Mock test #3 Solutions

Part A: Short-answer type questions

For this part, answers must be written without any explanation.

How many factors of $2^{5} 3^{6} 5^{2}$ are perfect squares?
A) 24
B) 20
C) 30
D) 36

Sol.
Ans: 24.
The factors are of the form $2^x3^y5^z$ where $x\leq 5,y \leq 6 ,z \leq 2 $ must be even.
Find the value of expression:
\[ \sum_{n=1}^{\infty} \frac{\sin n}{n} \]

Sol.
Ans: $ (\pi-1)/2 $
Power series of $ - \log (1-z) = \sum_{n=1}^{\infty} \frac{z^n}{n}$ and we can write $\sin x$ as $\dfrac{1}{2i}(e^{ix}-e^{-ix})$. The infinite sum is $\dfrac{1}{2i}(-\ln(1-e^{i})+\ln(1-e^{-i})=\dfrac{1}{2i}(-\ln(-e^{-i})=\dfrac{\pi - 1}{2}$.
Let $\omega \neq 1$ be a complex cube root of unity. If \[\left(3-3 \omega+2 \omega^{2}\right)^{4 n+3}+\left(2+3 \omega-3 \omega^{2}\right)^{4 n+3}+\left(-3+2 \omega+3 \omega^{2}\right)^{4 n+3}=0\] then the possible value(s) of $n$ is (are)
A) 1
B) 2
C) 4
D) 5

Sol.
Ans: A, B, C and D.
Let $a, b \in R$ and $f: R \rightarrow R$ be defined by $f(x)=a \cos \left(\left|x^{3}-x\right|\right)+b|x| \sin \left(\left|x^{3}+x\right|\right)$. Then $f$ is
A) differentiable at $x=0$ if $a=0$ and $b=1$
B) differentiable at $x=1$ if $a=1$ and $b=0$
C) NOT differentiable at $x=0$ if $a=1$ and $b=0$
D) NOT differentiable at $x=1$ if $a=1$ and $b=1$

Sol.
Ans: A and B.
What is value of the series given below: \[(20)^2+ \dfrac{(21)^2}{1!} + \dfrac{(22)^2}{2!} + \dfrac{(23)^2}{3!} + \cdots \]

Sol.
Ans: $442e$.
We know that \[e^x = \sum_{n=0}^{\infty}\dfrac{x^n}{n!}.\] Then we can do a pattern of differentiating and multiplying by $x$. \[x^{20}e^x=\sum_{n=0}^{\infty}\dfrac{x^{n+20}}{n!}\] \[(20x^{19} + x^{20})e^x = \sum_{n=0}^{\infty}\dfrac{(n+20)x^{n+19}}{n!}\] \[(20x^{20} + x^{2021})e^x = \sum_{n=0}^{\infty}\dfrac{(n+20)x^{n+20}}{n!}\] \[(20^2x^{19} + 41x^{20} + x^{21})e^x =\sum_{n=0}^{\infty}\dfrac{(n+20)^2x^{n+19}}{n!}\] So, our desired sum occurs when $x = 1$, and we obtain $442e$.
Let $g$ be a continuous function which is not differentiable at 0 and let $g(0)=8$. If $f(x)=x g(x)$, then $f^{\prime}(0)=$
A) 0
B) 4
C) 2
D) 8

Sol.
Ans: D.
If the circles $x^{2}+y^{2}=1$ and $(x-a)^{2}+(y-b)^{2}=1$ have exactly one point in common, then the point $(a, b)$ lies on
A) $x^{2}+y^{2}=1$
B) $(x-a)^{2}+(y-b)^{2}=1$
C) $x^{2}+y^{2}=2$
D) $x^{2}+y^{2}=4$.

Sol.
Ans: D.
If $x$ and $y$ are non-zero real numbers, then $x^{2}+x y+y^{2}$
A) is always negative
B) takes the value zero for some $x, y$
C) is always positive
D) takes both positive and negative values.

Sol.
Ans: C. $x^2+y^2 \geq |2xy| > |xy|$. So $x^2+y^2$ is always greater than $xy$.
The set of all real numbers $x$ such that || $3-x|-| x+2||=5$ is
A) $[3, \infty)$
В) $(-\infty,-2] \cup[3, \infty)$
C) $(-\infty,-2]$
D) $(-\infty,-3] \cup[2, \infty)$.

