The number of solutions of $x^{2} + | x - 1 | = 1 i s$

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Solution

The correct option is C
2

$x^{2} + | x - 1 | = x^{2} + x - 1, x \geq 1 x^{2} - x + 1, x < 1 (i i) x^{2} + x - 1 = 1 \Rightarrow x^{2} + x - 2 = 0 \Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 o r x - 1 = 0 \Rightarrow x = - 2 o r x = 1 Since -2 does not satisfy the condition x \geq 1. (i i) x^{2} - x + 1 = 1 \Rightarrow x^{2} - x = 0 \Rightarrow x^{2} - x = 0 \Rightarrow x (x - 1) = 0 \Rightarrow x = 0 o r (x - 1) = 0 \Rightarrow x = 0, x = 1 x = 1 does not satisfy the condition x < 1 So, there are two solutions.$

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