A)
Re(2iz1−z2)=Re(2i(cosθ+isinθ)1−(cosθ+isinθ)2)
=Re(2i(cosθ+isinθ)1−cos2θ−isin2θ)
=−1sinθ=cscθ
Range of cscθ is (−∞,−1]∪[1,∞]
So, Re(2iz1−z2)∈(−∞,−1]∪[1,∞]
B) Let y=sin−1(8(3)x−21−32(x−1))
⇒siny=8(3)x−21−32(x−1)
−1≤8(3)x−21−32(x−1)≤1
⇒−1≤8(3)x9−32x≤1
Let 3x=t
⇒−1≤8t9−t2≤1
⇒−1≤8t9−t2 and 8t9−t2≤1
⇒t2−8t−9t2−9≥0 ⇒t2+8t−9t2−9≥0
⇒(t−9)(t+1)(t−3)(t+3)≥0 ⇒(t+9)(t−1)(t−3)(t+3)≥0
⇒(t−9)(t−3)≥0 ⇒(t−9)(t−3)≥0
⇒t<3 andt≥9 ⇒t≤1 and t>3
⇒x≤1 and x>2 ⇒x≤0 and x>2
⇒x∈(−∞,1)∪[2,∞) ⇒x∈(−∞,0]∪(2,∞)
So, common solution is ⇒x∈(−∞,0]∪(2,∞)
C) f(θ)=∣∣
∣∣1tanθ1−tanθ1tanθ−1−tanθ1∣∣
∣∣
=2sec2θ
Since, sec2θ≥1
⇒f(θ)∈[2,∞)
D) f(x)=x32(3x−10)=3x52−10x32
⇒f′(x)=152x32−15x12
⇒f′(x)=152x12(x−2)
For increasing , f′(x)≥0
⇒x≥2
f(x) is increasing in [2,∞)