D.

R and S both are equivalence relations

B.

1

|z – 1| = |z + 1|
⇒ lies on y-axis (perpendicular bisector
of the line segment joining (0, 1) and (0,-1)].
|z + 1| = |z – 1|
⇒ lies on y = -x
hence (0 + oe) is the only solution.

C.

1

The quadratic equation ax2 + bx +c = 0 has roots α and β,
Then α + β = - b/a, α β = c/a
Also, if ax²+ bx +c = 0
Then, 
We know that 1,ω, ω² are cube roots of unity.
1+ω + ω² = 0  (ω² = 1)
and 
Since α and β are roots of the equations
x²-x+1 = 0
⇒ α + β = 1 , α β =1

x = ω², or -ω
α = -ω², then β =-ω
or α = -ω, then β =-ω², (where ω³ = 1)
Hence,  α²⁰⁰⁹ + β²⁰⁰⁹ =(-ω)²⁰⁰⁹ + (-ω²)²⁰⁰⁹
 = - [(ω³)⁶⁶⁹. ω² + (ω³)¹³³⁷.ω]
 = - [ω2 + ω]
 = -(-1) = 1

B.

34 min

Let the first term of an AP be a and common difference be d and number of terms be n, then 
t_n = a + (n-1)d and Sn = n/2 [ 2a + (n-1)d]
Number of notes that the person counts in 10 min = 10 x 150 = 1500
Since a10, a11, a12, .... are in AP in with common difference -2
Let n be the time has taken to count remaining 3000 notes than
n/2[2 x 148 + (n-1) x -2] = 3000
⇒ n2-149n +3000 = 0 
⇒ (n-24)n-125) = 0 
n = 24, 125
Then, the total time taken by the person to count all notes = 10 +24 = 34 min
n = 125 is discarded as putting n = 125
an = 148 + (125-1)(-2)
= 148 - 124 x 2 = 148-248 = -100
⇒ Number of notes cannot be negative.

D.

y= 3

We have, 

On differentiating w.r.t x, we get

since the tangent is parallel to X- axis, therefore
dy/dx = 0
⇒ x³ = 8
⇒ x = 2 abd y =3