If $f$ is defined on $R$ such that $f(x) f(-x) = 9$, then find the value of $\int_{-23}^{23} \frac{dx}{3+f(x)}$.

Read the following mathematical statements carefully:
$I.$ $A$ differentiable function $f$ with maximum at $x = c$ $\implies f''(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$,$\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx}$.
$IV.$ If $f(x)$ has a maxima at $x = c$,then $f$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \to 0$ for $h > 0$. Now indicate the correct alternative.

If $I_{m, n} = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$ for $m, n \geq 1$ and $\int_{0}^{1} \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} dx = \alpha I_{m, n}$,where $\alpha \in R$,then $\alpha$ equals .... .

$\int_0^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x=$

Show that $\int_{0}^{a} f(x) g(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$,if $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$.

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to: