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

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Solution

$y = | x - 1 | \Rightarrow y = x - 1$ and $y = - (x - 1)$
$\Rightarrow y = x - 1$ and $y = - x + a$
Also, $y = 3 - | x | \Rightarrow y = 3 - x$ and $y = 3 + x$
Solving $y = x - 1$ and $y = 3 - x$ , we have
$x - 1 = 3 - x \Rightarrow 2 x = 4 \Rightarrow x = 2$
and $y = 3 - 2 \Rightarrow y = 1$
${A B}^{2} = {(2 - 1)}^{2} + {(1 - 0)}^{2} = 1 + 1 = 2 \Rightarrow A B = \sqrt{2}$
and, ${B C}^{2} = {(2 - 0)}^{2} + {(1 - 3)}^{2} = 4 + 4 = 2 \sqrt{2}$
Area of rectangle $A B C D = A B \times B C = \sqrt{2} \times 2 \sqrt{2} = 4$ square unit

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