Let $f(x) = \int\limits_0^x {(t^2 + 2t + 2)dt}$ where $x$ is the set of real numbers satisfying the inequation $\log_{\sqrt{2}}(1 + \sqrt{6x - x^2 - 8}) \ge 0$. If the range of $f(x)$ is $[a, b]$,then $(a + b)$ is:

Let $\operatorname{Max} \limits _{0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\alpha$ and $\operatorname{Min} \limits _ {0 \leq x \leq 2}\left\{\frac{9-x^{2}}{5-x}\right\}=\beta$. If $\int\limits_{\beta-\frac{8}{3}}^{2 \alpha-1} \operatorname{Max}\left\{\frac{9- x ^{2}}{5- x }, x \right\} dx =\alpha_{1}+\alpha_{2} \log _{e}\left(\frac{8}{15}\right)$,then $\alpha_{1}+\alpha_{2}$ is equal to

If $x$ satisfies the equation $\left( \int_{0}^{1} \frac{dt}{t^2 + 2t \cos \alpha + 1} \right) x^2 - \left( \int_{-3}^{3} \frac{t^2 \sin 2t}{t^2 + 1} dt \right) x - 2 = 0$ for $0 < \alpha < \pi$,then the value of $x$ is

Let $f: R \rightarrow R$ be a function defined by $f(x)=\begin{cases} [x], & x \leq 2 \\ 0, & x>2 \end{cases}$,where $[x]$ is the greatest integer less than or equal to $x$. If $I=\int_{-1}^2 \frac{x f(x^2)}{2+f(x+1)} dx$,then the value of $(4I-1)$ is

If $f(x)$ and $g(x)$ are inverse functions of each other such that $f(1) = 3$ and $f(3) = 1$,then $\int_{1}^{3} \left( g(x) + \frac{x}{f'(g(x))} \right) dx$ is equal to -

Let $f(\alpha) = \int_{0}^{\alpha} x^{2} \left(1 - \frac{x}{\alpha}\right)^{\alpha} dx$ (where $\alpha > 0$),then $\sum_{\alpha=1}^{5} \frac{f(\alpha)}{\alpha^{3}}$ is equal to-