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43

We can define $x \oplus y=y$. Then $(x \oplus y) \oplus z =z= x \oplus (y \oplus z)$ but $y=x \oplus y \neq y \oplus x=x$


13

What you are looking for is the indicator function: $$ \mathbf{1}_{A}(x)=\begin{cases} 1 & \text{if }x\in A\\ 0 & \text{otherwise} \end{cases} $$ In your case, $A=(0,1]$ (I have used $x$ instead of $i$ and $A$ instead of $j$ since it is more customary to use lower case letters at the end of the alphabet for real numbers and upper case letters for ...


11

If you already know that matrix multiplication is associative, but not commutative, then you can just choose your favorite bijection $f:\mathbb R\rightarrow M_2(\mathbb R)$, since the two sets are equinumerous. Then, define $a\oplus b = f^{-1}(f(a)f(b))$ to get an operation on $\mathbb R$ which is associative, but not commutative. If you want to have ...


9

I think I have an answer. Look at $\frac{1}{3}$ in binary format and you will notice that it is $0.\overline{01}$. This tells us something, namely that $[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to: $$\sum \limits_{i=0}^{50} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]=\sum \limits_{i=0}^{49} [2^{2i}/3]+\sum \...


8

There are $\binom{7}{3}$ ways to choose the pre-image of $\{3\}$. Given that choice, you still need to decide where the other four elements of $\{1,2,3,4,5,6,7\}$ map. For each one, there are four choices of where it maps ($0,1,2,$ or $4$), so there are $4^4$ choices for where those four elements not in the pre-image of $\{3\}$ map. Thus there are $\binom{7}{...


7

Hint: $A\sin(x)+B\cos(x)=c\sin(x+x_0)$. In order to derive the new coefficients: $c\sin(x+x_0)=c\sin(x)\cos(x_0)+c\sin(x_0)\cos(x)=A\sin(x)+B\cos(x)$ Compare coefficients: $A=c\cos(x_0)$ and $B=c\sin(x_0)$. Divide the equations. $B/A=\tan(x_0)$ and square and add both equations $A^2+B^2=c^2$. Hence, $c=\sqrt{A^2+B^2}=\sqrt{5}$ and $x_0=\arctan(B/A)=\...


5

Here's a really neat geometric approach: Consider that $(x_1, x_2) = (\cos(x), \sin(x))$ traces a circle once, and that $2x_1+x_2=\sqrt{3}$ is a linear equation. The answers correspond to the intersections of a line with a circle. Since the line passes through the interior of the circle, but not through $(1, 0)$ (which is mapped twice by the parametric ...


5

The sequence of $[2^n/3]$ for $n\in \mathbb{N}$ and $n\geq 2$ would be like this: $$1,\ 2,\ 5,\ 10,\ 21,\ 42,\ 85,\ \cdots $$ So, comparing the numbers in the sequence successively, you may guess that it is a recursive sequence like $a_1=1$, and for $n>1$, we have $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \...


4

first line in i) is confusing since if $C\cap D=\emptyset$ then $x \in f^c(C \cap D)$ does not even exist. I would write something like this: assume ac. $f^c(C)\cap f^c(D)\neq \emptyset$ (so the NEGATION of what we are trying to prove) then $\exists_{x \in X} x \in f^c(C)\cap f^c(D)$ $\exists_{x \in X} x \in f^c(C) \wedge x \in f^c(D)$ $\exists_{f(x) \...


4

$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1=x(x-1)(x^2+x+1)(x^8+1)+1$$ We can see that this is positive for $x\ge 1$ or $x\le 0$. For $0\lt x\lt 1$, $$x^{12}-x^9+x^4-x+1=x^{12}+(1-x)+x^4(1-x^5)$$ is positive since $1-x^5\gt 0$. Therefore, $x^{12}-x^9+x^4-x+1$ is positive.


4

This function does not have an inverse as it is not injective: check that $f(x-2) = f(-x-2)$ whatever $x$ is - for example, $f(1) = f(-5)$, etc. We want to solve $y=2x^2+8x+13$ for $x$. Applying the quadratic formula for $2x^2+8x+(13-y)=0$ yields $$x = \frac{-8\pm \sqrt{64-4\cdot 2 \cdot (13-y)}}{4} = -2\pm \frac{\sqrt{16-26+2y}}{2} = -2\pm\frac{\sqrt{-10+...


4

This one is pretty straightforward. Intuitively, all we have to do is pick a function that maps set $A$ to the image of your initial fuction, and a function that "extends" the image of $f$ to set $B$, namely the identity function. Here, the functions are named "$map$" and "$ext$", respectively and are defined as follows: $map:A\rightarrow Imf\subseteq B, ...


