Engineering mathematics set 2

By | November 21, 2018

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Engineering mathematics questions answers

Question 1 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$\int_0^\infty \frac{\sin x} x \, dx =?$
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=?$
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} =?$
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx =?$
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$\int_0^1\ln\left(1+\frac{\ln^2x}{4\, \pi^2}\right)\frac{\ln(1-x)}x dx=?$
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
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 (Click on any option to know correct answer. सही उत्तर जानने के लिए किसी भी choice पर क्लिक करें।)
$1^{\infty}=?$
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.

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