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$\large {\red {\mathtt{\fcolorbox{yellow}{blue}{ \fcolorbox{blue}{black}{QUESTION}}}}}$
Solution of $\large \bold{ |1 + \frac{3}{x} | > 2}$ is

(a) (0, 3]
(b) [-1, 0)
(c) (-1, 0) U (0, 3)

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Answer 1

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Solution:-

First, factor the numerator and denominator.

First, factor the numerator and denominator.{k(x)=5+2x22−x−x2 =5+2x2(2+x)(1−x)

To find :-

the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:{(2+x)(1−x)=0 x=−2,1

the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:{(2+x)(1−x)=0 x=−2,1Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.

Answer 2

Answer:

$\large \bold{ |1 + \frac{3}{x} | > 2} \\ \\ \to \: |x + 3| > 2 |x| \\ \to |x + 3| - 2 |x| > 0$