Let f: ℝ → ℝ be a function defined by f(x) = 4 ^ x 4 ^ x + 2 and M...

Given,

$f\left(x\right) = \dfrac{{4}^{x}}{{4}^{x} + 2}$

$⇒ f\left(1 - x\right) = \dfrac{{4}^{1 - x}}{{4}^{1 - x} + 2}$

$⇒ f\left(1 - x\right) = \dfrac{4}{4 + 2 \cdot {4}^{x}} = \dfrac{2}{2 + {4}^{x}}$

$⇒ f\left(x\right) + f\left(1 - x\right) = 1$

Now, solving $M = \mathop{∫}_{f\left(a\right)}^{f\left(1 - a\right)}x \cdot \sin^{4}\left(x\left(1 - x\right)\right)dx$

Now, using the formula ${∫}_{a}^{b}f\left(x\right) d x = {∫}_{a}^{b}f\left(a + b - x\right) d x$ we get,

$⇒ M = \mathop{∫}_{f\left(a\right)}^{f\left(1 - a\right)}\left(f\left(a\right) + f\left(1 - a\right) - x\right) \cdot \sin^{4}\left(f\left(a\right) + f\left(1 - a\right) - x\right)\left(1 - \left(f\left(a\right) + f\left(1 - a\right) - x\right)\right)dx$

$⇒ M = \mathop{∫}_{f\left(a\right)}^{f\left(1 - a\right)}\left(1 - x\right) \cdot \sin^{4}\left(\left(1 - x\right)x\right)dx$

$⇒ M = \mathop{∫}_{f\left(a\right)}^{f\left(1 - a\right)}\sin^{4}\left(\left(1 - x\right)x\right)dx - \mathop{∫}_{f\left(a\right)}^{f\left(1 - a\right)}x \cdot \sin^{4}\left(\left(1 - x\right)x\right)dx$

$⇒ M = N - M$

$⇒ 2M = N$

So, on comparing we get, $\alpha = 2, \beta = 1$

Hence, ${\alpha}^{2} + {\beta}^{2} = {2}^{2} + {1}^{2} = 5$