| {"year": "2010", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Suppose that $p(x)$ is a polynomial and that $p(x)-p^{\\prime}(x)=x^{2}+2 x+1$. Compute $p(5)$.", "solution": "50 Observe that $p(x)$ must be quadratic. Let $p(x)=a x^{2}+b x+c$. Comparing coefficients gives $a=1, b-2 a=2$, and $c-b=1$. So $b=4, c=5, p(x)=x^{2}+4 x+5$ and $p(5)=25+20+5=50$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Let $f$ be a function such that $f(0)=1, f^{\\prime}(0)=2$, and\n\n$$\nf^{\\prime \\prime}(t)=4 f^{\\prime}(t)-3 f(t)+1\n$$\n\nfor all $t$. Compute the 4th derivative of $f$, evaluated at 0 .", "solution": "54 Putting $t=0$ gives $f^{\\prime \\prime}(0)=6$. By differentiating both sides, we get $f^{(3)}(t)=4 f^{\\prime \\prime}(t)-$ $3 f^{\\prime}(t)$ and $f^{(3)}(0)=4 \\cdot 6-3 \\cdot 2=18$. Similarly, $f^{(4)}(t)=4 f^{(3)}(t)-3 f^{\\prime \\prime}(t)$ and $f^{(4)}(0)=4 \\cdot 18-3 \\cdot 6=54$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let $p$ be a monic cubic polynomial such that $p(0)=1$ and such that all the zeros of $p^{\\prime}(x)$ are also zeros of $p(x)$. Find $p$. Note: monic means that the leading coefficient is 1 .", "solution": "$(x+1)^{3}$ A root of a polynomial $p$ will be a double root if and only if it is also a root of $p^{\\prime}$. Let $a$ and $b$ be the roots of $p^{\\prime}$. Since $a$ and $b$ are also roots of $p$, they are double roots of $p$. But $p$ can have only three roots, so $a=b$ and $a$ becomes a double root of $p^{\\prime}$. This makes $p^{\\prime}(x)=3 c(x-a)^{2}$ for some constant $3 c$, and thus $p(x)=c(x-a)^{3}+d$. Because $a$ is a root of $p$ and $p$ is monic, $d=0$ and $c=1$. From $p(0)=1$ we get $p(x)=(x+1)^{3}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Compute $\\lim _{n \\rightarrow \\infty} \\frac{\\sum_{k=1}^{n}|\\cos (k)|}{n}$.", "solution": "$\\frac{2}{\\pi}$ The main idea lies on the fact that positive integers are uniformly distributed modulo $\\pi$. (In the other words, if each integer $n$ is written as $q \\pi+r$ where $q$ is an integer and $0 \\leq r<\\pi$, the value of $r$ will distribute uniformly in the interval $[0, \\pi]$.) Using this fact, the summation is equivalent to the average value (using the Riemann summation) of the function $|\\cos (k)|$ over the interval $[0, \\pi]$. Therefore, the answer is $\\frac{1}{\\pi} \\int_{0}^{\\pi}|\\cos (k)|=\\frac{2}{\\pi}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let the functions $f(\\alpha, x)$ and $g(\\alpha)$ be defined as\n\n$$\nf(\\alpha, x)=\\frac{\\left(\\frac{x}{2}\\right)^{\\alpha}}{x-1} \\quad g(\\alpha)=\\left.