exam: HMMT

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.

solution_match: Answer:


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,

$$
\frac{t^{3}+t+16}{t}=3 t^{2}+1
$$

Simplifying, we get $t^{3}=8$, so $t=2$. It follows that the slope is $3(2)^{2}+1=13$.


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

$$
\int_{1}^{y} x \ln x d x=\left.\frac{1}{2} x^{2} \ln x\right|_{1} ^{y}-\frac{1}{2} \int_{1}^{y} x d x=\frac{1}{2} y^{2} \ln y-\frac{1}{4} y^{2}+\frac{1}{4}
$$

So $y^{2} \ln y=\frac{1}{2} y^{2}$. Since $y>1$, we obtain $\ln y=\frac{1}{2}$, and thus $y=\sqrt{e}$.


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

$$
1=\lim _{x \rightarrow 1} \frac{(\ln (2-x))^{2}}{x^{2}+a x+b}=\lim _{x \rightarrow 1} \frac{2 \ln (2-x)}{(x-2)(2 x+a)}
$$

and 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.


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:
| $\frac{3}{8}$ |
| :---: |
| We have |

$$
\begin{aligned}
\sin ^{6} x+\cos ^{6} x & =\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right) \\
& =1-3 \sin ^{2} x \cos ^{2} x=1-\frac{3}{4} \sin ^{2} 2 x=1-\frac{3}{4}\left(\frac{1-\cos 4 x}{2}\right) \\
& =\frac{5}{8}+\frac{3}{8} \cos 4 x
\end{aligned}
$$

It 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 $\frac{3}{8}$.


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

$$
S_{n}=\binom{n}{0}^{-1}+\binom{n}{1}^{-1}+\binom{n}{n-1}^{-1}+\binom{n}{n}^{-1}+\sum_{k=2}^{n-2}\binom{n}{k}^{-1} \leq 2+\frac{2}{n}+(n-3)\binom{n}{2}^{-1}
$$

which goes to 2 as $n \rightarrow \infty$. Therefore, $S_{n} \rightarrow 2$.


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

$$
\lim _{t \rightarrow 0} t^{-p}\left(\sqrt[3]{\frac{1}{t}+1}+\sqrt[3]{\frac{1}{t}-1}-2 \sqrt[3]{\frac{1}{t}}\right)=\lim _{t \rightarrow 0} t^{-p-\frac{1}{3}}(\sqrt[3]{1+t}+\sqrt[3]{1-t}-2)
$$

We need the degree of the first nonzero term in the MacLaurin expansion of $\sqrt[3]{1+t}+\sqrt[3]{1-t}-2$. We have

$$
\sqrt[3]{1+t}=1+\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right), \quad \sqrt[3]{1-t}=1-\frac{1}{3} t-\frac{1}{9} t^{2}+o\left(t^{2}\right)
$$

It 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 $p=\frac{5}{3}$.


solution:
$\frac{11}{4}$ Divide the top and bottom by $e^{x}$ to obtain that

$$
T=\int_{0}^{\ln 2} \frac{2 e^{2 x}+e^{x}-e^{-x}}{e^{2 x}+e^{x}-1+e^{-x}} d x
$$

Notice that $2 e^{2 x}+e^{x}-e^{-x}$ is the derivative of $e^{2 x}+e^{x}-1+e^{-x}$, and so

$$
T=\left[\ln \left|e^{2 x}+e^{x}-1+e^{-x}\right|\right]_{0}^{\ln 2}=\ln \left(4+2-1+\frac{1}{2}\right)-\ln 2=\ln \left(\frac{11}{4}\right)
$$

Therefore, $e^{T}=\frac{11}{4}$.


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

$$
-\frac{1}{2}\left(1+\frac{1}{n}\right) \ln n+\frac{1}{n^{2}} \sum_{k=1}^{n} k \ln k=\frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \ln \left(\frac{k}{n}\right)
$$

We 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

$$
\int_{0}^{1} x \ln x d x=\left.\frac{1}{2} x^{2} \ln x\right|_{0} ^{1}-\int_{0}^{1} \frac{x}{2} d x=-\frac{1}{4}
$$

Therefore, the original limit is $e^{-1 / 4}$.


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

$$
\int_{0}^{1} \ln x \ln (1-x) d x=-\int_{0}^{1} \ln x \sum_{n=1}^{\infty} \frac{x^{n}}{n} d x=-\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n} \ln x d x
$$

Using integration by parts, we get

$$
\int_{0}^{1} x^{n} \ln x d x=\left.\frac{x^{n+1} \ln x}{n+1}\right|_{0} ^{1}-\int_{0}^{1} \frac{x^{n}}{n+1} d x=-\frac{1}{(n+1)^{2}}
$$

(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

$$
\sum_{n=1}^{\infty} \frac{1}{n(n+1)^{2}}=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{(n+1)^{2}}\right)
$$

Telescoping 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 $2-\frac{\pi^{2}}{6}$.