Solving for a parameter $t$

To prove: $\displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx = \dfrac{b - a}{2}$

Let $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $u = a + b - x$

$\implies I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} du$

Since $u$ is a dummy variable we can replace $u$ by $x$ to obtain:

$I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

Adding $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

$\implies 2I = \displaystyle\int_{a}^{b} dx = b - a \implies \boxed{I = \dfrac{b - a}{2}}$ .

$(\sin(\theta) + \cos(\lambda))^2 = (4t + 6)^2$

$(\cos(\theta) + \sin(\lambda))^2 = (3t + 2)^2$

Adding the two equations above $\implies$

$2 + 2\sin(\theta + \lambda) = 25t^2 + 60t + 40 = (5t + 6)^2 + 4$

and $0 \leq 2 + 2\sin(\theta + \lambda) \leq 4 \implies 0 \leq (5t + 6)^2 + 4 \leq 4 \implies$

$0 \leq (5t + 6)^2 \leq 0 \implies (5t + 6)^2 = 0 \implies 5t + 6 = 0 \implies t = -\dfrac{6}{5}$

Now using $t = -\dfrac{6}{5} \implies I = \displaystyle\int_{-\frac{6}{5}}^{0} \dfrac{f(x)}{f(-\dfrac{6}{5} - x) + f(x)} dx = \dfrac{3}{5} = \dfrac{\alpha}{\beta}$

$\implies \alpha + \beta = \boxed{8}$