\\frac{d^{4} f}{d x^{4}}\\right|_{x=2}\n$$\n\nThen $g(\\alpha)$ is a polynomial in $\\alpha$. Find the leading coefficient of $g(\\alpha)$.", "solution": "$\\frac{1}{16}$ Write the first equation as $(x-1) f=\\left(\\frac{x}{2}\\right)^{\\alpha}$. For now, treat $\\alpha$ as a constant. From this equation, repeatedly applying derivative with respect to $x$ gives\n\n$$\n\\begin{aligned}\n(x-1) f^{\\prime}+f & =\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{x}{2}\\right)^{\\alpha-1} \\\\\n(x-1) f^{\\prime \\prime}+2 f^{\\prime} & =\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{\\alpha-1}{2}\\right)\\left(\\frac{x}{2}\\right)^{\\alpha-2} \\\\\n(x-1) f^{(3)}+3 f^{\\prime \\prime} & =\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{\\alpha-1}{2}\\right)\\left(\\frac{\\alpha-2}{2}\\right)\\left(\\frac{x}{2}\\right)^{\\alpha-3} \\\\\n(x-1) f^{(4)}+4 f^{(3)} & =\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{\\alpha-1}{2}\\right)\\left(\\frac{\\alpha-2}{2}\\right)\\left(\\frac{\\alpha-3}{2}\\right)\\left(\\frac{x}{2}\\right)^{\\alpha-4}\n\\end{aligned}\n$$\n\nSubstituting $x=2$ to all equations gives $g(\\alpha)=f^{(4)}(\\alpha, 2)=\\left(\\frac{\\alpha}{2}\\right)\\left(\\frac{\\alpha-1}{2}\\right)\\left(\\frac{\\alpha-2}{2}\\right)\\left(\\frac{\\alpha-3}{2}\\right)-4 f^{(3)}(\\alpha, 2)$. Because $f^{(3)}(\\alpha, 2)$ is a cubic polynomial in $\\alpha$, the leading coefficient of $g(\\alpha)$ is $\\frac{1}{16}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n5. [4]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Let $f(x)=x^{3}-x^{2}$. For a given value of $c$, the graph of $f(x)$, together with the graph of the line $c+x$, split the plane up into regions. Suppose that $c$ is such that exactly two of these regions have finite area. Find the value of $c$ that minimizes the sum of the areas of these two regions.", "solution": "$-\\frac{11}{27}$ Observe that $f(x)$ can be written as $\\left(x-\\frac{1}{3}\\right)^{3}-\\frac{1}{3}\\left(x-\\frac{1}{3}\\right)-\\frac{2}{27}$, which has $180^{\\circ}$ symmetry around the point $\\left(\\frac{1}{3},-\\frac{2}{27}\\right)$. Suppose the graph of $f$ cuts the line $y=c+x$ into two segments of lengths $a$ and $b$. When we move the line toward point $P$ with a small distance $\\Delta x$ (measured along the line perpendicular to $y=x+c)$, the sum of the enclosed areas will increase by $|a-b|(\\Delta x)$. As long as the line $x+c$ does not passes through $P$, we can find a new line $x+c^{*}$ that increases the sum of the enclosed areas. Therefore, the sum of the areas reaches its maximum when the line passes through $P$. For that line, we can find that $c=y-x=-\\frac{2}{27}-\\frac{1}{3}=-\\frac{11}{27}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Let $a_{1}, a_{2}$, and $a_{3}$ be nonzero complex numbers with non-negative real and imaginary parts. Find the minimum possible value of\n\n$$\n\\frac{\\left|a_{1}+a_{2}+a_{3}\\right|}{\\sqrt[3]{\\left|a_{1} a_{2} a_{3}\\right|}}\n$$", "solution": "$\\sqrt{3} \\sqrt[3]{2}$ Write $a_{1}$ in its polar form $r e^{i \\theta}$ where $0 \\leq \\theta \\leq \\frac{\\pi}{2}$. Suppose $a_{2}, a_{3}$ and $r$ are fixed so that the denominator is constant. Write $a_{2}+a_{3}$ as $s e^{i \\phi}$. Since $a_{2}$ and $a_{3}$ have non-negative real and imaginary