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 :

<p>$$\begin{aligned} & (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right)\left({ }^q \mathrm{C}_0-{ }^q \mathrm{C}_1 \mathrm{x}+{ }^q \mathrm{C}_2 \mathrm{x}^2+\ldots\right) \\ & \text { coff of } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_0=1 \\ & \mathrm{p}-\mathrm{q}=1 \\ & \text { coff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^q \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \\ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \\ & \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \\ & (\mathrm{p}-\mathrm{q})^2-(\mathrm{p}+\mathrm{q})=-4 \\ & \mathrm{p}+\mathrm{q}=5 \\ & \mathrm{p}=3 \\ & \mathrm{q}=2 \\ & \text { so } \mathrm{p}^2+\mathrm{q}^2=13 \end{aligned}$$</p>