Question

The number of solutions of $x^2+\left|x-1\right|=1$ is
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

(c) 2
$$\begin{gathered} x^2+|x-1|=x^2+x-1,x\ge 1 \\ =x^2-x+1,x<1 \end{gathered}$$

(i)
$$\begin{gathered} x^2+x-1=1 \\ \Rightarrow x^2+x-2=0 \\ \Rightarrow x^2+2x-x-2=0 \\ \Rightarrow x\left(x+2\right)-1\left(x+2\right)=0 \\ \Rightarrow \left(x+2\right)\left(x-1\right)=0 \\ \Rightarrow x+2=0\mathrm{or},x-1=0 \\ \Rightarrow x=-2\mathrm{or}x=1 \end{gathered}$$

Since $-$2 does not satisfy the condition $x\ge 1$

(ii)
$$\begin{gathered} x^2-x+1=1 \\ \Rightarrow x^2-x=0 \\ \Rightarrow x^2-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0\mathrm{or},(x-1)=0 \\ \Rightarrow x=0,x=1 \end{gathered}$$

x = 1 does not satisfy the condition x < 1

So, there are two solutions.