Is the xor of a prg and a function still a prg?

I have this couple of deterministic functions $G_1$ and $G_2$. Suppose at least I of them is a PRG. Take $G^*=G_1(x)$ xor $G_2(x)$ with the same $x$. I have to show whether this is still a PRG. I thought about this counterexample: if $G_1=G_2$ then $G^*$ cannot be a PRG but my teacher said then $G^*$ isn't interesting.

What should I do?

Take 2 bits $x1$ & $x2$ generateD by $G_1$ & $G_2$ respectively. Let $G_2$ be a good PRG i.e $x2$ generated by $G_2$ is perfectly random. So there is 50% probability that $x2$ is $0$ or $1$. Let $G_1$ be a terrible PRG - i.e. it always generates the same value.

Truth table of $x1$ XOR $x2$

Even with a fixed $x1$ (i.e. $x1$ is always 1), the output after XOR with a perfectly Random $x2$ has 50% probability of being 0 or 1. i.e. $G_1$ $XOR$ $G_2$ is a PRG.

More formal Proof

Let's have a biased $a$ and a perfectly unbiased $b$:

$P(a = 0) = x$

$P(a = 1) = 1 - x$

$P(b = 0) = 0.5$

$P(b = 1) = 0.5$

$P(a\text{ XOR } b = 0) = $

$= P(a = 0) * P(b = 0) + P(a = 1) * P(b = 1) =$

$= x * 0.5 + (1 - x) * 0.5$

$= 0.5x + 0.5 - 0.5x$

$= 0.5$

$P(a \text { XOR } b = 1)$

$= P(a = 1) * P(b = 0) + P(a = 0) * P(b = 1) $

$= (1 - x) * 0.5 + x * 0.5$

$= 0.5 - 0.5x + 0.5x $

$ = 0.5$