Question

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

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

Answer 1

Answer: (C)

$(-\infty ,4)$

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

for maximum or minimum $f^{\prime }\left(x\right)=0$

$\Rightarrow -2x^2+11x-12-t-3x+x^2=0$

$\Rightarrow -x^2+8x-(12+t)=0$

for maxima and minima to exits; the above equation must have to real and distinct roots.

$\therefore D>0\Rightarrow 64-4(12+t)>0\Rightarrow 16-12-1>0\Rightarrow 4>t\Rightarrow t<4$