exam: HMMT

problem:
Compute:

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

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

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

Remarks. Another solution, using L'Hôpital's rule, is possible: $\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos (x)}=\lim _{x \rightarrow 0} \frac{2 x}{\sin (x)}=2$.

solution_match: Answer:


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 $f(a)=a$.


problem:
Let $a$ be a positive real number. Find the value of $a$ such that the definite integral

$$
\int_{a}^{a^{2}} \frac{\mathrm{~d} x}{x+\sqrt{x}}
$$

achieves its smallest possible value.

solution:
$\sqrt[{3-2 \sqrt{2}}]{ }$ Let $F(a)$ denote the given definite integral. Then

$$
F^{\prime}(a)=\frac{\mathrm{d}}{\mathrm{~d} a} \int_{a}^{a^{2}} \frac{\mathrm{~d} x}{x+\sqrt{x}}=2 a \cdot \frac{1}{a^{2}+\sqrt{a^{2}}}-\frac{1}{a+\sqrt{a}} .
$$

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 $\sqrt{a}>0, a=(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$.


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 $\alpha=e^{x_{0}} /\left(x_{0}^{2}\right)=e^{2} / 4$.


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

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

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 $f(x)=e^{2 x-2}$.


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 $x_{0}=-1$.


problem:
Compute

$$
\sum_{n=1}^{\infty} \frac{1}{n \cdot(n+1) \cdot(n+1)!}
$$

solution:
3-e. We write

$$
\begin{gathered}
\sum_{n=1}^{\infty} \frac{1}{n \cdot(n+1) \cdot(n+1)!}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) \frac{1}{(n+1)!}=\sum_{n=1}^{\infty} \frac{1}{n \cdot(n+1)!}-\sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot(n+1)!} \\
\frac{1}{2}+\sum_{n=2}^{\infty} \frac{1}{n \cdot(n+1)!}-\sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot(n+1)!}=\frac{1}{2}+\sum_{n=1}^{\infty} \frac{1}{(n+1) \cdot(n+2)!}-\frac{1}{(n+1) \cdot(n+1)!} \\
\frac{1}{2}+\sum_{n=1}^{\infty} \frac{1-(n+2)}{(n+1) \cdot(n+2)!}=\frac{1}{2}-\left(\frac{1}{3!}+\frac{1}{4!}+\cdots\right)=3-\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots\right)=3-e .
\end{gathered}
$$

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

$$
\begin{aligned}
2-e & +\sum_{n=1}^{\infty} \frac{n(n+1)+1}{n \cdot(n+1) \cdot(n+1)!}=2-e+\sum_{n=1}^{\infty} \frac{(n+1)^{2}-n}{n \cdot(n+1) \cdot(n+1)!} \\
2 & -e+\sum_{n=1}^{\infty} \frac{1}{n \cdot n!}-\frac{1}{(n+1) \cdot(n+1)!}=3-e
\end{aligned}
$$


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

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 $z^{k}=1$.

solution:
$2005 \cdot \mathbf{2}^{2006}+1$. Note that

$$
\begin{aligned}
& \frac{1}{z-\omega}+\cdots+\frac{1}{z-\omega^{2006}}=\frac{\sum_{j=1}^{2006} \prod_{i \neq j}\left(z-\omega^{i}\right)}{(z-\omega) \cdots\left(z-\omega^{2006}\right)} \\
& \quad=\frac{\frac{\mathrm{d}}{\mathrm{~d} z}\left[z^{2006}+z^{2005}+\cdots+1\right]}{z^{2006}+z^{2005}+\cdots+1}=\frac{2006 z^{2005}+2005 z^{2004}+\cdots+1}{z^{2006}+z^{2005}+\cdots+1} \cdot \frac{z-1}{z-1} \\
& \quad=\frac{2006 z^{2006}-z^{2005}-z^{2004}-\cdots-1}{z^{2007}-1} \cdot \frac{z-1}{z-1}=\frac{2006 z^{2007}-2007 z^{2006}+1}{\left(z^{2007}-1\right)(z-1)} .
\end{aligned}
$$

Plugging in $z=2$ gives $\frac{2005 \cdot 2^{2006}+1}{2^{2007}-1}$; whence the answer.


problem:
g$ is a twice differentiable function over the positive reals such that

$$
\begin{aligned}
g(x)+2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x) & =0 \quad \text { for all positive reals } x . \\
\lim _{x \rightarrow \infty} x g(x) & =1
\end{aligned}
$$

Find 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,

$$
\begin{aligned}
0=g(x)+2 x^{3} g^{\prime}(x)+x^{4} g^{\prime \prime}(x) & =g\left(\frac{1}{y}\right)+2 y^{-3} g^{\prime}\left(\frac{1}{y}\right)+y^{-4} g^{\prime \prime}\left(\frac{1}{y}\right) \\
& =g\left(\frac{1}{y}\right)+\frac{\mathrm{d}}{\mathrm{~d} y}\left[-y^{-2} g^{\prime}\left(\frac{1}{y}\right)\right] \\
& =g\left(\frac{1}{y}\right)+\frac{\mathrm{d}^{2}}{\mathrm{~d} y^{2}}\left[g\left(\frac{1}{y}\right)\right]
\end{aligned}
$$

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

$$
1=\lim _{x \rightarrow \infty} c_{1} x+c_{2} \cdot \frac{\sin (1 / x)}{1 / x}=c_{2}+\lim _{x \rightarrow \infty} c_{1} x
$$

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 $\alpha=6 / \pi$.


problem:
Compute

$$
\int_{0}^{\infty} \frac{e^{-x} \sin (x)}{x} d x
$$

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

$$
\begin{aligned}
\int_{0}^{\infty} e^{-x}\left(\frac{\sin (x)}{x}\right) \mathrm{d} x & =\int_{0}^{\infty} e^{-x}\left(\int_{0}^{1} \cos (a x) \mathrm{d} a\right) \mathrm{d} x=\int_{0}^{1}\left(\int_{0}^{\infty} e^{-x} \cos (a x) \mathrm{d} x\right) \mathrm{d} a \\
& =\int_{0}^{1} \operatorname{Re}\left[\int_{0}^{\infty} e^{(-1+a i) x} \mathrm{~d} x\right] \mathrm{d} a=\int_{0}^{1} \operatorname{Re}\left[\left.\frac{e^{(-1+a i) x}}{-1+a i}\right|_{x=0} ^{\infty}\right] \mathrm{d} a \\
& =\int_{0}^{1} \operatorname{Re}\left[\frac{1}{1-a i}\right] \mathrm{d} a=\int_{0}^{1} \operatorname{Re}\left[\frac{1+a i}{1+a^{2}}\right] \mathrm{d} a \\
& =\int_{0}^{1} \frac{1}{1+a^{2}} \mathrm{~d} a=\left.\tan ^{-1}(a)\right|_{a=0} ^{1}=\frac{\pi}{4}
\end{aligned}
$$