SAT1000 - P159

We are given that $x - f(g(x)) = 0$ so we can conclude that $f(g(x)) = x$ . Using this, let's evaluate the options.

Option A - $g(f(x)) = x^{2} + x + \frac{1}{5}$ . Let x = g(u). Then $g(f(g(u))) = g(u)^{2} + g(u) + \frac{1}{5}$ . But $f(g(u)) = u$ so $g(u) = g(u)^{2} + g(u) + \frac{1}{5}$ or $g(u)^{2} + \frac{1}{5} = 0$ . There does not exist any real solution $g(u)$ to this equation so $g(f(x)) = x^{2} + x + \frac{1}{5}$ cannot satisfy the constraints provided.

We have already solved the problem, however, I will show that the other options satisfy the constraints.

Option B - $g(f(x)) = x^{2} + \frac{1}{5}$ . Let x = g(u). Then $g(f(g(u))) = g(u)^{2} + \frac{1}{5}$ . But $f(g(u)) = u$ so $g(u) = g(u)^{2} + \frac{1}{5}$ or $g(u)^{2} - g(u) + \frac{1}{5} = 0$ . There a real solution $g(u)$ to this equation so $g(f(x)) = x^{2} + \frac{1}{5}$ satisfies the constraints provided.

Option C - $g(f(x)) = x^{2} - \frac{1}{5}$ . Let x = g(u). Then $g(f(g(u))) = g(u)^{2} - \frac{1}{5}$ . But $f(g(u)) = u$ so $g(u) = g(u)^{2} - \frac{1}{5}$ or $g(u)^{2} - g(u) - \frac{1}{5} = 0$ . There a real solution $g(u)$ to this equation so $g(f(x)) = x^{2} - \frac{1}{5}$ satisfies the constraints provided.

Option D - $g(f(x)) = x^{2} + x - \frac{1}{5}$ . Let x = g(u). Then $g(f(g(u))) = g(u)^{2} - \frac{1}{5}$ . But $f(g(u)) = u$ so $g(u) = g(u)^{2} + g(u) - \frac{1}{5}$ or $g(u)^{2} - \frac{1}{5} = 0$ . There a real solution $g(u)$ to this equation so $g(f(x)) = x^{2} + x - \frac{1}{5}$ satisfies the constraints provided.

Nice problem!

Take $f(x)=x+a,g(x)=x^2+bx+c$ as an example. The result follows.