MCQ
વિધેય $f(x) = \frac{1}{{{{\log }_{10}}(1 - x)}} + \sqrt {x + 2} $ નો પ્રદેશ મેળવો.
- A$( - 3,\; - 2.5) \cup ( - 2.5,\; - 2)$
- ✓$( - 2,\;0) \cup (0,\;1)$
- C$(0, 1)$
- Dએકપણ નહી.
$\because {\log _{10}}(1 - x) \ne 0$==> $1 - x \ne 1$==> $x \ne 0$
Again $1 - x > 0$ ==> $1 > x$ ==> $x < 1$
All these can be combined as $ - 2 \le x < 0$ and $0 < x < 1$.
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