JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 6)
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_______.
Answer
5
Explanation
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$$
= 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
= $$\mathop {\lim }\limits_{x \to 0} \left( {{{\ln \left( {1 + 3x} \right)} \over x} - {{\ln \left( {1 - 2x} \right)} \over x}} \right)$$
= $$\mathop {\lim }\limits_{x \to 0} \left( {3{{\ln \left( {1 + 3x} \right)} \over {3x}} - \left( { - 2} \right){{\ln \left( {1 - 2x} \right)} \over { - 2x}}} \right)$$
= 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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