Let f and g be twice differentiable even func | vedclass

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(C) Define $h(x) = f(x)g'(x)$. The given equation is $h'(x) = 0$.Since $f(x)$ is an even function,$f(x) = f(-x)$. Given $f(\frac{1}{4}) = 0$ and $f(\f

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$4$

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(C) Define $h(x) = f(x)g'(x)$. The given equation is $h'(x) = 0$.Since $f(x)$ is an even function,$f(x) = f(-x)$. Given $f(\frac{1}{4}) = 0$ and $f(\frac{1}{2}) = 0$,we also have $f(-\frac{1}{4}) = 0$ and $f(-\frac{1}{2}) = 0$. Thus,$f(x)$ has at least $4$ roots in $(-1, 1)$,specifically at $\pm \frac{1}{4}$ and $\pm \frac{1}{2}$.Since $g(x)$ is an even function,$g'(x)$ is an odd function. Given $g(\frac{3}{4}) = 0$,we have $g(-\frac{3}{4}) = 0$. By Rolle's Theorem,$g'(x)$ must have at least one root in $(-\frac{3}{4}, \frac{3}{4})$,which is $x = 0$ (since $g'(x)$ is odd).Now,$h(x) = f(x)g'(x)$. The roots of $h(x)$ include $\pm \frac{1}{4}, \pm \frac{1}{2}$ (from $f(x)$) and $0$ (from $g'(x)$). Thus,$h(x)$ has at least $5$ roots in $(-1, 1)$.By Rolle's Theorem,if $h(x)$ has $5$ roots,then $h'(x)$ must have at least $4$ roots in $(-1, 1)$. Since the interval is $(-2, 2)$,the minimum number of solutions is $4$.

Let $f$ and $g$ be twice differentiable even functions on $(-2, 2)$ such that $f(\frac{1}{4}) = 0, f(\frac{1}{2}) = 0, f(1) = 1$ and $g(\frac{3}{4}) = 0, g(1) = 2$. Then,the minimum number of solutions of $f(x)g''(x) + f'(x)g'(x) = 0$ in $(-2, 2)$ is equal to

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