f-x-arctg-1-x-2-x-1-and-f-1-f-2-f-21-find-tg-

Question Number 156807 by amin96 last updated on 15/Oct/21

f (x) = a r c t g \frac{1}{x^{2} + x + 1} a n d α = f (1) + f (2) + \dots + f (21)

f i n d t g (α) = ?

Commented by amin96 last updated on 15/Oct/21

A) \frac{6}{11} B) \frac{7}{11} C) \frac{11}{21} D) \frac{1}{21} E) \frac{21}{23}

Commented by amin96 last updated on 15/Oct/21

n o p e s i r

Commented by ghimisi last updated on 15/Oct/21

a r c t g \frac{1}{x^{2} + x + 1} = a r c t g \frac{1}{x} - a r c t g \frac{1}{x + 1}

Answered by gsk2684 last updated on 16/Oct/21

f (x) = \tan^{- 1} \frac{1}{1 + (x + 1) x} = \tan^{- 1} \frac{(x + 1) - x}{1 + (x + 1) x} = \tan^{- 1} (x + 1) - \tan^{- 1} x

α = f (1) + f (2) + f (3) + \dots . + f (21)

α = (\tan^{- 1} 2 - \tan^{- 1} 1)

+ (\tan^{- 1} 3 - \tan^{- 1} 2)

+ (\tan^{- 1} 4 - \tan^{- 1} 3)

+ \dots

+ (\tan^{- 1} 22 - \tan^{- 1} 21)

α = \tan^{- 1} 22 - \tan^{- 1} 1 = \tan^{- 1} \frac{22 - 1}{1 + 22 \times 1} = \tan^{- 1} \frac{21}{23}

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