If sin -1 x- x 2 - 2 - x 3 - 4 - -cos -1 x 2 - x 4 - 2 - x 6 - 4 - - 2 and 0 - x - - 2 then x -

The correct option is B 1 x x ^ 2 / 2 + x ^ 3 / 4 + ……∞ = x ^ 2 x ^ 4 / 2 + x ^ 6 / 4 ……∞ x = x ^ 2 ⇒ x = 1 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If sin -1 x...

Question

If ${sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \infty) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . . . \infty) = \frac{π}{2}$ and $0 < x < \sqrt{2}$ then $x =$

\frac{1}{2}

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- \frac{1}{2}

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Solution

The correct option is B $1$
$x - \frac{x^{2}}{2} + \frac{x^{3}}{4} + \dots \dots \infty = x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots \dots \infty$
$x = x^{2} \Rightarrow x = 1$

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If sin−1(x−x22+x34−∞)+cos−1(x2−x42+x64−.........∞)=π2 and 0<x<√2 then x=

The correct option is B 1x−x22+x34+……∞=x2−x42+x64−……∞x=x2⇒x=1

If ${sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \infty) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . . . . \infty) = \frac{π}{2}$ and $0 < x < \sqrt{2}$ then $x =$

The correct option is B $1$
$x - \frac{x^{2}}{2} + \frac{x^{3}}{4} + \dots \dots \infty = x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots \dots \infty$
$x = x^{2} \Rightarrow x = 1$