GATE 2025 EE Q.41: solution and answer

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

1. With unity negative feedback, the open-loop transfer function is the forward block: $$K(s-1)G(s)=\frac{K(s-1)}{(s+1)(s+2)}$$

2. Characteristic equation $1+K(s-1)G(s)=0$. Multiply through by $(s+1)(s+2)$: $$(s+1)(s+2)+K(s-1)=s^2+(3+K)s+(2-K)=0$$

3. Routh array: row $s^2$: $1,2-K$; row $s^1$: $3+K,0$; row $s^0$: $2-K$.

4. For no sign change in the first column we need $3+K>0$ and $2-K>0$, so the closed loop is stable only for $-3<K<2$. (For a second-order polynomial this is the same as requiring every coefficient to be positive.)

5. At $K=2$ the characteristic polynomial becomes $s^2+5s=s(s+5)$, so there is a closed-loop pole at $s=0$ (marginally stable). For $K>2$ the first column reads $1,3+K>0,2-K<0$: one sign change, so exactly one closed-loop pole lies in the right half-plane. You can see the same thing from the roots: their product is $2-K<0$, so they are real and of opposite sign.

6. Check the options. For every $K>2$ the system is unstable, so (B) is true and (A) is false. At $K=1.5$ (which is greater than 1) the system is stable, so (C) is false. At $K=3$ (also greater than 1) it is unstable, so (D) is false. The answer is (B).