Exams › JEE Advanced › Maths

Find the area (in square units) of the region enclosed by the curve defined by |x + y| + |x - y| = 11.

1. 121
2. 144
3. 100
4. 169

Correct answer: 121

Solution

The relation |x + y| + |x - y| = 2*max(|x|, |y|) = 11 gives max(|x|, |y|) = 5.5, which is a square with side 11 centred at the origin. Its area is 11² = 121.

Related JEE Advanced Maths questions

• How many lattice points (points with both coordinates as integers) lie strictly inside the triangle with vertices at (0,0), (0,41), and (41,0)?
• In a chemistry class, there are 20 students, while a physics class has 30 students. If 10 students are enrolled in both classes, and the two classes are held at separate times, determine the value of k/8, where k represents the total number of students attending either class.
• Given that A represents the divisors of 15, B contains prime numbers less than 10, and C includes even numbers less than 9, how many elements are there in the intersection of (A ∪ C) and B?
• Identify the periodic function among the following:
• The function f(x) = √|x² - 5| x + 6 + √8 + 2|x| - |x|² is defined as a real number for values of x within which range?
• Let f(x) = 4x / (4x² + 2). If the sum of the integrals ∫(1/1997) + ∫(2/1997) +... + ∫(1196/1997) equals 499q, what is the value of q?

⚔️ Practice JEE Advanced Maths free + battle 1v1 →