begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 & | vedclass

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(C) Given function is $f(x, y) = 2(x-y)^2 - x^4 - y^4$. First, we find the partial derivatives: $f_x = \frac{\partial}{\partial x} [2(x-y)^2 - x^4 - y

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(C) Given function is $f(x, y) = 2(x-y)^2 - x^4 - y^4$. First, we find the partial derivatives: $f_x = \frac{\partial}{\partial x} [2(x-y)^2 - x^4 - y^4] = 4(x-y) - 4x^3$. $f_y = \frac{\partial}{\partial y} [2(x-y)^2 - x^4 - y^4] = -4(x-y) - 4y^3$. Now, we find the second-order partial derivatives: $f_{xx} = \frac{\partial}{\partial x} [4(x-y) - 4x^3] = 4 - 12x^2$. $f_{yy} = \frac{\partial}{\partial y} [-4(x-y) - 4y^3] = 4 - 12y^2$. $f_{xy} = \frac{\partial}{\partial y} [4(x-y) - 4x^3] = -4$. Evaluating these at $(0,0)$: $f_{xx}(0,0) = 4 - 12(0)^2 = 4$. $f_{yy}(0,0) = 4 - 12(0)^2 = 4$. $f_{xy}(0,0) = -4$. Finally, calculate the expression: $f_{xx}f_{yy} - f_{xy}^2 = (4)(4) - (-4)^2 = 16 - 16 = 0$.

Column-$I$ (Molecules)	Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$	$1$. Two
$B$. $H_2O$	$2$. Three
$C$. $XeF_2$	$3$. Zero
$D$. $CH_4$	$4$. Four
	$5$. One

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

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