Consider the following listsList − I List − IIA) f(x)= |x+2|x+2, x≠−2 I) [13, 1]B) g(x)=|[x]|, x∈R II) ZC) h(x)= |x−[x]|, x∈R III) WD) f(x)=12−sin3x, x∈R IV) {−1, 1}Match the function List –I with their ranges I List – II and choose the correct answer

A) f(x)=|x+2|x+2={(x+2)x+2=1ifx≥−2−(x+2)x+2=−1ifx< −2∴ Range of f(x)= {−1, 1}B) g(x)=|[x]|∀x∈R then [x] =z⇒|[x]|=z+ =WC) h(x)=|x−[x]|=|{x}|∈[0, 1)∴ Range of h(x)=[0, 1)D) f(x)=12− sin3xWe know that−1≤ sin3x≤1⇒ 1≥ −sin3x ≥ −1⇒ 3≥ 2− sin3x ≥ 1⇒ 113≤ 12− sin3x≤1∴ Range = [13, 1]A−V, B−III,C−IV, D−I

Consider the following listsList − I List − IIA) f(x)= |x+2|x+2, x≠−2 I) [13, 1]B) g(x)=|[x]|, x∈R II) ZC) h(x)= |x−[x]|, x∈R III) WD) f(x)=12−sin3x, x∈R IV) {−1, 1}Match the function List –I with their ranges I List – II and choose the correct answer

A) f(x)=|x+2|x+2={(x+2)x+2=1ifx≥−2−(x+2)x+2=−1ifx< −2∴ Range of f(x)= {−1, 1}B) g(x)=|[x]|∀x∈R then [x] =z⇒|[x]|=z+ =WC) h(x)=|x−[x]|=|{x}|∈[0, 1)∴ Range of h(x)=[0, 1)D) f(x)=12− sin3xWe know that−1≤ sin3x≤1⇒ 1≥ −sin3x ≥ −1⇒ 3≥ 2− sin3x ≥ 1⇒ 113≤ 12− sin3x≤1∴ Range = [13, 1]A−V, B−III,C−IV, D−I

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$\begin{array}{l} C o n s i d e r t h e f o l l o w i n g l i s t s \\ L i s t - I L i s t - I I \\ A) f (x) = \frac{| x + 2 |}{x + 2}, x \neq - 2 I) [\frac{1}{3}, 1] \\ B) g (x) = | [x] |, x \in R I I) Z \\ C) h (x) = | x - [x] |, x \in R I I I) W \\ D) f (x) = \frac{1}{2 - \sin 3 x}, x \in R I V) {- 1, 1} \end{array}$
Match the function List –I with their ranges I List – II and choose the correct answer

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

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

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} A) f (x) = \frac{| x + 2 |}{x + 2} \\ = {\begin{cases} \frac{(x + 2)}{x + 2} = 1 i f x \geq - 2 \\ - \frac{(x + 2)}{x + 2} = - 1 i f x < - 2 \end{cases} \\ ∴ R a n g e o f f (x) = {- 1, 1} \\ B) g (x) = | [x] | \\ \forall x \in R t h e n [x] = z \\ \Rightarrow | [x] | = z^{+} = W \\ C) h (x) = | x - [x] | = | {x} | \in [0, 1) \\ ∴ R a n g e o f h (x) = [0, 1) \\ D) f (x) = \frac{1}{2 - \sin 3 x} \\ W e k n o w t h a t \end{array}$

$\begin{array}{l} - 1 \leq \sin 3 x \leq 1 \\ \Rightarrow 1 \geq - \sin 3 x \geq - 1 \\ \Rightarrow 3 \geq 2 - \sin 3 x \geq 1 \\ \Rightarrow \frac{11}{3} \leq \frac{1}{2 - \sin 3 x} \leq 1 \\ ∴ R a n g e = [\frac{1}{3}, 1] \\ A - V, B - I I I, C - I V, D - I \end{array}$