26. MATHs Mock Test 26 for NDA

• 0 Questions answered correctly
• 0 Questions answered incorrect
• 10 Questions unattempted

• 1

  

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  Explanation

  Question:

  dy/dx = (ax+h)/(by+k)

  When does the general solution represent a parabola?

  Explanation:

  dy/dx = (ax+h)/(by+k)

  ⇒ (by+k)dy = (ax+h)dx

  Integrating:

  (b/2)y² + ky
  = (a/2)x² + hx + C

  General equation:

  ay² − bx² + linear terms + constant = 0

  This is a second-degree curve.

  For parabola:

  Coefficient of one squared term must vanish.

  Thus either

  a = 0, b ≠ 0

  or

  b = 0, a ≠ 0

  Among options only

  a = 0, b ≠ 0

  satisfies.

  Answer:
  (c)

• 2

  

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  Explanation

  Question:

  lim x→0

  [√(1+x²) − √(1+x)]
  -------------------
  [√(1+x³) − √(1+x)]

  Explanation:

  Use expansion:

  √(1+t)

  ≈ 1 + t/2

  Numerator:

  ≈ (1+x²/2) − (1+x/2)

  = (x²−x)/2

  Denominator:

  ≈ (1+x³/2) − (1+x/2)

  = (x³−x)/2

  Therefore

  Limit

  = (x²−x)/(x³−x)

  = x(x−1)/x(x²−1)

  = (x−1)/[(x−1)(x+1)]

  = 1/(x+1)

  Putting x=0

  = 1

  Answer:
  (b) 1

• 3

  

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  Explanation

  Question:

  Coefficient of x⁵ and x¹⁵ are equal in

  (x² + a/x³)¹⁰

  Find positive a.

  Explanation:

  General term:

  Tr+1

  = ¹⁰Cr (x²)^(10−r)
  (a/x³)^r

  = ¹⁰Cr a^r x^(20−5r)

  For x⁵:

  20−5r=5

  r=3

  Coefficient

  = ¹⁰C₃ a³

  For x⁻¹⁵:

  20−5r=−15

  r=7

  Coefficient

  = ¹⁰C₇ a⁷

  Since

  ¹⁰C₃=¹⁰C₇

  Therefore

  a³=a⁷

  a⁴=1

  Positive value:

  a=1

  Answer:
  (c) 1

• 4

  

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  Explanation

  Question:

  Find n if determinant is zero.

  Explanation:

  Using identity:

  ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁

  Observe:

  C(8,3)+C(8,4)=C(9,4)

  C(9,5)+C(9,6)=C(10,6)

  C(10,7)+C(10,8)=C(11,8)

  Hence

  R₃ becomes

  (C(9,n), C(10,n+2), C(11,n+4))

  To make determinant zero,

  R₃ must equal R₁+R₂

  Thus

  n=4

  because

  C(9,4), C(10,6), C(11,8)

  Answer:
  (c) 4

• 5

  

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  Explanation

  Question:

  Find inverse of

  | 3  -10  -1 |
  | -2   8   2 |
  |  2  -4  -2 |

  Explanation:

  Checking option (b):

  A × B

  =

  |1 0 0|
  |0 1 0|
  |0 0 1|

  Hence option (b)
  is the inverse matrix.

  Answer:
  (b)

  |1/2   0    1/2 |
  | 1   1/4   1/2 |
  |3/4  1/4  -1/4 |

• 6

  

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  Explanation

  Question:

  f(x)=a|x|²+b|x|+c

  When does f'(0) exist?

  Explanation:

  Since

  |x|² = x²

  f(x)

  = ax² + b|x| + c

  Derivative of ax² exists everywhere.

  Problematic term:

  b|x|

  LHD at 0 = −b

  RHD at 0 = +b

  For derivative to exist:

  −b = b

  b = 0

  Answer:
  (b)

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  Explanation

  Question:

  R on integers defined by

  (a,b)∈R ⇔ a²+b²=25

  Find domain.

  Explanation:

  Need all integers a for which
  some integer b exists.

  a²+b²=25

  Possible integer pairs:

  (±5,0)

  (0,±5)

  (±4,±3)

  (±3,±4)

  Hence possible a values:

  {0, ±3, ±4, ±5}

  Answer:
  (c)

• 8

  

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  Explanation

  Question:

  P(win)=1/3

  P(draw)=1/6

  P(loss)=1/2

  3 matches.

  Win=2 points
  Draw=1 point
  Loss=0 point

  Find probability of scoring 5 points.

  Explanation:

  Total 5 points possible only as:

  Win + Win + Draw

  Probability

  = 3 × (1/3)(1/3)(1/6)

  = 3/54

  = 1/18

  Answer:
  (d)

• 9

  

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  Explanation

  Question:

  3/4 − 5/4² + 3/4³ − 5/4⁴ + ...

  Explanation:

  Group terms:

  S

  = (3/4)(1+1/4²+1/4⁴+...)

  − (5/4²)(1+1/4²+1/4⁴+...)

  Common GP:

  1 + 1/16 + 1/256 + ...

  = 1/(1−1/16)

  = 16/15

  Therefore

  S

  = [3/4 − 5/16] × 16/15

  = (12−5)/16 ×16/15

  = 7/15

  Answer:
  (c) 7/15

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  Explanation

  Question:

  z=(√3+i)/2

  Find

  (z¹⁰¹ + i¹⁰³)¹⁰⁵

  Explanation:

  z

  = cos(π/6)+i sin(π/6)

  Therefore

  z¹⁰¹

  = z^(96+5)

  = z⁵

  = cos(5π/6)+i sin(5π/6)

  = (-√3+i)/2

  Also

  i¹⁰³

  = i³

  = -i

  Hence

  z¹⁰¹+i¹⁰³

  = (-√3+i)/2 - i

  = (-√3-i)/2

  = cos(7π/6)+i sin(7π/6)

  = z⁷

  Therefore

  (z¹⁰¹+i¹⁰³)¹⁰⁵

  = (z⁷)¹⁰⁵

  = z⁷³⁵

  = z^(735 mod 12)

  = z³

  Answer:
  (c) z³