If in the expansion of (1+x)ᵖ(1-x)^q, the coefficients of x and x² are 1 and -2 , respectively, then p²+q² is equal to :

Solution

Maths sir · 5 steps

Write each bracket up to x², multiply, and the x² condition turns into p + q.

1. Expand each bracket. Keep terms only up to x²: (1+x)ᵖ = 1 + px + p(p-1)/2x² + … and (1-x)^q = 1 - qx + q(q-1)/2x² - …
2. Coefficient of x. An x term comes from x in one bracket times 1 in the other. This gives p - q = 1.
3. Coefficient of x squared. There are three ways to get x²: x² times 1, 1 times x², and x times x. Rewrite p² + q² - 2pq as (p-q)²:

   p(p-1)/2 + q(q-1)/2 - pq = ((p-q)² - (p+q))/2 = -2

4. Solve. Put p - q = 1: (1 - (p+q))/2 = -2, so p + q = 5. Together with p - q = 1, this gives p = 3 and q = 2.
5. So. p² + q² = 9 + 4 = 13. Check: (1+3x+3x²)(1-2x+x²) has x-coefficient 3 - 2 = 1 and x²-coefficient 1 - 6 + 3 = -2.

The idea to keep. For a product of two binomials, the coefficient of x² comes from three pairs: x² with 1, 1 with x², and x with x. Never forget the cross term.

Answer from the NTA final answer key. Solution by PrepWiser.