question_answer

Consider a function \[f(n)=\frac{1}{1+{{n}^{2}}}.\]. Let \[{{\alpha }_{n}}=\frac{1}{n}\sum\limits_{r=1}^{n}{f\left( \frac{r}{n} \right)}\] and \[{{\beta }_{n}}=\frac{1}{n}\sum\limits_{r=0}^{n-1}{f\left( \frac{r}{n} \right)}\] for \[n=1,2,3,....\] Then which of the following inequalities is true?

A) \[{{\alpha }_{n}}<{{\beta }_{n}}<\frac{\pi }{4}\]

B)       \[{{\alpha }_{n}}>{{\beta }_{n}}>\frac{\pi }{4}\]

C) \[{{\alpha }_{n}}>\frac{\pi }{4}>{{\beta }_{n}}\]

D)        \[{{\alpha }_{n}}<\frac{\pi }{4}<{{\beta }_{n}}\]

Correct Answer: D

Solution :

[d] \[f(x)=\frac{1}{1+{{x}^{2}}}\] is decreasing function Clearly, \[{{\alpha }_{n}}\] represents the area of n rectangles below the graph of \[\frac{1}{1+{{x}^{2}}}\]as shown in the following diagram: and \[\underset{n\to \infty }{\mathop{Lim}}\,\,{{\alpha }_{n}}\] will be equal to the area bounded by curve \[y=\frac{1}{1+{{x}^{2}}},\,\,x=0,\,\,x=1\] and x-axis. \[{{\beta }_{n}}\] represents the area of n rectangles above the graph of \[\frac{1}{1+{{x}^{2}}}\] as shown in the following figure: and \[\underset{n\to \infty }{\mathop{Lim}}\,{{\beta }_{n}}\] will be equal to the area bounded by the curve \[y=\frac{1}{1+{{x}^{2}}},\,\,x=0,\,\,x=1\] and x-axis.             \[\int\limits_{0}^{1}{\frac{1}{1+{{x}^{2}}}}dx=[{{\tan }^{-1}}x]_{0}^{1}=\frac{\pi }{4}\] So,  \[{{\alpha }_{n}}<\int\limits_{0}^{1}{\frac{1}{1+{{x}^{2}}}}dx<{{\beta }_{n}}\] \[\Rightarrow \,\,\,\,\,\,\,{{\alpha }_{n}}<\frac{\pi }{4}<{{\beta }_{n}}\]