Let S = {(lambda, mu) in R times R : f(t) = ( | vedclass

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(A) Given $f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|)$.Since $\|t\|$ is involved,we analyze $f(t)$ for $t > 0$ and $t  0$,$f(t) = (

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$R 	imes [0, \infty)$

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(A) Given $f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|)$.Since $\|t\|$ is involved,we analyze $f(t)$ for $t > 0$ and $t For $t > 0$,$f(t) = (\|\lambda\|e^t - \mu) \sin(2t)$.For $t For $f(t)$ to be differentiable at $t=0$,it must be continuous at $t=0$.$f(0) = (\|\lambda\| - \mu) \sin(0) = 0$.Now,we check the derivative $f'(t)$ at $t=0$ using $LHD = RHD$.$RHD = \lim_{t 	o 0^+} \frac{f(t) - f(0)}{t} = \lim_{t 	o 0^+} (\|\lambda\|e^t - \mu) \frac{\sin(2t)}{t} = (\|\lambda\| - \mu) 	imes 2 = 2(\|\lambda\| - \mu)$.$LHD = \lim_{t 	o 0^-} \frac{f(t) - f(0)}{t} = \lim_{t 	o 0^-} -(\|\lambda\|e^{-t} - \mu) \frac{\sin(2t)}{t} = -(\|\lambda\| - \mu) 	imes 2 = -2(\|\lambda\| - \mu)$.For differentiability,$LHD = RHD \implies 2(\|\lambda\| - \mu) = -2(\|\lambda\| - \mu)$.$4(\|\lambda\| - \mu) = 0 \implies \|\lambda\| = \mu$.Since $\|\lambda\| \ge 0$,we must have $\mu \ge 0$.Thus,$S = \{(\lambda, \mu) : \mu = \|\lambda\|, \mu \ge 0, \lambda \in R\}$.This set $S$ is a subset of $R 	imes [0, \infty)$.

Let $S = \{(\lambda, \mu) \in R \times R : f(t) = (\|\lambda\|e^{\|t\|} - \mu) \sin(2\|t\|), t \in R\}$ be a differentiable function. Then $S$ is a subset of?

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