Mathematics - Functions Question with Solution | TestHub

$fog(x)=f(g(x))=f(4x(1-x))$

$$\Rightarrow \frac{1-4x(1-x)}{1+4x(1-x)}\text{ when }0\le 4x(1-x)\le 1\text{ and }0\le x\le 1$$

But $4x-4x^2\ge 0\Rightarrow 0\le x\le 1$

$$4x-4x^2\le 1\Rightarrow (2x-1{)}^2\ge 0\Rightarrow x\in R$$

Hence $fog(x)=\frac{1-4x+4x^2}{1+4x-4x^2},0\le x\le 1$

Let $y=\frac{4x^2-4x+1}{-\left(4x^2-4x\right)+1},0\le x\le 1$

put $0\le x\le 1t\in [-1,0]$

$$y=\frac{1+t}{1-t},\frac{dy}{dt}=\frac{1-t+1+t}{(1-t{)}^2}>0$$

Range of $fog(x)=[0,1]$

$$\Rightarrow \alpha +\beta =1$$