Q42 / 160 · AP EAMCET 2021 (25 Aug Shift 2) · Mathematics

What is the range the function $h(x)=\frac{x-2}{x+3}$ ?

  • A
    $(-\infty, 2) \cup(2, \infty)$
  • Correct answer
    B
    $(-\infty, 1) \cup(1, \infty)$
  • C
    $(-\infty,-3) \cup(-3, \infty)$
  • D
    $(-\infty,-1) \cup(-1, \infty)$

Solution

To find, range of the function $h(x)=\frac{x-2}{x+3}$ Let $h(x)=y$ i.e. set of possible values of $y$ will give the range of $h(x)$. $$ \begin{aligned} & \Rightarrow & \frac{x-2}{x+3} & =y \\ & & x-2 & =x y+3 y \\ \Rightarrow & & x-x y & =3 y+2 \\ \Rightarrow & & x & =\frac{3 y+2}{1-y} \end{aligned} $$ which is not defined for $y=1$ Hence, range $=R-\{1\}$ or $(-\infty,-1) \cup(1, \infty)$

Hence, the correct answer is option: 'B'.

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