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Existence of Limit

If x 3 - x ...

Question

If $x (3 - x) ⩾ 2$ then $S i n^{- 1} (x) + S i n^{- 1} (x^{2}) + . . . . . . + S i n^{- 1} (x^{10})$ is equal to

\frac{π}{2}

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2

π

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5

π

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10

π

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Solution

The correct option is A $5$ $π$
$x (3 - x) \geq 2$

$3 x - x^{2} \geq 2$

$x^{2} - 3 x + 2 \leq 0$

$(x - 1) (x - 2) \leq 0$

$x \in (1, 2)$

but for domain of ${sin}^{- 1} x, {sin}^{- 1} x^{2}, . . . {sin}^{- 1} x^{10}$

$x \in (- 1, 1)$

Intersection is $x = 1$

$= \frac{π}{2} + \frac{π}{2} + . . . . . + \frac{π}{2}$

$= 5 π$

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Similar questions

x (3 - x) \geq 2

s i n^{- 1} (x) + s i n^{- 1} (x^{2}) + \dots + s i n^{- 1} (x^{10}) =

sin^{- 1} (x - \frac{x}{2} + \frac{x^{3}}{4} - . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . .) = \frac{π}{2}

0 < | x | < \sqrt{2}

x

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . \infty) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . \infty) = \frac{π}{2}

0 < | x | < \sqrt{2}

x

^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} + . . . . . . . . . . . . . . \infty) +

^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . . . . . \infty)

= \frac{π}{2}

0 < x < \sqrt{2}

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \infty) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . . . \infty) = \frac{π}{2}

0 < x < \sqrt{2}

x =

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