Here I put the entrance exams of Kyoto University for this year, which is organised on 25/02/2023.
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amazon’s_hanging_cable_interview_question_.pdf | |
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Just uploaded the chapter about the Differential Equations to the maths books page.
You will find the basics of Laplace Transform too.
Good Luck.
You will find the basics of Laplace Transform too.
Good Luck.
I have just uploaded the Hyperbolic Functions chapter at the maths books page.
In the complex number world, you will see that the hyperbolic functions and the trigonometric functions are the same as $$\begin{split} \cosh x &= \dfrac12 ( e^x+x^{-x}) \\ &= \dfrac12 ( e^{-i(ix)}+e^{i(ix)}) \\ &= \cos (ix) \end{split}$$ and $$\begin{split} \sinh x &= \dfrac12 ( e^x-x^{-x})\\ &= \dfrac12 ( e^{-i(ix)}-e^{i(ix)}) \\ &= -i \left(\dfrac1{2i}(( e^{i(ix)}-e^{-i(ix)}) \right) \\ &=- i \sin (ix) \end{split}$$ Then, for example, the integral of the hyperbolic function is only the integral of the trigonometric function or vice versa. $$\begin{split} \int \cosh x\; dx &= \int \cos (ix) \; dx \\ &= \dfrac{1}{i} \sin (ix) + C \\ &=- i \sin (ix) + C \\ &= \sinh x + C \end{split}$$
I put the PDF files of exams of some famous Japanese Universities
(You can find them just click the text " Entrance Exams of Japanese Universities 2019"
on the bottom of the English web page.)
Original exams are written in Japanese, then I translated them into English,
and two of these universities questions ( Kyoto and Tokyo) I made answers too.
(Generally there are no official corrections for Japanese Entrance Exams.)
If you need answers for another universities,
please request me from the comments area or e-mail me.
(You can find them just click the text " Entrance Exams of Japanese Universities 2019"
on the bottom of the English web page.)
Original exams are written in Japanese, then I translated them into English,
and two of these universities questions ( Kyoto and Tokyo) I made answers too.
(Generally there are no official corrections for Japanese Entrance Exams.)
If you need answers for another universities,
please request me from the comments area or e-mail me.
Since $$1+i = \sqrt2 (\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4})$$ and $$1-i = \sqrt2 (\cos \dfrac{\pi}{4} - i \sin \dfrac{\pi}{4})$$ then $$\begin{split} (1+i)^n + (1-i)^n &= \left( \sqrt2 (\cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{4}) \right)^n + \left( \sqrt2 (\cos \dfrac{\pi}{4} - i \sin \dfrac{\pi}{4}) \right)^n \\ &= 2^{\frac{n}{2}} ( \cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4} ) + 2^{\frac{n}{2}} ( \cos \dfrac{n \pi}{4} - i \sin \dfrac{n \pi}{4} ) \\ &= 2^{\frac{n}{2}+1} \cos \dfrac{n \pi}{4} \end{split}$$ For \(\cos \dfrac{n \pi}{4}\) is positive, $n = 8k, \; n = 8k+1$ or $n=8k+7$, where $k$ is a non-negative integer.
-
When $n = 8k$,
\(\cos \dfrac{n \pi}{4} = \cos \dfrac{8k \pi}{4} = \cos 2k\pi = 1\).
Then $$(1+i)^n +(1-i)^n > 10^{10}$$ $$2^{\frac{n}{2}+1} \cos \dfrac{n \pi}{4} > 10^{10}$$ $$2^{\frac{8k}{2}+1} > 10^{10}$$ $$2^{4k+1} > 10^{10}$$ $$\log_{10} 2^{4k+1} > \log_{10}10^{10}$$ $$(4k+1) \log_{10}2 > 10$$ $$k > \dfrac14 \left( \dfrac{10}{\log_{10}2}-1 \right)$$ $$k > 8.05$$ Therefore the smallest integer $k$ is $k = 9$.
