JEE Mains · Maths · STD 11 - 12. limits
If the function \(\mathrm{f}\) defined on \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) by \(f(x)=\left\{\begin{array}{ll}{\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}\right)} & {, \text { when } x \neq 0} \\ {k} & {, \text { when } x=0}\end{array}\right.\) is continuous, then \(\mathrm{k}\) is equal to
- A \(4\)
- B \(5\)
- C \(6\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
\(\mathrm{k}=\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\ln (1+3 \mathrm{x})}{\mathrm{x}}-\frac{\ln (1-2 \mathrm{x})}{\mathrm{x}}\right)\) \(\mathrm{k}=3+2=5\)
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