Question:
The area enclosed by the curves \( |y + x| \le 1 \), \( |y - x| \le 1 \) and \( 2x^2 + 2y^2 = 1 \) is
The area enclosed by the curves \( |y + x| \le 1 \), \( |y - x| \le 1 \) and \( 2x^2 + 2y^2 = 1 \) is
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Convert modulus inequalities into geometric shapes to simplify area calculations.
Updated On: Apr 23, 2026
- \( \left(2 + \frac{\pi}{2}\right) \) sq units
- \( \left(2 - \frac{\pi}{2}\right) \) sq units
- \( \left(3 + \frac{\pi}{2}\right) \) sq units
- \( \left(3 - \frac{\pi}{4}\right) \) sq units
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The Correct Option is B
Solution and Explanation
Concept:
The given inequalities represent a rotated square, and the equation represents a circle.
Step 1: Interpret the inequalities. \[ |y+x| \le 1, \quad |y-x| \le 1 \Rightarrow \text{Square with area } 2 \]
Step 2: Convert circle equation: \[ 2x^2 + 2y^2 = 1 \Rightarrow x^2 + y^2 = \frac{1}{2} \] \[ \text{Area of circle} = \pi \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2} \]
Step 3: Required area: \[ \text{Area} = 2 - \frac{\pi}{2} \]
Step 1: Interpret the inequalities. \[ |y+x| \le 1, \quad |y-x| \le 1 \Rightarrow \text{Square with area } 2 \]
Step 2: Convert circle equation: \[ 2x^2 + 2y^2 = 1 \Rightarrow x^2 + y^2 = \frac{1}{2} \] \[ \text{Area of circle} = \pi \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2} \]
Step 3: Required area: \[ \text{Area} = 2 - \frac{\pi}{2} \]
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