MHT CET · Maths · Functions
The domain of definition of \(\mathrm{f}(x)=\frac{\log _2(x+3)}{x^2+3 x+2}\) is
- A \(\mathrm{R}-\{1,2\}\)
- B \((-2, \infty)\)
- C \(\mathrm{R}-\{-1,-2,-3\}\)
- D \((-3, \infty)-\{-1,-2\}\)
Answer & Solution
Correct Answer
(D) \((-3, \infty)-\{-1,-2\}\)
Step-by-step Solution
Detailed explanation
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, \(x+3\gt0\).
\(x\gt-3\)
\(\therefore \quad x \in(-3, \infty)-\{-1,-2\}\)
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, \(x+3\gt0\).
\(x\gt-3\)
\(\therefore \quad x \in(-3, \infty)-\{-1,-2\}\)
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