Question

The number of solutions of `x^2 + |x - 1| = 1` is ______. 

Options

• 0

• 1

• 2

• 3

Answer 1

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.