The value of ' $a$ ' for which $a x^{2} + {sin}^{- 1} (x^{2} - 2 x + 2) + {cos}^{- 1} (x^{2} - 2 x + 2) = 0$ has a real solution, is

- \frac{2}{π}

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\frac{2}{π}

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- \frac{π}{2}

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\frac{π}{2}

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Solution

The correct option is A $- \frac{π}{2}$
Here, $x^{2} - 2 x + 2 = (x - 1)^{2} + 1 \geq 1$
But $- 1 \leq (x^{2} - 2 x + 2) \leq 1$
which is possible only when
$x^{2} - 2 x + 2 = 1$
$∴ x = 1$
Then, $a (1)^{2} + {sin}^{- 1} (1) + {cos}^{- 1} (1) = 0$
$\Rightarrow a + \frac{π}{2} + 0 = 0$
$∴ a = - \frac{π}{2}$

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