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Question
The number of solutions of `x^2 + |x - 1| = 1` is ______.
Options
0
1
2
3
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Solution
The number of solutions of `x^2 + |x - 1| = 1` is 2.
\[x^2 + |x - 1| = x^2 + x - 1 , x \geq 1\]
\[ = x^2 - x + 1 , x < 1\]
\[x^2 + x - 1 = 1\]
\[ \Rightarrow x^2 + x - 2 = 0\]
\[ \Rightarrow x^2 + 2x - x - 2 = 0\]
\[ \Rightarrow x\left( x + 2 \right) - 1\left( x + 2 \right) = 0\]
\[ \Rightarrow \left( x + 2 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x + 2 = 0 \text { or }, x - 1 = 0\]
\[ \Rightarrow x = - 2 \text { or } x = 1\]
Since
\[-\] 2 does not satisfy the condition
\[x \geq 1\]
(ii)
\[x^2 - x + 1 = 1\]
\[ \Rightarrow x^2 - x = 0\]
\[ \Rightarrow x^2 - x = 0\]
\[ \Rightarrow x ( x - 1) = 0\]
\[ \Rightarrow x = 0 \text { or }, (x - 1) = 0\]
\[ \Rightarrow x = 0, x = 1\]
x = 1 does not satisfy the condition x < 1
So, there are two solutions.
