Question

Number of ordered pair(s) of (x, y) satisfying the system of simultaneous equations $|x^2-2x|+y=1$ and $x^2+|y|=1$ is $(x,yϵR);$

• A. $1$
• B. $2$
• C. $4$
• D. infinitely many

Answer 1

Answer: (C)

$4$

Given: Simultaneous equations $|x^2-2x|+y=1$ and $x^2+\left|y\right|=1$, and $x,y\in R$

To find the number of ordered pairs of $(x,y)$

Solution:

The equation $|x^2-2x|+y=1$ can be written without modulus as

(i) $x^2-2x+y=1⟹y=1-x^2+2x$.

This is quadratic where $a=-1,b=2,c=1$. Solving for the roots of the equation we get $x=\frac{2\pm \sqrt{8}}{-2}⟹x=-2.4,0.4$. Hence this equation has two roots therefore it will have 2 ordered pair of $(x,y)$, i.e., $(-2.4,-9.56),(0.4,0.2)$

(ii) $-(x^2-2x)+y=1⟹y=1+x^2-2x⟹y=(x-1{)}^2$.

This gives $x=1$. Hence in this case it will have one ordered pair of $(x,y)$ i.e., $(1,0)$

The equation $x^2+\left|y\right|=1$ can be written without modulus as

(i) $x^2+y=1⟹y=1-x^2$. This gives $x=\pm 1$. Hence in this case we get $(x,y)$ pair as $(1,0),(-1,0)$.

(ii) $x^2-y=1⟹y=x^2-1$. This gives $x=\pm 1$. Hence in this case we get $(x,y)$ pair as $(1,0),(-1,0)$.

So the ordered pairs so obtained are $(-2.4,-9.56),(0.4,0.2),(1,0),(-1,0)$