Then $n = 8k = 72$. -
When $n = 8k+1$,
\(\cos \dfrac{n \pi}{4} = \cos \dfrac{(8k+1)\pi}{4} = \cos (2k\pi+\dfrac{\pi}{4} )= \dfrac{1}{\sqrt2}\).
Then $$(1+i)^n +(1-i)^n > 10^{10}$$ $$2^{\frac{n}{2}+1} \cos \dfrac{n \pi}{4} > 10^{10}$$ $$2^{\frac{8k+1}{2}+1} 2^{-\frac12} > 10^{10}$$ $$2^{4k+1} > 10^{10}$$ $$\log_{10} 2^{4k+1} > \log_{10}10^{10}$$ $$(4k+1) \log_{10}2 > 10$$ $$k > \dfrac14 \left( \dfrac{10}{\log_{10}2}-1 \right)$$ $$k > 8.05$$ Therefore the smallest integer $k$ is $k = 9$.
Then $n = 8k+1 = 73$. -
When $n = 8k+7$,
\(\cos \dfrac{n \pi}{4} = \cos \dfrac{(8k+7)\pi}{4} = \cos (2k\pi+\dfrac{7\pi}{4}) = \dfrac{1}{\sqrt2}\).
Then $$(1+i)^n +(1-i)^n > 10^{10}$$ $$2^{\frac{n}{2}+1} \cos \dfrac{n \pi}{4} > 10^{10}$$ $$2^{\frac{8k+7}{2}+1} 2^{-\frac12} > 10^{10}$$ $$2^{4k+4} > 10^{10}$$ $$\log_{10} 2^{4k+4} > \log_{10}10^{10}$$ $$(4k+4) \log_{10}2 > 10$$ $$k > \dfrac14 \left( \dfrac{10}{\log_{10}2}-4 \right)$$ $$k > 7.30$$ Therefore the smallest integer $k$ is $k = 8$.
Then $n = 8k+7 = 71$.
Hence the smallest positive integer $n$ is $$n = 71$$
Let $x$ be the distance from the center $O$ of the sphere and the square $B_1B_2B_3B_4$.
Then the length of the diagonal of the square is given by $2 \sqrt{1-x^2}$.
The area of the square $B_1B_2B_3B_4$ is $$\dfrac12 (2 \sqrt{1-x^2})^2 = 2(1-x^2)$$ The height of the pyramid $AB_1B_2B_3B_4$ is $1+x$.
Then the volume $V$ of the pyramid $AB_1B_2B_3B_4$ is $$V = \dfrac13 \cdot 2(1-x^2) \cdot (1+x) = \dfrac23 ( 1+x-x^2-x^3)$$ $$\dfrac{dV}{dx} = \dfrac23 ( 1-2x-3x^2) = -\dfrac23 (3x-1)(x+1)$$ Then \(\dfrac{dV}{dx} = 0\), when \(x = -1, \; \dfrac13\)
Since $0<x<1$, the variation of $V$ is $$\begin{array}{c|ccccc} x&0& &\frac13& &1 \\\hline \frac{dV}{dx}& &+&0&-& \\\hline V& & \nearrow&\frac{64}{81}&\searrow& \end{array}$$ Hence the maximum volume of a pyramid $AB_1B_2B_3B_4$ is \(\dfrac{64}{81}\).
The condition (I) means:
Before $k$, every number of the face of die are less than or equal to $4$, and after $k$ , once $X_{i+1}$ is less than or equal to $4$, after that every number of the face of die must be less than or equal to $4$.
Then $$X_0=0,$$ $$X_1 \leq 4, \; X_2 \leq 4, \; \cdots, \; X_{k-1} \leq 4,\qquad \qquad \text{probability} = \left( \dfrac46 \right)^k$$ $$X_k \geq 5,x_{k+1} \geq 5, \; \cdots \; X_{i} \geq 5, \qquad \qquad \text{probability} = \left( \dfrac26 \right)^{i-k+1}$$ $$X_{i+1} \leq 4, \; X_{i+1} \leq 4 , \; \cdots \; X_n \leq 4 \qquad \qquad \text{probability} = \left( \dfrac46 \right)^{n-i}$$ where $k \leq i \leq n$.
