"3-kx" and "\\frac{x^2}{4}-3" are continuous all over their domains because these are polynomials, thus "F" is continuous all over "[1;2[\\cup]2;\\infty[" for an arbitrary "k".
So we need to find "\\lim\\limits_{x\\to 2-0}3-kx" and "\\lim\\limits_{x\\to 2+0}\\left(\\frac{x^2}{4}-3\\right)" and set them equal to achieve "F" to be continuous at "x=2" . So
"\\lim\\limits_{x\\to 2-0}3-kx=3-2k"
"\\lim\\limits_{x\\to 2+0}\\left(\\frac{x^2}{4}-3\\right)=\\frac{2^2}{4}-3=1-3=-2" , so
"3-2k=-2"
"k=\\frac52"
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