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

Hint: 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 \[D={{b}^{2}}-4ac\]