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 the greatest value of $t$ is .........

Answer 1

For function to be maximum or minimum, the necessary

condition is its $1^{st}$ derivative should be zero.

$\therefore \frac{df\left(x\right)}{dx}=0$

$\Rightarrow \frac{(x-4)(3-2x)-(t+3x-x^2)}{(x-4{)}^2}=0$

$\therefore 3x-2x^2-12+8x-t-3x+x^2=0$

$-x^2+8x-12-t=0$

Now this function has maximum & minimum

$\therefore b^2-4ac>0$ $(\because \Delta >0)$

$\therefore 64-4(-1)(-12-t)>0$

$64-48-4t>0$

$16>4t$

$4>t$