KCET · Maths · Continuity and Differentiability
If \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & ; x \neq 1 \\ k & ; x=1\end{array}\right. \) is continuous at \( x=1 \), then the value of \( k \) is
- A e
- B \( 11 \)
- C \( -1 \)
- D \( 00 \)
Answer & Solution
Correct Answer
(B) \( 11 \)
Step-by-step Solution
Detailed explanation
Given that, \( f(x)=\left\{\begin{array}{cc}\frac{\log _{e} x}{x-1} & x \neq 1 \\ k & x=1\end{array}\right. \)
For continuous at \( x=1 \), we have
\[
\begin{array}{l}
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\
k=\lim _{x \rightarrow 1} \frac{\log _{e} x}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=1 \\
\text { So, } k=1
\end{array}
\]
For continuous at \( x=1 \), we have
\[
\begin{array}{l}
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \\
k=\lim _{x \rightarrow 1} \frac{\log _{e} x}{x-1}=\lim _{x \rightarrow 1} \frac{1 / x}{1}=1 \\
\text { So, } k=1
\end{array}
\]
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