Question

The set of values of $k$ for which roots of the quadratic equation $-x^2-2(k-1)x-(k+5)=0$ are less than \(1)\ is

• A. $R-1$
• B. $[4,\infty )$
• C. $[2,\frac{3+\sqrt{27}}{2}]$
• D. $R$

Answer 1

The correct option is B $[4,\infty )$

$\text{Let}f\left(x\right)=-x^2-2(k-1)x-(k+5)$
On comparing with standard quadratic equation $y=a{x}^2+bx+c$ we get
$a=-1,b=-2(k-1),c=-(k+5).$

We can see that $a<0,$ so the given expression will be a downward opening parabola as shown below

$\therefore$ Required conditions are
$\left(i\right)D\ge 0$
$D=b^2-4ac\ge 0$
$D=(-2(k-1){)}^2-4.(-1).(-(k+5)\ge 0$
$\Rightarrow 4k^2-12k-16\ge 0$
$\Rightarrow k^2-3k-4\ge 0$
$\Rightarrow (k-4)(k+1)\ge 0$
$\Rightarrow k\in (-\infty ,-1]\cup [4,\infty )$

$\left(ii\right)f\left(1\right)<0$
$-(1{)}^2-2(k-1\left)\right(1)-(k+5)<0$
$\Rightarrow 3k+4>0$
$\Rightarrow k\in (\frac{-4}{3},\infty )\cdots \left(2\right)$

$\left(iii\right)x$ coordinate of the vertex will be less than $1$
$\therefore \frac{-b}{2a}<1$
$\frac{-(-2(k-1\left)\right)}{2(-1)}<1$
$\Rightarrow k>0$
$\Rightarrow k\in (0,\infty )\cdots \left(3\right)$

Final set of solution will be the intersection of $\left(1\right),\left(2\right),\left(3\right).$
$k\in ((-\infty ,-1]\cup [4,\infty ))\cap (\frac{-4}{3},\infty )\cap (0,\infty )$

$\Rightarrow k\in [4,\infty )$