ISRO EC 2025 Q.31: solution and answer

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

1. Write the Routh array for $s^4+2s^3+3s^2+4s+K=0$. The first two rows take the coefficients alternately: $s^4$: $1,3,K$ and $s^3$: $2,4$.

2. Row $s^2$: $b_1=\frac{2\cdot 3-1\cdot 4}{2}=1$ and $b_2=\frac{2\cdot K-1\cdot 0}{2}=K$.

3. Row $s^1$: $c_1=\frac{b_1\cdot 4-2\cdot b_2}{b_1}=\frac{4-2K}{1}=4-2K$.

4. Row $s^0$: $d_1=K$.

5. For stability every entry in the first column must be positive: $1,2,1,4-2K,K$.

6. $4-2K>0$ gives $K<2$, and the last row gives $K>0$. So $0<K<2$.

7. Check at $K=2$: the $s^1$ row is zero, and the auxiliary polynomial from the $s^2$ row, $s^2+2=0$, gives $s=\pm j\sqrt{2}$ on the imaginary axis. Indeed $s^4+2s^3+3s^2+4s+2=(s^2+2)(s^2+2s+1)$. The system is only marginally stable there, so the range is strict.