JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 2)
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 :
$${4 \over 3}$$
$${2 \over 3}$$
$${3 \over 4}$$
$${2 \over {\sqrt 3 }}$$
Explanation
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 = $${4 \over {2\sqrt 3 }} \times {2 \over {\sqrt 3 }}$$ = $${4 \over 3}$$
$$ \Rightarrow $$ 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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