If in the expansion of \(\ (1+x)^p(1-x)^q\) coefficient of \(x \& x^2\) is \(1 \&-2\) then find \(p^2+q^2\)

\(\begin{aligned}
&\begin{aligned}
& \text { coeff. of } \mathrm{x} \Rightarrow \mathrm{p}-\mathrm{q}=1 \Rightarrow \mathrm{p}=1+\mathrm{q} \\
& \text { coeff. of } \mathrm{x}^2 \Rightarrow{ }^q C_2-p q+{ }^p C_2=-2 \\
& \frac{p(p-1)}{2}-p q+\frac{(q-1) q}{2}=-2 \\
& \Rightarrow \mathrm{p}^2-\mathrm{p}-2 \mathrm{pq}+ \mathrm{q}^2-\mathrm{q}=-4 \\
& \Rightarrow \mathrm{q}=2 \\
& \mathrm{p}=3
\end{aligned}\\
&\text { So, } p^2+q^2=3^2+2^2=13
\end{aligned}\)