Question

If we have $x\in R$ and $k=\dfrac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}$, then
A. $x\le 0$
B. $\dfrac{1}{3}\le k\le 3$
C. $k\ge 5$
D. None of these

Answer 1

First we will cross multiply ${{x}^{2}}+x+1$ , and then equation will look like $({{x}^{2}}+x+1)k={{x}^{2}}-x+1$ , now making a quadratic in x which we look like $(k-1){{x}^{2}}+(k+1)x+k-1=0$, given $x\in R$ so applying $D\ge 0$
Which gives quadratic in k as $-3{{k}^{2}}+10k-3\ge 0$ now finding the range of k

Complete step-by-step solution:
We are given $x\in R$ and $k=\dfrac{{{x}^{2}}-x+1}{{{x}^{2}}+x+1}$, then we have to find the range of k
For that we will cross multiply ${{x}^{2}}+x+1$ and get $({{x}^{2}}+x+1)k={{x}^{2}}-x+1$
On solving we get equation $(k-1){{x}^{2}}+(k+1)x+k-1=0$ now we have a quadratic in x and given a condition $x\in R$ so we can say that discriminant of this quadratic is greater than equals to zero
$D\ge 0$, where $D$ of equation $a{{x}^{2}}+bx+c=0$ is $D={{b}^{2}}-4ac$
So, calculating $D$ value for equation $(k-1){{x}^{2}}+(k+1)x+k-1=0$
Using $D={{b}^{2}}-4ac$, it will be
$D={{(k+1)}^{2}}-4(k-1)(k-1)$ which on simplifying equals to
Writing $D\ge 0$ gives expression as
$({{k}^{2}}+2k+1)-4({{k}^{2}}-2k+1)\ge 0$
Which on further solving gives equation
$-3{{k}^{2}}+10k-3\ge 0$
Now as coefficient of ${{k}^{2}}$ is negative so we will multiply with -1 both sides and it will change direction of sign and our equation will look like
$3{{k}^{2}}-10k+3\le 0$
Now to find the range of k we can write this equation as
$3{{k}^{2}}-9k-k+3\le 0$
Now taking common we can write it as
$3k(k-3)-1(k-3)\le 0$
Finally, it will look like $(3k-1)(k-3)\le 0$
Hence on looking to it we can say that range of k is $\dfrac{1}{3}\le k\le 3$
The answer is the option (B).

Note: Most of the students have doubt that why we take condition $D\ge 0$, it is because For a quadratic equation we can write its solution as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Now It is given that $x\in R$ it means value of $\sqrt{{{b}^{2}}-4ac}$ should be real and for that we can write ${{b}^{2}}-4ac\ge 0$ and $D={{b}^{2}}-4ac$