Engineering mathematics set 2

By | November 21, 2018

Select your preferred language

Engineering mathematics questions answers

Question 1
The only prime followed by a cube?
A
2
B
3
C
5
D
7
Question 1 Explanation: 
$ x^3 - 1 = \underbrace{(x-1)(x^2+x+1)}. $

Being a product of two numbers, the expression over the $\underbrace{\text{underbrace}}$ is composite UNLESS $(x-1)=1$. That happens only if $x=2$, so $x^3=8$.

Question 2
What is $\infty - \infty$?
A
$\infty$
B
0
C
Indeterminate
D
None of the above
Question 2 Explanation: 
Answer is taken from this link

Imagine that I have an infinite number of hotel rooms, each numbered 1, 2, 3, 4, ... Then I give you all of them. I would have none left, so $\infty - \infty = 0$ On the other hand, if I give you all of the odd-numbered ones, then I still have an infinite number left. So $\infty - \infty = \infty$. Now suppose that I give you all of them except for the first seven. Then $\infty - \infty = 7$. While this doesn't explain why this is indeterminate, hopefully you can agree that it is indeterminate!

Question 3
$\int_0^\infty \frac{\sin x} x \, dx = ?$
A
$\frac \pi 2$
B
1
C
$\pi$
D
0
Question 3 Explanation: 
Here's another way of finishing off Derek's argument. He proves $$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$ Let $$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx= \int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$ Let $$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$ where $$f(x)=\frac1{\sin x}-\frac1x.$$ We need the fact that if we define $f(0)=0$ then $f$ has a continuous derivative on the interval $[0,\pi/2]$. Integration by parts yields $$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$ Hence $I_n\to\pi/2$ and we conclude that $$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$

Answers from link

Question 4
$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=?$
A
1/2
B
$\pi$
C
$\pi/2$
D
$\pi^2$
Question 4 Explanation: 
This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$. By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$. Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.

Reference

Question 5
$\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} =?$
A
$\frac{15\pi^{2}}{96}$
B
$\frac{\pi^{2}}{96}$
C
$\frac{5\pi^{2}}{96}$
D
$\frac{3\pi^{2}}{96}$
Question 5 Explanation: 
Check this link for discussion
Question 6
$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx =?$
A
$ \frac{\sqrt \pi}{2}$
B
$ \pi$
C
1
D
$\infty$
Question 6 Explanation: 
Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$ Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$ $$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$ Now change to polar coordinates $$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$ The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$ $$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$ So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry

Reference

Question 7
$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=?$
A
$-\pi^2\left(3\,\zeta'(-1)+\frac23\right).$
B
$-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$
C
$\pi^2\left(4\,\zeta'(-1)+\frac23\right).$
D
$-\pi^2\left(4\,\zeta'(-1)+\frac25\right).$
Question 7 Explanation: 
Here is a solution: Let $I$ denote the integral. Then \begin{align*} I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\ &= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\ &= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\ &= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\ &= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du. \end{align*} Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$: $$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1. $$ Then it follows that \begin{align*} I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\ &= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}. \end{align*} Now we consider the exponential of the partial sum: \begin{align*} &\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\ &= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\ &= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\ &= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}. \end{align*} In view of the definition of Glaisher-Kinkelin constant $A$, we have $$ \lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}. $$ This, together with the identity $ \log A = \frac{1}{12} - \zeta'(-1)$, yields $$ I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right) $$ as desired.

Reference

Question 8
How many ways can you draw this starting from the center of the petals, without lifting the pen.
A
3072
B
2071
C
3065
D
3987
Question 8 Explanation: 
First you have to draw the petals. There are $4!=24$ ways to choose the order of the petals and $2^4=16$ ways to choose the direction you go around each petal. Then you go down the stem to the leaves. There are $2! \cdot 2^2=8$ ways to draw the leaves. Finally you draw the lower stem. $24 \cdot 16 \cdot 8=3072$ At the beginning you could go 8 different ways, then you could go 6 different ways, then you could go 4 and 2 different ways but in the down of the picture you could go at first 4 different ways and 2 at the end. $8\cdot6\cdot4\cdot2\cdot4\cdot2 = 3072$,

Reference

Question 9
$1^{\infty}=?$
A
1
B
$\infty$
C
$\pi^\pi$
D
Indeterminate
Question 9 Explanation: 
$1^\infty$ can be roughly rewritten as: $1^{\frac 10}=\sqrt[0]{1}$ Now just think to the zeroth root of 1: every number raised to 0 is one so the zeroth root of 1 could be every number! This is why $1^\infty$ is an indeterminate form.
There are 9 questions to complete.
Please report wrong answer

Leave a Reply

Your email address will not be published. Required fields are marked *