Question

If $x$ is real and $k=\frac{x^2-x+1}{x^2+x+1}$, then

• A. none of these
• B. $k\in [\frac{1}{3},3]$
• C. $k\le \frac{1}{3}$
• D. $k\ge 3$

Answer 1

The correct option is B $k\in [\frac{1}{3},3]$

We have,

$k=\frac{x^2-x+1}{x^2+x+1}$

$\Rightarrow k(x^2+x+1)=x^2-x+1$

$\Rightarrow (k-1)x^2+(k+1)x+(k-1)=0$

As $x\in R$, so the equation has real roots when $D\ge 0$

$\Rightarrow b^2-4ac\ge 0$

On comparing with general form of Quadratic Equation $a{x}^2+bx+c=0$

We get $a=k-1,b=k+1,c=k-1$

$\Rightarrow (k+1{)}^2-4(k-1\left)\right(k-1)\ge 0$

$\Rightarrow (k+1{)}^2-(2(k-1){)}^2\ge 0$

$\Rightarrow [k+1+2(k-1\left)\right][k+1-2(k-1\left)\right]\ge 0$

$[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$\Rightarrow (3k-1)(-k+3)\ge 0$

$\Rightarrow -3(k-\frac{1}{3})(k-3)\ge 0$

$\Rightarrow (k-\frac{1}{3})(k-3)\le 0$

$\therefore k\in [\frac{1}{3},3]$

Hence, Option (A) is correct.