HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl

{"year": "2007", "tier": "T4", "problem_label": "1", "problem_type": null, "exam": "HMMT", "problem": "Compute:\n\n$$\n\\lim _{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos (x)}\n$$", "solution": "2. Since $\\sin ^{2}(x)=1-\\cos ^{2}(x)$, we multiply the numerator and denominator by $1+\\cos (x)$ and use the fact that $x / \\sin (x) \\rightarrow 1$, obtaining\n\n$$\n\\lim _{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos (x)}=\\lim _{x \\rightarrow 0} \\frac{x^{2}(1+\\cos (x))}{1-\\cos ^{2}(x)}=\\lim _{x \\rightarrow 0}\\left(\\frac{x}{\\sin (x)}\\right)^{2} \\cdot 2=2\n$$\n\nRemarks. Another solution, using L'Hôpital's rule, is possible: $\\lim _{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos (x)}=\\lim _{x \\rightarrow 0} \\frac{2 x}{\\sin (x)}=2$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n1. [3]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "2", "problem_type": null, "exam": "HMMT", "problem": "Determine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=$ $x^{4}-x^{3}-x^{2}+a x+1$.", "solution": "1. Being a relative minimum, we have $0=f^{\\prime}(a)=4 a^{3}-3 a^{2}-2 a+a=a(4 a+1)(a-1)$. Then $a=0,1,-1 / 4$ are the only possibilities. However, it is easily seen that $a=1$ is the only value satisfying $f(a)=a$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n2. [3]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "3", "problem_type": null, "exam": "HMMT", "problem": "Let $a$ be a positive real number. Find the value of $a$ such that the definite integral\n\n$$\n\\int_{a}^{a^{2}} \\frac{\\mathrm{~d} x}{x+\\sqrt{x}}\n$$\n\nachieves its smallest possible value.", "solution": "$\\sqrt[{3-2 \\sqrt{2}}]{ }$ Let $F(a)$ denote the given definite integral. Then\n\n$$\nF^{\\prime}(a)=\\frac{\\mathrm{d}}{\\mathrm{~d} a} \\int_{a}^{a^{2}} \\frac{\\mathrm{~d} x}{x+\\sqrt{x}}=2 a \\cdot \\frac{1}{a^{2}+\\sqrt{a^{2}}}-\\frac{1}{a+\\sqrt{a}} .\n$$\n\nSetting $F^{\\prime}(a)=0$, we find that $2 a+2 \\sqrt{a}=a+1$ or $(\\sqrt{a}+1)^{2}=2$. We find $\\sqrt{a}= \\pm \\sqrt{2}-1$, and because $\\sqrt{a}>0, a=(\\sqrt{2}-1)^{2}=3-2 \\sqrt{2}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n3. [4]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "4", "problem_type": null, "exam": "HMMT", "problem": "Find the real number $\\alpha$ such that the curve $f(x)=e^{x}$ is tangent to the curve $g(x)=\\alpha x^{2}$.", "solution": "$\\mathbf{e}^{\\mathbf{2} / 4}$. Suppose tangency occurs at $x=x_{0}$. Then $e^{x_{0}}=\\alpha x_{0}^{2}$ and $f^{\\prime}\\left(x_{0}\\right)=2 \\alpha x_{0}$. On the other hand, $f^{\\prime}(x)=f(x)$, so $\\alpha x_{0}^{2}=2 \\alpha x_{0}$. Clearly, $\\alpha=0$ and $x_{0}=0$ are impossible, so it must be that $x_{0}=2$. Then $\\alpha=e^{x_{0}} /\\left(x_{0}^{2}\\right)=e^{2} / 4$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n4. [4]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "5", "problem_type": null, "exam": "HMMT", "problem": "The function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $f\\left(x^{2}\\right) f^{\\prime \\prime}(x)=f^{\\prime}(x) f^{\\prime}\\left(x^{2}\\right)$ for all real $x$. Given that $f(1)=1$ and $f^{\\prime \\prime \\prime}(1)=8$, determine $f^{\\prime}(1)+f^{\\prime \\prime}(1)$.", "solution": "6. Let $f^{\\prime}(1)=a$ and $f^{\\prime \\prime}(1)=b$. Then setting $x=1$ in the given equation, $b=a^{2}$. Differentiating the given yields\n\n$$\n2 x f^{\\prime}\\left(x^{2}\\right) f^{\\prime \\prime}(x)+f\\left(x^{2}\\right) f^{\\prime \\prime \\prime}(x)=f^{\\prime \\prime}(x) f^{\\prime}\\left(x^{2}\\right)+2 x f^{\\prime}(x) f^{\\prime \\prime}\\left(x^{2}\\right)\n$$\n\nPlugging $x=1$ into this equation gives $2 a b+8=a b+2 a b$, or $a b=8$. Then because $a$ and $b$ are real, we obtain the solution $(a, b)=(2,4)$.