GATE 2025 EE Q.21: solution and answer

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

1. Sifting property of the impulse: ${\int }_{-\infty }^{\infty }x(t)\delta (t-t_0)dt=x(t_0)$ whenever $x(t)$ is continuous at $t=t_0$. Here $\delta (t+3)=\delta (t-(-3))$, so $t_0=-3$.

2. The window: $u(t+4)-u(t-4)$ equals $1$ for $-4<t<4$ and $0$ for $|t|>4$. Since $-4<-3<4$, the window equals $1$ at $t=-3$ and throughout a neighbourhood of it. So near $t=-3$ the signal is simply $x(t)=-t^2$, which is continuous there, and the sifting property applies. (The jumps at $t=\pm 4$ are far from $t=-3$.)

3. Evaluate: $$x(-3)=-(-3{)}^2\cdot 1=-9$$

4. Therefore ${\int }_{-\infty }^{\infty }x(t)\delta (t+3)dt=-9$, which is option (A).