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Step-by-step Solution
Step 1: Define an auxiliary function
Let us define the function
h(x) = f(x)\,g'(x).
Our goal is to analyze the zeros of
h'(x) = f'(x)\,g'(x) + f(x)\,g''(x),
which is the given expression
f(x)\,g''(x) + f'(x)\,g'(x) = 0.
Step 2: Identify the roots of f(x) and g'(x)
Since f(x) is an even function and we know
f\left(\tfrac{1}{4}\right) = 0 and f\left(\tfrac{1}{2}\right) = 0,
by the property of even functions, it follows:
f\left(-\tfrac{1}{4}\right) = 0 \quad \text{and} \quad f\left(-\tfrac{1}{2}\right) = 0.
Hence, f(x) has at least four distinct roots in the interval (-2,\,2).
Next, g(x) is also an even function. For an even function g(x), its derivative
g'(x) is an odd function. An odd function g'(x) must satisfy
g'(0) = 0.
Additionally, since g(1) = 2, there must be at least one zero of g'(x) in the interval
(0,\,1) (by the Mean Value Theorem), and being odd implies a reflection root in (-1,\,0).
Thus, g'(x) has at least the following distinct roots: x=0, x=a, and x=-a (for some a in (0,1) ).
Step 3: Count the roots of h(x) = f(x)\,g'(x)
Because f(x) has four distinct roots (at \pm \tfrac{1}{4} and \pm \tfrac{1}{2} ) and
g'(x) has at least three distinct roots (at 0, \pm a ), the product h(x) = f(x)\,g'(x)
will have at least a combined total of five distinct roots in (-2,2).
(Some of these roots might coincide, but the minimal distinct count still reaches five.)
Step 4: Relate the zeros of h(x) to the zeros of h'(x)
By Rolle's Theorem, if a function has n distinct zeros in an interval, then its derivative
has at least n - 1 distinct zeros in that interval. Here, h(x) has at least five distinct zeros,
so h'(x) must have at least 5 - 1 = 4 distinct zeros.
But recall that
h'(x) = f'(x)\,g'(x) + f(x)\,g''(x),
which matches the expression we need to analyze. Thus, there are at least four distinct solutions
to
f(x)\,g''(x) + f'(x)\,g'(x) = 0
in the interval (-2,2).
Answer
The minimum number of solutions to the equation
f(x)\,g''(x) + f'(x)\,g'(x) = 0
in (-2,2) is
4.
