Fourier Transform

x(t)\rightleftharpoons X(j\omega )=\frac{1}{(10+j\omega )^2}
By taking inverse Fourier transform,
\begin{aligned} x(t)&=te^{-10t}u(t) \\ \text{Now, } x(t)_{t=1}&= 1 \times e^{-10} \times 1=e^{-10}\\ \text{Thus, }|\ln (x(t))| &= |\ln (e^{-10})|=|-10|=10 \end{aligned}

The Fourier transform of
\begin{aligned} &\frac{2 \sin (t\tau /2)}{t}\rightarrow 2 \pi \text{rect}\left ( \frac{\omega }{\tau } \right ) \\ &\frac{\sin (2 \pi t)}{\pi t}\rightarrow \text{rect} \left ( \frac{\omega }{4 \pi} \right )\\ &\text{So, }\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} e^{-j\omega t}dt=\text{rect}\left ( \frac{\omega }{4 \pi} \right )\\ &\text{Putting }\omega =0 \text{ in the above equation} \\ & \int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=1\\ &2\int_{-\infty }^{\infty } \frac{\sin (2 \pi t)}{\pi t} dt=2 \end{aligned}

Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz

By observing X_1(j \omega ) and X_2(j \omega ), we can say that they are not conjugate symmetric. Since, the fourier transform is not conjugate symmetric the signal will not be real. So, x_1(t), x_2(t) are not real. Now the fourier transform of x_1(t) \cdot x_2(t) will be \frac{1}{2\pi}X_1(j \omega )*X_2(j \omega ) and by looking at X_1(j \omega ) and X_2(j \omega ), we can say that X_1(j \omega ) *X_2(j \omega ) will be conjugate symmetric and thus x_1(t) \cdot x_2(t) will be real.
By observing X_1(j \omega ) and X_2(j \omega ), we can say,
X_2( \omega )=X_1(- \omega )\overset{IFT}{\rightarrow}x_2(t)=x_1(-t)\;\;...(i)
Now, X_1(j \omega ) is real. Therefore, x_1(t) will be conjugate symmetric.
\begin{aligned} \text{i.e., } x_1(t)&=x_1^*(-t) \\ \downarrow t&= -t\\ x_1(-t)&= x_1^*(t)\\ \text{Now, }x_1(t)\cdot x_2(t) &= x_1(t)\cdot x_1(-t) \;\;(from \;(i))\\ &= x_1(t)\cdot x_1^*(t)\\ &= |x_1(t)|^2=\text{real function} \end{aligned}

\begin{aligned} \text{Since, }x(t) &=\left\{\begin{matrix} e^{j10t} &\text{for }|t|\leq 1 \\ 0& \text{for }|t|\gt 1 \end{matrix}\right. \\ \text{as, } X(\omega )&=\int_{-\infty }^{\infty } x(t)e^{-j\omega t}dt\\ \text{so, } X(\omega )&=\int_{-1}^{1 } e^{j10t}e^{-j\omega t}dt\\ &=\int_{-1}^{1 } e^{jt(10-\omega )}dt\\ &= \frac{1}{j(10-\omega )}[e^{j(10-\omega )}- e^{-j(10-\omega )}]\\ X(\omega ) &=\frac{2 \sin (\omega -10)}{(\omega -10)} \end{aligned}

For real even function x(t), the Fourier transform X(\omega ) is always real even. y(t) is a derivative of x(t) which is a real odd function because derivative of even function is an odd function and hence, Fourier transform Y(\omega ) is imaginary odd.

\Rightarrow \;\; Symmetrically located pole and zero.
\Rightarrow \;\; All pass filter.
\Rightarrow \;\; Constant magnitude
(-\infty \leq \omega \leq \infty )

Given signal can be drawn as

Therefore,
\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}
Convolution in time domain multiplication in frequency domain
\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}

Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .

Given that,
\begin{aligned} f(t)&\rightleftharpoons F(\omega )\\ g(t)&=\int_{-\infty }^{\infty }F(u)e^{-jut}du\\ \text{Now, }f(t)&=\frac{1}{2\pi}\int_{-\infty }^{\infty }F(\omega)e^{j \omega t}d\omega \\ \Rightarrow \; 2 \pi f(t)&=\int_{-\infty }^{\infty }F(\omega)e^{j \omega t}d\omega \\ \omega &\rightarrow u,\\ 2 \pi f(t)&=\int_{-\infty }^{\infty }F(u)e^{j u t}du \\ t &\rightarrow -t,\\ 2 \pi f(-t)&=\int_{-\infty }^{\infty }F(u)e^{-j u t}du \\ &=g(t)\\ \Rightarrow \; g(t)&=2 \pi f(-t)\\ &= 2\pi f(t) \end{aligned}
[If f(t) is even then f(t)=f(-t)
\Rightarrow \;\; g(t) \propto f(t) \text{ if }f(t) \text{ is even.}