parts, the angle $\\phi$ lies between 0 and $\\frac{\\pi}{2}$. Consider the function\n\n$$\nf(\\theta)=\\left|a_{1}+a_{2}+a_{3}\\right|^{2}=\\left|r e^{i \\theta}+s e^{i \\phi}\\right|^{2}=r^{2}+2 r s \\cos (\\theta-\\phi)+s^{2}\n$$\n\nIts second derivative is $\\left.f^{\\prime \\prime}(\\theta)=-2 \\operatorname{rs}(\\cos (\\theta-\\phi))\\right)$. Since $-\\frac{\\pi}{2} \\leq(\\theta-\\phi) \\leq \\frac{\\pi}{2}$, we know that $f^{\\prime \\prime}(\\theta)<0$ and $f$ is concave. Therefore, to minimize $f$, the angle $\\theta$ must be either 0 or $\\frac{\\pi}{2}$. Similarly, each of $a_{1}, a_{2}$ and $a_{3}$ must be either purely real or purely imaginary to minimize $f$ and the original fraction.\nBy the AM-GM inequality, if $a_{1}, a_{2}$ and $a_{3}$ are all real or all imaginary, then the minimum value of the fraction is 3 . Now suppose only two of the $a_{i}$ 's, say, $a_{1}$ and $a_{2}$ are real. Since the fraction is homogenous, we may fix $a_{1}+a_{2}$ - let the sum be 2 . The term $a_{1} a_{2}$ in the denominator acheives its maximum only when $a_{1}$ and $a_{2}$ are equal, i.e. when $a_{1}=a_{2}=1$. Then, if $a_{3}=k i$ for some real number $k$, then the expression equals\n\n$$\n\\frac{\\sqrt{k^{2}+4}}{\\sqrt[3]{k}}\n$$\n\nSquaring and taking the derivative, we find that the minimum value of the fraction is $\\sqrt{3} \\sqrt[3]{2}$, attained when $k=\\sqrt{2}$. With similar reasoning, the case where only one of the $a_{i}$ 's is real yields the same minimum value.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n7. [6]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let $f(n)=\\sum_{k=2}^{\\infty} \\frac{1}{k^{n} \\cdot k!}$. Calculate $\\sum_{n=2}^{\\infty} f(n)$.", "solution": "$3-e$\n\n$$\n\\begin{aligned}\n\\sum_{n=2}^{\\infty} f(n) & =\\sum_{k=2}^{\\infty} \\sum_{n=2}^{\\infty} \\frac{1}{k^{n} \\cdot k!} \\\\\n& =\\sum_{k=2}^{\\infty} \\frac{1}{k!} \\sum_{n=2}^{\\infty} \\frac{1}{k^{n}} \\\\\n& =\\sum_{k=2}^{\\infty} \\frac{1}{k!} \\cdot \\frac{1}{k(k-1)} \\\\\n& =\\sum_{k=2}^{\\infty} \\frac{1}{(k-1)!} \\cdot \\frac{1}{k^{2}(k-1)}\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n& =\\sum_{k=2}^{\\infty} \\frac{1}{(k-1)!}\\left(\\frac{1}{k-1}-\\frac{1}{k^{2}}-\\frac{1}{k}\\right) \\\\\n& =\\sum_{k=2}^{\\infty}\\left(\\frac{1}{(k-1)(k-1)!}-\\frac{1}{k \\cdot k!}-\\frac{1}{k!}\\right) \\\\\n& =\\sum_{k=2}^{\\infty}\\left(\\frac{1}{(k-1)(k-1)!}-\\frac{1}{k \\cdot k!}\\right)-\\sum_{k=2}^{\\infty} \\frac{1}{k!} \\\\\n& =\\frac{1}{1 \\cdot 1!}-\\left(e-\\frac{1}{0!}-\\frac{1}{1!