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29. MATHs Mock Test 29 for NDA

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Explanation

Question No.    281
Topic    Sets
Given
A∪B = A∪C
A∩B = A∩C
Let x ∈ B
Case 1: x ∈ A
Then x ∈ A∩B
⇒ x ∈ A∩C
⇒ x ∈ C
Case 2: x ∉ A
Since x ∈ B
⇒ x ∈ A∪B
⇒ x ∈ A∪C
⇒ x ∈ C
Hence B ⊆ C
Similarly C ⊆ B
Therefore
B = C
Answer    (b) B = C
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Explanation

Question No.    282
Topic    Vectors
Given
a = λi - 3j - k
b = 2λi + λj - k
Condition 1
Angle between a and b is acute
a·b > 0
(λ)(2λ)+(-3)(λ)+(-1)(-1) > 0
2λ²-3λ+1 > 0
(2λ-1)(λ-1) > 0
⇒ λ < 1/2 or λ > 1
Condition 2
Angle between b and Y-axis lies between π/2 and π
⇒ component along j-axis is negative
λ < 0
Combining both conditions
λ < 0
Answer    (d) λ < 0
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Explanation

Question No.    283
Topic    Permutation & Combination
Required
4-digit numbers between 5000 and 10000
Digits available
1,2,3,4,5,6,7,8,9
First digit
Must be 5,6,7,8 or 9
Number of choices = 5
Remaining three places
Choose and arrange 3 digits from remaining 8 digits
= 8P3
Total numbers
= 5! × 8P3
Answer    (a) 5! × 8P3
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Explanation

Question No.    284
Topic    Determinants
Let cube roots of unity be
1, ω, ω²
Using
1+ω+ω² = 0
and
ω³ = 1
Substituting values and simplifying determinant
Δ = -3√3 i
Therefore
Re(Δ) = 0
Im(Δ) ≠ 0
Answer    (a) Re(Δ) = 0
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Explanation

Question No.    285
Topic    Inverse Trigonometry
Given
sin⁻¹(1-x) - 2sin⁻¹x = π/2
Let
sin⁻¹x = θ
Then
sin⁻¹(1-x) = π/2 + 2θ
LHS can equal π/2 only when
θ = 0
⇒ x = 0
Verification
sin⁻¹(1) - 0
= π/2
RHS = π/2
Satisfied
Answer    (c) 0
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Explanation

Question No.    286
Topic    Circle
Circle 1
(x-1)² + (y-3)² = r²
Centre C₁ = (1,3)
Radius = r
Circle 2
x²+y²-8x+2y+8=0
⇒ (x-4)²+(y+1)²=9
Centre C₂=(4,-1)
Radius=3
Distance between centres
d = √[(4-1)²+(-1-3)²]
= √(9+16)
= 5
For two distinct intersections
|r-3| < 5 < r+3
Solving
r < 8
and
r > 2
Hence
2 < r < 8
Answer    (a) 2 < r < 8
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Explanation

Question No.    287
Topic    Matrices
Given
A =
[1 2 2]
[2 1 2]
[2 2 1]
Observe
A = 2J - I
where J is matrix of all ones
Using
J² = 3J
A²
=(2J-I)²
=4J²-4J+I
=12J-4J+I
=8J+I
Now
A²-4A
=(8J+I)-4(2J-I)
=8J+I-8J+4I
=5I
Answer    (d) 5I
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Explanation

Question No.    288
Topic    Determinants
D =
|a b c|
|b c a|
|c a b|
Using standard determinant expansion
D
= -(a³+b³+c³-3abc)
= 3abc-a³-b³-c³
Answer    (b)
3abc - a³ - b³ - c³
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Explanation

Question No.    289
Topic    Vectors
For unit vectors a and b
|a+b|²
= |a|²+|b|²+2a·b
= 1+1+2cosθ
= 2(1+cosθ)
= 4cos²(θ/2)
Therefore
|a+b| = 2cos(θ/2)
Hence
cos(θ/2)
= |a+b|/2
Answer    (b) |a+b|/2
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Explanation

Question No.    290
Topic    Permutation & Combination
MISSISSIPPI
M = 1
I = 4
S = 4
P = 2
Arrange all letters except S
Letters:
M, I,I,I,I, P,P
Number of arrangements
= 7!/(4!2!)
= 105
Available gaps around these letters
_ M _ I _ I _ I _ I _ P _ P _
Total gaps = 8
Choose 4 gaps for 4 S's
= 8C4
Required arrangements
= [7!/(4!2!)] × 8C4
= 105 × 70
= 7350
Equivalent option
= 7·6 × 8C4
Answer    (d)

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