If `f(x)=(t+3x-x^2)/(x-4),` where `t` is a parameter that has minimum and maximum, then the range of values of `t` is `(0,4)` (b) `(0,oo)` `(-oo,4)` (d) `(4,oo)`

Options

A. `(0,4)`
B. `(0,oo)`
C. `-(oo,4)`
D. `(4,oo)`

Answer

3

Explanation

`f(x)=(t+3x-x^(2))/(x-4),f(x)=(x-4)(3x-2x)-(t+3x-x^(2))/(x-4)^(2)`
for maximum or minimum f(X)=0
`-2x^(2)+11x-12-t-3x+x^(2)=0`
`-x^(2)+8x-(12-t)=0`
For one maxima and minima
`64-4(12+t)gt0`
`16-12-tgt0, i.e 4gtt or tlt4`