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 #10;^ 3 / 4 ....∞ #160; = x ^ 2 x #10;^ 4 / 2 + x ^ 6 / 4 ....∞ #160; x= x ^ 2 ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 1

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 $1$
$x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . \infty$ = $x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . \infty$
$x = x^{2} \Rightarrow x = 1$

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Similar questions

Q. 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 =

Q. 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

\leq \sqrt{2}

then

x =

If $s i n^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + c o s^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for 0 < 1x < $\sqrt{2}$ , then x equals

Q. Assertion :

s i n^{- 1} [x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . .] = π / 2 - c o s^{- 1} [x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . .]

for

0 < | x | < \sqrt{2}

has a unique solution. Reason:

t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = π / 2

has no solution for

- \sqrt{2} < x < 0

Q.
(a) Prove that

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

(b) If

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

for

0 < | x | < \sqrt{2}

, then x equals

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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 $1$
$x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . \infty$ = $x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . \infty$
$x = x^{2} \Rightarrow x = 1$