QJEEM20Ma197b: JEE Main 2020 Mathematics - Class 12 JEEM20Ma197b Solution: If the function ƒ defined on by f(x) = { 1 x log e ( 1 + 3 x 1…

Question

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If the function ƒ defined on

\left( { - {1 \over 3},{1 \over 3}} \right)

by

f(x) =

{\begin{matrix} \frac{1}{x} \log_{e} (\frac{1 + 3 x}{1 - 2 x}), & w h e n x \neq 0 \\ k, & w h e n x = 0 \end{matrix}

is continuous, then k is equal to_______.

Accepted Answer

\mathop {\lim }\limits_{x \to 0} f\left( x \right)

=

\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)

=

\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)

= 3 + 2 = 5

f(x) is continuous

\therefore

\mathop {\lim }\limits_{x \to 0} f\left( x \right)

= f(0)

So f(0) = 5 = k

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