Consider the quadratic equation ax^2 + bx + c | vedclass

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(A) Given $g(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$. Note that $g'(x) = ax^2 + bx + c$.Calculate $g(0) = 0$ and $g(1) = \frac{a}{3} + \frac{b}{2}

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true. Statement-$2$ is the correct explanation for Statement-$1$.

Vedclass · Answer

(A) Given $g(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$. Note that $g'(x) = ax^2 + bx + c$.Calculate $g(0) = 0$ and $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.Since $2a + 3b + 6c = 0$,we have $g(1) = 0$.Since $g(0) = 0$ and $g(1) = 0$,and $g(x)$ is a polynomial (which is continuous and differentiable everywhere),Rolle's Theorem applies to $g(x)$ on $[0, 1]$.Thus,Statement-$2$ is true.By Rolle's Theorem,there exists at least one $c_0 \in (0, 1)$ such that $g'(c_0) = 0$.Since $g'(x) = ax^2 + bx + c$,this means $ac_0^2 + bc_0 + c = 0$.Therefore,the quadratic equation $ax^2 + bx + c = 0$ has at least one root in $(0, 1)$.Thus,Statement-$1$ is true and Statement-$2$ is the correct explanation for Statement-$1$.

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