GATE 2025 IN Q.16: solution and answer

Official answer: BSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Factor out the mean frequency $e^{j7t}$ (half of $9+5$): $$x(t)=e^{j7t}\left(e^{j2t}+e^{-j2t}\right)=2\cos (2t)e^{j7t}$$

2. The magnitude of a product is the product of the magnitudes, and $|e^{j7t}|=1$, so $$|x(t)|=2|\cos (2t)|$$

3. Check with $|x{|}^2=x{x}^{\ast }$: $$|x(t){|}^2=\left(e^{j9t}+e^{j5t}\right)\left(e^{-j9t}+e^{-j5t}\right)=2+e^{j4t}+e^{-j4t}=2+2\cos (4t)=4{\cos }^2(2t)$$ This agrees with step 2.

4. $T=\frac{\pi }{2}$ is a period: $$\left|\cos \left(2\left(t+\frac{\pi }{2}\right)\right)\right|=|\cos (2t+\pi )|=|-\cos (2t)|=|\cos (2t)|$$ In words, $\cos (2t)$ has period $\frac{2\pi }{2}=\pi$, and taking the absolute value flips the negative half-cycle up, so $|\cos (2t)|$ repeats after half of that.

5. No shorter period exists: $|\cos (2t)|=1$ at $t=0$, so for any period $T$ it must also equal 1 at $t=T$. But $|\cos (2t)|=1$ only at $t=\frac{k\pi }{2}$, so the smallest positive period is $\frac{\pi }{2}$. The fundamental period of $|x(t)|$ is $\frac{\pi }{2}\text{s}$, option (B).