

(A) f(A) ⊂ B
(B) f(A) = B
(C) f(A) ⊃ B
(D) None of these

Answer: (B)

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78. What type of relation is “less than” in the set of real numbers?

(A) Only symmetric
(B) Only transitive
(C) Only reflexive
(D) Equivalence

Answer: (B)

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79. | 3i + 4j + 7k | =

(A) √14
(B) √74
(C) √61
(D) √94

Answer: (D)

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80. i · ( j × k ) =

(A) 1
(B) 0
(C) −1
(D) i

Answer: (A)

81. If | a⃗ + b⃗ | = | a⃗ − b⃗ |, then

(A) |a⃗| = |b⃗|
(B) a⃗ ∥ b⃗
(C) a⃗ ⟂ b⃗
(D) None of these

Answer: (C)

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82. a⃗ · (a⃗ × a⃗) =

(A) 1
(B) 0
(C) a⃗
(D) −1

Answer: (B)

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83. If 3i⃗ + j⃗ − 2k⃗ and i⃗ + λj⃗ − 3k⃗ are perpendicular to each other, then the value of λ is

(A) −3
(B) −6
(C) −9
(D) −1

Answer: (D)

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84. The projection of the vector i⃗ + 3j⃗ + 7k⃗ on the vector 2i⃗ − 3j⃗ + 6k⃗ is

(A) 5
(B) 25
(C) 6
(D) None of these

Answer: (A)

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85. If |a⃗| = 2, |b⃗| = 3 and a⃗ · b⃗ = 4, then |a⃗ − b⃗| =

(A) 5
(B) √5
(C) 4
(D) 2

Answer: (B)

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86. If a⃗ = i⃗ − j⃗ + 2k⃗ and b⃗ = 2i⃗ + 3j⃗ − 4k⃗, then |a⃗ × b⃗| =

(A) √174
(B) √87
(C) √93
(D) None of these

Answer: (B)

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87. If |a⃗ + b⃗| = |a⃗ − b⃗|, then

(A) |a⃗| = |b⃗|
(B) a⃗ ∥ b⃗
(C) a⃗ ⟂ b⃗
(D) None of these

Answer: (C)

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88. 2j⃗ · (−3k⃗) =

(A) 6
(B) −6
(C) 0
(D) −6i⃗

Answer: (C)

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89. The direction ratios of a straight line are 2, 6, −3. Then its direction cosines are

(A) 1/7, 2/7, 3/7
(B) 2/7, −6/7, 3/7
(C) 2/7, 6/7, −3/7
(D) None of these

Answer: (C)

90. If a line makes angles α, β and γ with the positive directions of x, y and z axes respectively, then

(A) cos²α + cos²β + cos²γ = 1
(B) sin²α + sin²β + sin²γ = 4
(C) cos²α + cos²β + cos²γ = 2
(D) sin²α + sin²β + sin²γ = 1

Answer: (A)

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91. The angle between the straight lines (x − 2)/2 = (y − 1)/7 = (z + 3)/(−3) and (x + 2)/(−1) = (y − 4)/2 = (z − 5)/4 is

(A) π/2
(B) 0
(C) π/6
(D) π/4

Answer: (C)

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92. The distance of the plane x + 2y − 2z = 9 from the point (2, 3, −5) is

(A) 1
(B) 2
(C) 3
(D) 4

Answer: (C)

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93. If two planes x − 4y + λz + 3 = 0 and 2x + 2y + 3z = 5 are perpendicular to each other, then λ =

(A) 1
(B) 2
(C) 3
(D) 4

Answer: (C)

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94. If the line (x − x₁)/a₁ = (y − y₁)/b₁ = (z − z₁)/c₁ is parallel to the plane a₂x + b₂y + c₂z + d = 0, then

(A) a₁/a₂ = b₁/b₂ = c₁/c₂
(B) a₁x + b₁y + c₁z + d = 0
(C) a₁a₂ + b₁b₂ + c₁c₂ = 0
(D) None of these

Answer: (C)

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95. The equation of the plane whose intercepts on the x, y and z axes are −2, 3 and 4 respectively is

(A) 6x − 4y − 3z + 12 = 0
(B) 6x + 4y + 3z + 12 = 0
(C) 6x − 4y − 3z = 0
(D) None of these

Answer: (A)

96. If n(A) = 4 and n(B) = 2, then n(A × B) =

(A) 6
(B) 8
(C) 16
(D) None of these

Answer: (B) 8

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97. The solution of the differential equation x (dy/dx) = cot y is

(A) x cos y = k
(B) x tan y = k
(C) x sec y = k
(D) x sin y = k

Answer: (C)

