HarvardMIT/md/en-102-2007-feb-calc-solutions.md

$10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament
Saturday 24 February 2007

Individual Round: Calculus Test

1. [3] Compute:

$$\underset{x\to 0}{\lim }\frac{x^2}{1-\cos (x)}\backslash lim\; \_\{x\; \backslash rightarrow\; 0\}\; \backslash frac\{x^\{2\}\}\{1-\backslash cos\; (x)\}$$

Answer: 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

$$\underset{x\to 0}{\lim }\frac{x^2}{1-\cos (x)}=\underset{x\to 0}{\lim }\frac{x^2(1+\cos (x))}{1-{\cos }^2(x)}=\underset{x\to 0}{\lim }{\left(\frac{x}{\sin (x)}\right)}^2\cdot 2=2\backslash lim\; \_\{x\; \backslash rightarrow\; 0\}\; \backslash frac\{x^\{2\}\}\{1-\backslash cos\; (x)\}=\backslash lim\; \_\{x\; \backslash rightarrow\; 0\}\; \backslash frac\{x^\{2\}(1+\backslash cos\; (x))\}\{1-\backslash cos\; ^\{2\}(x)\}=\backslash lim\; \_\{x\; \backslash rightarrow\; 0\}\backslash left(\backslash frac\{x\}\{\backslash sin\; (x)\}\backslash right)^\{2\}\; \backslash cdot\; 2=2$$

Remarks. Another solution, using L'Hôpital's rule, is possible: $\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos (x)}=\lim _{x \rightarrow 0} \frac{2 x}{\sin (x)}=2$. 2. [3] Determine the real number $a$ having the property that $f(a)=a$ is a relative minimum of $f(x)=$ $x^{4}-x^{3}-x^{2}+a x+1$. Answer: 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 $f(a)=a$. 3. [4] Let $a$ be a positive real number. Find the value of $a$ such that the definite integral

$${\int }_a^{a^2}\frac{\mathrm{d}x}{x+\sqrt{x}}\backslash int\_\{a\}^\{a^\{2\}\}\; \backslash frac\{\backslash mathrm\{~d\}\; x\}\{x+\backslash sqrt\{x\}\}$$

achieves its smallest possible value. Answer: $\sqrt[{3-2 \sqrt{2}}]{ }$ Let $F(a)$ denote the given definite integral. Then

$$F^{\mathrm{\prime }}(a)=\frac{\mathrm{d}}{\mathrm{d}a}{\int }_a^{a^2}\frac{\mathrm{d}x}{x+\sqrt{x}}=2a\cdot \frac{1}{a^2+\sqrt{a^2}}-\frac{1}{a+\sqrt{a}}\mathrm{.}F^\{\backslash prime\}(a)=\backslash frac\{\backslash mathrm\{d\}\}\{\backslash mathrm\{~d\}\; a\}\; \backslash int\_\{a\}^\{a^\{2\}\}\; \backslash frac\{\backslash mathrm\{~d\}\; x\}\{x+\backslash sqrt\{x\}\}=2\; a\; \backslash cdot\; \backslash frac\{1\}\{a^\{2\}+\backslash sqrt\{a^\{2\}\}\}-\backslash frac\{1\}\{a+\backslash sqrt\{a\}\}\; .$$

Setting $F^{\prime}(a)=0$, 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 $\sqrt{a}>0, a=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$. 4. [4] Find the real number $\alpha$ such that the curve $f(x)=e^{x}$ is tangent to the curve $g(x)=\alpha x^{2}$.

Answer: $\mathbf{e}^{\mathbf{2} / 4}$. 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 $\alpha=e^{x_{0}} /\left(x_{0}^{2}\right)=e^{2} / 4$. 5. [5] 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 $f^{\prime}(1)+f^{\prime \prime}(1)$. Answer: 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

$$2x{f}^{\mathrm{\prime }}\left(x^2\right)f^{\mathrm{\prime }\mathrm{\prime }}(x)+f\left(x^2\right)f^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(x)=f^{\mathrm{\prime }\mathrm{\prime }}(x)f^{\mathrm{\prime }}\left(x^2\right)+2x{f}^{\mathrm{\prime }}(x)f^{\mathrm{\prime }\mathrm{\prime }}\left(x^2\right)2\; x\; f^\{\backslash prime\}\backslash left(x^\{2\}\backslash right)\; f^\{\backslash prime\; \backslash prime\}(x)+f\backslash left(x^\{2\}\backslash right)\; f^\{\backslash prime\; \backslash prime\; \backslash prime\}(x)=f^\{\backslash prime\; \backslash prime\}(x)\; f^\{\backslash prime\}\backslash left(x^\{2\}\backslash right)+2\; x\; f^\{\backslash prime\}(x)\; f^\{\backslash prime\; \backslash prime\}\backslash left(x^\{2\}\backslash right)$$

