Let $$f$$ and $$g$$ be twice differentiable even functions on $$(-2, 2)$$ such that $$f\left(\frac{1}{4}\right) = 0$$, $$f\left(\frac{1}{2}\right) = 0$$, $$f(1) = 1$$ and $$g\left(\frac{3}{4}\right) = 0$$, $$g(1) = 2$$. Then, the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$(-2, 2)$$ is equal to ______.

We are given that $$f$$ and $$g$$ are twice differentiable even functions on $$(-2, 2)$$ satisfying $$f(1/4) = 0$$, $$f(1/2) = 0$$, $$f(1) = 1$$, $$g(3/4) = 0$$, and $$g(1) = 2$$, and we want to determine the minimum number of solutions of $$f(x)g''(x) + f'(x)g'(x) = 0$$ in $$(-2, 2)$$. Noting that $$f(x)g''(x) + f'(x)g'(x) = \frac{d}{dx}[f(x)g'(x)]$$, we set $$h(x) = f(x)g'(x)$$ and focus on the zeros of $$h'(x)$$.

Since $$f$$ is even, its zeros at $$x=1/4$$ and $$x=1/2$$ also occur at $$x=-1/4$$ and $$x=-1/2$$. The evenness of $$g$$ makes $$g'$$ odd and hence $$g'(0)=0$$. In addition, $$g(3/4) = 0$$ and $$g(1) = 2$$ together with Rolle’s theorem on the interval $$[-3/4,3/4]$$ confirm at least one zero of $$g'$$ in $$(-3/4,3/4)$$, which we already have at $$x=0$$.

Therefore, the zeros of $$h(x)=f(x)g'(x)$$ occur at least at $$x=-1/2,-1/4,0,1/4,1/2$$, giving five zeros. By Rolle’s theorem, between each pair of consecutive zeros of $$h$$ there is at least one zero of $$h'(x)$$, namely in each of the intervals $$(-1/2,-1/4)$$, $$(-1/4,0)$$, $$(0,1/4)$$, and $$(1/4,1/2)$$. Hence there are at least four zeros of $$h'(x)$$ in $$(-2,2)$$, and so the minimum number of solutions of the given equation is 4.