Question

Let $S=\left\{\right(\lambda ,\mu )ϵR\times R:f(t)=(|\lambda |e^{|t|}-\mu ).\sin (2|t|),tϵR,\text{is a differentiable function}\}$.

Then $S$ is a subset of?

• A. $R\times [0,\infty )$
• B. $(-\infty ,0)\times R$
• C. $[0,\infty )\times R$
• D. $R\times (-\infty ,0)$

Answer 1

The correct option is A $R\times [0,\infty )$

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

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

$=\{\begin{array}{cc}(|\lambda |e^t-\mu )\sin et& t>0\\ (|\lambda |e^{-t}-\mu )(-\sin 2t)& t<0\end{array}$

$f^{\prime }\left(t\right)=\{\begin{array}{cc}\left(|\lambda |e^t\right)\sin 2t+(|\lambda |e^t-\mu )\left(2\cos et\right)& t>0\\ +|\lambda |e^{-t}\sin 2t+(|\lambda |e^{-t}-\mu )(-\cos 2t)& t<0\end{array}$

Given $f\left(t\right)$ is differentiable

$\therefore$ LHD$=$RHD at $t=0$

$|\lambda |\cdot \sin 2\left(0\right)+(|\lambda |e^o-\mu )2\cos \left(\infty \right)$

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

$0+(|\lambda |-\mu )2=0-2(|\lambda |e-\mu )$

$4(|\lambda |-\mu )=0$

$|\lambda |=\mu$

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

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