Plugging $x=1$ 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 $(a, b)=(2,4)$. Remarks. A priori, the function needn't exist, but one possibility is $f(x)=e^{2 x-2}$. 6. [5] 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 $x_{0}$. Answer: $\frac{\mathbf{1}}{\mathbf{3}}$. 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 $x_{0}=-1$. 7. [5] Compute

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\cdot (n+1)\cdot (n+1)!}\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{n\; \backslash cdot(n+1)\; \backslash cdot(n+1)!\}$$

Answer: 3-e. We write

$$\begin{array}{c}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\cdot (n+1)\cdot (n+1)!}=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})\frac{1}{(n+1)!}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\cdot (n+1)!}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+1)\cdot (n+1)!}}\\ {\displaystyle \frac{1}{2}+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\cdot (n+1)!}-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+1)\cdot (n+1)!}=\frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+1)\cdot (n+2)!}-\frac{1}{(n+1)\cdot (n+1)!}}\\ {\displaystyle \frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{1-(n+2)}{(n+1)\cdot (n+2)!}=\frac{1}{2}-(\frac{1}{3!}+\frac{1}{4!}+\cdots )=3-(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots )=3-e\mathrm{.}}\end{array}\backslash begin\{gathered\}\; \backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{n\; \backslash cdot(n+1)\; \backslash cdot(n+1)!\}=\backslash sum\_\{n=1\}^\{\backslash infty\}\backslash left(\backslash frac\{1\}\{n\}-\backslash frac\{1\}\{n+1\}\backslash right)\; \backslash frac\{1\}\{(n+1)!\}=\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{n\; \backslash cdot(n+1)!\}-\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{(n+1)\; \backslash cdot(n+1)!\}\; \backslash \backslash \; \backslash frac\{1\}\{2\}+\backslash sum\_\{n=2\}^\{\backslash infty\}\; \backslash frac\{1\}\{n\; \backslash cdot(n+1)!\}-\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{(n+1)\; \backslash cdot(n+1)!\}=\backslash frac\{1\}\{2\}+\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1\}\{(n+1)\; \backslash cdot(n+2)!\}-\backslash frac\{1\}\{(n+1)\; \backslash cdot(n+1)!\}\; \backslash \backslash \; \backslash frac\{1\}\{2\}+\backslash sum\_\{n=1\}^\{\backslash infty\}\; \backslash frac\{1-(n+2)\}\{(n+1)\; \backslash cdot(n+2)!\}=\backslash frac\{1\}\{2\}-\backslash left(\backslash frac\{1\}\{3!\}+\backslash frac\{1\}\{4!\}+\backslash cdots\backslash right)=3-\backslash left(\backslash frac\{1\}\{0!\}+\backslash frac\{1\}\{1!\}+\backslash frac\{1\}\{2!\}+\backslash cdots\backslash right)=3-e\; .\; \backslash end\{gathered\}$$

Alternatively, but with considerably less motivation, we can induce telescoping by adding and subtracting $e-2=1 / 2!+1 / 3!+\cdots$, obtaining

8. [6] Suppose that $\omega$ is a primitive $2007^{\text {th }}$ root of unity. Find $\left(2^{2007}-1\right) \sum_{j=1}^{2006} \frac{1}{2-\omega^{j}}$.

For 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 $z^{k}=1$. Answer: $2005 \cdot \mathbf{2}^{2006}+1$. Note that

Plugging in $z=2$ gives $\frac{2005 \cdot 2^{2006}+1}{2^{2007}-1}$; whence the answer. 9. $[7] g$ is a twice differentiable function over the positive reals such that

Find the real number $\alpha>1$ such that $g(\alpha)=1 / 2$. Answer: $\frac{6}{\pi}$. 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,

Thus $g\left(\frac{1}{y}\right)=c_{1} \cos (y)+c_{2} \sin (y)$ or $g(x)=c_{1} \cos (1 / x)+c_{2} \sin (1 / x)$. Now the second condition gives

$$1=\underset{x\to \mathrm{\infty }}{\lim }{c}_{1}x+c_2\cdot \frac{\sin (1\mathrm{/}x)}{1\mathrm{/}x}=c_2+\underset{x\to \mathrm{\infty }}{\lim }{c}_{1}x1=\backslash lim\; \_\{x\; \backslash rightarrow\; \backslash infty\}\; c\_\{1\}\; x+c\_\{2\}\; \backslash cdot\; \backslash frac\{\backslash sin\; (1\; /\; x)\}\{1\; /\; x\}=c\_\{2\}+\backslash lim\; \_\{x\; \backslash rightarrow\; \backslash infty\}\; c\_\{1\}\; x$$

It 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 $\alpha=6 / \pi$. 10. [8] Compute

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x}\sin (x)}{x}dx\backslash int\_\{0\}^\{\backslash infty\}\; \backslash frac\{e^\{-x\}\; \backslash sin\; (x)\}\{x\}\; d\; x$$

Answer: $\frac{\pi}{4}$. We can compute the integral by introducing a parameter and exchanging the order of integration: