Question

If $f\left(x\right)=\frac{t+3x-x^2}{x-4},$ where $t$ is a parameter and $f\left(x\right)$ has exactly one minimum and one maximum, then the range of values of $t$ is

• A. $(0,\infty )$
• B. $(4,\infty )$
• C. $(-\infty ,4)$
• D. $(0,4)$

Answer 1

The correct option is C $(-\infty ,4)$

$f\left(x\right)=\frac{t+3x-x^2}{x-4}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x-4)(3-2x)-(t+3x-x^2)\cdot 1}{(x-4{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{-x^2+8x-12-t}{(x-4{)}^2}$
Given that $f\left(x\right)$ has a maximum and minimum. It means $f^{\prime }\left(x\right)=0$ have two different and real roots.
$f^{\prime }\left(x\right)=0$
$\Rightarrow -x^2+8x-(12+t)=0.$
$D>0\Rightarrow 8^2-4(-1)(-12-t)>0$
$\Rightarrow 16-12-t>0\Rightarrow t<4$