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