Question
The domain of $$F\left( x \right) = \frac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}}$$ is :
A.
$$R - \left\{ { - 1,\, - 2} \right\}$$
B.
$$\left( { - 2,\,\infty } \right)$$
C.
$$R - \left\{ { - 1,\, - 2 - 3} \right\}$$
D.
$$\left( { - 3,\,\infty } \right) - \left\{ { - 1,\, - 2} \right\}$$
Answer :
$$\left( { - 3,\,\infty } \right) - \left\{ { - 1,\, - 2} \right\}$$
Solution :
$$\eqalign{
& {\text{We have, }}F\left( x \right) = \frac{{{{\log }_2}\left( {x + 3} \right)}}{{{x^2} + 3x + 2}} \cr
& \therefore \,F\left( x \right){\text{ is defined if }}x + 3 > 0{\text{ and }}{x^2} + 3x + 2 \ne 0 \cr
& \Rightarrow F\left( x \right){\text{ is defined if }}x > - 3{\text{ and }}x \ne - 1,\, - 2 \cr
& \Rightarrow {\text{Domain of }}F\left( x \right) = \left( { - 3,\,\infty } \right) - \left\{ { - 1,\, - 2} \right\} \cr} $$