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).

The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = -1$,$f'(\log 2) = 31$ and $\int_0^{\log 4} [f(x) - Rx] \, dx = \frac{39}{2}$ are given by

$A$ polynomial function $f(x)$ satisfying the conditions $f(x) = [f'(x)]^2$ and $\int_{0}^{1} f(x) dx = \frac{19}{12}$ can be:

The tangent to the graph of the function $y = f(x)$ at the point with abscissa $x = a$ forms with the $x$-axis an angle of $\pi/3$ and at the point with abscissa $x = b$ at an angle of $\pi/4$. Then the value of the integral $\int_{a}^{b} f(x) \cdot f''(x) \, dx$ is equal to (assume $f''(x)$ to be continuous).

Let $I_n = \int_0^1 (\log x)^n dx$,where $n$ is a non-negative integer. Then,$I_{2011} + 2011 I_{2010}$ is equal to

Let $f: R \rightarrow R$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1$,$f^{\prime}(\log _e 2)=21$,and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$,then the value of $|a+b+c|$ equals: