Question

A function $f$ is defined as $f\left(\right.x\left.\right)=\frac{k - x^{2} + 3 x}{- 4 + x}$ where $k$ is a parameter, Then the greatest integral value of $k$ so that function $f$ has a relative minimum and a relative maximum is:

Answer 1

$f(x)=\frac{-x^2+3 x+k}{x-4} $
$f^{\prime}(x)=\frac{(-2 x+3)(x-4)--x^2+3 x+k}{(x-4)^2} $
$f^{\prime}(x)=0 \Leftrightarrow-x^2+8 x-12-k=0 $
$\Leftrightarrow x^2-8 x+12+k=0 ...(i)$
$f$ has a relative minimum and a relative maximum
$\Rightarrow $ Equation (i) has real and distinct roots
$\Leftrightarrow$ Discriminant of $ i>0 $
$\Leftrightarrow 64-4(12+k)>0 $
$\Leftrightarrow k< 4$
$\Rightarrow$ The greatest integral value of $k=3$.