JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $a_1 = 1, a_2, a_3, a_4, \dots$ be consecutive natural numbers. Then $\tan^{-1}\left(\frac{1}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{1}{1 + a_2 a_3}\right) + \dots + \tan^{-1}\left(\frac{1}{1 + a_{2021} a_{2022}}\right)$ is equal to:

Choose the correct answer:

Explanation

Solution:

Chunki $a_n$ consecutive natural numbers hain, isliye $a_{n+1} - a_n = 1$ .
General Term ( $T_n$ ):

$T_n = \tan^{-1}\left(\frac{a_{n+1} - a_n}{1 + a_n a_{n+1}}\right) = \tan^{-1}(a_{n+1}) - \tan^{-1}(a_n)$
Series ka Sum ( $S$ ):

$S = \sum_{n=1}^{2021} [\tan^{-1}(a_{n+1}) - \tan^{-1}(a_n)]$

$S = [\tan^{-1}(a_2) - \tan^{-1}(a_1)] + [\tan^{-1}(a_3) - \tan^{-1}(a_2)] + \dots + [\tan^{-1}(a_{2022}) - \tan^{-1}(a_{2021})]$
Simplification:

Sare middle terms cancel ho jayenge:

$S = \tan^{-1}(a_{2022}) - \tan^{-1}(a_1)$

Chunki $a_1 = 1$ aur $a_{2022} = 2022$ :

$S = \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{\pi}{4}$

Sahi Vikalp: (3)