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If x3-x ≥ 2, ...

Question

If $x (3 - x) \geq 2$ , then the value of ${sin}^{- 1} (x) + {sin}^{- 1} (x^{2}) + {sin}^{- 1} (x^{3}) + \dots + {sin}^{- 1} (x^{10})$ is

A

\frac{π}{2}

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B

2 π

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C

5 π

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D

10 π

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Solution

The correct option is C $5 π$
$x (3 - x) \geq 2$
$\Rightarrow 3 x - x^{2} \geq 2$
$\Rightarrow x^{2} - 3 x + 2 \leq 0$
$\Rightarrow (x - 1) (x - 2) \leq 0$
$\Rightarrow x \in [1, 2]$
But the domain of ${sin}^{- 1} x$ is $[- 1, 1]$
$∴ x = 1$

${sin}^{- 1} (x) + {sin}^{- 1} (x^{2}) + {sin}^{- 1} (x^{3}) + \dots + {sin}^{- 1} (x^{10})$
$= \frac{π}{2} + \frac{π}{2} + \dots + \frac{π}{2} (10 times)$
$= 5 π$

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