Question

The area of the region bounded by the curves $y=|x-1|$ and $=3-|x|$ is

• A. 2 sq units
• B. 3 sq units
• C. 4 sq units
• D. 6 sq units

Answer 1

Answer: (C)

Since, $y=|x-1|=\{\begin{array}{ll}x-1,& x>1\\ -x+1,& x\le 1\end{array}$
and
$y=3-|x|=\{\begin{array}{ll}3+x,& x\le 0\\ 3-x,& x>0\end{array}$

On solving $y=x-1$ and $y=3-x$, we get
$x-1=3-x$
$\Rightarrow x=2$
and $y=3-2\Rightarrow y=1$
Now, $A{B}^2=(2-1{)}^2+(1-0{)}^2=1+1=2$
$\Rightarrow AB=\sqrt{2}$
and $B{C}^2=(0-2{)}^2+(3-1{)}^2$
$=4+4=8$
$\Rightarrow BC=2\sqrt{2}$
Area of rectangle $ABCD=AB\times BC$
$=\sqrt{2}\times 2\sqrt{2}=4sq$ units
Note The above approach is objective approach, for board (school) exams find area using integration.