Boiling point elevation: $\Delta T_b = K_b \cdot m$

$2 = K_b \times 1 \Rightarrow K_b = 2\ \text{K kg mol}^{-1}$

Freezing point depression: $\Delta T_f = K_f \cdot m$

$2 = K_f \times 2 \Rightarrow K_f = 1\ \text{K kg mol}^{-1}$

Therefore $K_b = 2 = 2K_f$, i.e., $\mathbf{K_b = 2\,K_f}$

Wait — that gives option D. $K_b=2$ and $K_f=1$, so $K_b=2K_f$. But let me recheck: $\Delta T_b=K_b\cdot m_1$: $2=K_b\cdot1$, $K_b=2$. $\Delta T_f=K_f\cdot m_2$: $2=K_f\cdot2$, $K_f=1$. So $K_b=2K_f$. Official JEE answer is $K_b=K_f$ — this would require $K_b=K_f=2$. In that case $\Delta T_f=2\times2=4\neq2$. The correct computation gives $K_b=2K_f$ (index 3).