JEE Main 2025 Jan 23 Shift 2, Mathematics Q1: Introduction to Binomial Expansion

JEE Main2025Jan 23, Shift 2Mathematics

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ , the coefficients of $x$ and $x^{2}$ are $1$ and $- 2$ , respectively, then $p^{2} + q^{2}$ is equal to :

A
8
B
18
C
13
D
20

Solution

Expand each factor as a binomial series and collect the coefficients.

Coefficient of $x$ : $(1 p) - (1 q) = p - q = 1$ .

Coefficient of $x^{2}$ : $(2 p) - (1 p) (1 q) + (2 q) = \frac{p ( p - 1 )}{2} - pq + \frac{q ( q - 1 )}{2} = - 2$ .

Simplify the $x^{2}$ relation: $p^{2} + q^{2} - p - q - 2 pq = - 4$ , i.e. $(p - q)^{2} - (p + q) = - 4$ . With $p - q = 1$ : $1 - (p + q) = - 4 \Rightarrow p + q = 5$ .

Solving $p - q = 1$ , $p + q = 5$ gives $p = 3$ , $q = 2$ . Hence $p^{2} + q^{2} = 9 + 4 = 13$ .

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q1: Introduction to Binomial Expansion

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