The domain of definition of $\mathrm{f}(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is

The domain of definition of $\mathrm{f}(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
  1. $\mathrm{R}-\{1,2\}$
  2. $(-2, \infty)$
  3. $\mathrm{R}-\{-1,-2,-3\}$
  4. $\quad(-3, \infty)-\{-1,-2\}$

Solution

Given, $\mathrm{f}(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ Now, $\mathrm{f}(x)$ is defined if $x^2+3 x+2 \neq 0$ $\Rightarrow(x+1)(x+2) \neq 0$ $\Rightarrow x \neq-1,-2$. Also, $x+3\gt0$. $x\gt-3$ $\therefore \quad x \in(-3, \infty)-\{-1,-2\}$

Asked in: MHT CET 2024 (03 May Shift 2)

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