\nRemarks. A priori, the function needn't exist, but one possibility is $f(x)=e^{2 x-2}$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n5. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "6", "problem_type": null, "exam": "HMMT", "problem": "The elliptic curve $y^{2}=x^{3}+1$ is tangent to a circle centered at $(4,0)$ at the point $\\left(x_{0}, y_{0}\\right)$. Determine the sum of all possible values of $x_{0}$.", "solution": "$\\frac{\\mathbf{1}}{\\mathbf{3}}$. Note that $y^{2} \\geq 0$, so $x^{3} \\geq-1$ and $x \\geq-1$. Let the circle be defined by $(x-4)^{2}+y^{2}=c$ for some $c \\geq 0$. Now differentiate the equations with respect to $x$, obtaining $2 y \\frac{\\mathrm{~d} y}{\\mathrm{~d} x}=3 x^{2}$ from the given and $2 y \\frac{\\mathrm{~d} y}{\\mathrm{~d} x}=-2 x+8$ from the circle. For tangency, the two expressions $\\frac{\\mathrm{d} y}{\\mathrm{~d} x}$ must be equal if they are well-defined, and this is almost always the case. Thus, $-2 x_{0}+8=3 x_{0}^{2}$ so $x_{0}=-2$ or $x_{0}=4 / 3$, but only the latter corresponds to a point on $y^{2}=x^{3}+1$. Otherwise, $y_{0}=0$, and this gives the trivial solution $x_{0}=-1$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n6. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "7", "problem_type": null, "exam": "HMMT", "problem": "Compute\n\n$$\n\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot(n+1) \\cdot(n+1)!}\n$$", "solution": "3-e. We write\n\n$$\n\\begin{gathered}\n\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot(n+1) \\cdot(n+1)!}=\\sum_{n=1}^{\\infty}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\frac{1}{(n+1)!}=\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot(n+1)!}-\\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot(n+1)!} \\\\\n\\frac{1}{2}+\\sum_{n=2}^{\\infty} \\frac{1}{n \\cdot(n+1)!}-\\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot(n+1)!}=\\frac{1}{2}+\\sum_{n=1}^{\\infty} \\frac{1}{(n+1) \\cdot(n+2)!}-\\frac{1}{(n+1) \\cdot(n+1)!} \\\\\n\\frac{1}{2}+\\sum_{n=1}^{\\infty} \\frac{1-(n+2)}{(n+1) \\cdot(n+2)!}=\\frac{1}{2}-\\left(\\frac{1}{3!}+\\frac{1}{4!}+\\cdots\\right)=3-\\left(\\frac{1}{0!}+\\frac{1}{1!}+\\frac{1}{2!}+\\cdots\\right)=3-e .\n\\end{gathered}\n$$\n\nAlternatively, but with considerably less motivation, we can induce telescoping by adding and subtracting $e-2=1 / 2!+1 / 3!+\\cdots$, obtaining\n\n$$\n\\begin{aligned}\n2-e & +\\sum_{n=1}^{\\infty} \\frac{n(n+1)+1}{n \\cdot(n+1) \\cdot(n+1)!}=2-e+\\sum_{n=1}^{\\infty} \\frac{(n+1)^{2}-n}{n \\cdot(n+1) \\cdot(n+1)!} \\\\\n2 & -e+\\sum_{n=1}^{\\infty} \\frac{1}{n \\cdot n!}-\\frac{1}{(n+1) \\cdot(n+1)!}=3-e\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n7. [5]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "8", "problem_type": null, "exam": "HMMT", "problem": "Suppose that $\\omega$ is a primitive $2007^{\\text {th }}$ root of unity. Find $\\left(2^{2007}-1\\right) \\sum_{j=1}^{2006} \\frac{1}{2-\\omega^{j}}$.\n\nFor this problem only, you may express your answer in the form $m \\cdot n^{k}+p$, where $m, n, k$, and $p$ are positive integers. Note that a number $z$ is a primitive $n^{\\text {th }}$ root of unity if $z^{n}=1$ and $n$ is the smallest number amongst $k=1,2, \\ldots, n$ such that $z^{k}=1$.", "solution": "$2005 \\cdot \\mathbf{2}^{2006}+1$. Note that\n\n$$\n\\begin{aligned}\n& \\frac{1}{z-\\omega}+\\cdots+\\frac{1}{z-\\omega^{2006}}=\\frac{\\sum_{j=1}^{2006} \\prod_{i \\neq j}\\left(z-\\omega^{i}\\right)}{(z-\\omega) \\cdots\\left(z-\\omega^{2006}\\right)} \\\\\n& \\quad=\\frac{\\frac{\\mathrm{d}}{\\mathrm{~d} z}\\left[z^{2006}+z^{2005}+\\cdots+1\\right]}{z^{2006}+z^{2005}+\\cdots+1}=\\frac{2006 z^{2005}+2005 z^{2004}+\\cdots+1}{z^{2006}+z^{2005}+\\cdots+1} \\cdot \\frac{z-1}{z-1} \\\\\n& \\quad=\\frac{2006 z^{2006}-z^{2005}-z^{2004}-\\cdots-1}{z^{2007}-1} \\cdot \\frac{z-1}{z-1}=\\frac{2006 z^{2007}-2007 z^{2006}+1}{\\left(z^{2007}-1\\right)(z-1)} .