}\\right) \\\\\n& =3-e\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Let $x(t)$ be a solution to the differential equation\n\n$$\n\\left(x+x^{\\prime}\\right)^{2}+x \\cdot x^{\\prime \\prime}=\\cos t\n$$\n\nwith $x(0)=x^{\\prime}(0)=\\sqrt{\\frac{2}{5}}$. Compute $x\\left(\\frac{\\pi}{4}\\right)$.", "solution": "$\\frac{\\sqrt[4]{450}}{5}$ Rewrite the equation as $x^{2}+2 x x^{\\prime}+\\left(x x^{\\prime}\\right)^{\\prime}=\\cos t$. Let $y=x^{2}$, so $y^{\\prime}=2 x x^{\\prime}$ and the equation becomes $y+y^{\\prime}+\\frac{1}{2} y^{\\prime \\prime}=\\cos t$. The term $\\cos t$ suggests that the particular solution should be in the form $A \\sin t+B \\cos t$. By substitution and coefficient comparison, we get $A=\\frac{4}{5}$ and $B=\\frac{2}{5}$. Since the function $y(t)=\\frac{4}{5} \\sin t+\\frac{2}{5} \\cos t$ already satisfies the initial conditions $y(0)=x(0)^{2}=\\frac{2}{5}$ and $y^{\\prime}(0)=2 x(0) x^{\\prime}(0)=\\frac{4}{5}$, the function $y$ also solves the initial value problem. Note that since $x$ is positive at $t=0$ and $y=x^{2}$ never reaches zero before $t$ reaches $\\frac{\\pi}{4}$, the value of $x\\left(\\frac{\\pi}{4}\\right)$ must be positive. Therefore, $x\\left(\\frac{\\pi}{4}\\right)=+\\sqrt{y\\left(\\frac{\\pi}{4}\\right)}=\\sqrt{\\frac{6}{5} \\cdot \\frac{\\sqrt{2}}{2}}=\\frac{\\sqrt[4]{450}}{5}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}} |
| {"year": "2010", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Let $f(n)=\\sum_{k=1}^{n} \\frac{1}{k}$. Then there exists constants $\\gamma, c$, and $d$ such that\n\n$$\nf(n)=\\ln (n)+\\gamma+\\frac{c}{n}+\\frac{d}{n^{2}}+O\\left(\\frac{1}{n^{3}}\\right)\n$$\n\nwhere the $O\\left(\\frac{1}{n^{3}}\\right)$ means terms of order $\\frac{1}{n^{3}}$ or lower. Compute the ordered pair $(c, d)$.", "solution": "$\\left(\\frac{1}{2},-\\frac{1}{12}\\right)$ From the given formula, we pull out the term $\\frac{k}{n^{3}}$ from $O\\left(\\frac{1}{n^{4}}\\right)$, making $f(n)=$ $\\log (n)+\\gamma+\\frac{c}{n}+\\frac{d}{n^{2}}+\\frac{k}{n^{3}}+O\\left(\\frac{1}{n^{4}}\\right)$. Therefore,\n$f(n+1)-f(n)=\\log \\left(\\frac{n+1}{n}\\right)-c\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right)-d\\left(\\frac{1}{n^{2}}-\\frac{1}{(n+1)^{2}}\\right)-k\\left(\\frac{1}{n^{3}}-\\frac{1}{(n+1)^{3}}\\right)+O\\left(\\frac{1}{n^{4}}\\right)$.\nFor the left hand side, $f(n+1)-f(n)=\\frac{1}{n+1}$. By substituting $x=\\frac{1}{n}$, the formula above becomes\n\n$$\n\\frac{x}{x+1}=\\log (1+x)-c x^{2} \\cdot \\frac{1}{x+1}-d x^{3} \\cdot \\frac{x+2}{(x+1)^{2}}-k x^{4} \\cdot \\frac{x^{2}+3 x+3}{(x+1)^{3}}+O\\left(x^{4}\\right)\n$$\n\nBecause $x$ is on the order of $\\frac{1}{n}, \\frac{1}{(x+1)^{3}}$ is on the order of a constant. Therefore, all the terms in the expansion of $k x^{4} \\cdot \\frac{x^{2}+3 x+3}{(x+1)^{3}}$ are of order $x^{4}$ or higher, so we can collapse it into $O\\left(x^{4}\\right)$. Using the Taylor expansions, we get\n\n$$\nx\\left(1-x+x^{2}\\right)+O\\left(x^{4}\\right)=\\left(x-\\frac{1}{2} x^{2}+\\frac{1}{3} x^{3}\\right)-c x^{2}(1-x)-d x^{3}(2)+O\\left(x^{4}\\right) .\n$$\n\nCoefficient comparison gives $c=\\frac{1}{2}$ and $d=-\\frac{1}{12}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-132-2010-feb-calc-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}} |
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