Question:
If the function ƒ defined on \( \left( { - {1 \over 3},{1 \over 3}} \right) \) by
f(x) = \( \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right. \)
is continuous, then k is equal to_______.
If the function ƒ defined on \( \left( { - {1 \over 3},{1 \over 3}} \right) \) by
f(x) = \( \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right. \)
is continuous, then k is equal to_______.
f(x) = \( \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + 3x} \over {1 - 2x}}} \right),} & {when\,x \ne 0} \cr {k,} & {when\,x = 0} \cr } } \right. \)
is continuous, then k is equal to_______.
Show Hint
For continuity at 0, k must equal the limit of f(x) as x tends to 0; use \(\ln(1+u)/u\to1\).
Updated On: Oct 31, 2026
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Correct Answer: 5
Solution and Explanation
Step 1: Continuity at 0 requires \(k=\lim_{x\to0}\dfrac1x\ln\dfrac{1+3x}{1-2x}\).
Step 2: Write \(\ln\dfrac{1+3x}{1-2x}=\ln(1+3x)-\ln(1-2x)\).
Step 3: \(\lim\dfrac{\ln(1+3x)}{x}=3\) and \(\lim\dfrac{\ln(1-2x)}{x}=-2\).
Step 4: \(k=3-(-2)=5\).
Answer: 5
Step 2: Write \(\ln\dfrac{1+3x}{1-2x}=\ln(1+3x)-\ln(1-2x)\).
Step 3: \(\lim\dfrac{\ln(1+3x)}{x}=3\) and \(\lim\dfrac{\ln(1-2x)}{x}=-2\).
Step 4: \(k=3-(-2)=5\).
Answer: 5
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