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

Answer: 4

Let $h(x)=f(x)g^{\prime }(x)\to 5$ 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