JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Application of Derivatives

The given equation can be rewritten as $\frac{d}{dx}[f(x)g^{\prime }(x)]=0$, so $h(x)=f(x)g^{\prime }(x)$ is a function whose derivative has roots.

Since $f$ is even with $f(\frac{1}{4})=f(\frac{1}{2})=0$, by symmetry $f(-\frac{1}{4})=f(-\frac{1}{2})=0$ as well. So $f(x)$ has at least 4 zeros in $(-2,2)$.

Since $g$ is even with $g(\frac{3}{4})=0$, we have $g(-\frac{3}{4})=0$. By Rolle's theorem, $g^{\prime }(x)$ has at least one zero in $(-\frac{3}{4},\frac{3}{4})$, and since $g$ is even, $g^{\prime }(0)=0$.

The function $h(x)=f(x)g^{\prime }(x)$ has at least 5 zeros (from the zeros of $f$ and $g^{\prime }$). By Rolle's theorem, $h^{\prime }(x)=f(x)g^{\prime \prime }(x)+f^{\prime }(x)g^{\prime }(x)$ has at least 4 zeros between consecutive zeros of $h(x)$.

Therefore, the minimum number of solutions is 4.