Graphical Method of Solving Linear Programming Problems
The area of t...
Question
The area of the region bounded by y=|x−1| and y=3−|x| is:
A
less than 5
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B
greater than 3
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C
greater than or equal to 4
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D
less than or equal to 4
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Solution
The correct options are A less than 5 B greater than or equal to 4 C less than or equal to 4 D greater than 3 y=|x−1|={1−xx<1x−1x≥1} Similarly y=3−|x|={3+xx<03−xx≥0} The required region is a rectangle with length 2√2 and breadth √2. Hence, the area is 4. Hence, all the options are correct