The domain of definition of the function f(x)=1+loge(1-x)

f(x)=1+loge(1−x) value of f(x) is real when 1+loge(1−x)≥0 and 1−x>0⇒loge(1−x)≥−1 and x<1⇒loge(1−x)≥−1 and x<1⇒1−x≥1e and x<1⇒x≤e−1e and x<1 .

The domain of definition of the function f(x)=1+loge(1-x)

f(x)=1+loge(1−x) value of f(x) is real when 1+loge(1−x)≥0 and 1−x>0⇒loge(1−x)≥−1 and x<1⇒loge(1−x)≥−1 and x<1⇒1−x≥1e and x<1⇒x≤e−1e and x<1 .

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$The domain of definition of the$ function $f (x) = \sqrt{1 + \log_{e} (1 - x)}$

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$- \infty < x \leq 0$

$- \infty < x \leq \frac{e - 1}{e}$

$- \infty < x \leq 1$

$x \geq 1 - e$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{1 + \log_{e} (1 - x)} value of f (x) is real when \\ 1 + \log_{e} (1 - x) \geq 0 and 1 - x > 0 \\ \Rightarrow \log_{e} (1 - x) \geq - 1 and x < 1 \\ \Rightarrow \log_{e} (1 - x) \geq - 1 and x < 1 \\ \Rightarrow 1 - x \geq \frac{1}{e} and x < 1 \Rightarrow x \leq \frac{e - 1}{e} and x < 1 . \end{array}$