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98. | a + ib  c + id |
    | −c + id a − ib | =

(A) a² + b² + c² + d²
(B) a² − b² − c² − d²
(C) a² − b² + c² + d²
(D) a² + b² + c² − d²

Answer: (A)

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99. The maximum value of Z = 4x + y subject to the constraints x + y ≤ 50, x ≥ 0, y ≥ 0 is

(A) 50
(B) 250
(C) 0
(D) None of these

Answer: (B) 250

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100. If n(A) = 4 and n(B) = 2, then n(A × B) =

(A) 6
(B) 8
(C) 16
(D) None of these

Answer: (B) 8

Section B Bihar Board 12th Math 2026 Question

Short Answer Type Question

1. Prove that tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π

Solution:
Using the identity tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1−ab))

tan⁻¹1 + tan⁻¹2 = tan⁻¹(3 / −1) = tan⁻¹(−3)

tan⁻¹(−3) + tan⁻¹3 = π

Hence proved.

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2. Find the value of
2tan⁻¹(1/3) + tan⁻¹(1/7)

Solution:
2tan⁻¹x = tan⁻¹(2x / (1 − x²))

2tan⁻¹(1/3) = tan⁻¹(3/4)

tan⁻¹(3/4) + tan⁻¹(1/7) = tan⁻¹(1)

= π/4

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3. Prove that
sec²(tan⁻¹2) + cosec²(cot⁻¹3) = 15

Solution:
Let tanθ = 2 ⇒ sec²θ = 1 + tan²θ = 5

Let cotφ = 3 ⇒ cosec²φ = 1 + cot²φ = 10

∴ Required sum = 5 + 10 = 15

Hence proved.

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4. Prove that the value of the determinant is zero:

| 102  18  36 |
|  1   3   4 |
| 17   3   6 |

Solution:
First row = 6 × Third row

Since two rows are proportional, the determinant is zero.

Hence proved.

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5. Prove that

| 0  a  b |
| c  0  b | = 2abc
| c  a  0 |

Solution:
Expanding along the first row:

= −a(−bc) + b(ca) = abc + abc = 2abc

Hence proved.

6. Find the value of x, when

| 2x  8 |
| 4   x | = 0

Solution:
Determinant = (2x)(x) − (8)(4)

2x² − 32 = 0
x² = 16
x = ±4

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7. Find the slope at the point (1, √2) of the curve
x² + y² = 3

Solution:
Differentiate implicitly:

2x + 2y (dy/dx) = 0
dy/dx = −x / y

At (1, √2):
dy/dx = −1 / √2

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8. Find dy/dx when
x = at², y = 2at

Solution:
dx/dt = 2at
dy/dt = 2a

dy/dx = (dy/dt)/(dx/dt) = 2a / (2at) = 1/t

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9. If y = sec(tan√x), find dy/dx

Solution:
Let u = tan√x

dy/dx = sec(u)tan(u) · du/dx

du/dx = sec²√x · (1 / 2√x)

∴ dy/dx = sec(tan√x) · tan(tan√x) · sec²√x / (2√x)

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10. Find dy/dx when
x^{sin y} = y^{sin x}

Solution:
Taking log on both sides:

sin y · log x = sin x · log y

Differentiate implicitly:

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

Collecting dy/dx terms and simplifying gives the required dy/dx.

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11. Find the inverse matrix of

[ 4  1 ]
[ 2  3 ]

Solution:
Determinant = (4)(3) − (1)(2) = 10

Inverse matrix = (1/10)

[ 3  −1 ]
[ −2  4 ]

12. If
A = [ 1  2
−3  4 ], B = [ 0  7
2  −4 ], find the value of (4A − B).

Solution:
4A = [ 4  8
−12  16 ]

4A − B = [ 4−0  8−7
−12−2  16−(−4) ]

= [ 4  1
−14  20 ]

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13. Integrate:
∫ dx / (1 + cos x)

Solution:
Multiply numerator and denominator by (1 − cos x):

∫ (1 − cos x) / (1 − cos²x) dx = ∫ (1 − cos x) / sin²x dx

= ∫ (cosec²x − cot x cosec x) dx

= −cot x + cosec x + C

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14. Integrate:
∫ sec⁴x dx

Solution:
sec⁴x = sec²x · sec²x

∫ sec²x (1 + tan²x) dx

= ∫ (1 + tan²x) d(tan x)

= tan x + (tan³x)/3 + C

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15. Integrate:
∫ sin⁻¹x / √(1 − x²) dx

Solution:
Let t = sin⁻¹x

dt = dx / √(1 − x²)

∴ Integral = ∫ t dt = t²/2 + C

= (sin⁻¹x)² / 2 + C

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16. Integrate:
∫ dx / (x² + 6x + 13)

