Question

The function $f(x)=\frac{ax+b}{(x-1)(x-4)}$ has a local maxima at $(2,-1)$ , then

• A. $b=1,a=0$
• B. $a=1,b=0$
• C. $b=-1,a=0$
• D. $a=-1,b=0$

Answer 1

Answer: (B)

Clearly, $f(2)=-1$
$\Rightarrow -1=\frac{2a+b}{(2-1)(2-4)}$
$\Rightarrow 2a+b=2$
Now, $f^{\prime }(x)=\frac{4a+5b+2bx-a{x}^2}{(x-1{)}^2(x-4{)}^2}$
$f^{\prime }(2)=0$
$\Rightarrow b=0$
$\Rightarrow a=1$
$\Rightarrow f^{\prime }(x)=-\frac{(x-2)(x+2)}{(x-1{)}^2(x-4{)}^2}$
Clearly, for $x>2,f^{\prime }(x)<0$ and for $x<2,f^{\prime }x)>0$.
Thus, $x=2$ is indeed the point of local maxima for $y=f(x)$