माना $S =\left\{(\lambda, \mu) \in R \times R : f( t )=\left(|\lambda| e ^{| t |}-\mu\right)\right.. \sin (2|t|), t \in R$ एक अवकलनीय फलन है $\}$ तो $S$ जिसका उपसम्च्चय है, वह है
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Answer
$S = \left\{ {\lambda ,\mu } \right\} \in R \times R:f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right),$$t \in R$
$f\left( t \right) = \left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$
$\left\{ \begin{array}{l} \left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t\,\,\,\,\,\,\,\,t > 0\\ \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t\,} \right)\,\,\,\,\,\,\,t < 0 \end{array} \right.$
$f'(t)$
$ = \left\{ \begin{array}{l} \left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right)\,\,\,t > 0\\ \left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - 2\cos 2t} \right)\,\,\,\,t < 0 \end{array} \right.$
As, $f(t)$ is differentiable
$\therefore \text{LHD = RHD}$ at $t=0$
$\left| \lambda \right|.\sin 2\left( 0 \right) + \left( {\left| \lambda \right|{e^0} - \mu } \right)2\cos \left( 0 \right)$
$ = \left| \lambda \right|{e^0}.\sin 2\left( 0 \right) - 2\cos \left( 0 \right)\left( {\left| \lambda \right|{e^0} - \mu } \right)$
$ \Rightarrow 0 + \left( {\left| \lambda \right| - \mu } \right)2 = 0 - 2\left( {\left| \lambda \right| - \mu } \right)$
$4\left( {\left| \lambda \right| - \mu } \right) = 0$
$\left| \lambda \right| = \mu $
So, $S \equiv \left( {\lambda ,\mu } \right) = \left\{ {\lambda \in R \text{ and } \mu \in \left[ {0,\infty } \right)} \right\}$
Therefore set $S$ is subset of $R \times \left[ {0,\infty } \right)$
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