Solution:
x² + 6x + 13 = (x + 3)² + 4

∫ dx / [(x + 3)² + 2²]

= (1/2) tan⁻¹((x + 3)/2) + C

17. Find the value of
∫₀^a x / √(a² + x²) dx

Solution:
Let u = a² + x² ⇒ du = 2x dx

∫ x / √(a² + x²) dx = ½ ∫ du / √u = √u

= √(a² + x²)

Applying limits 0 to a:

= √(a² + a²) − √(a²) = a(√2 − 1)

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18. Solve:
log_e(dy/dx) = ax + by

Solution:
dy/dx = e^ax+by

e^−by dy = e^ax dx

∫ e^−by dy = ∫ e^ax dx

(−1/b)e^−by = (1/a)e^ax + C

This is the required solution.

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19. Solve:
∫₁² (log x)/x dx

Solution:
Let t = log x ⇒ dt = dx/x

Integral = ∫ t dt = t²/2

Applying limits:

= (log 2)²/2 − (log 1)²/2 = (log 2)² / 2

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20. Find the value of
∫₀^π/2 √cos x / (√cos x + √sin x) dx

Solution:
Let I = ∫ √cos x / (√cos x + √sin x) dx

Replace x by (π/2 − x):

I = ∫ √sin x / (√sin x + √cos x) dx

Adding both:

2I = ∫ dx = π/2

∴ I = π/4

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21. Solve:
dy/dx + 2y tan x = sin x

Solution:
This is a linear differential equation

IF = e^{∫2 tan x dx} = sec²x

Multiplying throughout by IF:

d/dx (y sec²x) = sin x sec²x

Integrating:

y sec²x = sec x + C

y = cos x + C cos²x

22. If a + b + c = 0, prove that a × b = b × c = c × a.

Solution:
Given: a + b + c = 0 ⇒ a = −(b + c)

a × b = −(b + c) × b = −(b × b + c × b) = −(0 + c × b) = b × c

Similarly, b × c = c × a

∴ a × b = b × c = c × a. ✔

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23. Find the area of the parallelogram whose adjacent sides are
(i + 2j + 3k) and (−3i − 2j + k).

Solution:
Area = |a × b|

a × b = | i  j  k | | 1  2  3 | | −3 −2  1 |

= i(2×1 − 3×(−2)) − j(1×1 − 3×(−3)) + k(1×(−2) − 2×(−3))

= 8i − 10j + 4k

|a × b| = √(8² + (−10)² + 4²) = √180 = 6√5

∴ Area = 6√5 square units.

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24. Find the sine of the angle between the vectors
3i + j + 2k and 2i − 2j + 4k.

Solution:
sinθ = |a × b| / (|a||b|)

a × b = | i  j  k | | 3  1  2 | | 2 −2  4 |

= 8i − 8j − 8k

|a × b| = 8√3 |a| = √14, |b| = √24 = 2√6

sinθ = (8√3) / (√14 × 2√6) = 2 / √7

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25. If P(A)=1/4, P(B)=1/3 and P(A∪B)=1/2, prove that A and B are independent.

Solution:
P(A∪B) = P(A) + P(B) − P(A∩B)

1/2 = 1/4 + 1/3 − P(A∩B)

⇒ P(A∩B) = 1/12

P(A)·P(B) = (1/4)(1/3) = 1/12

∴ P(A∩B) = P(A)·P(B) Hence, A and B are independent events. ✔

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26. Two dice are thrown. Find the probability that the sum is 8 if the second die always shows 4.

Solution:
Given: second die = 4

Sum = 8 ⇒ first die must be 4

Total possible outcomes = 6 Favourable outcomes = 1

Probability = 1/6

27. Maximize Z = 3x + 2y
subject to:
3x + y ≤ 15,
x ≥ 0, y ≥ 0

Solution:
The feasible region is bounded by the axes and the line 3x + y = 15.

Corner points are:
(0,0), (5,0), (0,15)

Evaluate Z at these points:

Z(0,0) = 0
Z(5,0) = 15
Z(0,15) = 30

∴ Maximum value of Z is 30 at point (0,15).

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28. Find the distance between the planes
2x − y + 3z + 4 = 0 and 6x − 3y + 9z − 3 = 0

Solution:
Divide the second equation by 3:

2x − y + 3z − 1 = 0

Distance between parallel planes is:

|d₂ − d₁| / √(a² + b² + c²)

= |−1 − 4| / √(2² + (−1)² + 3²)

= 5 / √14

∴ Required distance = 5/√14.

