# Happy e day!

This Quiz was uploaded on 7 February celebrating e day.

Only $7$ responses were recorded (Unfortunately). Question in order were:

Q1. The number e is also called as __(Euclid's number/ Euler's number/ Euler-Mascheroni number/ Archimedes number/ None of the above)

Q2. e is __ (an Irrational Number/ a Rational number/ a non real-complex number/ None of the above)

Q3. There exists non zero polynomial $p(x)$ with rational coefficients such that p(e)=0 (True/ False)

Q4. Note : ⌊ ⋅ ⌋ denotes the floor function. $\lfloor e\rfloor=?$

$Q5. \lim_{x\to\infty}\frac{1}{e}\times(1+x^{-1})^x=?$

$Q6. \sum_{n=0}^\infty\frac{(-1)^n}{n!}=?$

Q7. If $e^x=-1$, then $x=?$

$Q8.\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=?$

$Q9. \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}+ i\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!},\space i^2=-1$

Q10. $f(x)=y$ is a straight line which is tangent to $y=e^x$ at $x=0.5$, find $f(x)$

$Q11. \forall n\in\mathbb{W}\frac{b_{n+1}}{b_n}=\frac{1}{a}-n,b_0=1$ $\lim_{a\to 0}\sum_{n=0}^\infty\frac{b_na^n}{n!}=?$

Q12. Note : $x(t):\mathbb{R}\to\mathbb{R}$ $a\frac{d^2x}{dt^2}+cx+d=0$ Find $x(t)$ if it satisfies above differential equation.

Attempts

Note by Zakir Husain
4 months, 2 weeks ago

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## Comments

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- 4 months, 1 week ago

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You may post your answers without the solutions below this comment. I will mark it.

- 4 months, 2 weeks ago

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- 4 months, 1 week ago

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Interesting !

- 4 months, 1 week ago

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I shall make a note with all the solutions :)

- 4 months, 2 weeks ago

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I did it:

$\frac{8}{27}$

- 4 months, 2 weeks ago

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I don’t quite understand Q12 though... :(

- 4 months, 1 week ago

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Substitute $z:\mathbb{C}\to\mathbb{C}$ and $z(t)=Ae^{i(\alpha t +\beta)}+\gamma$ $\therefore \frac{d^2z}{dz^2}=-\alpha^2 Ae^{i(\alpha t +\beta)}$ Putting this in the equation $-a\alpha^2 Ae^{i(\alpha t +\beta)}+cAe^{i(\alpha t +\beta)}+c\gamma=-d$ $\Rightarrow Ae^{i(\alpha t +\beta)}(-a\alpha^2+c)=-(d+c\gamma)$ Because this must be true for all $t\in\mathbb{R}\therefore -(d+c\gamma)=0$ $\Rightarrow \boxed{\gamma=\frac{-d}{c}}$ $\Rightarrow Ae^{i(\alpha t +\beta)}(-a\alpha^2+c)=0$ Now either $A=0$ or $-a\alpha^2+c=0$

Because $z$ is a non zero function. $\therefore A\cancel{=}0$ $\Rightarrow -a\alpha^2+c=0$ $\Rightarrow \boxed{\alpha=\sqrt{\frac{c}{a}}}$

Now let $x(t)=$ real part of $z(t)$ $\Rightarrow x(t)=A\cos(\alpha t +\beta)+\gamma$

- 4 months, 1 week ago

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Oh I understand now! Thanks :)

- 4 months, 1 week ago

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