Sol.
Ans: B
If $f(x)=nx , g(x) = e^{2x},$ and $h(x)=g(f(x))$. If $h'(0)=100$, find $n$.

Sol.
Ans: 50.
If $f(x)=n x, g(x)=e^{2 x}$ and $h(x)=g(f(x))$, then $h(x)=e^{2(n x)}$. The derivative of $h(x)$ is $2 n \cdot e^{2 n x}$, by the chain rule. Then, plugging in 0 for $x$ gets us $2 n=100$, so $n=50$.

Part B: Subjective questions

Submission files: Each question in this part must be answered on a page of its own. Name the files as B1.jpg, B2.jpg, etc. In case you have multiple files for the same question, say B4, name the corresponding files as B4-1.jpg, B4-2.jpg, etc.

Clearly explain your entire reasoning. No credit will be given without reasoning. Partial solutions may get partial credit.

B1. Five people each choose an integer between $1$ and $3$, inclusive. What is the probability that all three numbers get chosen?

Sol.

Answer: $50/81$.
Use complementary counting and principle of inclusion and exclusion. Call $E_n$ the event where $n$ is not chosen. Then, using the PIE, $(E_1 \cup E_2 \cup E_3)=(E_1) \cup (E_2) \cup (E_3)-((E_1 \cap E_2) + (E_2 \cap E_3) + (E_1 \cap E_3)) + (E_1 \cap E_2 \cap E_3)$. $(E_1) = 2^5$, since there are two numbers are left over for 5 people to choose and similarly, $(E_1 \cup E_2) = 1^5$, and $(E_1 \cup E_2 \cup E_3) = 0$. Thus there are $ 3 * 2^5 - 3 = 93$ ways that one of the 3 numbers is not selected. Our final answer is then $P = 1 - \dfrac{93}{3^5} = \dfrac{50}{81}$.

B2. In a convex quadrilateral $ABCD$ ,$\angle ABC$ and $\angle BCD$ are $\geq 120^\circ$. Prove that $|AC|$ + $|BD| \geq |AB|+|BC|+|CD|$. Notation. $|XY|$ denotes the length of the segment $XY$.

Sol.

By the cosine theorem: \[ AC^2=AB^2+BC^2-2AB.BC cos(\angle ABC) \geq AB^2+BC^2-2AB.BC (-\frac{1}{2}) = AB^2+BC^2+AB.BC \] It's obvious $ cos(\angle ABC) \leq -\frac{1}{2}$ because $ \angle ABC \geq 120^{o} $. On the other hand: \[ AB^2+BC^2+AB.BC > AB^2 + \frac{1}{4}BC^2 + AB.BC = (AB + \frac{1}{2}BC)^2 \] So $ AC > AB + \dfrac{1}{2}BC $. In the same way we show that : $ BD > CD + \frac{1}{2}BC $ Adding these inequalities we obtain: $ AC + BD \geq AB+BC+CD$.

B3. Consider a $3 \times 3$ grid with the first $9$ positive integers placed in the grid. Take the greatest integer in each row and let $r$ be the smallest of those numbers. Take the smallest integer in each column and let $c$ be the greatest of those numbers.
(a) True or false: Regardless of the arrangement of the numbers, $r \leq c$ always. Give explanation if true or a counterexample, if false.
(b) How many arrangements are there such that $r \leq c \leq 4$?

Sol.

(a) False. The matrix below is a counterexample. \[ \begin{bmatrix} 8 & 1 & 4 \\ 2 & 5 & 9 \\ 3 & 7 & 6 \end{bmatrix} \] (b) Answer is $38880$.
We first note that $r$ and $c$ are have minimum value $3$. We proceed using casework. If $r = c = 3$, then we need to split $1,2,3$ into different columns for $c = 3$. Then they necessarily need to be in the same row for $r = 3$. There are $9$ places to choose where the $3$ is, $2$ more ways to arrange $1,2$ in the row and $6!$ ways to arrange the other numbers. Thus, there are $9 \cdot 2 \cdot 6! = 720 \cdot 18 = 12960$ arrangements. If $r = 3, c = 4$. Then again $1,2,3$ need to be in the same row, forcing them to be in different columns. But we need 4 to be the smallest integer in some column, hence causing a problem. So, there are $0$ arrangements. Finally if $r = 4, c = 4$, we can choose $9$ places for the $4$ to be. Then we are forced to place two of the numbers $1,2,3$ in the row, giving us $6$ ordered ways. Then the final number can not be placed in the same column as the $4$, hence there are $4$ places to put it. Finally there are $5!$ ways to place the final numbers. So there are $9 \cdot 6 \cdot 4 \cdot 5! = 25920$. Thus there are $12960 + 25920 = 38880$ arrangements.

