Question

SOLVE THIS QUESTION OF IIT JEE 2008 ​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle S_{n} = \sum _{k = 0}^{n} \dfrac{n}{n^{2} + kn + k^{2} }$

$\sf \implies \displaystyle T_{n} = \sum _{k = 0}^{n - 1} \dfrac{n}{n^{2} + kn + k^{2} }$

⇒ n = 1, 2, 3, . . . . .  then.

As we know that,

This question is from limits of the sum topic.

We can write equation as,

$\sf \implies \displaystyle S_{n} = \sum _{k = 0}^{n} \dfrac{n}{n^{2} + kn + k^{2} }$

Divide numerator and denominator by n², we get.

$\sf \implies \displaystyle \sum _{k = 0}^{n} \dfrac{\bigg(\dfrac{n}{n^{2} }\bigg) }{\bigg( \dfrac{n^{2} }{n^{2} } + \dfrac{kn}{n^{2} } + \dfrac{k^{2} }{n^{2} } \bigg)}$

$\sf \implies \displaystyle \sum _{k = 0}^{n} \dfrac{1}{n} \bigg(\dfrac{1}{1 + k/n + k^{2} /n^{2} } \bigg) < \lim_{n \to \infty} \sum _{k = 0}^{n} \dfrac{1}{n} \bigg(\dfrac{1}{1 + k/n + (k/n)^{2} } \bigg)$

As we know that,

Now we put k/n = x and integrate from 0 to 1, we get.

$\sf \implies \displaystyle \int\limits^1_0 \dfrac{1}{1 + x + x^{2} } dx$

As we know that,

We can write equation as,

$\sf \implies \displaystyle \int\limits^1_0 \dfrac{1}{\bigg(x^{2} + x + \dfrac{1}{4} \bigg) + \dfrac{3}{4} } dx.$

$\sf \implies \displaystyle \int\limits^1_0 \dfrac{1}{\bigg( x + \dfrac{1}{2}\bigg)^{2} + \bigg(\dfrac{\sqrt{3} }{2}\bigg)^{2} } dx.$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{1}{x^{2} + a^{2} } dx = \dfrac{1}{a} tan^{-1} \bigg(\dfrac{x}{a} \bigg) + C.$

Using this formula in the equation, we get.

$\sf \implies \displaystyle \bigg[ \dfrac{1}{(\sqrt{3} /2)} tan^{-1} \bigg(\dfrac{(x + 1/2)}{(\sqrt{3}/2)} \bigg)\bigg]_{0}^{1}$

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2}{\sqrt{3} } \bigg(x + \dfrac{1}{2} \bigg) \bigg) \bigg]_{0}^{1}$

As we know that,

In definite integration first we put upper limit then we put lower limit, we get.

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg( \dfrac{2}{\sqrt{3} }\bigg( 1 + \dfrac{1}{2} \bigg) \bigg) \bigg] \ - \ \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg(\dfrac{2}{\sqrt{3} } \bigg( 0 + \dfrac{1}{2} \bigg) \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg( \dfrac{2}{\sqrt{3} } \times \dfrac{3}{2} \bigg) \bigg] \ - \ \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg( \dfrac{2}{\sqrt{3} } \times \dfrac{1}{2} \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1}(\sqrt{3} ) \bigg] \ - \bigg[ \dfrac{2}{\sqrt{3} } tan^{-1} \bigg( \dfrac{1}{\sqrt{3} } \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } \bigg(\dfrac{\pi}{3} - \dfrac{\pi}{6} \bigg) \bigg]$

$\sf \implies \displaystyle \bigg[ \dfrac{2}{\sqrt{3} } \bigg( \dfrac{2 \pi - \pi}{6} \bigg) \bigg]$

$\sf \implies \displaystyle \dfrac{2}{\sqrt{3} } \times \dfrac{\pi}{6} = \dfrac{\pi}{3\sqrt{3} }$

We can write equation as,

$\sf \implies \displaystyle S_{n} < \dfrac{\pi}{3\sqrt{3} }$

Similarly, we can write as.

$\sf \implies \displaystyle T_{n} > \dfrac{\pi}{3\sqrt{3} }$

Option [A, D] is correct answer.