If local maximum of f(x) = frac{ax + b}{(x - | vedclass

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(C) Given $f(x) = \frac{ax + b}{x^2 - 5x + 4}$. Since the local maximum exists at $(2, -1)$, we have $f(2) = -1$. Substituting $x = 2$ into the functi

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(C) Given $f(x) = \frac{ax + b}{x^2 - 5x + 4}$. Since the local maximum exists at $(2, -1)$, we have $f(2) = -1$. Substituting $x = 2$ into the function: $f(2) = \frac{2a + b}{(2 - 1)(2 - 4)} = \frac{2a + b}{(1)(-2)} = \frac{2a + b}{-2} = -1$. This implies $2a + b = 2$. Also, for a local maximum at $x = 2$, the derivative $f'(2) = 0$. Using the quotient rule $f'(x) = \frac{a(x^2 - 5x + 4) - (ax + b)(2x - 5)}{(x^2 - 5x + 4)^2}$. Setting $f'(2) = 0$: $a(4 - 10 + 4) - (2a + b)(4 - 5) = 0$. $a(-2) - (2a + b)(-1) = 0 \implies -2a + 2a + b = 0 \implies b = 0$. Substituting $b = 0$ into $2a + b = 2$, we get $2a = 2$, so $a = 1$. Thus, $a + b = 1 + 0 = 1$.

If local maximum of $f(x) = \frac{ax + b}{(x - 1)(x - 4)}$ exists at $(2, -1)$, then $a + b =$

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