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QJEEM21Mad4f5: JEE Main 2021 Mathematics - Class 12 JEEM21Mad4f5 Solution: If is equal to L, then the value of (6L + 1) is

If lim⁡x→0sin⁡−1x−tan⁡−1x3x3\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

Multiple Choice Options

  • A 16{1 \over 6}
  • B 12{1 \over 2}
  • C 6
  • D 2

Step-by-Step Verified Solution

L=lim⁡x→0(x+x36+.....)−(x−x33.....)3x3L = \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=13(16+13)=16L = {1 \over 3}\left( {{1 \over 6} + {1 \over 3}} \right) = {1 \over 6}

∴\therefore 6L+1=6.16+1=26L + 1 = 6.{1 \over 6} + 1 = 2

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