S = {($$\lambda $$, $$\mu $$) $$ \in $$ R $$ \times $$ R : f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$$ sin $$\left( {2\left| t \right|} \right),$$ t $$ \in $$ R

f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$$

= $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$$

f'(t) = $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$$

As, f(t) is differentiable

$$ \therefore $$  LHD = RHD at t = 0

$$ \Rightarrow $$   $$\left| \lambda \right|$$ . sin2(0) + $$\left( {\left| \lambda \right|{e^0} - \mu } \right)$$2cos(0)

      = $$\left| \lambda \right|{e^{ - 0}}\,$$ . sin 2(0) $$-$$ 2 cos (0) $$\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$$

$$ \Rightarrow $$$$\,\,\,$$ 0 + $$\left( {\left| \lambda \right| - \mu } \right)$$2 = 0 $$-$$ 2$$\left( {\left| \lambda \right| - \mu } \right)$$

$$ \Rightarrow $$  4$$\left( {\left| \lambda \right| - \mu } \right)$$ = 0

$$ \Rightarrow $$  $$\left| \mu \right|$$ = $$\mu $$

So, S $$ \equiv $$ ($$\lambda $$, $$\mu $$) = {$$\lambda $$ $$ \in $$ R & $$\mu $$ $$ \in $$ [0, $$\infty $$)}

Therefore set S is subset of R $$ \times $$ [0, $$\infty $$)