Discussion

Explanation:

Let X = $\left[1+\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\mathrm{ }\dots \dots \dots .\mathrm{ }\frac{1}{\sqrt{100}+\sqrt{99}}\right]$

Consider the 2nd term here i.e., $\frac{1}{\sqrt{2}+\sqrt{1}}$.

Upon rationalization, $\frac{1}{\sqrt{2}+\sqrt{1}}$ = √2 - √1.

Similarly, when we rationalize all the terms of the given expression it becomes

X = [1 + (√2 - √1) + (√3 - √2) + (√4 - √3) + … + (√100 - √99)]

⇒ X = [√100] = 10

We next find √X = √10.

Hence, option (d).

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