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_______.
Answer (integer)
5
Solution
$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$$
<br><br>= 3 + 2 = 5
<br><br>f(x) is continuous
<br><br>$\therefore$ $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ = f(0)
<br><br>So f(0) = 5 = k
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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