Q103 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

The set of values of $x$ for which the function $f(x)=\log \left(\frac{x-1}{x+2}\right)$ is continuous, is

  • A
    $R$
  • B
    $(-\infty,-2) \cup(0, \infty)$
  • Correct answer
    C
    $(-\infty,-2) \cup(1, \infty)$
  • D
    $(-2,-1)$

Solution

We have, $$ f(x)=\log \left(\frac{x-1}{x+2}\right) $$ $f(x)$ is continuous of $$ \begin{aligned} \frac{x-1}{x+2}>0 \Rightarrow(x-1)(x+2)>0 \\ \therefore x \in(-\infty,-2) \cup(1, \infty) \end{aligned} $$

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

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