Consider the quadratic equation ax^2+bx+c=0,w | vedclass

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(D) Given $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Calculating $g(0)$ and $g(1)$: $g(0) = 0$. $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a+3

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Statement-$I$ is true,Statement-$II$ is true and Statement-$II$ is a correct explanation of Statement-$I$

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(D) Given $g(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$. Calculating $g(0)$ and $g(1)$: $g(0) = 0$. $g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a+3b+6c}{6}$. Since $2a+3b+6c=0$,we have $g(1) = 0$. Since $g(x)$ is a polynomial,it is continuous on $[0,1]$ and differentiable on $(0,1)$. Since $g(0) = g(1) = 0$,by Rolle's theorem,there exists at least one $c_1 \in (0,1)$ such that $g'(c_1) = 0$. Note that $g'(x) = ax^2+bx+c$. Thus,$g'(c_1) = ac_1^2+bc_1+c = 0$. This implies that the quadratic equation $ax^2+bx+c=0$ has at least one root $c_1 \in (0,1)$. Therefore,Statement-$I$ is true. Statement-$II$ is also true because $g(x)$ satisfies all conditions of Rolle's theorem on $[0,1]$,and it is the correct explanation for Statement-$I$.

Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.
Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.
Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.
Then

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Consider the quadratic equation $ax^2+bx+c=0$,where $2a+3b+6c=0$ and let $g(x)=\frac{ax^3}{3}+\frac{bx^2}{2}+cx$.Statement-$I$ : The given quadratic equation $ax^2+bx+c=0$ has at least one root in $(0,1)$.Statement-$II$ : Rolle's theorem is applicable to $g(x)$ on $[0,1]$.Then