Question

If \( 4x^2 + y^2<52 \), where \( x, y \in \mathbb{Z} \), then the number of ordered pairs \( (x,y) \) is:

• A. \(67\)
• B. \(87\)
• C. \(77\)
• D. \(38\)

Hint:

For inequalities involving integers:
First restrict one variable using bounds
Then count valid integers for the other variable Symmetry about axes simplifies counting.

Answer 1

The Correct Option is C

To solve the problem and find the number of ordered pairs \((x, y)\) such that the inequality \(4x^2 + y^2 < 52\) holds and \(x, y \in \mathbb{Z}\) (integers), we will analyze the constraints systematically.

1.  First, let's consider the range for \(x\). Since \(4x^2 < 52\), we can derive:
   • \(x^2 < \frac{52}{4} = 13\).
2. This implies:
   • \(-\sqrt{13} < x < \sqrt{13}\), which gives:
   • \(-3.605 < x < 3.605\).
3. Since \(x\) is an integer, possible integer values for \(x\) are: \(-3, -2, -1, 0, 1, 2, 3\).

Next, we determine the values of \(y\) for each integer value of \(x\).

1. For each integer \(x\), compute the maximum value of \(y^2\):
   • \(y^2 < 52 - 4x^2\).
2. We calculate the possible \(y\) values for each \(x\):
   • \(x = 0\): \(4(0)^2 + y^2 < 52 \Rightarrow y^2 < 52\) gives \(y = -7, -6, \ldots, 0, \ldots, 6, 7\), yielding 15 values.
   • \(x = 1\) or \(-1\): \(4(1)^2 + y^2 < 52 \Rightarrow y^2 < 48\) gives \(y = -6, -5, \ldots, 0, \ldots, 5, 6\), yielding 13 values each.
   • \(x = 2\) or \(-2\): \(4(2)^2 + y^2 < 52 \Rightarrow y^2 < 36\) gives \(y = -5, -4, \ldots, 0, \ldots, 4, 5\), yielding 11 values each.
   • \(x = 3\) or \(-3\): \(4(3)^2 + y^2 < 52 \Rightarrow y^2 < 16\) gives \(y = -3, -2, -1, 0, 1, 2, 3\), yielding 7 values each.

Finally, sum these values to find the total number of integer solutions:

• For \(x = 0\): 15 values
• For \(x = \pm 1\): 13 + 13 = 26 values
• For \(x = \pm 2\): 11 + 11 = 22 values
• For \(x = \pm 3\): 7 + 7 = 14 values

Total number of ordered pairs is \(15 + 26 + 22 + 14 = 77\).

Therefore, the number of ordered pairs \((x, y)\) that satisfy the inequality is 77.