If the abscissa of the vertex of the parabola | vedclass

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(B) The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Given $- \frac{b}{2a} = 1$,we have $b = -2a$.Since $a, b, c > 0$,and $b

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$2$

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(B) The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Given $- \frac{b}{2a} = 1$,we have $b = -2a$.Since $a, b, c > 0$,and $b = -2a$,this implies $a$ must be negative,which contradicts $a > 0$. However,assuming the condition $a, b, c > 0$ refers to the coefficients of the parabola and the vertex abscissa is $1$,we proceed with $b = -2a$.Given $f(x) = \int_0^x (3at^2 + bt + c) dt$,then $f'(x) = 3ax^2 + bx + c$.For $f(x)$ to be strictly increasing,$f'(x) > 0$ for all $x \in R$.This requires $3a > 0$ (which is true as $a > 0$) and the discriminant $D $D = b^2 - 4(3a)(c) = b^2 - 12ac Substituting $b = -2a$,we get $(-2a)^2 - 12ac Dividing by $4a$ (since $a > 0$),we get $a - 3c Thus,$\frac{a}{c} The greatest integer value $[\frac{a}{c}]$ can take is $2$.

If the abscissa of the vertex of the parabola $y = ax^2 + bx + c$ is $1$ $(a, b, c > 0)$ and $f(x) = \int_0^x (3at^2 + bt + c) dt$ is a strictly increasing function $\forall x \in R$,then the maximum possible value of $[\frac{a}{c}]$ is (where $[.]$ denotes the greatest integer function).

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