Question

If the inequality $\frac{x^2-kx-2}{x^2-3x+4}>-1$ for every $x\in R$ and $S$ is the sum of all integral values of $k$ then find the value of $S^2$ ?

Answer 1

$\frac{x^2-kx-2}{x^2-3x+4}>-1\Rightarrow x^2-kx-2>-(x^2-3x+4)\Rightarrow x^2-\left(\frac{k+3}{2}\right)x+1>0$

for real value of $x$, the discriminant should be greater than (or) equal to zero

${[-\left(\frac{k+3}{2}\right)]}^2-\mathrm{4.1.1}\ge 0\Rightarrow k^2+6k-7\ge 0\Rightarrow (k-1)(k+7)\ge 0\therefore k\ge 1ork\ge -7fork\ge 1S=1+2+3+4+....+n=\frac{n(n+1)}{2}\therefore S^2=\frac{n^2(n+1{)}^2}{4}andfork\ge -7S=(-7-6-5-4-.......)+0+(1+2+3+.....+n)=-28+\frac{n(n+1)}{2}=\frac{n^2+n-56}{2}{S}^2=\frac{{(n^2+n-56)}^2}{4}$