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:

Let the function $f: R \rightarrow R$ be defined by
$f(t)=\begin{cases} (-1)^{n+1} 2, & \text{if } t=2n-1, n \in N \\ \frac{(2n+1-t)}{2} f(2n-1) + \frac{(t-(2n-1))}{2} f(2n+1), & \text{if } 2n-1 < t < 2n+1, n \in N \end{cases}$
Define $g(x) = \int_1^x f(t) dt, x \in (1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x) = 0$ in the interval $(1, 8]$ and $\beta = \lim_{x \rightarrow 1^+} \frac{g(x)}{x-1}$. Then the value of $\alpha + \beta$ is equal to.

The number of solutions of the equation $6 \int_{0}^{|x|} ((t^2-1) \ln t) dt = 5|x|$ for $x \in R \setminus \{0\}$ is

If ${I_n} = \int_{ - n}^n {{{\tan }^2}\{x\}dx} $ then (where $\{.\}$ denotes the fractional part function and $n \in N$):

Let $r_k = \frac{\int_0^1 (1-x^7)^k dx}{\int_0^1 (1-x^7)^{k+1} dx}$,$k \in N$. Then the value of $\sum_{k=1}^{10} \frac{1}{7(r_k-1)}$ is equal to ...........

Let $\{x\}$ and $[x]$ denote the fractional part of $x$ and the greatest integer $\leq x$ respectively of a real number $x$. If $\int_{0}^{n}\{x\} dx$,$\int_{0}^{n}[x] dx$,and $10(n^{2}-n)$ $(n \in N, n > 1)$ are three consecutive terms of a $G.P.$,then $n$ is equal to