JEE Main 2025 — 23 January, Evening Shift — Question 1

$(1+x)^{p}(1-x)^{q}=\left({ }^{p} C_{0}+{ }^{p} C_{1} x+{ }^{p} C_{2} x^{2}+\ldots\right)\left({ }^{q} C_{0}-{ }^{q} C_{1} x+{ }^{q} C_{2} x^{2}+\ldots\right)$

coff of $\mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_{0}{ }^{\mathrm{q}} \mathrm{C}_{1}+{ }^{\mathrm{p}} \mathrm{C}_{1}{ }^{\mathrm{q}} \mathrm{C}_{0}=1$

$\mathrm{p}-\mathrm{q}=1$

coff of $\mathrm{x}^{2} \equiv{ }^{\mathrm{p}} \mathrm{C}_{0}{ }^{q} \mathrm{C}_{2}-{ }^{\mathrm{p}} \mathrm{C}_{1}{ }^{q} \mathrm{C}_{1}+{ }^{\mathrm{p}} \mathrm{C}_{2}{ }^{q} \mathrm{C}_{0}=-2$

$\frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2$

$q^{2}-q-2 p q+p^{2}-p=-4$

$1-(p+q)=-4$

$\mathrm{p}+\mathrm{q}=5$

$\mathrm{p}=3$ $\mathrm{q}=2$

so $\boxed{{p}^{2}+\mathrm{q}^{2}=13}$