Question

{"given":"The function $f(x)$ is continuous, twice differentiable, and even on $\\mathbb{R}$. We are given $f(4) = -4$, $f(-2) = 4$, $f(0) = -4$, and $\\lim\\limits_{x \\to \\infty} f(x) = 8 = \\lim\\limits_{x \\to -\\infty} f(x)$.","key_observation":"Since $f(x)$ is an even function, $f(-x) = f(x)$. We can find the values of $f(x)$ at negative points and use the Intermediate Value Theorem to find the minimum number of roots of $f(x)$. Then, by constructing a function $\\phi(x) = f(x)e^x$ and applying Rolle's Theorem repeatedly, we can find the minimum number of roots for the given differential equation.","formula_steps":[""],"option_analysis":[{"label":"(A)","text":"$2$","verdict":"incorrect","explanation":"Applying Rolle's Theorem twice on the 6 roots of $f(x)$ yields at least 4 roots, not 2."},{"label":"(B)","text":"$3$","verdict":"incorrect","explanation":"The minimum number of roots is 4, so 3 is incorrect."},{"label":"(C)","text":"$4$","verdict":"correct","explanation":"\nSince $f(x)$ is an even function, we have:\n$f(-4) = f(4) = -4$\n$f(2) = f(-2) = 4$\nLet's observe the signs of $f(x)$ at various points:\nAs $x \\to -\\infty$, $f(x) \\to 8 > 0$\nAt $x = -4$, $f(-4) = -4 < 0$\nAt $x = -2$, $f(-2) = 4 > 0$\nAt $x = 0$, $f(0) = -4 < 0$\nAt $x = 2$, $f(2) = 4 > 0$\nAt $x = 4$, $f(4) = -4 < 0$\nAs $x \\to \\infty$, $f(x) \\to 8 > 0$\n\nBy the Intermediate Value Theorem, $f(x)$ must have at least one root in each of the following 6 intervals: $(-\\infty, -4)$, $(-4, -2)$, $(-2, 0)$, $(0, 2)$, $(2, 4)$, and $(4, \\infty)$.\nThus, $f(x)$ has at least 6 roots.\n\nLet $\\phi(x) = f(x)e^x$.\nThe roots of $\\phi(x)$ are the same as the roots of $f(x)$, so $\\phi(x)$ has at least 6 roots.\nBy Rolle's Theorem, between any two roots of $\\phi(x)$, there is at least one root of $\\phi'(x)$.\nSo, $\\phi'(x)$ has at least $6 - 1 = 5 roots$.\n$\\phi'(x) = f'(x)e^x + f(x)e^x = (f'(x) + f(x))e^x$\n\nApplying Rolle's Theorem again to $\\phi'(x)$, we find that $\\phi''(x)$ has at least $5 - 1 = 4$ roots.\n$\\phi''(x) = \\frac{d}{dx}[(f'(x) + f(x))e^x] = (f''(x) + f'(x))e^x + (f'(x) + f(x))e^x = (f''(x) + 2f'(x) + f(x))e^x$\n\nSince $e^x \\neq 0$ for all $x \\in \\mathbb{R}$, the equation $f''(x) + 2f'(x) + f(x) = 0$ has at least 4 roots."},{"label":"(D)","text":"$5$","verdict":"incorrect","explanation":"We can only guarantee 4 roots for the second derivative expression using Rolle's Theorem, so 5 is incorrect."}],"answer":"(C)"}