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29. Find the value of p so that the straight lines
(x − 1)/(-1) = (y + 3)/p = (z − 7)/2
and
(x + 1)/p = y/2 = (z + 6)/(-3)
are perpendicular.

Solution:
Direction ratios of first line: (−1, p, 2)

Direction ratios of second line: (p, 2, −3)

For perpendicular lines, dot product = 0:

(−1)(p) + (p)(2) + (2)(−3) = 0

−p + 2p − 6 = 0 ⇒ p = 6

∴ Value of p = 6.

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30. If f : R → R, f(x) = (3 − x³)^1/3, prove that (f ∘ f)(x) = x.

Solution:
Given f(x) = (3 − x³)^1/3

(f ∘ f)(x) = f(f(x))

= [3 − {(3 − x³)^1/3}³]^1/3

= (3 − (3 − x³))^1/3

= (x³)^1/3 = x

∴ (f ∘ f)(x) = x. ✔

Long Answer Type Question

31. Prove that

| x  y  z |
| x² y² z² | = (x − y)(y − z)(z − x)(xy + yz + zx)
| yz zx xy |

Solution:

Let Δ =

| x  y  z |
| x² y² z² |
| yz zx xy |

Apply the operation:
R₂ → R₂ − xR₁
R₃ → R₃ − yzR₁

Then the determinant simplifies and common factors (x−y), (y−z), (z−x) appear.

After simplification, we get:

Δ = (x − y)(y − z)(z − x)(xy + yz + zx)

Hence proved.

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32. If sin⁻¹x + sin⁻¹y + sin⁻¹z = π, prove that

x√(1 − x²) + y√(1 − y²) + z√(1 − z²) = 2xyz

Solution:

Let sin⁻¹x = A, sin⁻¹y = B, sin⁻¹z = C

Then A + B + C = π

So, sin(A + B) = sin(π − C) = sin C

Using identity:

sin(A + B) = sinA cosB + cosA sinB

⇒ x√(1 − y²) + y√(1 − x²) = z

Multiplying both sides appropriately and adding cyclic terms, we get:

x√(1 − x²) + y√(1 − y²) + z√(1 − z²) = 2xyz

Hence proved.

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33. Solve:

(x² − y²) dy/dx = 2xy

Solution:

Rewriting the equation:

(x² − y²) dy = 2xy dx

Separate variables:

(x² − y²)/y dy = 2x dx

⇒ (x²/y − y) dy = 2x dx

Integrate both sides:

∫(x²/y − y) dy = ∫2x dx

⇒ x² ln|y| − y²/2 = x² + C

This is the required solution.

34. Prove that

∫₀^π/2 ( √tan x + √cot x ) dx = π√2

Solution:

Let I = ∫₀^π/2 ( √tan x + √cot x ) dx

Using the property:
∫₀^π/2 f(x) dx = ∫₀^π/2 f(π/2 − x) dx

⇒ I = ∫₀^π/2 ( √cot x + √tan x ) dx

Adding both expressions:

2I = 2 ∫₀^π/2 √tan x dx

⇒ I = ∫₀^π/2 √tan x dx

Using standard result:

∫₀^π/2 √tan x dx = π/√2

∴ I = π√2

Hence proved.

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35. Find dy/dx, when x^y + y^x = 1.

Solution:

Differentiate both sides w.r.t. x:

d/dx (x^y) + d/dx (y^x) = 0

x^y(y’/ln x + y/x) + y^x(ln y + x·y’/y) = 0

Collect y’:

y’ [ x^y ln x + y^x x/y ] = − [ x^y y/x + y^x ln y ]

∴ dy/dx = − ( x^y y/x + y^x ln y ) / ( x^y ln x + y^x x/y )

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36. Prove that

| a × b |² + (a · b)² = a²b²

Solution:

We know:

|a × b| = ab sinθ
a · b = ab cosθ

Squaring and adding:

|a × b|² + (a · b)²
= a²b²(sin²θ + cos²θ)

= a²b²

Hence proved.

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37. Maximize Z = x + y

subject to:
x − y ≤ −1
−x + y ≤ 0
x ≥ 0, y ≥ 0

Solution:

The feasible region is bounded by the lines:
x − y = −1, y = x, x = 0

Corner points are:

(0,1), (0,0), (1,1)

Evaluate Z:

Z(0,1) = 1
Z(0,0) = 0
Z(1,1) = 2

∴ Maximum value of Z is 2 at point (1,1).

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38. 10 coins are tossed. Find the probability that exactly 5 heads appear.

Solution:

Total number of outcomes = 2¹⁰

Number of favourable outcomes = ¹⁰C₅

Probability = ¹⁰C₅ / 2¹⁰

= 252 / 1024 = 63 / 256