Question:
The number of solutions of $x^{2}+|x-1|=1$ is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2
$x^{2}+|x-1|=x^{2}+x-1, x \geq 1$
$=x^{2}-x+1 \quad, x<1$
(i) $x^{2}+x-1=1$
$\Rightarrow x^{2}+x-2=0$
$\Rightarrow x^{2}+2 x-x-2=0$
$\Rightarrow x(x+2)-1(x+2)=0$
$\Rightarrow(x+2)(x-1)=0$
$\Rightarrow x+2=0$ or, $x-1=0$
$\Rightarrow x=-2$ or $x=1$
Since $-2$ does not satisfy the condition $x \geq 1$
(ii) $x^{2}-x+1=1$
$\Rightarrow x^{2}-x=0$
$\Rightarrow x^{2}-x=0$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0$ or, $(x-1)=0$
$\Rightarrow x=0, \quad x=1$
$x=1$ does not satisfy the condition $x<1$
So, there are two solutions.
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