Question

If the range of the real valued function $f(x) = \frac{x^2 + x + k}{x^2 - x + k}$ is $[\frac{1}{3}, 3]$, then $k =$

• A. -2
• B. -1
• C. 1
• D. 2

Hint:

For a rational function $y = \frac{ax^2+bx+c}{dx^2+ex+f}$, to find the range, rearrange it into a quadratic equation in $x$. Then, use the condition that the discriminant ($\Delta$) must be non-negative ($\Delta \geq 0$) since $x$ is real. The resulting inequality in $y$ will give the range. The boundary values of the range are the roots of the equation $\Delta = 0$.

Answer 1

The Correct Option is C

Let $y = f(x) = \frac{x^2 + x + k}{x^2 - x + k}$.
Given that the range of the function is $[\frac{1}{3}, 3]$. This means $\frac{1}{3} \leq y \leq 3$.
We can write the equation as $y(x^2 - x + k) = x^2 + x + k$.
Rearranging the terms to form a quadratic equation in $x$:
$yx^2 - yx + yk = x^2 + x + k$
$(y - 1)x^2 - (y + 1)x + (yk - k) = 0$.
Since $x$ is a real number, the discriminant ($\Delta$) of this quadratic equation must be greater than or equal to zero ($\Delta \geq 0$).
$\Delta = b^2 - 4ac = (-(y + 1))^2 - 4(y - 1)(k(y - 1)) \geq 0$.
$(y + 1)^2 - 4k(y - 1)^2 \geq 0$.
The extreme values of the range of $y$, which are $1/3$ and $3$, are the roots of the equation $(y + 1)^2 - 4k(y - 1)^2 = 0$.
Let's substitute one of the boundary values, $y = 3$, into the equation:
$(3 + 1)^2 - 4k(3 - 1)^2 = 0$.
$4^2 - 4k(2^2) = 0$.
$16 - 4k(4) = 0$.
$16 - 16k = 0$.
$16k = 16$.
$k = 1$.
Let's verify this using the other boundary value, $y = 1/3$:
$(\frac{1}{3} + 1)^2 - 4k(\frac{1}{3} - 1)^2 = 0$.
$(\frac{4}{3})^2 - 4k(-\frac{2}{3})^2 = 0$.
$\frac{16}{9} - 4k(\frac{4}{9}) = 0$.
$\frac{16}{9} - \frac{16k}{9} = 0$.
$16 - 16k = 0$.
$k = 1$.
Both boundary values yield $k = 1$. Therefore, the correct value of $k$ is 1.