Question

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

• A. $(0,4)$
• B. $(0,\infty )$
• C. $(-\infty ,11)$
• D. $(4,\infty )$

Answer 1

The correct option is C $(-\infty ,11)$

$f^{\prime }\left(x\right)=\frac{(x-4)(3-2x)-(t+3x-x^2)\cdot 1}{{(x-4)}^2}=\frac{(x-12-2x^2)-(t+3x-x^2)}{{(x-4)}^2}=\frac{-(12+t)-2x-x^2}{{(x-4)}^2}=\frac{-(x^2+2x+(12+t\left)\right)}{{(x-4)}^2}$

$f^{\prime }\left(x\right)=0$ has two roots

$\Rightarrow -(x^2+2x+12+t)=0$ has two roots

$\Rightarrow$ Discriminant $>0$

$D$$=2^2-4(12+t)>0\Rightarrow 4(1-(12+t\left)\right)>0\Rightarrow -(t+11)>0\Rightarrow t+11<0\Rightarrow t<-11$

Hence, correct answer is $t\in (-\infty ,-11)$