KCET · Maths · Continuity and Differentiability
\(f(x)=\left\{\begin{array}{cc}\frac{\log x}{x-1}, & \text { if } x \neq 1 \\ k, & \text { if } x=1\end{array}\right.\)
is continuous at
\(x=1\), then the value of \(k\) is
- A 0
- B \(-1\)
- C 1
- D \(e\)
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{cl}\frac{\log x}{x-1}, & \text { if } x \neq 1 \\ k & , \text { if } x=1\end{array}\right.\) at \(x=1\) Since, the function is continuous at \(x=1\). Then, \(f(1)=\lim _{x \rightarrow 1} f(x)\)
\[
k=\lim _{x \rightarrow 1} \frac{\log x}{x-1} \quad\left(\frac{0}{0} \text { from }\right)
\]
Apply L'hopital rule,
\[
\begin{aligned}
&k=\lim _{x \rightarrow 1} \frac{1 / x}{1}=\frac{1 / 1}{1} \\
&k=1
\end{aligned}
\]
\[
\Rightarrow \quad k=1
\]
\[
k=\lim _{x \rightarrow 1} \frac{\log x}{x-1} \quad\left(\frac{0}{0} \text { from }\right)
\]
Apply L'hopital rule,
\[
\begin{aligned}
&k=\lim _{x \rightarrow 1} \frac{1 / x}{1}=\frac{1 / 1}{1} \\
&k=1
\end{aligned}
\]
\[
\Rightarrow \quad k=1
\]
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