Question

The coefficient of \( (y-2) \) in the Taylor's series expansion of \( f(x, y) = x^2 + xy + y^2 \) in powers of \( (x-1) \) and \( (y-2) \) is:

• A. 5
• B. 3
• C. 2
• D. 1

Hint:

In the Taylor series, the coefficient of \( (y - b) \) is the partial derivative of the function with respect to \( y \), evaluated at the point \( (a, b) \).

Answer 1

The Correct Option is A

The function given is: \[ f(x, y) = x^2 + xy + y^2 \] We are tasked with finding the coefficient of \( (y - 2) \) in the Taylor series expansion of this function about \( x = 1 \) and \( y = 2 \). The Taylor series expansion of a function \( f(x, y) \) around a point \( (a, b) \) is given by: \[ f(x, y) = f(a, b) + \frac{\partial f}{\partial x} (a, b) (x - a) + \frac{\partial f}{\partial y} (a, b) (y - b) + \cdots \] We expand this for \( f(x, y) \) around \( (1, 2) \), so we need to compute the derivatives of \( f(x, y) \) at \( (1, 2) \). First, we calculate \( f(1, 2) \): \[ f(1, 2) = 1^2 + 1 \times 2 + 2^2 = 1 + 2 + 4 = 7 \] Next, compute the partial derivative of \( f(x, y) \) with respect to \( y \): \[ \frac{\partial f}{\partial y} = x + 2y \] At \( (x, y) = (1, 2) \), we have: \[ \frac{\partial f}{\partial y} (1, 2) = 1 + 2 \times 2 = 1 + 4 = 5 \] Thus, the coefficient of \( (y - 2) \) in the Taylor expansion is 5.