JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 15)
If in the expansion of $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$, the coefficients of $x$ and $x^2$ are 1 and -2 , respectively, then $\mathrm{p}^2+\mathrm{q}^2$ is equal to :
8
20
13
18
Explanation
$$\begin{aligned}
& (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right)\left({ }^q \mathrm{C}_0-{ }^q \mathrm{C}_1 \mathrm{x}+{ }^q \mathrm{C}_2 \mathrm{x}^2+\ldots\right) \\
& \text { coff of } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_0=1 \\
& \mathrm{p}-\mathrm{q}=1 \\
& \text { coff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^q \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \\
& \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \\
& \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \\
& (\mathrm{p}-\mathrm{q})^2-(\mathrm{p}+\mathrm{q})=-4 \\
& \mathrm{p}+\mathrm{q}=5 \\
& \mathrm{p}=3 \\
& \mathrm{q}=2 \\
& \text { so } \mathrm{p}^2+\mathrm{q}^2=13
\end{aligned}$$
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