Let $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}$. Then the value of $8\left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right)$ is equal to
(1) 92
(2) 118
(3) 102
(4) 108

The value of $\int _ { e ^ { 2 } } ^ { e ^ { 4 } } \frac { 1 } { x } \left( \frac { e ^ { \left( \left( \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } } { e ^ { \left( \left( \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } + e ^ { \left( \left( 6 - \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } } \right) d x$ is
(1) 2
(2) $\log _ { e } 2$
(3) 1
(4) $e ^ { 2 }$

If $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \frac { 96 x ^ { 2 } \cos ^ { 2 } x } { \left( 1 + e ^ { x } \right) } \mathrm { d } x = \pi \left( \alpha \pi ^ { 2 } + \beta \right) , \alpha , \beta \in \mathbb { Z }$, then $( \alpha + \beta ) ^ { 2 }$ equals
(1) 64
(2) 196
(3) 144
(4) 100

For a continuous function $f$ defined on the set of real numbers and the function $g(x) = 2x + 2$ defined as,
$$\begin{aligned} & \int_{-1}^{1} f(g(x))\, dx = 18 \\ & \int_{2}^{4} g(f(x))\, dx = 18 \end{aligned}$$
are satisfied. Accordingly, what is the value of the integral $\int_{0}^{2} f(x)\, dx$?
A) 20 B) 23 C) 26 D) 29 E) 32