Let $h(x)=f(x) \cdot g^{\prime}(x)$ As $f(x)$ is even $f\left(\frac{1}{2}\right)=\left(\frac{1}{4}\right)=0$ $\Rightarrow f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{4}\right)=0$ and $g(x)$ is even $\Rightarrow g^{\prime}(x)$ is odd and $g(1)=2$ ensures one root of $g^{\prime}(x)$ is 0 . So, $h(x)=f(x) \cdot g^{\prime}(x)$ has minimum five zeroes $\therefore h^{\prime}(x)=f^{\prime}(x) \cdot g^{\prime}(x)+f(x) \cdot g^{\prime \prime}(x)=0$, has minimum 4 zeroes