Therefore the required probability $P$ is $$\begin{split} P &= \sum_{k=1}^n \sum_{i = k}^n \left( \dfrac46 \right)^{k-1} \left( \dfrac26 \right)^{i-k+1}\left( \dfrac46 \right)^{n-i} \\ &= \sum_{k=1}^n \sum_{i = k}^n \dfrac{2^{n+k-i-1}}{3^n}\\ &= \sum_{k=1}^n \left( \dfrac{2^{n+k-1}}{3^n} \sum_{i = k}^n 2^{-i} \right) \\ &= \sum_{k=1}^n \left( \dfrac{2^{n+k-1}}{3^n} \dfrac{2^{-k}(1-2^{-(n-k+1)})}{1-2^{-1}} \right) \\ &= \sum_{k=1}^n \left( \dfrac{2^{n}(1-2^{-n+k-1})}{3^n} \right) \\ &= \sum_{k=1}^n \left( \dfrac{2^n -2^{k-1}}{3^n} \right) \\ &= n \left(\dfrac23 \right)^n - \left( \dfrac13 \right)^n \dfrac{(1-2^n)}{1-2} \\ &= \dfrac{n \cdot 2^n - 2^{n}+1}{3^n} \\ &= \dfrac{(n-1)2^n +1}{3^n} \end{split}$$
Today I put rest of the problems of entrance exams of Kyoto Univeristy.
I am afraid that they are not presented normally on some browser, then I will put PDF file on the main page of this WEB site 'Entrance Exams of Japanese University 2019'.
We fix the coordinates system such that $A(v, w), B(0,0)$ and $C(u,0)$ where $u,v$ and $w$ are positive numbers.
Then $$\vec{OQ} = \vec{OA} + \vec{AQ} = \vec{OA} + t\vec{AC} = \vec{OA} +t(\vec{OC} - \vec{OA}) = (1-t) \vec{OA} + \vec{OC}$$ Therefore $$\vec{OQ} = (1-t) \begin{pmatrix} v \\ w \end{pmatrix} + \begin{pmatrix} u \\ 0 \end{pmatrix} = \begin{pmatrix} (1-t)v+u \\ (1-t)w \end{pmatrix}$$ And $$\vec{OP} = t \vec{OQ}$$ Then $$\vec{OP} = t \begin{pmatrix} (1-t)v+u \\ (1-t)w \end{pmatrix} = \begin{pmatrix} t(1-t)v+tu \\ t(1-t)w \end{pmatrix}$$ Therefore the equation of the locus of the point $P$ is given as $$\left\{ \begin{array}{l} x = t(1-t)v+tu = (u+v)t - vt^2 \\ y = t(1-t)w = wt - wt^2 \end{array} \right.$$ $$\dfrac{dx}{dt} = (u+v) -2vt, \qquad dx = ((u+v) -2vt)\; dt$$ Hence the required area $A$ is $$\begin{split} A &= \int_0^u y \; dx \\ &= \int_0^1 (wt-wt^2) ((u+v)-2vt)\; dt \\ &= \int_0^1 (w(u+v)t+(-w(u+v)-2vw)t^2 + 2vwt^3)\; dt \\ &= \int_0^1 (w(u+v)t-w(u+3v)t^2 + 2vwt^3)\; dt \\ &= w \Big[ \dfrac12(u+v)t^2 -\dfrac13(u+3v)t^3 + \dfrac12vt^4 \Big]_0^1\\ &= w \left( \dfrac12(u+v) -\dfrac13(u+3v) + \dfrac12v \right)\\ &= \dfrac16uw \end{split}$$
Since \(S =\dfrac12uw\), $$A = \dfrac13S$$
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