35. MATHs Mock Test 35 for NDA

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  Explanation

  Question No.    341
  Topic    Definite Integrals

  Given

  I = ∫₀^(π/2) log(tan x) dx

  Let

  x = π/2 - t

  Then

  I = ∫₀^(π/2) log(cot t) dt

  = -∫₀^(π/2) log(tan t) dt

  = -I

  Therefore

  2I = 0

  I = 0

  Answer    (a) 0

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  Explanation

  Question No.    342
  Topic    Limits

  Given

  lim(x→0) [sin²(ax)]/(bx)

  Using

  sin(ax) ~ ax

  Therefore

  sin²(ax) ~ a²x²

  Limit

  = lim(x→0) a²x²/(bx)

  = (a²/b) lim(x→0)x

  = 0

  Answer    (a) 0

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  Explanation

  Question No.    343
  Topic    Differentiation

  Given

  f(x)=tanx + e^(-2x) - 7x³

  Differentiate

  f'(x)

  = sec²x - 2e^(-2x) - 21x²

  At x=0

  f'(0)

  = 1 - 2(1) - 0

  = -1

  Answer    (b) -1

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  Explanation

  Question No.    344
  Topic    Differential Equations

  Family of rectangular hyperbolas

  xy = c

  Differentiate

  x(dy/dx) + y = 0

  Highest derivative

  dy/dx

  Order = 1

  Power of derivative = 1

  Degree = 1

  Answer    (a) 1,1

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  Explanation

  Question No.    345
  Topic    Definite Integrals

  I = ∫₀¹ x(1-x)ⁿ dx

  Using Beta Function

  I = B(2,n+1)

  = Γ(2)Γ(n+1)/Γ(n+3)

  = 1!·n!/(n+2)!

  = 1/[(n+1)(n+2)]

  Answer    (b) 1/[(n+1)(n+2)]

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  Question No.    346
  Topic    Application of Derivatives

  Given

  P(x)

  = -3500 + (400-x)x

  = -x² + 400x -3500

  Differentiate

  P'(x)

  = -2x + 400

  For maximum profit

  P'(x)=0

  -2x+400=0

  x=200

  Since coefficient of x² is negative,
  profit is maximum.

  Answer    (c) 200

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  Explanation

  Question No.    347
  Topic    Application of Derivatives

  Given

  s = 64t -16t²

  Differentiate

  ds/dt

  = 64 -32t

  At maximum height

  ds/dt = 0

  64-32t=0

  t=2

  Answer    (b) 2 s

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  Question No.    348
  Topic    Functions

  Given

  f(x)=cosx

  g(x)=logx

  y=(g∘f)(x)

  = log(cosx)

  Differentiate

  dy/dx

  = 1/cosx × (-sinx)

  = -tanx

  At x=0

  dy/dx = 0

  Answer    (a) 0

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  Explanation

  Question No.    349
  Topic    Continuity

  Given

  f(x)=3x-4 , 0≤x≤2

  f(x)=2x+λ , 2<x≤3

  For continuity at x=2

  Left value

  f(2)=3(2)-4

  =2

  Right limit

  =2(2)+λ

  =4+λ

  Equate

  2=4+λ

  λ=-2

  Answer    (d) -2

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  Explanation

  Question No.    350
  Topic    Conic Sections

  Property of Ellipse

  Sum of distances of any point
  from the two foci

  = 2a

  where 2a is length of major axis.

  Therefore

  Sum of focal radii

  = Length of major axis

  Answer    (b) Length of major-axis