If y = ( a x - b ) ( x - 1 ) ( x - 4 ) has a turning point P 2 , - 1 , then the values of a and b are

Given, $y=\frac{ax-b}{\left(x-1\right)\left(x-4\right)} \dots \left(\mathrm{i}\right)$ has turning point $P\left(2,-1\right)$

$\therefore$ Point $P$ satisfy the equation $\left(\mathrm{i}\right)$

$\Rightarrow -1=\frac{2a-b}{\left(2-1\right)\left(2-4\right)}=\frac{2a-b}{-2}$

$\Rightarrow 2a-b=2 \dots \left(\mathrm{ii}\right)$

As we know turning point of function is a point where $f^{\text{'}}\left(x\right)=0$.

So, let's find $\frac{dy}{dx}$

From equation $\left(\mathrm{i}\right)$, $y\left(x-1\right)\left(x-4\right)=ax-b$

$\Rightarrow y\left(x^2-5x+4\right)=ax-b$

Differentiating the above equation w.r.t $x,$we get

$\Rightarrow y\left(2x-5\right)+\left(x^2-5x+4\right)\frac{dy}{dx}=a$

$\Rightarrow \frac{dy}{dx}=\frac{a-y\left(2x-5\right)}{\left(x^2-5x+4\right)}$

$\because {\left( \frac{dy}{dx}\right)}_{\left(2,-1\right)}=0$

$\Rightarrow \frac{a-\left(-1\right)\left(-1\right)}{4-10+4}=0$

$\Rightarrow a=1$

Put $a=1$ in the equation $\left(\mathrm{ii}\right)$, we get

$\Rightarrow b=0$