Binomial Theorem – Coefficient of Terms – JEE Main 23 January 2025 Shift 2

Solution:

The given expression is $(1+x)^{p}(1-x)^{q}$.

Expanding using binomial theorem:
$$(1+x)^{p}(1-x)^{q} = \left(1 + p x + \frac{p(p-1)}{2}x^2 + \dots \right) \left(1 – q x + \frac{q(q-1)}{2}x^2 – \dots \right)$$

The coefficient of $x$ is given as 1.
$$1 \cdot (-q) + p \cdot 1 = 1$$
$$p – q = 1 \quad \dots (1)$$

The coefficient of $x^2$ is given as -2.
$$1 \cdot \frac{q(q-1)}{2} + p(-q) + \frac{p(p-1)}{2} \cdot 1 = -2$$
$$\frac{q^2-q}{2} – pq + \frac{p^2-p}{2} = -2$$
$$q^2 – q – 2pq + p^2 – p = -4$$
$$(p^2 – 2pq + q^2) – (p + q) = -4$$
$$(p – q)^2 – (p + q) = -4$$

Substituting $(p – q) = 1$ from equation (1):
$$(1)^2 – (p + q) = -4$$
$$1 – (p + q) = -4$$
$$p + q = 5 \quad \dots (2)$$

Solving equations (1) and (2):
$$(p – q) + (p + q) = 1 + 5 \Rightarrow 2p = 6 \Rightarrow p = 3$$
$$q = 5 – 3 \Rightarrow q = 2$$

We need to find the value of $p^2 + q^2$:
$$p^2 + q^2 = (3)^2 + (2)^2$$
$$= 9 + 4 = 13$$

Ans. (13)