Question

The number of elements in the relation \[ R=\{(x,y): 4x^2+y^2<52,\; x,y\in\mathbb{Z}\} \] is

• A. 67
• B. 89
• C. 86
• D. 77

Hint:

When counting integer solutions of inequalities, first restrict one variable, then count valid values of the other variable for each case.

Answer 1

The Correct Option is D

We are required to count the number of integer ordered pairs \((x,y)\) satisfying \[ 4x^2 + y^2<52. \]
Step 1: Find possible integer values of \(x\).
\[ 4x^2<52 \Rightarrow x^2<13 \Rightarrow x = -3,-2,-1,0,1,2,3 \]
Step 2: Count possible integer values of \(y\) for each \(x\).

For \(x=0\): \[ y^2<52 \Rightarrow |y|\le 7 \Rightarrow 15 \text{ values} \]
For \(x=\pm1\): \[ y^2<48 \Rightarrow |y|\le 6 \Rightarrow 13 \text{ values each} \]
For \(x=\pm2\): \[ y^2<36 \Rightarrow |y|\le 5 \Rightarrow 11 \text{ values each} \]
For \(x=\pm3\): \[ y^2<16 \Rightarrow |y|\le 3 \Rightarrow 7 \text{ values each} \]
Step 3: Add all valid ordered pairs.
\[ 15 + 2(13) + 2(11) + 2(7) = 15 + 26 + 22 + 14 = 77 \]
Final Answer: \[ \boxed{77} \]