Question

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 $f(x) g^{\prime \prime}(x)+f^{\prime}(x) g^{\prime}(x)=0$ in $(-2,2)$ is equal to______.

Answer 1

Let $h(x)=f(x) g^{\prime}(x) \rightarrow 5$ roots
$f ( x )$ is even $\Rightarrow$
$f \left(\frac{1}{4}\right)= f \left(\frac{1}{2}\right)= f \left(-\frac{1}{2}\right)= f \left(\frac{1}{4}\right)=0$
$g ( x )$ is even $\Rightarrow g \left(\frac{3}{4}\right)= g \left(-\frac{3}{4}\right)=0$
$g ^{\prime}( x )=0$ has minimum one root
$h^{\prime}( x )$ has at last $4$ roots