Question

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

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

Answer 1

The correct option is A

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

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

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

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

$\text{For real values of x, the discriminant of}(k-1)x^2+(k+1)x+k-1=0\text{should be greater than or equal to zero.}$

$\therefore ifk\ne 1$

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

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

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

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

$\Rightarrow (3k-1)(k-3)\le 0$

$\Rightarrow \frac{1}{3}\le k\le 3i.e.k\in [\frac{1}{3},3]\left\{1\right\}...\left(i\right)$

And if k = 1, then, x = 0, which is real ...(ii) So, from (i) and (ii), we get $k\in [\frac{1}{3},3]$