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Sol:--(1)Given that,F\left(f^{\prime}\right)=\left\{\begin{array}{ll}i\left(3 \omega+2 \omega^{2}\right) & |\omega|<4, \\0 & |\omega|>4\end{array}\right.(a) To find an expression for \hat{f}(\omega) in terms of unit step function 4 ?Now let us take the given function,\begin{array}{l}F\left(f^{\prime}\right)=\left\{\begin{array}{ll}i\left(3 \omega+2 \omega^{2}\right) & |\omega|<4, \\0 & |\omega|>4 .\end{array}\right. \\\Rightarrow(i \omega) F(f)=\left\{\begin{array}{cc}i \omega(3+2 \omega), & |\omega|<4 \\0, & |\omega|>4\end{array}\right. \\\end{array}we know that (iw) F(f)=\hat{f}(\omega).\Rightarrow \hat{f}(\omega)=\left\{\begin{array}{clc}0, & -\infty & <\omega<-4 \\3+2 \omega, & -4<\omega<4 \\0, & 4<\omega<\infty\end{array}\right.The unit step function in terms of u is,u(\omega-a)=\left\{\begin{array}{ll}0, & \omega<a \\1, & \omega>a\end{array}\right.Now,\begin{aligned}& \hat{f}(\omega)=(3+2 \omega-0) u(\omega-(-4))+(0-(3+2 \omega)) u(\omega-4) \\\Rightarrow & \hat{f}(\omega)=(3+2 \omega) u(\omega+4)+[-(3+2 \omega) u(\omega-4)] \\\Rightarrow & \hat{f}(\omega)=(3+2 \omega)[u(\omega+4)-u(\omega-4)] \\\Rightarrow & \hat{f}(\omega)=(3+2 \omega)[u(\omega+4 u-4(\omega+4 u]\end{aligned}\begin{array}{l}\Rightarrow \hat{f}(\omega)=(3+2 \omega) 8 u \\\quad \therefore \hat{f}(\omega)=8 u(3+2 \omega)\end{array}(2)(b) To find the inverse fourier transform of \hat{f}(\omega) ?Here given the function as,F^{-1}(\hat{f}(\omega))=\sqrt{\frac{2}{\pi}}\{F(x) \sin (4 x)+G(x) \cos (4 x)\} .We know that,\begin{array}{l} F^{-1}(\hat{f}(\omega))=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega x} d \omega \\=\frac{1}{\sqrt{2 \pi}} \int_{-4}^{4}(3+2 \omega) e^{i \omega x} d \omega \\=\frac{1}{\sqrt{2 \pi}}\left[\frac{-(i x(2 \omega+3)-2) \cdot e^{i \omega x}}{x^{2}}\right]+c \\=\frac{-2 i x(2 \omega+3) e^{i \omega x}}{\sqrt{2 \pi} x^{2}}+c \\=\frac{-2 i(x \omega+i) i x \omega}{x^{2}}-\frac{3 i e^{i \omega x}}{x} \\=\frac{(11 x+2 i) \sin 4 x+(2-1) i x) \cos 4 x}{x^{2}}-\frac{(5 x-2 i) \sin 4 x+(5 x i+2) \cos 4 x}{x^{2}} \\\frac{\sqrt{2 \pi}}{\sqrt{2 \pi}}\end{array}(3)\begin{array}{l}=[\sin 4 x][11 x+2 i-5 x+2 i]+[\cos 4 x][\not-11 i x-5 i x-\not 2] \\\sqrt{2 \pi} x^{2} \\=\frac{(6 x+4 i) \sin 4 x-16 i x \cos 4 x}{\sqrt{2 \pi} x^{2}} \\F^{+}\{\hat{f}(\omega)\}=\frac{1}{\sqrt{2 \pi} x^{2}}\{F(x) \sin 4 x+G(x) \cos 4 x\} \\\end{array}Here, F(x)=\frac{6 x+4 i}{\sqrt{2 \pi} x^{2}} and\begin{array}{l}G(x)=\frac{-16 i x}{\sqrt{2 \pi} x^{2}}=\frac{-16 i}{\sqrt{2 \pi} x} \\\Rightarrow G(x)=\frac{-16 i}{\sqrt{2 \pi} x}\end{array} ...