JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 6)
If $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x} \over {3{x^3}}}$$ is equal to L, then the value of (6L + 1) is
$${1 \over 6}$$
$${1 \over 2}$$
6
2
Explanation
$$L = \mathop {\lim }\limits_{x \to 0} {{\left( {x + {{{x^3}} \over 6} + .....} \right) - \left( {x - {{{x^3}} \over 3}.....} \right)} \over {3{x^3}}}$$
$$L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}$$
$$ \therefore $$ $$ 6L + 1 = 6.{1 \over 6} + 1 = 2$$
$$L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}$$
$$ \therefore $$ $$ 6L + 1 = 6.{1 \over 6} + 1 = 2$$
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