If $\left|\frac{x^{2}+k x+1}{x^{2}+x+1}\right| < 3$ for all real numbers $x$, then the range of the…
If $\left|\frac{x^{2}+k x+1}{x^{2}+x+1}\right| < 3$ for all real numbers $x$, then the range of the parameter $k$ is
- \((0,4)\)
- \((-1,5)\)
- \((-4,0)\)
- \((-5,1)\)
Solution
It is given, for all $x \in R$
$\begin{aligned}
&\left|\frac{x^{2}+k x+1}{x^{2}+x+1}\right| < 3 \\
& \Rightarrow -3 < \frac{x^{2}+k x+1}{x^{2}+x+1} < 3 \\
& \Rightarrow -3 x^{2}-3 x-3 < x^{2}+k x+1 < 3 x^{2}+3 x+3 \\
& \left[\because x^{2}+x+1 > 0, \forall x \in R\right]
\end{aligned}$
So, $4 x^{2}+(k+3) x+4 > 0, \forall x \in R$
Then $D < 0 \Rightarrow(k+3)^{2}-4(4)(4) < 0$
$\begin{aligned}
& \Rightarrow (k+3)^{2}-8^{2} < 0 \\
& \Rightarrow (k+3-8)(k+3+8) < 0 \\
& \Rightarrow k \in(-11,5) \quad \ldots (i)
\end{aligned}$
and \(2 x^2+(3-k) x+2 > 0, \forall x \in R\)
Then \(D < 0 \Rightarrow(3-k)^2-16 < 0\)
\(\Rightarrow \quad(k-3-4)(k-3+4) < 0\)
\(\Rightarrow \quad k \in(-1,7) \quad \ldots (ii)\)
From intervals (i) and (ii), we get
\(k \in(-1,5)\)
Hence, option (b) is correct.
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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