A) \[R\times [0,\infty )\]
B) \[(-\infty ,0)\times R\]
C) \[[0,\infty )\times R\]
D) \[R\times (-\infty ,0)\]
Correct Answer: A
Solution :
\[S=\{(\lambda ,\mu )\in R\times R:f(t)=(|\lambda |{{e}^{|t|}}-\mu )\sin 2|t|,t\in R\] \[f(t)=(|\lambda |{{e}^{|t|}}-\mu )\sin (2|t|)\] \[=\left\{ \begin{matrix} (|\lambda |{{e}^{t}}-\mu )\sin et & t>0 \\ (|\lambda |{{e}^{-t}}-\mu )(-sin2t) & t<0 \\ \end{matrix} \right.\] \[f'(t)=\left\{ \begin{matrix} (|\lambda |{{e}^{t}})sin2t+(|\lambda |{{e}^{t}}-\mu )(2\cos et) & t>0 \\ +|\lambda |{{e}^{-t}}sin2t+(|\lambda |{{e}^{-t}}-\mu )(-\cos 2t) & t<0 \\ \end{matrix} \right.\] Given \[f(t)\] is differentiable \[\therefore LHD=RHD\]at \[t=0\] \[|\lambda |\cdot \sin 2(0)+(|\lambda |{{e}^{o}}-\mu )2\cos (\infty )\] \[=|\lambda |{{e}^{-0}}\sin 2(\infty )-2cos(0)(\lambda |{{e}^{-0}}-\mu )\] \[0+(|\lambda |-\mu )2=0-2(|\lambda |e-\mu )\] \[4(|\lambda |-\mu )=0\] \[|\lambda |=\mu \] \[S\equiv (\lambda ,\mu )=\{\lambda \in R\And \mu \in (0,\infty )\}\] Set S is subset of \[R\times [0,\infty ]\]
You need to login to perform this action.
You will be redirected in
3 sec
