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Unit step function is given by\begin{array}{l}u(t-a)=\left\{\begin{array}{ll}0 & t<a \\1 & t>a\end{array} \quad\right. \\\therefore u(t-\pi)=\left\{\begin{array}{ll}0 & t<\pi \\1 & t>\pi\end{array} \text { and } u(t-3 \pi)=\left\{\begin{array}{ll}0 & t<3 \pi \\1 & t>3 \pi\end{array}\right.\right.\end{array}L( 6Considerf(t)=-\sin (t-\pi) u(t-\pi)+\sin (t-3 \pi) u(t-3 \pi) \text {. }\therefore If t<\pi thenf(t)=-\sin (t-\pi)(0)+\sin (t-3 \pi)(0) \ldots(\text { by }(1) f(2))\Rightarrow f(t)=0 if t<\piIf \pi<t<3 \pi then\begin{aligned}f(t) & =-\sin (t-\pi)(1)+\sin (t-3 \pi)(0) \quad \ldots\left(b_{y}(1) t(2)\right) \\& =-\sin (t-\pi)=-\sin (-(\pi-t)) \\& =\sin (\pi-t) \quad(\because \sin (-\theta)=-\sin \theta) \\\Rightarrow f(t) & =\sin (t) \text { if } \pi<t<3 \pi \quad(\because \sin (\pi-\theta)=\sin \theta)\end{aligned}If t>3 \pi, then\begin{aligned}f(t) & =-\sin (t-\pi)(1)+\sin (t-3 \pi) \\& =-\sin (t-\pi)+\sin (t-\pi-2 \pi) \\& =-\sin (t-\pi)+\sin (t-\pi) \quad(0 \sin (\theta-2 \pi)=\sin \theta) \\\Rightarrow f(t) & =0 \text { if } t>3 \pi\end{aligned}\therefore Given function f can be expressed asf(t)=-\sin (t-\pi) u(t-\pi)+\sin (t-3 \pi) u(t-3 \pi) \text {. }\Rightarrow option (5) - correct.\begin{array}{l} L\{f(t)\}=\int_{0}^{\infty} e^{-s t} f(t) d t \\=\int_{0}^{\pi} e^{-s t} \cdot(0) d t+\int_{\pi}^{3 \pi} e^{-s t} \sin t d t+\int_{3 \pi}^{\infty} e^{-s t}(0) d t \\I=\int_{\pi}^{3 \pi} e^{-s t} \sin t d t \\=-\left.\frac{e^{-s t} \sin t}{s}\right|_{\pi} ^{3 \pi}+\int_{\pi}^{3 \pi} \cos t \cdot \frac{e^{-s t}}{s} d t \\=-\frac{1}{s}(0)+\frac{1}{s}\left[\left.\frac{e^{-s t}}{-s} \cos t\right|_{\pi} ^{3 \pi}-\int_{\pi}^{3 \pi}-\sin t \cdot-\frac{e^{-s t}}{s^{2}}\right] \\=-\frac{1}{s^{2}}\left[-e^{-3 \pi s}+e^{-\pi s}\right]-\frac{I}{s^{2}} \\I+\frac{I}{s^{2}}=\frac{1}{s^{2}}\left[e^{-3 \pi s}-e^{-\pi s}\right] \\I=\frac{1}{s^{2}+1}\left[e^{-3 \pi s}-e^{-\pi s}\right] \\\therefore L\{f(t)\}=\frac{e^{-3 \pi s}-e^{-\pi s}}{s^{2}+1}\end{array}Note that, if 2\{f(t)\}=F(s) then\begin{array}{c}L\left\{e^{a t} f(t)\right\}=F(s-a) \\\therefore L\left\{e^{t} f(t)\right\}=\frac{e^{-3 \pi(s-1)}-e^{-\pi(s-1)}}{(s-1)^{2}+1} \quad(\because a=1)\end{array} ...