The equation 2 ${cos}^{- 1} x = {cos}^{- 1} (2 x^{2} - 1)$ is satisfied by

- 1 \leq x \leq 1

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0 \leq x \leq 1

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x \geq 1

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x \leq 1

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Solution

The correct option is A $- 1 \leq x \leq 1$
$2 c o s^{- 1} x = c o s^{- 1} (2 x^{2} - 1)$
domain $x ϵ [- 1, 1]$
and
$- 1 \leq 2 x^{2} - 1 \leq 1$
$0 \leq 2 x^{2} \leq 2 \Rightarrow 0 \leq x^{2} \leq 1$
$\Rightarrow - 1 \leq x \leq 1$
take $c o s$ on both sides
$c o s (2 c o s^{- 1} x) = 2 x^{2} - 1 [c o s c o s^{- 1} x = x]$
$2 (c o s c o s^{- 1} x^{2}) - 1 = 2 x^{2} - 1$
$2 x^{2} - 1 = 2 x^{2} - 1$ valid for any x
$∴ x ϵ [- 1, 1]$

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Similar questions

{cos}^{- 1} x = {\begin{matrix} a π - b {cos}^{- 1} (2 x^{2} - 1), i f - 1 \leq x < 0 c {cos}^{- 1} (2 x^{2} - 1), i f 0 \leq x \leq 1 \end{matrix}

a + b + c

c o s^{- 1} x = 2 s i n^{- 1} \sqrt{\frac{1 - x}{2}} = 2 c o s^{- 1} \sqrt{\frac{1 + x}{2}}

s i n^{- 1} (- x) = - s i n^{- 1} x, c o s^{- 1} (- x) = π - c o s^{- 1} x (- 1 \leq x \leq 1)

{tan}^{- 1} (\frac{1}{x - 1} + \frac{1}{x - 2} + \frac{1}{x - 3} + \frac{1}{x - 4}) + {cos}^{- 1} (x) = \frac{3 π}{4} - {sin}^{- 1} (x)

s i n^{- 1} x + s i n^{- 1} (1 - x) = c o s^{- 1}

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