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Mathematics Question on Relations and functions

Consider the following lists. List-IList-II
(A)$f(x) = \frac{
(B)$(x)=
(C)$h(x) =
(D)f(x)=12−sin⁡3x,x∈[Rf(x) = \frac{1}{2 - \sin 3x} , x \in [R
(V)
A

(A)-(V), (B)-(III) , (C)-(II), (D)-(I)

B

(A)-(III), (B)-(II) , (C)-(IV), (D)-(I)

C

(A)-(V), (B)-(III) , (C)-(IV), (D)-(I)

D

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Answer

(A)-(V), (B)-(III) , (C)-(IV), (D)-(I)

Explanation

Solution

f(x)=∣x+2∣x+2,x≠−2f(x)=\frac{|x+2|}{x+2}, x \neq-2

So, range of f(x)f(x) is −1,1\\{-1,1\\}.
(B) ∵g(x)=∣[x]∣,x∈R\because g(x)=|[x]|, x \in R
As [x]∈I⇒∣[x]∣∈W[x] \in I \Rightarrow|[x]| \in W
So, range of g(x)g(x) is WW.
(C) ∵h(x)=∣x−[x]∣,x∈R=∣x∣∈[0,1)\because h(x)=|x-[x]|, x \in R=|\\{x\\}| \in[0,1)
[∵x=x−[x][\because\\{x\\}=x-[x] and x∈[0,1)]\\{x\\} \in[0,1)]
So, range of h(x)h(x) is [0,1)[0,1).
(D) ∵f(x)=12−sin⁡3x,x∈R\because f(x)=\frac{1}{2-\sin 3 x}, x \in R
∵−1≤sin⁡3x≤1,∀x∈R\because -1 \leq \sin 3 x \leq 1, \forall x \in R
⇒−1≤−sin⁡3x≤1\Rightarrow -1 \leq-\sin 3 x \leq 1
⇒2−1≤2−sin⁡3x≤2+1\Rightarrow 2-1 \leq 2-\sin 3 x \leq 2+1
⇒13≤12−sin⁡3x≤11\Rightarrow \frac{1}{3} \leq \frac{1}{2-\sin 3 x} \leq \frac{1}{1}
So, range of f(x) is [13,1]f(x) \text { is }\left[\frac{1}{3}, 1\right]