B4. Three squares have average area $\bar{A}=100 \mathrm{~m}^{2}$. The average of the lengths of their sides is $\bar{l}=10 \mathrm{~m}$. Find all possible values of the size of the largest of the three squares. The answer must be given along with reasoning.

Sol.

Solution 1.
Using Cauchy-Schwarz inequality: See Colin Maclaurin's solution.
Solution 2.
Let $x$ be the length of the side of a square, and let the probability of $x$ be $1 / 3,1 / 3,1 / 3$ over the three lengths $l_{1}, l_{2}, l_{3}$. Let $\mathcal{E}[x]$ denote the expected length. Then the information that we have is that $\mathcal{E}[x]=10$ and $\mathcal{E}[f(x)]=100$, where $f(x)=x^{2}$ is the function mapping lengths to areas. This is a strictly convex function. We notice that the equality $\mathcal{E}[f(x)]=f(\mathcal{E}[x])$ holds, therefore $x$ is a constant, and the three lengths must all be equal. The area of the largest square is $100 \mathrm{m}^{2}$.

B5. Let $t$ be a randomly chosen positive divisor of 20!. What is the probability that $t$ can be written as $a^{2}+b^{2}$ for some integers $a$ and $b$?

Sol.

Ans. $\frac{5}{54}$
The choice $k$ can be written as the sum of two squares if and only if every prime $p \equiv 3(\bmod 4)$ appears to an even power in the prime factorization of $k$. The primes $p \leq 20$ with $p \equiv 3(\bmod 4)$ are $3,7,11$, and 19. Also, $20 !=3^{8} \cdot 7^{2} \cdot 11^{1} \cdot 19^{1} \cdot k$ where $k$ is not divisible by any of these primes. Therefore, a randomly positive divisor of 20 ! has prime factorization $3^{a} \cdot 7^{b} \cdot 11^{c} \cdot 19^{d} \cdot l$ where $l$ is not divisible by any of these primes, $a \in[0,8], a \in[0,2], a \in[0,1]$, $a \in[0,1]$. Furthermore $a, b, c$, and $d$ take on the possible values with equally likelihood and independently of each other. It suffices to compute the probability that $a, b, c$, and $d$ are all even which is $\left(\frac{5}{9}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{5}{54}$.

B6. Compute \[ \int_{\frac{1}{2}}^{2} \frac{\tan ^{-1} x}{x^{2}-x+1} dx \]

Sol.

Take $y=1 / x$, then $\frac{d x}{x^{2}-x+1}=-\frac{d y}{y^{2}-y+1}$. Note furthermore by the tangent addition formula that $\tan ^{-1}(x)+\tan ^{-1}(y)=\pi / 2$. The original integral is equal to the average of these two integrals: \[ \frac{1}{2}\left(\int_{\frac{1}{2}}^{2} \frac{\tan ^{-1} x}{x^{2}-x+1} d x+\int_{\frac{1}{2}}^{2} \frac{\frac{\pi}{2}-\tan ^{-1} y}{y^{2}-y+1} d y\right)=\frac{\pi}{4} \int_{1 / 2}^{2} \frac{d x}{x^{2}-x+1} \] Substitute $x=\frac{\sqrt{3}}{2} \theta+1 / 2$, then \[ \frac{\pi}{4} \int_{1 / 2}^{2} \frac{d x}{x^{2}-x+1}=\frac{\pi}{4} \frac{4}{3} \frac{\sqrt{3}}{2} \int_{0}^{\sqrt{3}} \frac{1}{\theta^{2}+1} d \theta=\frac{\pi^{2} \sqrt{3}}{18} \]