Two reactions $\mathrm{A} \rightarrow$ products and $\mathrm{B} \rightarrow$ products have rate constant, $\mathrm{k_a}$ and $\mathrm{k_b}$, respectively at temperature T and activation energies are $\mathrm{E_a}$ and $\mathrm{E_b}$, respectively. If $\mathrm{k_a} > \mathrm{k_b}$ and $\mathrm{E_a} < \mathrm{E_b}$ and assuming the frequency factor A in both the reactions are same then

Using Arrhenius equation for both reactions, $$k = A e^{-E/RT}$$ Since the frequency factor is same, $$k_a = A e^{-E_a/RT}$$ $$k_b = A e^{-E_b/RT}$$ Therefore, $$\frac{k_a}{k_b} = e^{(E_b-E_a)/RT}$$ Given $E_b>E_a$, so the exponent is positive and hence $k_a>k_b$ at any temperature. Thus option $A$ is correct. Now as temperature increases, the quantity $\frac{E_b-E_a}{RT}$ decreases, so $$\frac{k_a}{k_b} \to 1$$ This means $k_a$ and $k_b$ become closer in magnitude at higher temperature. Thus option $C$ is correct. At lower temperature, $RT$ is smaller, so $\frac{E_b-E_a}{RT}$ is larger and hence the ratio differs more from $1$. Therefore the two rate constants differ more at lower temperature. Thus option $B$ is also correct. Since $A$, $B$, and $C$ are all correct, the correct choice is $D$.