The domain of the real valued function $f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}$ is
The domain of the real valued function $f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}$ is
- $(-3, \infty)$
- $(-3,-1) \cup(-1, \infty)$
- $(-3,-2) \cup(-2,-1) \cup(-1, \infty)$
- $(-3,-2) \cup(-1, \infty)$
Solution
$
\begin{aligned}
& \text { Given, } f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}} \\
& \\
& \quad x^2+3 x+2>0 \\
& \Rightarrow \quad x^2+2 x+x+2>0 \\
& \Rightarrow \quad x(x+2)+1(x+2)>0 \\
& \Rightarrow \quad(x+2)(x+1)>0 \\
& \Rightarrow \quad x \in(-\infty,-2) \cup(-1, \infty) \\
& \text { But } x+3>0 \\
& \quad x>-3
\end{aligned}
$
Hence, $x \in(-3,-2) \cup(-1, \infty)$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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