\n\\end{aligned}\n$$\n\nPlugging in $z=2$ gives $\\frac{2005 \\cdot 2^{2006}+1}{2^{2007}-1}$; whence the answer.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n8. [6]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "9", "problem_type": null, "exam": "HMMT", "problem": "g$ is a twice differentiable function over the positive reals such that\n\n$$\n\\begin{aligned}\ng(x)+2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x) & =0 \\quad \\text { for all positive reals } x . \\\\\n\\lim _{x \\rightarrow \\infty} x g(x) & =1\n\\end{aligned}\n$$\n\nFind the real number $\\alpha>1$ such that $g(\\alpha)=1 / 2$.", "solution": "$\\frac{6}{\\pi}$. In the first equation, we can convert the expression $2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x)$ into the derivative of a product, and in fact a second derivative, by writing $y=1 / x$. Specifically,\n\n$$\n\\begin{aligned}\n0=g(x)+2 x^{3} g^{\\prime}(x)+x^{4} g^{\\prime \\prime}(x) & =g\\left(\\frac{1}{y}\\right)+2 y^{-3} g^{\\prime}\\left(\\frac{1}{y}\\right)+y^{-4} g^{\\prime \\prime}\\left(\\frac{1}{y}\\right) \\\\\n& =g\\left(\\frac{1}{y}\\right)+\\frac{\\mathrm{d}}{\\mathrm{~d} y}\\left[-y^{-2} g^{\\prime}\\left(\\frac{1}{y}\\right)\\right] \\\\\n& =g\\left(\\frac{1}{y}\\right)+\\frac{\\mathrm{d}^{2}}{\\mathrm{~d} y^{2}}\\left[g\\left(\\frac{1}{y}\\right)\\right]\n\\end{aligned}\n$$\n\nThus $g\\left(\\frac{1}{y}\\right)=c_{1} \\cos (y)+c_{2} \\sin (y)$ or $g(x)=c_{1} \\cos (1 / x)+c_{2} \\sin (1 / x)$. Now the second condition gives\n\n$$\n1=\\lim _{x \\rightarrow \\infty} c_{1} x+c_{2} \\cdot \\frac{\\sin (1 / x)}{1 / x}=c_{2}+\\lim _{x \\rightarrow \\infty} c_{1} x\n$$\n\nIt must be that $c_{1}=0, c_{2}=1$. Now since $0<1 / \\alpha<1$, the value of $\\alpha$ such that $g(\\alpha)=\\sin (1 / \\alpha)=1 / 2$ is given by $1 / \\alpha=\\pi / 6$ and so $\\alpha=6 / \\pi$.", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n9. $[7]", "solution_match": "\nAnswer: "}}
{"year": "2007", "tier": "T4", "problem_label": "10", "problem_type": null, "exam": "HMMT", "problem": "Compute\n\n$$\n\\int_{0}^{\\infty} \\frac{e^{-x} \\sin (x)}{x} d x\n$$", "solution": "$\\frac{\\pi}{4}$. We can compute the integral by introducing a parameter and exchanging the order of integration:\n\n$$\n\\begin{aligned}\n\\int_{0}^{\\infty} e^{-x}\\left(\\frac{\\sin (x)}{x}\\right) \\mathrm{d} x & =\\int_{0}^{\\infty} e^{-x}\\left(\\int_{0}^{1} \\cos (a x) \\mathrm{d} a\\right) \\mathrm{d} x=\\int_{0}^{1}\\left(\\int_{0}^{\\infty} e^{-x} \\cos (a x) \\mathrm{d} x\\right) \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\operatorname{Re}\\left[\\int_{0}^{\\infty} e^{(-1+a i) x} \\mathrm{~d} x\\right] \\mathrm{d} a=\\int_{0}^{1} \\operatorname{Re}\\left[\\left.\\frac{e^{(-1+a i) x}}{-1+a i}\\right|_{x=0} ^{\\infty}\\right] \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\operatorname{Re}\\left[\\frac{1}{1-a i}\\right] \\mathrm{d} a=\\int_{0}^{1} \\operatorname{Re}\\left[\\frac{1+a i}{1+a^{2}}\\right] \\mathrm{d} a \\\\\n& =\\int_{0}^{1} \\frac{1}{1+a^{2}} \\mathrm{~d} a=\\left.\\tan ^{-1}(a)\\right|_{a=0} ^{1}=\\frac{\\pi}{4}\n\\end{aligned}\n$$", "metadata": {"resource_path": "HarvardMIT/segmented/en-102-2007-feb-calc-solutions.jsonl", "problem_match": "\n10. [8]", "solution_match": "\nAnswer: "}}