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The domain of f(x)=\frac{\log _{(x+1)}(x-2)}{e^{2 \log _e x-(2 x+3)}}, x \in \mathbb{R} is

 

Option: 1

\mathbb{R}-\{3\}
 


Option: 2

(-1, \infty)-\{3\}
 


Option: 3

(2, \infty)-\{3\}
 


Option: 4

\mathbb{R}-\{-1,3\}


Answers (1)

best_answer

f(x)=\frac{\log _{(x+1)}(x-2)}{e^{2 \ln x}-(2 x+3)}
case (i)       x-2>0 \Rightarrow x>2

                  \mathrm{ x \in(2, \infty) }

case (ii)    \mathrm{x+1> 0} and \mathrm{x+1\neq1}
                       \mathrm{x> -1}    \mathrm{x\neq0}
\therefore \mathrm{x}_{\mathrm{t}}(-1,0) \cup(0, \infty)
\begin{aligned} & \text { case (iii) } \quad x>0 \Rightarrow x_t(0, \infty) \\ & \text { case (iv) } \quad \mathrm{e}^{2<n}-(2 \mathrm{x}+3) \neq 0 \\ \end{aligned}

\begin{aligned} & \Rightarrow \quad x^2-2 \mathrm{x}+3 \neq 0 \\ & \quad(\mathrm{x}-3)(\mathrm{x}+1) \neq 0 \\ \end{aligned}

\mathrm{\begin{aligned} & \Rightarrow \quad x \neq 3, \mathrm{x} \neq-1 \\ & \therefore \text { from (i) } n(\text { ii) } n \text { (iii)n (iv) } \\ & \quad x_t(2, \infty)-\{3\} \end{aligned} }
 

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