JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 5)
If the function $$f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.$$
is continuous at $$x = {\pi \over 2}$$, then $$9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$$ is equal to
11
10
8
2e$$^4$$ + 8
Explanation
$$
\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^ \frac{\lambda}{|\cos x|}=e^\lambda
$$
And,
$$ \begin{aligned} &\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}\\\\ &= \mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} {e^{{{\sin 4x.\cos 6x} \over {\sin 6x.\cos 4x}}}} \\\\ & =e^{\frac{2}{3}} \end{aligned} $$
Also, $$ f(\pi / 2)=\mu $$
For continuous function,
$ \mathrm{e}^{2 / 3}=\mathrm{e}^\lambda=\mu$
$$ \lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3} $$
$$ \therefore $$ $9 \lambda+6 \ln \mu+\mu^{6}-e^{6 \lambda}$
$=6+4+e^{4}-e^{4}=10$
And,
$$ \begin{aligned} &\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}\\\\ &= \mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} {e^{{{\sin 4x.\cos 6x} \over {\sin 6x.\cos 4x}}}} \\\\ & =e^{\frac{2}{3}} \end{aligned} $$
Also, $$ f(\pi / 2)=\mu $$
For continuous function,
$ \mathrm{e}^{2 / 3}=\mathrm{e}^\lambda=\mu$
$$ \lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3} $$
$$ \therefore $$ $9 \lambda+6 \ln \mu+\mu^{6}-e^{6 \lambda}$
$=6+4+e^{4}-e^{4}=10$
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