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27. MATHs Mock Test 27 for NDA

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Question No.    261
Topic    Functions
Given    f(x) = (x-1)/(x+1)
Find    f(f(x))
Step 1    Let y = f(x) = (x-1)/(x+1)
Step 2    f(f(x))
    = (y-1)/(y+1)
Step 3    = [((x-1)/(x+1))-1] / [((x-1)/(x+1))+1]
Step 4    = [(-2)/(x+1)] / [(2x)/(x+1)]
Step 5    = -1/x
Answer    (d) -1/x
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Explanation

Question No.    262
Topic    Complex Numbers
Given    z = (√3+i)^(4n+1)/(1-i√3)^(4n)
Step 1    √3+i = 2(cosπ/6 + i sinπ/6)
Step 2    1-i√3 = 2(cos(-π/3)+i sin(-π/3))
Step 3    Arg(z)
    = (4n+1)(π/6) - 4n(-π/3)
Step 4    = (4n+1)π/6 + 8nπ/6
Step 5    = (12n+1)π/6
Step 6    = 2nπ + π/6
Principal Argument    π/6
Answer    (a) π/6
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Explanation

Question No.    263
Topic    3D Geometry
Plane 1    2x-y+z=6
Plane 2    x+y+2z=7
Normal Vector 1    n₁=(2,-1,1)
Normal Vector 2    n₂=(1,1,2)
n₁·n₂    =2(1)+(-1)(1)+1(2)=3
|n₁|    =√6
|n₂|    =√6
cosθ    =3/(√6×√6)=1/2
θ    =60°
Answer    (a) 60°
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Explanation

Question No.    264
Topic    Vectors
Given    |a|=4, |b|=4, |c|=5
Condition 1    a·(b+c)=0
    ⇒ a·b+a·c=0
Condition 2    b·(c+a)=0
    ⇒ b·c+a·b=0
Condition 3    c·(a+b)=0
    ⇒ a·c+b·c=0
Solving    a·b=a·c=b·c=0
Hence    Vectors are mutually perpendicular
|a+b+c|²
    =|a|²+|b|²+|c|²
    =16+16+25
    =57
|a+b+c|
    =√57
Answer    (d) √57
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Explanation

Question No.    265
Topic    Differentiation
Given    y = Σ tan⁻¹[1/(n²+n+1)]
Identity
    tan⁻¹(n+1)-tan⁻¹(n)
    = tan⁻¹[1/(n²+n+1)]
Therefore
y    = [tan⁻¹2-tan⁻¹1]
    +[tan⁻¹3-tan⁻¹2]
    +...
    +[tan⁻¹(x+1)-tan⁻¹x]
Telescoping Sum
y    = tan⁻¹(x+1)-tan⁻¹1
Differentiate
dy/dx    =1/[1+(x+1)²]
Answer    (b) 1/[1+(x+1)²]
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Explanation

Question No.    266
Topic    Matrices
Coefficient Matrix
| α 1 1 |
| 1 α 1 |
| 1 1 α |
Determinant
Δ=(α-1)²(α+2)
For no unique solution
Δ=0
⇒ α=1 or α=-2
Checking consistency
At α=1
0=1 (inconsistent)
At α=-2
System inconsistent
Hence No Solution
Answer    (d) α = -2 or 1
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Explanation

Question No.    267
Topic    Matrices
Given
A = | 0 1 |
|-1 0 |
A²=-I
Given
A=(αI+βA)²
Expanding
(αI+βA)²
=(α²-β²)I+2αβA
Comparing coefficients
α²-β²=0
2αβ=1
From α²=β²
α=β
2α²=1
α=±1/√2
β=±1/√2
Answer    (c) α = β = ±1/√2
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Explanation

Question No.    268
Topic    Application of Derivatives
f(x)=sin⁴x+cos⁴x
Using Identity
f(x)
=(sin²x+cos²x)²
-2sin²xcos²x
=1-(1/2)sin²2x
Differentiate
f'(x)
=-sin4x
For Increasing Function
f'(x)>0
⇒ sin4x<0
This occurs in
(3π/8 , 5π/8)
Answer    (c) (3π/8 , 5π/8)
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Explanation

Question No.    269
Topic    Sequence & Series
Given    a,b,c are in G.P.
Therefore
b²=ac
Expression
1/(a²-b²)+1/b²
Taking LCM
=[b²+a²-b²] / [b²(a²-b²)]
=a²/[b²(a²-b²)]
Using c=b²/a
Simplifying
=1/(b²-c²)
Answer    (b) 1/(b²-c²)
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Explanation

Question No.    270
Topic    Inverse Function
Given
y=(10^x-10^(-x)) /(10^x+10^(-x))
Let
t=10^x
Then
y=(t²-1)/(t²+1)
Cross Multiply
y(t²+1)=t²-1
yt²+y=t²-1
t²(1-y)=1+y
t²=(1+y)/(1-y)
Taking log
x=(1/2)log₁₀[(1+y)/(1-y)]
Replacing y by x
f⁻¹(x)
=(1/2)log₁₀[(1+x)/(1-x)]
Answer    (d) ½log₁₀[(1+x)/(1-x)]

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