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

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Binomial Theorem

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 :

Answer: (C)

Write

(1 + x)^{p} (1 - x)^{q}

as a product of binomial series:

(1 + x)^{p} (1 - x)^{q} = (1 + p x + (2 p) x^{2} + \dots) (1 - q x + (2 q) x^{2} - \dots)

Coefficient of

x

p - q = 1

.

Coefficient of

x^{2}

(2 p) + (2 q) - pq = - 2

, i.e.

\frac{p ( p - 1 ) + q ( q - 1 )}{2} - pq = - 2 \Rightarrow (p - q)^{2} - (p + q) = - 4

Since

p - q = 1

, this gives

1 - (p + q) = - 4

, so

p + q = 5

.

Solving

p - q = 1

and

p + q = 5

gives

p = 3

q = 2

p^{2} + q^{2} = 9 + 4 = 13

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option