Let \(\mathrm{a}=\sqrt{6}-\sqrt{5}, \mathrm{~b}=\sqrt{5}-2, \mathrm{c}=2-\sqrt{3}\). Then point out the correct alternative among the four alternatives given below. (a) a \(<\) b \(<\) c (b) b \(<\) a \(<\) c (c) a \(<\) c \(<\) b (d) b \(<\) c \(<\) a;

Given the values for \(a\), \(b\), and \(c\) as:

– \(a = \sqrt{6} – \sqrt{5}\)
– \(b = \sqrt{5} – 2\)
– \(c = 2 – \sqrt{3}\)

And using the approximate square root values:

– \(\sqrt{6} \approx 2.45\)
– \(\sqrt{5} \approx 2.24\)
– \(\sqrt{3} \approx 1.73\)

We can approximate the values of \(a\), \(b\), and \(c\) as:

– \(a \approx 2.45 – 2.24 = 0.21\)
– \(b \approx 2.24 – 2 = 0.24\)
– \(c \approx 2 – 1.73 = 0.27\)

Thus, we have the order:

\[
a < b < c
\]

Therefore, the correct alternative among the given options is:

(a) \(a < b < c\)