Question

Find the coefficient of y in the expansion f (x, y) = (y + x – 1)^2.

Answer 1

Answer:

coefficient of y = - 2

Step-by-step explanation:

f (x, y) = (y + x – 1)^2.

=> f(x , y) = ( y + x - 1)²

=> f(x , y) = ( y + (x - 1))²

Using (a + b)² = a² + b² + 2ab

=> f(x , y) = y²  + (x - 1)²  + 2y(x - 1)

=>  f(x , y) = y²  + x² -2x + 1  + 2xy - 2y

=> f(x , y) = x² + y² + 2xy - 2x - 2y + 1

coefficient of y = - 2

Answer 2

The coefficient of y in the expansion f (x, y) = (y + x – 1)^2 is - 2

Given function

f(x, y) = ( y + x - 1)^2

Since it's a square of trinomial, We can directly expand and find the coefficient of any required term.

We use the expansion

(a + b + c)^2 = a^2+b^2+c^2+2(ab +bc +ca)

Expanding the given function ;

f ( x, y) = (y + x – 1)^2

f( x, y) = y² +x² + (-1)² + 2(xy)+2(y)(-1)+2(x)(-1)

f(x,y) = x² + y² + 1 + 2xy - 2y - 2x

Coefficient of x² = 1

Coefficient of y² = 1

Constant = 1

Coefficient of xy = 2

Coefficient of x = - 2

Coefficient of y = - 2

Therefore, The required coefficient of y in the expansion of f(x, y) is - 2