JEE Main 2023 (Online) 10th April Evening ShiftQuestion 1
If the coefficients of $x$ and $x^{2}$ in $(1+x)^{p}(1-x)^{q}$ are $4$ and $-5$ respectively, then $2p+3q$ is equal to
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Expand each factor up to $x^2$ and multiply. Formulas used. (1) $(1+x)^p=1+px+{}^{p}C_2\,x^2+\cdots$ and $(1-x)^q=1-qx+{}^{q}C_2\,x^2-\cdots$Step 1 - Coefficient of $x$. $p-q=4$. Step 2 - Coefficient of $x^2$. $${}^{p}C_2-pq+{}^{q}C_2=\frac{(p-q)^2-(p+q)}{2}=-5.$$ With $p-q=4$: $16-(p+q)=-10$, so $p+q=26$. Step 3 - Solve. $p=15$, $q=11$. Step 4 - Value. $2p+3q=30+33=63$. Answer: (B)