Sets and Relations – Integral Solutions – JEE Main 22 Jan 2026 Shift 2

Solution:

The inequality is $4x^2 + y^2 < 52$ where $x, y \in Z$. We analyze the possible integer values of $x$. Since $y^2 \ge 0$, we must have $4x^2 < 52$, which implies $x^2 < 13$.
Possible values for $x$ are $0, \pm 1, \pm 2, \pm 3$.

**Case 1:** $x = 0$
$$ 0 + y^2 < 52 \Rightarrow y^2 < 52 $$ $$ -\sqrt{52} < y < \sqrt{52} \Rightarrow -7.2 < y < 7.2 $$ $$ y \in \{0, \pm 1, \pm 2, ..., \pm 7\} $$ Total values = $15$
**Case 2:** $x = \pm 1$
$$ 4(1) + y^2 < 52 \Rightarrow y^2 < 48 $$ $$ -\sqrt{48} < y < \sqrt{48} \Rightarrow -6.9 < y < 6.9 $$ $$ y \in \{0, \pm 1, \pm 2, ..., \pm 6\} $$ Total values = $13$ for each $x$. Total for $x = \pm 1$ is $13 \times 2 = 26$.
**Case 3:** $x = \pm 2$
$$ 4(4) + y^2 < 52 \Rightarrow 16 + y^2 < 52 \Rightarrow y^2 < 36 $$ $$ -6 < y < 6 $$ $$ y \in \{0, \pm 1, \pm 2, ..., \pm 5\} $$ Total values = $11$ for each $x$. Total for $x = \pm 2$ is $11 \times 2 = 22$.
**Case 4:** $x = \pm 3$
$$ 4(9) + y^2 < 52 \Rightarrow 36 + y^2 < 52 \Rightarrow y^2 < 16 $$ $$ -4 < y < 4 $$ $$ y \in \{0, \pm 1, \pm 2, \pm 3\} $$ Total values = $7$ for each $x$. Total for $x = \pm 3$ is $7 \times 2 = 14$.
**Total number of elements:**
$$ \text{Total} = 15 + 26 + 22 + 14 = 77 $$

Ans. (1)