When x = 1.5, s^{2} = 0 and 0 - 0 = 0.

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Answer: A

$r = \frac{\partial^2 f}{\partial x^2} = 2y$

$s = \frac{\partial^2 f}{\partial x \partial y} = 2x - 3$

$t = \frac{\partial^2 f}{\partial y^2} = 0$

Since, $rt - s^2 \leq 0,$ (if < 0 then we have no maxima or minima, if = 0 then we can't say anything).

Maxima will exist when $rt - s^2 > 0$ and $r < 0.$

Minima will exist when $rt - s2 > 0$ and $r > 0.$

Since, $rt - s^2$ is never $> 0$ so we have no local extremum.

$r = \frac{\partial^2 f}{\partial x^2} = 2y$

$s = \frac{\partial^2 f}{\partial x \partial y} = 2x - 3$

$t = \frac{\partial^2 f}{\partial y^2} = 0$

Since, $rt - s^2 \leq 0,$ (if < 0 then we have no maxima or minima, if = 0 then we can't say anything).

Maxima will exist when $rt - s^2 > 0$ and $r < 0.$

Minima will exist when $rt - s2 > 0$ and $r > 0.$

Since, $rt - s^2$ is never $> 0$ so we have no local extremum.