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