We know $$x^2-9 \leq 0 \& y^2-16>0$$; we need to check from the options that which of them is true.

The equation $$x^2-9 \leq 0$$ is same as $$x^2 \leq 9 \Rightarrow-3<x<3$$ whereas the second equation $$y^2-16>0 \Rightarrow y>4$$ or $$y<-4$$

The minimum positive integral value of the product of $$x$$ and $$y$$ will come if we take either $x$ and $y$ both negative or both positive. The minimum possible negative values of $$x$$ and $$y$$ are -2 and -3 , so we get the product $$\mathrm{xy}=-2 \times-3=6 \&$$ the minimum possible positive values of x and y are 1 and 5 respectively, so we get the product $$x y=1 \times 5=5$$.

So, the minimum positive integral value of the product of x and y is 5 .

Similarly to find the maximum negative integral value of $$x$$ and $$y$$, we would consider exactly one of $$x$$ and $$y$$ negative and take their minimum possible values. If we take $$x=-1$$ and $$y=5$$, we get the product $$\mathrm{xy}=-1 \times 5=-5 \&$$ if we take $$\mathrm{x}=1$$ and $$\mathrm{y}=-5$$, we get $$\mathrm{xy}=1 \times-5=-5$$

So, the maximum possible negative integral value of the product of x and y is -5 .
Hence options (B) \& (C) are correct.