If left|frac{x^2+k x+1}{x^2+x+1}right|

Question

(B) Given that for all $x \in \mathbb{R}$,$\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$. This implies $-3 < \frac{x^2+k x+1}{x^2+x+1} < 3$. Since $x^2+

Vedclass · Accepted Answer

$(-1,5)$

Vedclass · Answer

(B) Given that for all $x \in \mathbb{R}$,$\left|\frac{x^2+k x+1}{x^2+x+1}\right| This implies $-3 Since $x^2+x+1 > 0$ for all $x \in \mathbb{R}$,we can multiply by the denominator: $-3(x^2+x+1) Case $1$: $x^2+k x+1  0$. For this to hold for all $x$,the discriminant $D_1 $(3-k)^2 - 4(2)(2) $-4 Case $2$: $x^2+k x+1 > -3x^2-3x-3 \Rightarrow 4x^2+(k+3)x+4 > 0$. For this to hold for all $x$,the discriminant $D_2 $(k+3)^2 - 4(4)(4) $-8 Taking the intersection of both intervals $(-1, 7)$ and $(-11, 5)$,we get $k \in (-1, 5)$. Thus,the correct option is $B$.

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

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