The domain of f ( x ) = log ( x + 1 ) ( x - 2 ) e 2 log e x - ( 2 x + 3 ) , x is [2023]

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Solution

Q.
The domain of $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)}, x \in ℝ$ is [2023]

1 $(2, \infty) - {3}$

2 $ℝ - {3}$

3 $ℝ - {- 1, 3}$

4 $(- 1, \infty) - {3}$

Ans.
(1)

Given, $f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)} = \frac{\log_{(x + 1)} (x - 2)}{x^{2} - 2 x - 3}$

Logarithmic function will be defined

$x - 2 > 0 \Rightarrow x > 2 and x + 1 > 0 \Rightarrow x > - 1$

$\Rightarrow x + 1 \neq 1 \Rightarrow x \neq 0 and x > 0$

In denominator, $x^{2} - 2 x - 3 \neq 0$

$\Rightarrow (x - 3) (x + 1) \neq 0 \Rightarrow x \neq - 1, 3 .$

So, $x \in (2, \infty) - {3}$

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