AP TET Paper 2 Math and Science Model Paper 2

Concept:Each fraction can be simplified by rationalizing the denominator using the identity $\frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$.Explanation:Take a general term of the series: $\frac{1}{\sqrt{n} + \sqrt{n-1}}$.Multiply the numerator and the denominator by $\sqrt{n} - \sqrt{n-1}$:$$\frac{1}{\sqrt{n} + \sqrt{n-1}} = \frac{\sqrt{n} - \sqrt{n-1}}{(\sqrt{n} + \sqrt{n-1})(\sqrt{n} - \sqrt{n-1})}$$Using $(a+b)(a-b) = a^2 - b^2$, the denominator simplifies to $n - (n-1) = 1$.Hence, $\frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$.Apply this to each term of the given sum:$\frac{1}{\sqrt{10} + \sqrt{9}} = \sqrt{10} - \sqrt{9}$$\frac{1}{\sqrt{11} + \sqrt{10}} = \sqrt{11} - \sqrt{10}$$\frac{1}{\sqrt{12} + \sqrt{11}} = \sqrt{12} - \sqrt{11}$Continuing this pattern, the last term becomes $\frac{1}{\sqrt{196} + \sqrt{195}} = \sqrt{196} - \sqrt{195}$.So the whole sum is:$$(\sqrt{10} - \sqrt{9}) + (\sqrt{11} - \sqrt{10}) + (\sqrt{12} - \sqrt{11}) + \dots + (\sqrt{196} - \sqrt{195})$$This is a telescoping series, so all intermediate terms cancel out.Only $-\sqrt{9} + \sqrt{196}$ remains.Since $\sqrt{9} = 3$ and $\sqrt{196} = 14$, the value is $14 - 3 = 11$.Answer:The value of the expression is $11$, which is option C.