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.

Let $f: \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ be a continuous function such that $f(0)=1$ and $\int_0^{\frac{\pi}{3}} f(t) dt = 0$. Then which of the following statements is (are) $TRUE$?
$(A)$ The equation $f(x) - 3 \cos 3x = 0$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(B)$ The equation $f(x) - 3 \sin 3x = -\frac{6}{\pi}$ has at least one solution in $\left(0, \frac{\pi}{3}\right)$
$(C)$ $\lim_{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{1 - e^{x^2}} = -1$
$(D)$ $\lim_{x \rightarrow 0} \frac{\sin x \int_0^x f(t) dt}{x^2} = -1$

If $n(2n+1) \int_{0}^{1}(1-x^n)^{2n} dx = 1177 \int_{0}^{1}(1-x^n)^{2n+1} dx$,then $n \in N$ is equal to $\dots\dots$

Let $u = \int_0^1 \frac{\ln(x + 1)}{x^2 + 1} \, dx$ and $v = \int_0^{\frac{\pi}{2}} \ln(\sin 2x) \, dx$,then:

Let $f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x$ for $x \in R$ be a function which satisfies $f(x)=x+\int \limits_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to $............$

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).