If sin -1x-sin -1 y-sin -1 z-3 -2 and -x2-y4x4-y8-z16- then -k-1-k-

Given: nbsp;sin ^ 1x+sin ^ 1 y+sin ^ 1 z= 3 π2 As sin ^ 1 x ∈[ π2, π2] So the given condition is possibe only if sin ^ 1 x=sin ^ 1 y=sin ^ 1 z= π2 ⇒ x=y=z= 1 ...

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Standard XII

Mathematics

Modulus of a Complex Number

If sin -1x+si...

Question

If ${sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{- 3 π}{2}$ and $λ = \frac{x^{2} + y^{4}}{x^{4} + y^{8} + z^{16}},$ then $\infty \sum k = 1 λ^{k} =$

\frac{2}{3}

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3

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\frac{3}{2}

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Solution

The correct option is C $2$
Given: ${sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{- 3 π}{2}$
As ${sin}^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}]$
So the given condition is possibe only if
${sin}^{- 1} x = {sin}^{- 1} y = {sin}^{- 1} z = - \frac{π}{2} \Rightarrow x = y = z = - 1 λ = \frac{x^{2} + y^{4}}{x^{4} + y^{8} + z^{16}} \Rightarrow λ = \frac{(- 1)^{2} + (- 1)^{4}}{(- 1)^{4} + (- 1)^{8} + (- 1)^{16}} = \frac{2}{3}$
$∴ \infty \sum k = 1 λ^{k} = λ + λ^{2} + λ^{3} + \dots = \frac{λ}{1 - λ} = 2$

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Modulus of a Complex Number

Standard XII Mathematics

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If sin−1x+sin−1y+sin−1z=−3π2 and λ=x2+y4x4+y8+z16, then ∞∑k=1λk=

The correct option is C 2Given: sin−1x+sin−1y+sin−1z=−3π2 As sin−1x∈[−π2,π2] So the given condition is possibe only if sin−1x=sin−1y=sin−1z=−π2⇒x=y=z=−1λ=x2+y4x4+y8+z16⇒λ=(−1)2+(−1)4(−1)4+(−1)8+(−1)16=23 ∴∞∑k=1λk=λ+λ2+λ3+⋯=λ1−λ=2

If ${sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = \frac{- 3 π}{2}$ and $λ = \frac{x^{2} + y^{4}}{x^{4} + y^{8} + z^{16}},$ then $\infty \sum k = 1 λ^{k} =$