{"year": "2008", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Let $f(x)=1+x+x^{2}+\\cdots+x^{100}$. Find $f^{\\prime}(1)$.", "solution": "5050 Note that $f^{\\prime}(x)=1+2 x+3 x^{2}+\\cdots+100 x^{99}$, so $f^{\\prime}(1)=1+2+\\cdots+100=\\frac{100 \\cdot 101}{2}=$ 5050.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Let $\\ell$ be the line through $(0,0)$ and tangent to the curve $y=x^{3}+x+16$. Find the slope of $\\ell$.", "solution": "13 Let the point of tangency be $\\left(t, t^{3}+t+16\\right)$, then the slope of $\\ell$ is $\\left(t^{3}+t+16\\right) / t$. On the other hand, since $d y / d x=3 x^{2}+1$, the slope of $\\ell$ is $3 t^{2}+1$. Therefore,\n\n$$\n\\frac{t^{3}+t+16}{t}=3 t^{2}+1\n$$\n\nSimplifying, we get $t^{3}=8$, so $t=2$. It follows that the slope is $3(2)^{2}+1=13$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Find all $y>1$ satisfying $\\int_{1}^{y} x \\ln x d x=\\frac{1}{4}$.", "solution": "$\\sqrt{\\sqrt{e}}$ Applying integration by parts with $u=\\ln x$ and $v=\\frac{1}{2} x^{2}$, we get\n\n$$\n\\int_{1}^{y} x \\ln x d x=\\left.\\frac{1}{2} x^{2} \\ln x\\right|_{1} ^{y}-\\frac{1}{2} \\int_{1}^{y} x d x=\\frac{1}{2} y^{2} \\ln y-\\frac{1}{4} y^{2}+\\frac{1}{4}\n$$\n\nSo $y^{2} \\ln y=\\frac{1}{2} y^{2}$. Since $y>1$, we obtain $\\ln y=\\frac{1}{2}$, and thus $y=\\sqrt{e}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Let $a, b$ be constants such that $\\lim _{x \\rightarrow 1} \\frac{(\\ln (2-x))^{2}}{x^{2}+a x+b}=1$. Determine the pair $(a, b)$.", "solution": "$(-2,1)$ When $x=1$, the numerator is 0 , so the denominator must be zero as well, so $1+a+b=0$. Using l'Hôpital's rule, we must have\n\n$$\n1=\\lim _{x \\rightarrow 1} \\frac{(\\ln (2-x))^{2}}{x^{2}+a x+b}=\\lim _{x \\rightarrow 1} \\frac{2 \\ln (2-x)}{(x-2)(2 x+a)}\n$$\n\nand by the same argument we find that $2+a=0$. Thus, $a=-2$ and $b=1$. This is indeed a solution, as can be seen by finishing the computation.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "Let $f(x)=\\sin ^{6}\\left(\\frac{x}{4}\\right)+\\cos ^{6}\\left(\\frac{x}{4}\\right)$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (i.e., $f$ differentiated 2008 times and then evaluated at $x=0$ ).", "solution": "| $\\frac{3}{8}$ |\n| :---: |\n| We have |\n\n$$\n\\begin{aligned}\n\\sin ^{6} x+\\cos ^{6} x & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{3}-3 \\sin ^{2} x \\cos ^{2} x\\left(\\sin ^{2} x+\\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x=1-\\frac{3}{4} \\sin ^{2} 2 x=1-\\frac{3}{4}\\left(\\frac{1-\\cos 4 x}{2}\\right) \\\\\n& =\\frac{5}{8}+\\frac{3}{8} \\cos 4 x\n\\end{aligned}\n$$\n\nIt follows that $f(x)=\\frac{5}{8}+\\frac{3}{8} \\cos x$. Thus $f^{(2008)}(x)=\\frac{3}{8} \\cos x$. Evaluating at $x=0$ gives $\\frac{3}{8}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n5. [4]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "Determine the value of $\\lim _{n \\rightarrow \\infty} \\sum_{k=0}^{n}\\binom{n}{k}^{-1}$.", "solution": "2 Let $S_{n}$ denote the sum in the limit. For $n \\geq 1$, we have $S_{n} \\geq\\binom{ n}{0}^{-1}+\\binom{n}{n}^{-1}=2$. On the other hand, for $n \\geq 3$, we have\n\n$$\nS_{n}=\\binom{n}{0}^{-1}+\\binom{n}{1}^{-1}+\\binom{n}{n-1}^{-1}+\\binom{n}{n}^{-1}+\\sum_{k=2}^{n-2}\\binom{n}{k}^{-1} \\leq 2+\\frac{2}{n}+(n-3)\\binom{n}{2}^{-1}\n$$\n\nwhich goes to 2 as $n \\rightarrow \\infty$. Therefore, $S_{n} \\rightarrow 2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Find $p$ so that $\\lim _{x \\rightarrow \\infty} x^{p}(\\sqrt[3]{x+1}+\\sqrt[3]{x-1}-2 \\sqrt[3]{x})$ is some non-zero real number.", "solution": "$\\sqrt{\\frac{5}{3}}$ Make the substitution $t=\\frac{1}{x}$. Then the limit equals to\n\n$$\n\\lim _{t \\rightarrow 0} t^{-p}\\left(\\sqrt[3]{\\frac{1}{t}+1}+\\sqrt[3]{\\frac{1}{t}-1}-2 \\sqrt[3]{\\frac{1}{t}}\\right)=\\lim _{t \\rightarrow 0} t^{-p-\\frac{1}{3}}(\\sqrt[3]{1+t}+\\sqrt[3]{1-t}-2)\n$$\n\nWe need the degree of the first nonzero term in the MacLaurin expansion of $\\sqrt[3]{1+t}+\\sqrt[3]{1-t}-2$. We have\n\n$$\n\\sqrt[3]{1+t}=1+\\frac{1}{3} t-\\frac{1}{9} t^{2}+o\\left(t^{2}\\right), \\quad \\sqrt[3]{1-t}=1-\\frac{1}{3} t-\\frac{1}{9} t^{2}+o\\left(t^{2}\\right)\n$$\n\nIt follows that $\\sqrt[3]{1+t}+\\sqrt[3]{1-t}-2=-\\frac{2}{9} t^{2}+o\\left(t^{2}\\right)$. By consider the degree of the leading term, it follows that $-p-\\frac{1}{3}=-2$. So $p=\\frac{5}{3}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n7. [5]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Let $T=\\int_{0}^{\\ln 2} \\frac{2 e^{3 x}+e^{2 x}-1}{e^{3 x}+e^{2 x}-e^{x}+1} d x$. Evaluate $e^{T}$.", "solution": "$\\frac{11}{4}$ Divide the top and bottom by $e^{x}$ to obtain that\n\n$$\nT=\\int_{0}^{\\ln 2} \\frac{2 e^{2 x}+e^{x}-e^{-x}}{e^{2 x}+e^{x}-1+e^{-x}} d x\n$$\n\nNotice that $2 e^{2 x}+e^{x}-e^{-x}$ is the derivative of $e^{2 x}+e^{x}-1+e^{-x}$, and so\n\n$$\nT=\\left[\\ln \\left|e^{2 x}+e^{x}-1+e^{-x}\\right|\\right]_{0}^{\\ln 2}=\\ln \\left(4+2-1+\\frac{1}{2}\\right)-\\ln 2=\\ln \\left(\\frac{11}{4}\\right)\n$$\n\nTherefore, $e^{T}=\\frac{11}{4}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n8. $[7]$", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "Evaluate the limit $\\lim _{n \\rightarrow \\infty} n^{-\\frac{1}{2}\\left(1+\\frac{1}{n}\\right)}\\left(1^{1} \\cdot 2^{2} \\cdots \\cdots n^{n}\\right)^{\\frac{1}{n^{2}}}$.", "solution": "$e^{-1 / 4}$ Taking the logarithm of the expression inside the limit, we find that it is\n\n$$\n-\\frac{1}{2}\\left(1+\\frac{1}{n}\\right) \\ln n+\\frac{1}{n^{2}} \\sum_{k=1}^{n} k \\ln k=\\frac{1}{n} \\sum_{k=1}^{n} \\frac{k}{n} \\ln \\left(\\frac{k}{n}\\right)\n$$\n\nWe can recognize this as the as Riemann sum expansion for the integral $\\int_{0}^{1} x \\ln x d x$, and thus the limit of the above sum as $n \\rightarrow \\infty$ equals to the value of this integral. Evaluating this integral using integration by parts, we find that\n\n$$\n\\int_{0}^{1} x \\ln x d x=\\left.\\frac{1}{2} x^{2} \\ln x\\right|_{0} ^{1}-\\int_{0}^{1} \\frac{x}{2} d x=-\\frac{1}{4}\n$$\n\nTherefore, the original limit is $e^{-1 / 4}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n9. [7]", "solution_match": "\nAnswer: "}}
{"year": "2008", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Evaluate the integral $\\int_{0}^{1} \\ln x \\ln (1-x) d x$.", "solution": "$2-\\frac{\\pi^{2}}{6}$ We have the MacLaurin expansion $\\ln (1-x)=-x-\\frac{x^{2}}{2}-\\frac{x^{3}}{3}-\\cdots$. So\n\n$$\n\\int_{0}^{1} \\ln x \\ln (1-x) d x=-\\int_{0}^{1} \\ln x \\sum_{n=1}^{\\infty} \\frac{x^{n}}{n} d x=-\\sum_{n=1}^{\\infty} \\frac{1}{n} \\int_{0}^{1} x^{n} \\ln x d x\n$$\n\nUsing integration by parts, we get\n\n$$\n\\int_{0}^{1} x^{n} \\ln x d x=\\left.\\frac{x^{n+1} \\ln x}{n+1}\\right|_{0} ^{1}-\\int_{0}^{1} \\frac{x^{n}}{n+1} d x=-\\frac{1}{(n+1)^{2}}\n$$\n\n(We used the fact that $\\lim _{x \\rightarrow 0} x^{n} \\ln x=0$ for $n>0$, which can be proven using l'Hôpital's rule.) Therefore, the original integral equals to\n\n$$\n\\sum_{n=1}^{\\infty} \\frac{1}{n(n+1)^{2}}=\\sum_{n=1}^{\\infty}\\left(\\frac{1}{n}-\\frac{1}{n+1}-\\frac{1}{(n+1)^{2}}\\right)\n$$\n\nTelescoping the sum and using the well-known identity $\\sum_{n=0}^{\\infty} \\frac{1}{n^{2}}=\\frac{\\pi^{2}}{6}$, we see that the above sum is equal to $2-\\frac{\\pi^{2}}{6}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-112-2008-feb-calc-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}