Let f : ℝ → ℝ be a function defined by f x = 4 x 4 x + 2 and M = ∫ f a f 1 − a x sin 4 x 1 − x d x , N = ∫ f…

Given,

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

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

$\Rightarrow f\left(1-x\right)=\frac{4}{4+2·4^x}=\frac{2}{2+4^x}$

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

Now, solving $M=\underset{f\left(a\right)}{\overset{f\left(1-a\right)}{\int }}x·{\sin }^4\left(x\left(1-x\right)\right)dx$

Now, using the formula ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get,

$\Rightarrow M=\underset{f\left(a\right)}{\overset{f\left(1-a\right)}{\int }}\left(f\left(a\right)+f\left(1-a\right)-x\right)·{\sin }^4\left(f\left(a\right)+f\left(1-a\right)-x\right)\left(1-\left(f\left(a\right)+f\left(1-a\right)-x\right)\right)dx$

$\Rightarrow M=\underset{f\left(a\right)}{\overset{f\left(1-a\right)}{\int }}\left(1-x\right)·{\sin }^4\left(\left(1-x\right)x\right)dx$

$\Rightarrow M=\underset{f\left(a\right)}{\overset{f\left(1-a\right)}{\int }}{\sin }^4\left(\left(1-x\right)x\right)dx-\underset{f\left(a\right)}{\overset{f\left(1-a\right)}{\int }}x·{\sin }^4\left(\left(1-x\right)x\right)dx$

$\Rightarrow M=N-M$

$\Rightarrow 2M=N$

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

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