The domain of f(x)=log(x+1)⁡(x−2)e2loge⁡x−(2x+3),x∈ℝ is

f(x)=log(x+1)⁡(x−2)e2loge⁡x−(2x+3)=logx+1⁡(x−2)x2−2x−3, if x>0f (x) is real⇒x−2>0,x+1>0,x+1≠1,x2−2x−3≠0⇒x>2,(x−3)(x+1)≠0⇒x>2,x≠3 Domain =(2,∞)−{3}

The domain of f(x)=log(x+1)⁡(x−2)e2loge⁡x−(2x+3),x∈ℝ is

f(x)=log(x+1)⁡(x−2)e2loge⁡x−(2x+3)=logx+1⁡(x−2)x2−2x−3, if x>0f (x) is real⇒x−2>0,x+1>0,x+1≠1,x2−2x−3≠0⇒x>2,(x−3)(x+1)≠0⇒x>2,x≠3 Domain =(2,∞)−{3}

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

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

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

$ℝ - {3}$

$ℝ - {- 1, 3}$

answer is B.

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Detailed Solution

$f (x) = \frac{\log_{(x + 1)} (x - 2)}{e^{2 \log_{e} x} - (2 x + 3)} = \frac{\log_{x + 1} (x - 2)}{x^{2} - 2 x - 3}, if x > 0$
f (x) is real
$\begin{array}{l} \Rightarrow x - 2 > 0, x + 1 > 0, x + 1 \neq 1, x^{2} - 2 x - 3 \neq 0 \\ \Rightarrow x > 2, (x - 3) (x + 1) \neq 0 \\ \Rightarrow x > 2, x \neq 3 \\ Domain = (2, \infty) - {3} \end{array}$