If y = frac{ax - b}{(x - 1)(x - 4)} has a tur | vedclass

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(B) Given,$y = \frac{ax - b}{(x - 1)(x - 4)} . . . . . . (i)$ has a turning point $P(2, -1)$.Since point $P$ lies on the curve,it must satisfy equatio

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$a = 1, b = 0$

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(B) Given,$y = \frac{ax - b}{(x - 1)(x - 4)} . . . . . . (i)$ has a turning point $P(2, -1)$.Since point $P$ lies on the curve,it must satisfy equation $(i)$:$-1 = \frac{2a - b}{(2 - 1)(2 - 4)} = \frac{2a - b}{-2}$$2a - b = 2 . . . . . . (ii)$At a turning point,the derivative $\frac{dy}{dx} = 0$.From equation $(i)$,$y(x^2 - 5x + 4) = ax - b$.Differentiating with respect to $x$ using the product rule:$y(2x - 5) + (x^2 - 5x + 4) \frac{dy}{dx} = a$.At $x = 2$ and $y = -1$,$\frac{dy}{dx} = 0$:$-1(2(2) - 5) + (2^2 - 5(2) + 4)(0) = a$$-1(4 - 5) = a$$a = 1$.Substituting $a = 1$ into equation $(ii)$:$2(1) - b = 2$$b = 0$.Thus,$a = 1$ and $b = 0$.

If $y = \frac{ax - b}{(x - 1)(x - 4)}$ has a turning point $P(2, -1)$,then the values of $a$ and $b$ are

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