Question

If $y=\frac{ax-b}{\left(x-1\right)\left(x-4\right)}$ has a turning point $P(2,-1)$, then find the value of $a$ and $b$ respectively.

• A. $1,2$
• B. $2,1$
• C. $0,1$
• D. $1,0$

Answer 1

Answer: (D)

We have,
$y=\frac{ax-b}{\left(x-1\right)\left(x-4\right)}$
$=\frac{ax-b}{x^2-5x+4}...\left(i\right)$
$\Rightarrow \frac{dy}{dx}=\frac{\left(x^2-5x+4\right)a-\left(ax-b\right)\left(2x-5\right)}{{\left(x^2-5x+4\right)}^2}...\left(ii\right)$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{P\left(2,-1\right)}=\frac{\left(4-10+4\right)a-\left(2a-b\right)\left(4-5\right)}{\left(4-10+4^2\right)}$
$=-\frac{b}{4}$
Since $P$ is a turning point of the curve $\left(i\right)$. Therefore,
$\Rightarrow {\left(\frac{dy}{dx}\right)}_P=0$
$\Rightarrow -\frac{b}{4}=0$
$\Rightarrow b=0...\left(iii\right)$
Since $P\left(2,-1\right)$ lies on $y=\frac{ax-b}{\left(x-1\right)\left(x-4\right)}$. Therefore,
$-1=\frac{2a-b}{\left(2-1\right)\left(2-4\right)}$
$\Rightarrow -1=\frac{2a-b}{-2}$
$\Rightarrow 2a-b-2...\left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right)$, we get $a=1$, $b=0$.