Question:

Step 1: Establish a relationship for $g(-x)$ using the first given condition. We are given that $f(-x) = g(x) - 1$. By replacing $x$ with $-x$ in this equation: $$f(x) = g(-x) - 1$$ Therefore: $$g(-x) = f(x) + 1$$ Step 2: Express $\frac{f(x)}{g(x)}$ in terms of $f(x)$. We are given that $f(x) = \frac{g(x)}{g(-x)}$. Substituting $g(-x) = f(x) + 1$: $$f(x) = \frac{g(x)}{f(x) + 1}$$ Multiplying both sides by $(f(x) + 1)$: $$f(x)(f(x) + 1) = g(x)$$ Therefore: $$\frac{f(x)}{g(x)} = \frac{f(x)}{f(x)(f(x) + 1)} = \frac{1}{f(x) + 1}$$ Step 3: Find a key relationship by adding $\frac{f(x)}{g(x)}$ and $\frac{f(-x)}{g(-x)}$. We need to compute $\frac{f(-x)}{g(-x)}$. Using $f(-x) = g(x) - 1$ and $g(-x) = f(x) + 1$: $$\frac{f(-x)}{g(-x)} = \frac{g(x) - 1}{f(x) + 1}$$ Adding the two fractions: $$\frac{f(x)}{g(x)} + \frac{f(-x)}{g(-x)} = \frac{1}{f(x) + 1} + \frac{g(x) - 1}{f(x) + 1}$$ Combining over a common denominator: $$= \frac{1 + g(x) - 1}{f(x) + 1} = \frac{g(x)}{f(x) + 1}$$ Since $g(x) = f(x)(f(x) + 1)$: $$= \frac{f(x)(f(x) + 1)}{f(x) + 1} = f(x)$$ Step 4: Integrate both sides over the symmetric interval $[-20, 20]$. Integrating the relationship from Step 3: $$\int_{-20}^{20} \frac{f(x)}{g(x)}\,dx + \int_{-20}^{20} \frac{f(-x)}{g(-x)}\,dx = \int_{-20}^{20} f(x)\,dx = 2020$$ Step 5: Use substitution to show the two integrals on the left are equal. Let $I = \int_{-20}^{20} \frac{f(x)}{g(x)}\,dx$. In the second integral, substitute $u = -x$ (so $du = -dx$): $$\int_{-20}^{20} \frac{f(-x)}{g(-x)}\,dx = \int_{20}^{-20} \frac{f(u)}{g(u)}\,(-du) = \int_{-20}^{20} \frac{f(u)}{g(u)}\,du = I$$ Step 6: Solve for the desired integral. From Step 4: $$I + I = 2020$$ $$2I = 2020$$ $$I = 1010$$ Therefore, the value of $\int_{-20}^{20} \frac{f(x)}{g(x)}\,dx$ is $\boxed{1010}$. The answer is **Option 1: 1010**.