Let a, b, c in mathbb{R} be such that 2a + 3b | vedclass

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(B) The slope of the tangent to the curve $y = g(x)$ at any point $x$ is given by $g'(x)$.Since $g(x)$ is the antiderivative of $f(x)$,we have $g'(x)

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$\frac{a}{6} = \frac{b}{-18} = \frac{c}{7}$

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(B) The slope of the tangent to the curve $y = g(x)$ at any point $x$ is given by $g'(x)$.Since $g(x)$ is the antiderivative of $f(x)$,we have $g'(x) = f(x) = ax^2 + bx + c$.The slopes of the tangents at $x = 1$ and $x = 2$ are equal,so $g'(1) = g'(2)$.This implies $f(1) = f(2)$.Substituting the values,we get $a(1)^2 + b(1) + c = a(2)^2 + b(2) + c$.$a + b + c = 4a + 2b + c$.$3a + b = 0 \implies b = -3a$.We are given the equation $2a + 3b + 6c = 0$.Substituting $b = -3a$ into this equation: $2a + 3(-3a) + 6c = 0$.$2a - 9a + 6c = 0 \implies -7a + 6c = 0 \implies 6c = 7a$.Thus,$a = \frac{a}{1}$,$b = -3a = \frac{a}{-1/3}$,and $c = \frac{7a}{6} = \frac{a}{6/7}$.To find the ratio,we can write $a : b : c = a : -3a : \frac{7a}{6}$.Multiplying by $6$,we get $6 : -18 : 7$.Therefore,$\frac{a}{6} = \frac{b}{-18} = \frac{c}{7}$.

Let $a, b, c \in \mathbb{R}$ be such that $2a + 3b + 6c = 0$ and $g(x)$ be the antiderivative of $f(x) = ax^2 + bx + c$. If the slopes of the tangents drawn to the curve $y = g(x)$ at $(1, g(1))$ and $(2, g(2))$ are equal,then

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