Lf x 3- x - 2- then sin-1 x-sin-1 x2-sin-1 x 10-

The correct option is C 5π If x(3 x) ≥ 2 x ^ 2 3x+2 ≤ 0 (x 1)(x 2) #160;≤0 x belongs to [ 1,2 ] #160; #160; ........(i) ...

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Complex Numbers

lf x 3- x ≥...

Question

lf $x (3 - x) \geq 2$ , then $s i n^{- 1} (x) + s i n^{- 1} (x^{2}) + \dots + s i n^{- 1} (x^{10}) =$

\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

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Similar questions

x (3 - x) ⩾ 2

S i n^{- 1} (x) + S i n^{- 1} (x^{2}) + . . . . . . + S i n^{- 1} (x^{10})

{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 =

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

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