Let a, b, c be real numbers such that 2a + 3b | vedclass

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(A) Given $2a + 3b + 6c = 0$ ... $(i)$ Since $f(x)$ is a primitive of $g(x) = ax^2 + bx + c$,we have $f(x) = \int (ax^2 + bx + c) dx = \frac{a}{3}x^3

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$x^3 - 3x^2 + 2x$

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(A) Given $2a + 3b + 6c = 0$ ... $(i)$ Since $f(x)$ is a primitive of $g(x) = ax^2 + bx + c$,we have $f(x) = \int (ax^2 + bx + c) dx = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx + K$. For Rolle's Theorem to hold on $[1, 2]$,we must have $f(1) = f(2)$. $f(1) = \frac{a}{3} + \frac{b}{2} + c + K$ $f(2) = \frac{8a}{3} + 2b + 2c + K$ Equating $f(1) = f(2)$,we get $\frac{a}{3} + \frac{b}{2} + c = \frac{8a}{3} + 2b + 2c$,which simplifies to $\frac{7a}{3} + \frac{3b}{2} + c = 0$,or $14a + 9b + 6c = 0$ ... (ii) Subtracting $(i)$ from (ii): $(14a + 9b + 6c) - (2a + 3b + 6c) = 0$ $\Rightarrow 12a + 6b = 0$ $\Rightarrow b = -2a$. Substituting $b = -2a$ into $(i)$: $2a + 3(-2a) + 6c = 0$ $\Rightarrow 2a - 6a + 6c = 0$ $\Rightarrow 6c = 4a$ $\Rightarrow c = \frac{2}{3}a$. Choosing $a = 3$,we get $b = -6$ and $c = 2$. Thus,$f(x) = \frac{3}{3}x^3 - \frac{6}{2}x^2 + 2x + K = x^3 - 3x^2 + 2x + K$. Assuming $K = 0$,$f(x) = x^3 - 3x^2 + 2x$.

Let $a, b, c$ be real numbers such that $2a + 3b + 6c = 0$ and $g(x) = ax^2 + bx + c = 0$ has at least one root in the interval $(1, 2)$. If a function $f: [1, 2] \rightarrow \mathbb{R}$ for which Rolle's Theorem holds is such that $f(x)$ is a primitive of $g(x)$,then $f(x) = $

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