QJEEAdv15Ma90b5: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma90b5 Solution: Let m and n be two positive integers greater than 1. If then…

Given, $\mathop {\lim }\limits_{\alpha \to 0} \left[ {{{{e^{\cos ({\alpha ^n})}} - e} \over {{\alpha ^m}}}} \right] = - {e \over 2}$

$\Rightarrow \mathop {\lim }\limits_{\alpha \to 0} {{e\{ {e^{\cos ({\alpha ^n}) - 1}} - 1\} } \over {\cos ({\alpha ^n}) - 1}}.{{\cos ({\alpha ^n}) - 1} \over {{\alpha ^m}}} = {{ - e} \over 2}$

$\Rightarrow \mathop {\lim }\limits_{\alpha \to 0} e\left\{ {{{{e^{\cos ({\alpha ^n}) - 1}} - 1} \over {\cos ({\alpha ^n}) - 1}}} \right\}.\mathop {\lim }\limits_{\alpha \to 0} {{ - 2{{\sin }^2}{{{\alpha ^n}} \over 2}} \over {{\alpha ^m}}} = - e/2$

$\Rightarrow e \times 1 \times ( - 2)\mathop {\lim }\limits_{\alpha \to 0} {{{{\sin }^2}\left( {{{{\alpha ^n}} \over 2}} \right)} \over {{{{\alpha ^{2n}}} \over 4}}}.{{{\alpha ^{2n}}} \over {4{\alpha ^m}}} = {{ - e} \over 2}$

$\Rightarrow e \times 1 \times - 2 \times 1 \times \mathop {\lim }\limits_{\alpha \to 0} {{{\alpha ^{2n - m}}} \over 4} = {{ - e} \over 2}$

For this to be exists,

$2n - m = 0 \Rightarrow {m \over n} = 2$