ISRO EC 2025 Q.60: solution and answer

Official answer: BSource: ISRO's final answer key of 15 May 2026

1. For a sinusoid $\cos (\omega t)$ or $\sin (\omega t)$ the frequency is $f=\omega /2\pi$.

2. First term: $5\cos (100\pi t)$ has ${\omega }_1=100\pi \text{rad/s}$, so $f_1=\frac{100\pi }{2\pi }=50\text{Hz}$.

3. Second term: $2\sin (200\pi t)$ has ${\omega }_2=200\pi \text{rad/s}$, so $f_2=\frac{200\pi }{2\pi }=100\text{Hz}$.

4. The spectrum of a sum of sinusoids has lines only at the component frequencies, so the signal is band-limited to $f_{\max }=\max (50,100)=100\text{Hz}$.

5. The Nyquist rate is twice the highest frequency: $f_N=2f_{\max }=2\times 100=200\text{Hz}$.

6. Option (a) is $f_{\max }$ itself, option (d) is the lower component, and option (c) is twice the Nyquist rate. Only option (b) is $200\text{Hz}$.