= Solution
The signal is continuous at zero with $f(0)=1$. Its envelope is $e^{-\lambda|t|}$, but its <angular frequency> falls from $2$ on the incoming side to $1$ on the outgoing side. Its one-sided slopes are $f'(0^-)=\lambda$ and $f'(0^+)=-\lambda$, producing a cusp at the peak. Both tails decay to zero.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2-q5-signal.png]
{title=Decaying signal with a frequency change at closest approach}
{description=The illustration uses $\lambda=0.3$ to make the incoming and outgoing oscillations visible. Dashed curves show the envelopes $\pm e^{-\lambda|t|}$; the incoming oscillations have twice the outgoing <angular frequency>.}
Use $\widetilde f(\omega)=\int_{\mathbb R}e^{-i\omega t}f(t)\,dt$. Expanding each cosine into two exponentials and integrating its half-line gives
$$
\begin{aligned}
\widetilde f(\omega)
&=\frac12\left[\frac1{\lambda-i(\omega-2)}+\frac1{\lambda-i(\omega+2)}\right]\\
&\quad+\frac12\left[\frac1{\lambda+i(\omega-1)}+\frac1{\lambda+i(\omega+1)}\right].
\end{aligned}
$$
Equivalently,
$$
\boxed{\widetilde f(\omega)=\frac{\lambda-i\omega}{(\lambda-i\omega)^2+4}+\frac{\lambda+i\omega}{(\lambda+i\omega)^2+1}.}
$$
The positive $\lambda$ guarantees convergence of the <Fourier transform> integrals.
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