Consider the following lists. List - I List - II A) f (x) = ≤ft|x+2|/x+2 , x ≠ - 2 I) ≤ft[1/3,1] B) g (x) = |[x]| , x ∊ R II) Z C) h (x) = |x - [x]| , x ∊ R III) W D) f (x) = 1/2- 3x , x ∊ R IV) [0, 1) V) {- 1 , 1}

Correct Answer is : A - V , B - III , C - IV , D - I

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

Correct Answer is : A - V , B - III , C - IV , D - I

Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 2 of 160

Marks: +1, -0

Solution:

For

A,

Consider the function,

f(x)=\;\;\frac{\left|x+2\right|}{x+2}, x\neq -2

Then

f(x)=\begin{cases} \;\;\frac{x+2}{x+2}, & x>-2 \\ -\;\;\frac{x+2}{x+2}, & x<-2 . \end{cases}

=\begin{cases} 1, & x>-2 \\ -1, & x<-2 . \end{cases}

This implies, the range of

f(x)

\{-1,1\}

For

B

g(x)=\left|\lfloor x\rfloor\right|, x\in R

\text{since } \lfloor x\rfloor \in I

[|x]| \in W

This implies the range of

g(x)

W

For

C

h(x)=\left|x-\lfloor x\rfloor\right|, x\in R

=\left|\{x\}\right| \in [0,1)

This implies the range is [0,1)

For

D

f(x)=\;\;\frac{1}{2-\sin 3x}, x\in R

Then

-1 \leq \sin 3x \leq 1

2-1 \leq 2-\sin 3x \leq 2+1

\;\;\frac{1}{3} \leq \;\;\frac{1}{2-\sin 2x} \leq \;\;\frac{1}{1}

This implies the range of

f(x)

\left[\;\;\frac{1}{3}, 1\right]

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