The number of ordered pairs of the form $(a, b)$ for which $a {(x - 1)}^{2} + b (x^{2} - 3 x + 2) + x - a^{2} = 0$ becomes an identity

0

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Infinite

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Solution

The correct option is B $1$
Given,
$a (x - 1)^{2} + b (x^{2} - 3 x + 2) + x - a^{2} = 0$
becomes an identity
$∴ a (x^{2} - 2 x + 1) + b (x^{2} - 3 x + 2) + x - a^{2} = 0$
$x^{2} (a + b) + x (- 2 a - 3 b + 1) + a + 2 b - a^{2} = 0$
becomes an identity
$\Rightarrow a + b = 0$
$\Rightarrow a = - b$
$- 2 a - 3 a + 1 = 0$
$a = - 1$
$\Rightarrow b = 1$
$a + 2 b - a^{2} = - 1 + 2 - 1 = 0$
Hence, there exists only one ordered pair $(a, b) = (- 1, 1)$
$∴$ Option B is correct.

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