Question

If $y=\frac{ax+b}{(x-1)(x-4)}$ has a turning value at $(2,-1)$ find $a$ & $b$ .

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

Answer 1

The correct option is B $a=1,b=0$

Given, $y=\frac{ax+b}{(x-1)(x-4)}$
Taking log both side, $\log y=\log (ax+b)-\log (x-1)-\log (x-4)$
Differentiating w.r.t $x$ both side,
$\frac{dy}{y.dx}=\frac{a}{ax+b}-\frac{1}{x-1}-\frac{1}{x-4}$
Given turning value at $(2,-1)$
$\Rightarrow y^{\prime }\left(2\right)=0\Rightarrow \frac{a}{2a+b}-\frac{1}{2}=0\Rightarrow b=0$
Also the given curve will pass through $(2,-1)$
$\Rightarrow -1=\frac{2a}{1.-2}\Rightarrow a=1$