If $x^{2} + x + 1, 2 x + 1, x^{2} - 1$ are the sides of a triangle, then its largest angle is

π / 3

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π / 4

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π / 6

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None of these

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Solution

The correct option is A $π / 3$
From given, we have,

$(x^{2} + x + 1)^{2} = (2 x + 1)^{2} + (x^{2} - 1)^{2} - 2 (2 x + 1) (x^{2} - 1) cos ϕ$

$(x^{2} + x + 1)^{2} - (x^{2} - 1)^{2} = (2 x + 1)^{2} - 2 (2 x + 1) (x^{2} - 1) cos ϕ$

$(2 + x) (2 x^{2} + x) = (2 x + 1)^{2} - 2 (2 x + 1) (x^{2} - 1) cos ϕ$

$x (2 + x) (2 x + 1) = (2 x + 1)^{2} - 2 (2 x + 1) (x^{2} - 1) cos ϕ$

$x (2 + x) = 2 x + 1 - 2 (x^{2} - 1) cos ϕ$

$x^{2} - 1 = - 2 (x^{2} - 1) cos ϕ$

$1 = - 2 cos ϕ \Rightarrow cos ϕ = - \frac{1}{2}$

$ϕ = \frac{π}{3}$

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x^{2} + x + 1, 2 x + 1

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3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

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