Q28 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
The number of elements in the relation $\mathrm{R}=\left\{(x, y): 4 x^{2}+y^{2}<52, x, y \in \mathbf{Z}\right\}$ is
- Correct answerA77
- B67
- C86
- D89
Solution
We need $4x^2 + y^2 < 52$ with $x, y \in \mathbb{Z}$. This requires $x^2 < 13$, so $x \in \{-3,-2,-1,0,1,2,3\}$.
$x=0$: $y^2<52 \Rightarrow y \in \{-7,...,7\}$ → 15 values
$x=\pm1$: $y^2<48 \Rightarrow |y|\leq 6$ → 13 each → 26 values
$x=\pm2$: $y^2<36 \Rightarrow |y|\leq 5$ → 11 each → 22 values
$x=\pm3$: $y^2<16 \Rightarrow |y|\leq 3$ → 7 each → 14 values
Total $= 15+26+22+14 = 77$
$x=0$: $y^2<52 \Rightarrow y \in \{-7,...,7\}$ → 15 values
$x=\pm1$: $y^2<48 \Rightarrow |y|\leq 6$ → 13 each → 26 values
$x=\pm2$: $y^2<36 \Rightarrow |y|\leq 5$ → 11 each → 22 values
$x=\pm3$: $y^2<16 \Rightarrow |y|\leq 3$ → 7 each → 14 values
Total $= 15+26+22+14 = 77$
Hence, the correct answer is option: 'A'.