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(B) Given the curve equation: $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) =

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$1$

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(B) Given the curve equation: $xy + ax + by = 0$. Since the point $(1, 1)$ lies on the curve,we substitute $x = 1$ and $y = 1$: $1(1) + a(1) + b(1) = 0 \implies 1 + a + b = 0 \implies a + b = -1$ ... $(i)$ Now,differentiate the equation with respect to $x$: $y + x \frac{dy}{dx} + a + b \frac{dy}{dx} = 0$ $\frac{dy}{dx}(x + b) = -(y + a)$ $\frac{dy}{dx} = -\frac{y + a}{x + b}$ Given the slope at $(1, 1)$ is $2$: $2 = -\frac{1 + a}{1 + b}$ $2(1 + b) = -(1 + a)$ $2 + 2b = -1 - a $$a + 2b = -3 ... (ii)$ Subtracting $(i)$ from $(ii)$: $(a + 2b) - (a + b) = -3 - (-1)$ $b = -2$ Substituting $b = -2$ into $(i)$: $a - 2 = -1 \implies a = 1$ Finally, calculate $3a + b$: $3(1) + (-2) = 3 - 2 = 1$.

If the slope of the tangent to the curve $xy + ax + by = 0$ at the point $(1, 1)$ is $2$,then the value of $3a + b$ is:

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