Correct option is (d) \(\frac{e^{-aλ}}{2a} \)
The best explanation: R (λ) = \(\int_{-∞}^∞ x(t) x(t±λ) \,dt\)
= \(\int_{-∞}^∞ e^{-at} u(t) e^{-a(t-λ)} u(t-λ) \,dt\)
= \(\int_λ^∞ e^{-2at} e^{aλ} \,dt\)
= \(\frac{e^{aλ}}{-2a}\)[0-e^-2aλ]
= \(\frac{e^{-aλ}}{2a} \).