Question

If the function $f\left(x\right)=\frac{ax+b}{(x-1)(x-4)}$is monotonic decreasing at $x=2$, then the possible values of $a$ and $b$ are:

• A. $a<0,b<0$
• B. $a<0,b>0$
• C. $a>0,b\in R$
• D. $a\ge 0,b<0$

Answer 1

Answer: (A), (D)

$a\ge 0,b<0$

Let $y=f\left(x\right)=\frac{ax+b}{(x-1)(x-4)}$
$\Rightarrow \frac{dy}{dx}=\frac{\left[a\right(x^2-5x+4)-(ax+b\left)\right(2x-5\left)\right]}{(x^2-5x+4{)}^2}\Rightarrow \frac{dy}{dx}=\frac{[a{x}^2-5ax+4a-2a{x}^2+5ax-2bx+5b]}{(x^2-5x+4{)}^2}\Rightarrow \frac{dy}{dx}=\frac{[-a{x}^2-2bx+4a+5b]}{(x^2-5x+4{)}^2}$
$\therefore \frac{dy}{dx}\text{at}x=2$ is:
$\frac{dy}{dx}=\frac{-4a-4b+4a+5b}{4}$
$\Rightarrow \frac{dy}{dx}=\frac{b}{4}$
For $f\left(x\right)$ to be monotonic decreasing at $x=2$ we need to have $\frac{b}{4}<0$
$\Rightarrow b<0,a\in R$