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Let $f\left ( x,y \right )=\dfrac{ax^{2}+by^{2}}{xy},$ where $a$ and $b$ are constants. If $\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}$ at $x = 1$ and $y = 2$, then

the relation between $a$ and $b$ is

  1. $a=\dfrac{b}{4}$
  2. $a=\dfrac{b}{2}$
  3. $a=2b$
  4. $a=4b$
in Differential Equations by (1.5k points)
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