Question

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

Options

• k ∈ [1/3,3]

•  k ≥ 3

•  k ≤ 1/3

•  none of these

Answer 1

k ∈ [1/3,3]

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

\[ \Rightarrow k x^2 + kx + k = x^2 - x + 1\]

\[ \Rightarrow \left( k - 1 \right) x^2 + \left( k + 1 \right)x + k - 1 = 0\]

For real values of x, the discriminant of

\[\left( k - 1 \right) x^2 + \left( k + 1 \right)x + k - 1 = 0\] should be greater than or equal to zero.

\[\therefore if k \neq 1\]

\[ \left( k + 1 \right)^2 - 4\left( k - 1 \right)\left( k - 1 \right) \geq 0 \]

\[ \Rightarrow \left( k + 1 \right)^2 - \left\{ 2\left( k - 1 \right) \right\}^2 \geq 0\]

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

\[ \Rightarrow \left( 3k - 1 \right)\left( - k + 3 \right) \geq 0\]

\[ \Rightarrow \left( 3k - 1 \right)\left( k - 3 \right) \leq 0\]

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

And if k=1, then,
x=0, which is real       ...(ii)
So, from (i) and (ii), we get,

\[k \in \left[ \frac{1}{3}, 3 \right]\]