Question

Find the values of k for which $\left|\frac{x^2+kx+1}{x^2+x+1}\right|<2\forall x\in R$

• A. 0 < k < 4
• B. 0 < k < 2
• C. 0 < k < 3
• D. 0 < k < 1

Answer 1

Answer: (A)

$|x|<a\Rightarrow -a<x<+a$
$\therefore$ the given inequality implies
$-2<\frac{x^2+kx+1}{x^2+x+1}<2$ ...(i)
Now $x^2+x+1={\left(x+\frac{1}{2}\right)}^2+\frac{3}{4}$ is positive for all values of x.
Multiplying (i) $by{x}^2+x+1-2(x^2+x+1)<x^2+kx+1<2(x^2+x+1)$
This yields two inequations $3x^2+(2+k)x+3>0$ and $x^2+(2-k)x+1>0$
For these quadratic expressions to be positive for all values of x, their discriminants must be negative
i.e., $(2+k{)}^2-36<0$ and $(2-k{)}^2-4<0$ ... (ii)
$(k+8)(k-4)<0$ and $k(k-4)<0$ ... (iii)
$\therefore$ - 8 < k < 4 and 0 < k < 4
For both these conditions to be satisfied, 0 < k < 4.