QJEEM21Ma0775: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0775 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}

is :

Accepted Answer

Let L =

\mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}

\Rightarrow

L =

\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} \right) - {1 \over 2}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}}} \right\}

\Rightarrow

L =

\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} \right) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)}}} \right\}

\Rightarrow

L =

\mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}

\Rightarrow

L =

{4 \over {2\sqrt 3 }}\mathop {\lim }\limits_{h \to 0} \left\{ {{{\sin \left( h \right)} \over {h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}

\Rightarrow

L =

{4 \over {2\sqrt 3 }} \times {2 \over {\sqrt 3 }}

=

{4 \over 3}

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