Arrange the following in descending order: A. $\sqrt{401}-\sqrt{399}$ B. $\sqrt{301}-\sqrt{299}$ C. $\sqrt{101}-\sqrt{99}$ D. $\sqrt{201}-\sqrt{199}$ Choose the correct answer from the options given below:

Each option is of the form $\sqrt{\text{bigger}}-\sqrt{\text{smaller}}$, which is painful to compare directly. Rationalise each one by multiplying and dividing by its conjugate, using the identity $(A-B)(A+B)=A^2-B^2$. For any number $x$, $\sqrt{x+1}-\sqrt{x-1}=\dfrac{(x+1)-(x-1)}{\sqrt{x+1}+\sqrt{x-1}}=\dfrac{2}{\sqrt{x+1}+\sqrt{x-1}}$ | Option | After rationalising | Size of denominator | |---|---|---| | A | $\dfrac{2}{\sqrt{401}+\sqrt{399}}$ | largest | | B | $\dfrac{2}{\sqrt{301}+\sqrt{299}}$ | third | | C | $\dfrac{2}{\sqrt{101}+\sqrt{99}}$ | smallest | | D | $\dfrac{2}{\sqrt{201}+\sqrt{199}}$ | second smallest | Every option now has the same numerator 2, so the smaller the denominator, the bigger the value. Descending order: $C>D>B>A$