Let S = { x: cos – 1 x = + sin – 1 x + sin – 1 ( 2 x + 1 ) } . Then x S ( 2 x – 1 ) 2 is equal to __________. [2025]

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Solution

Q.
Let $S = {x : \cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)}$ . Then $\sum_{x \in S} {(2 x - 1)}^{2}$ is equal to __________. [2025]

Ans.
(5)

We have, $\cos^{- 1} x = π + \sin^{- 1} x + \sin^{- 1} (2 x + 1)$

$\Rightarrow 2 \cos^{- 1} x - \sin^{- 1} (2 x + 1) = \frac{3 π}{2}$

$\Rightarrow 2 α - β = \frac{3 π}{2}, where \cos^{- 1} x = α, \sin^{- 1} (2 x + 1) = β$

$\Rightarrow 2 α = \frac{3 π}{2} + β \Rightarrow \cos 2 α = \sin β \Rightarrow 2 \cos^{2} α - 1 = \sin β$

$\Rightarrow 2 x^{2} - 1 = 2 x + 1$ [ $∵ x = \cos α and 2 x + 1 = \sin β$ ]

$\Rightarrow x^{2} - x - 1 = 0$

$\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$

$\Rightarrow x = \frac{1 - \sqrt{5}}{2}$ ( $∵ x = \frac{1 + \sqrt{5}}{2}$ rejected as $\cos α \leq 1$ )

Now, $\sum_{x \in S} {(2 x - 1)}^{2} = \sum_{x \in S} (4 x^{2} + 1 - 4 x) = 5$

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