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The set of real values of $x$ for which ${\log _{0.2}}{{x + 2} \over x} \le 1$ is
$\left( { - \infty ,\,\, - {5 \over 2}} \right] \cup (0, + \infty )$
$\left[ {{5 \over 2}, + \,\infty } \right)$
$( - \infty ,\, - 2) \cup (0, + \,\infty )$
None of these
Solution
(a) ${\log _{0.2}}{{x + 2} \over x} \le 1$…..$(i)$
For log to be defined, ${{x + 2} \over x} > 0$ $ \Rightarrow $$x > 0$ or $x < – 2$
Now from $(i),$ ${\log _{0.2}}{{x + 2} \over x} \le {\log _{0.2}}0.2$
$ \Rightarrow $${{x + 2} \over x} \ge 0.2$ …..$(ii)$
Case $(i)$ $x > 0$
From $(ii),$ $x + 2 \ge 0.2x$
$ \Rightarrow $ $0.8x \ge – 2$
$ \Rightarrow $$x \ge – {5 \over 2}$.
Case $(ii)$ $x < – 2$
From $(ii),$ $x + 2 \le 0.2x$$ \Rightarrow $$0.8x \le – 2$$ \Rightarrow $$x \le – {5 \over 2}$
$ \Rightarrow $$x \in (0,\,\infty )\, \cup \,\left( { – \infty ,\, – {5 \over 2}} \right]$;
$\therefore x \in \left( { – \infty ,\, – {5 \over 2}} \right] \cup (0,\,\infty )$.
