If $y = \frac{a x - b}{(x - 1) (x - 4)}$ has a turning point $P (2, - 1)$ , then find the value of $a + b$

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$1$

$0$

$- 1$

$- 2$

answer is A.

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Detailed Solution

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