Question

On differentiation if we get f (x,y)dy - g(x,y)dx = 0 from 2x²-3xy+y²+x+2y-8 = 0 then g(2,2)/f(1,1) =

• A. \(\frac{11}{7}\)
• B. -3
• C. \(\frac{-1}{3}\)
• D. 7

Answer 1

The Correct Option is B

To solve the problem, we need to find $\frac{g(2, 2)}{f(1, 1)}$ for functions derived from the implicit differentiation of the equation $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.

1. Differentiate the Equation:
Differentiate $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$ with respect to $x$:
$4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$.
Rearrange terms:
$4x - 3y + 1 = (3x - 2y - 2) \frac{dy}{dx}$.
Thus, $\frac{dy}{dx} = \frac{4x - 3y + 1}{3x - 2y - 2}$.

2. Identify $f(x, y)$ and $g(x, y)$:
We want the differential form $f(x, y) dy - g(x, y) dx = 0$, so $\frac{dy}{dx} = \frac{g(x, y)}{f(x, y)}$.
Comparing with $\frac{dy}{dx} = \frac{4x - 3y + 1}{3x - 2y - 2}$, we get:
$g(x, y) = 4x - 3y + 1$ and $f(x, y) = 3x - 2y - 2$.

3. Compute $\frac{g(2, 2)}{f(1, 1)}$:
Evaluate $g(2, 2) = 4(2) - 3(2) + 1 = 8 - 6 + 1 = 3$.
Evaluate $f(1, 1) = 3(1) - 2(1) - 2 = 3 - 2 - 2 = -1$.
Thus, $\frac{g(2, 2)}{f(1, 1)} = \frac{3}{-1} = -3$.

Final Answer:
The value of $\frac{g(2, 2)}{f(1, 1)}$ is $-3$.