Question

If the graph of $f\left(x\right)=x^2+(3-k)x+k,(wherek\in R)$ lies above and below $x-axis$, then $k$ cannot be

• A. $-1$
• B. $5$
• C. $1$
• D. $8$

Answer 1

The correct option is A $-1$

$f\left(x\right)=x^2+(3-k)x+k$ lies above and below x-axis

$D<0$

${(3-k)}^2-4k<0$

$9+k^2-6k-4k<0$

$k^2-10k+9<0$

$(k-1)(k-9)<0$

$k\in (1,9)$