If $α$ and $β$ are two real values of x which satisfy the equation $s i n^{- 1} x + s i n^{- 1} (1 - x) = c o s^{- 1} x$ , then

α + β = 1 / 2

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α β = 0

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α + β = 0

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α β = - 1 / 4

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Solution

The correct options are
A $α + β = 1 / 2$
C $α β = 0$
${sin}^{- 1} (x) + {sin}^{- 1} (1 - x) = {cos}^{- 1} (x)$
Now, $x ϵ [- 1, 1]$
and $- 1 \leq (1 - x) \leq 1$
$x ϵ [0, 2]$
Hence, $x ϵ [0, 1]$
Now we get two solutions of the above equation as $0$ and $\frac{1}{2}$
Thus, the two roots are
$x = 0$ and $x = \frac{1}{2}$
$α + β = \frac{1}{2}$
and $α β = 0$

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