Rozwiązanie:
Pytanie 1.
[tex]f(x)=x^{2}+kx+3[/tex]
Warunek:
[tex]\Delta<0\\\Delta=k^{2}-12\\k^{2}-12<0\\(k-2\sqrt{3})(k+2\sqrt{3})<0\\k \in (-2\sqrt{3},2\sqrt{3})[/tex]
Pytanie 2.
[tex]-x^{2}+6x+4k-2=0[/tex]
Warunek:
[tex]\Delta>0\\\Delta=36-4*(-1)*(4k-2)=36+4(4k-2)=36+16k-8=16k+28\\16k+28>0\\k>-\frac{7}{4}\\k \in (-\frac{7}{4},\infty)[/tex]
Pytanie 3.
[tex]x^{2}-3kx+2k^{2}-1=0[/tex]
Warunki:
[tex]\Delta>0\\\Delta=9k^{2}-4(2k^{2}-1)=9k^{2}-8k^{2}+4=k^{2}+4\\k^{2}+4>0\\k \in R[/tex]
[tex]x_{1}+x_{2}>x_{1}x_{2}\\3k>2k^{2}-1\\2k^{2}-3k-1<0\\\Delta_{k}=9-4*2*(-1)=17\\k_{1}=\frac{3-\sqrt{17} }{4} \\k_{2}=\frac{3+\sqrt{17} }{4} \\k \in (\frac{3-\sqrt{17} }{4}, \frac{3+\sqrt{17} }{4})[/tex]
Ostatecznie otrzymamy więc:
[tex]k \in (\frac{3-\sqrt{17} }{4}, \frac{3+\sqrt{17} }{4})[/tex]
