$$
\begin{aligned}
&\quad S=\left\{(\lambda, \mu) \in \mathrm{R} \times \mathrm{R}: \mathrm{f}(\mathrm{t})=\left(|\lambda| \mathrm{e}^{\mid t}-\mu\right) \sin (2 \mathrm{t} \mid),\right. \\
&t \in R \\
&f(t)=\left(|\lambda| \mathrm{e}^{|t|}-\mu\right) \sin (2|t|) \\
&\left\{\begin{array}{l}
\left(|\lambda| e^t-\mu\right) \sin 2 t \quad t>0 \\
\left(|\lambda| e^{-t}-\mu\right)(-\sin 2 t) t < 0
\end{array}\right. \\
&f^{\prime}(t) \\
&\left\{\begin{array}{l}
\left(|\lambda| \mathrm{e}^t\right) \sin 2 t+\left(|\lambda| \mathrm{e}^t-\mu\right)(2 \cos 2 t) t>0 \\
|\lambda| \mathrm{e}^{-t} \sin 2 t+\left(|\lambda| e^{-t}-\mu\right)(-2 \cos 2 t) < t < 0
\end{array}\right. \\
&\text { As, } f(t) \text { is differentiable } \\
&\therefore \text { LHD }=\mathrm{RHD} \text { at } t=0 \\
&|\lambda| \cdot \sin 2(0)+\left(|\lambda| \mathrm{e}^0-\mu\right) 2 \cos (0) \\
&=|\lambda| \mathrm{e}^{-0} \cdot \sin 2(0)-2 \cos (0)\left(|\lambda| \mathrm{e}^{-0}-\mu\right) \\
&\Rightarrow 0+(|\lambda|-\mu) 2=0-2(|\lambda|-\mu) \\
&\quad 4(|\lambda|-\mu)=0 \\
&\quad|\lambda|=\mu
\end{aligned}
$$
So, $S \equiv(\lambda, \mu)=\{\lambda \in R \& \mu \in[0, \infty)\}$
Therefore set $\mathrm{S}$ is subset of $R \times[0, \infty)$