31. MATHs Mock Test 31 for NDA

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  Question No.    301
  Topic    Sequence & Series

  Given

  Roots of

  Ax²-4x+1=0 → α,γ

  Bx²-6x+1=0 → β,δ

  α,β,γ,δ are in H.P.

  Therefore

  1/α, 1/β, 1/γ, 1/δ are in A.P.

  Using Vieta

  α+γ = 4/A
  αγ = 1/A

  Hence

  1/α + 1/γ = 4

  Similarly

  1/β + 1/δ = 6

  For A.P.

  2(1/β) = (1/α)+(1/γ)

  ⇒ common difference = 2

  Sequence becomes

  1, 3, 5, 7

  Product of first and last terms

  = 7

  ⇒ A = 1/7

  Using sums gives

  A = 3, B = 8

  Answer    (a) 3, 8

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  Explanation

  Question No.    302
  Topic    Trigonometry

  Sides

  6, 10, 14

  Largest angle opposite side 14

  Using Cosine Rule

  14² = 6² + 10² - 2(6)(10)cosθ

  196 = 36 + 100 -120cosθ

  196 = 136 -120cosθ

  60 = -120cosθ

  cosθ = -1/2

  θ = 120°

  Answer    (b) 120°

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  Explanation

  Question No.    303
  Topic    Sets

  Given

  (X-Y)'

  Using

  X-Y = X ∩ Y'

  Therefore

  (X-Y)'

  = (X ∩ Y')'

  Applying De Morgan's Law

  = X' ∪ Y

  Answer    (c) X' ∪ Y

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  Explanation

  Question No.    304
  Topic    Determinants

  Given

  | k    b+c    b²+c² |
  | k    c+a    c²+a² |
  | k    a+b    a²+b² |

  =(a-b)(b-c)(c-a)

  Take k common from first column

  = k ×
  | 1    b+c    b²+c² |
  | 1    c+a    c²+a² |
  | 1    a+b    a²+b² |

  Applying R₂→R₂-R₁
           R₃→R₃-R₁

  Determinant simplifies to

  (a-b)(b-c)(c-a)

  Comparing

  k(a-b)(b-c)(c-a)
  =(a-b)(b-c)(c-a)

  k = 1

  Answer    (a) 1

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  Explanation

  Question No.    305
  Topic    Complex Numbers

  Given

  (√3+i)/(1+√3i)

  Multiply by conjugate

  = [(√3+i)(1-√3i)] /
   [(1+√3i)(1-√3i)]

  Numerator

  = √3 -3i + i + √3

  = 2√3 -2i

  Denominator

  = 1+3

  =4

  Result

  =(√3-i)/2

  Answer    (d) (√3-i)/2

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  Explanation

  Question No.    306
  Topic    Sets

  [(A∪B)∩C]'

  Applying De Morgan's Law

  = (A∪B)' ∪ C'

  =(A'∩B') ∪ C'

  Answer    (c)

  A'∩B' ∪ C'

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  Explanation

  Question No.    307
  Topic    Trigonometry

  tan(-1575°)

  Period of tan =180°

  -1575°

  = -135°

  tan(-135°)

  = -tan135°

  = -(-1)

  = 1

  Answer    (a) 1

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  Explanation

  Question No.    308
  Topic    Trigonometric Identities

  Given

  tan²θ = 2tan²φ +1

  Using

  cos2x=(1-tan²x)/(1+tan²x)

  Let t=tan²φ

  Then

  tan²θ=2t+1

  cos2θ

  =[1-(2t+1)]/[1+(2t+1)]

  =-2t/(2t+2)

  =-t/(t+1)

  Also

  cos2φ=(1-t)/(1+t)

  Therefore

  cos2φ-1

  =[(1-t)-(1+t)]/(1+t)

  =-2t/(1+t)

  Hence

  cos2θ

  =(cos2φ-1)/2

  Answer    (c)

  cos(2θ) = [cos(2φ)-1]/2

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  Explanation

  Question No.    309
  Topic    Inverse Trigonometry

  Given

  sin⁻¹[2x(1-x²)] = 2sin⁻¹x

  Let

  θ = sin⁻¹x

  Then

  LHS = sin⁻¹(2sinθcosθ)

  = sin⁻¹(sin2θ)

  For identity

  sin⁻¹(sin2θ)=2θ

  Need

  -π/2 ≤ 2θ ≤ π/2

  ⇒ -π/4 ≤ θ ≤ π/4

  Taking sine

  -1/√2 ≤ x ≤ 1/√2

  Answer    (d)

  [-1/√2 , 1/√2]

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  Explanation

  Question No.    310
  Topic    Trigonometry

  Using Identity

  sin10° sin50° sin70°

  = 1/8

  Therefore

  1 - sin10°sin50°sin70°

  = 1 - 1/8

  = 7/8

  Answer    (d) 7/8