Question

The area of the region satisfying the inequalities $|x|-y≤1,y≥0$ and $y≤1$ is [This Question was asked as TITA]

• A. 3 Square units
• B. 2 Square units
• C. 5 Square units
• D. 4 Square units

Answer 1

Answer: (A)

3 Square units

Answer 2

The graph of $|x|-y≤1,y≥0$ and $y≤1$ is as follows:

Area of ABCD = Area of EFCD - Area of EAD - Area of BFC
= $EF×FC-\frac{1}{2}×EA×ED-\frac{1}{2}×BF×FC$
= $4×1-\frac{1}{2}×1×1-\frac{1}{2}×1×1$
= $4-1 = 3$ Square units