4

should be $$\left( \log _{ 3 }{ x } -1 \right) \left( \log _{ 3 }{ x } -2 \right) =0$$ $$\log _{ 3 }{ x } =1\Rightarrow \quad x=3\\ \log _{ 3 }{ x } =2\Rightarrow x=9$$


4

$\sin x=0$ is rejected because when $\sin x=0$, the value $\csc x$ does not exist, so $(\sin^2 x)(1+\csc x)$ does not exist, so it can't be equal to $0$ or to anything else either.


3

Take the function $x^{12}-x^9+x^4-x=x(x^3-1)(x^8+1)$ This has 2 real roots at $0$ and at $1$ and increases on both sides . So the minima is in between them. Find out the minima and check the function's value, ..it will come out to be $>-1$. So adding $+1$ to the entire function pushes up the lowest point above the $X$ axis ,resulting in no real roots..


3

Your argument for (1) is correct. For (2), how many divisors does $2^n$ have?


3

You can't find the inverse of a degree 2 polynomial defined on $R$ since it is not injective. If you want an inverse on invervals, write $2x^2+8x+13=y$, $2x^2+8x+13-y=0$, $\Delta = 64-8(13-y)$ and $x={{-8+\sqrt{8(13-y)}}\over 4}$ or $x={{-8-\sqrt{8(13-y)}}\over 4}$


3

Let $\sqrt{1+2x^2}=u\ge1,\implies1+x^2=\dfrac{1+u^2}2$ $$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2u}{1+u^2}=\dfrac2{u+\dfrac1u}$$ Now $u+\dfrac1u\ge2\sqrt{u\cdot\dfrac1u}=2$ Alternatively, let $\sqrt{1+2x^2}=\tan v$ Clearly, $\tan v\ge1+2\cdot0=1$ WLOG we can choose $\dfrac\pi4\le v<\dfrac\pi2\iff\dfrac\pi2\le2v<\pi$ Now $$\dfrac{\sqrt{1+2x^2}}{1+x^...


3

\begin{align} & \frac{\sqrt{1+2{{x}^{2}}}}{1+{{x}^{2}}}>0 \\ & f'(x)=\frac{-2{{x}^{3}}}{{{(1+{{x}^{2}})}^{2}}\sqrt{1+2{{x}^{2}}}} \\ & f(0)=1 \\ \end{align} $$\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{\sqrt{1+2{{x}^{2}}}}{1+{{x}^{2}}}=0$$ $$R_f=(0,1]$$


2

For the limit to exist we need \begin{align*} &\lim_{x\rightarrow 0} \ln \Delta(0) = 0\\ \implies &\lim_{x\rightarrow 0} \Delta(0) = 1\\ \implies &\lim_{x\rightarrow 0} f(x)/x^3+1 = 1\\ \implies &\lim_{x\rightarrow 0} f(x)/x^3 = 0. \end{align*} Since $f(x)$ is continuous this is only possible if $f(0) = 0$. By L'hopital you also get $f'(0) =...


2

\begin{align*} \int : \mathcal R([a,b],\mathbb R)&\longrightarrow \mathbb R\\ f&\longmapsto \int_a^b f \end{align*} where $\mathcal R([a,b],\mathbb R)=\{f:[a,b]\longrightarrow \mathbb R\mid f\text{ is Riemann integrable}\}$


2

Since $y^2=9-x^2$ on the set $S=\{(x,y):x^2+y^2=9\}$, your function is $$f(x,y)=4x^2+3(9-x^2)-5x=x^2-5x+27,\quad\text{on $S$}.$$ Now you just have to find the maximum and minimum of the quadratic $$x^2-5x+27,$$ for $-3\leq x\leq3$. (We have to restrict to $-3\leq x\leq3$ since otherwise there are no $y$ such that $(x,y)\in S$.)


2

No, the convention is that the Kronecker delta tests for equality and not for some other relation such as set membership. Also the elements tested for equality are most often numbers, and even integers, but in a stretch you could use it to test equality of objects of other types. So using the Kronecker delta with one index a number and the other an interval ...


2

The procedure is the same as always when you have to find an inverse: Set $u = \frac{x}{x^2+y^2}, \quad v = \frac{y}{x^2+y^2}$ and solve for $x,y$. You can easily see immediately that $u^2+v^2 = \frac{x^2+y^2}{(x^2+y^2)^2} = \frac{1}{x^2+y^2}$ and therefore $x^2+y^2 = \frac{1}{u^2+v^2}$. From the equations above we then get $$x = u\cdot (x^2+y^2) = \frac{...


2

You have $p+7=\sqrt{\frac{4000}{Q}}$. Increasing $Q$ by 1 decreases the price that can be charged (on all sales) by $\Delta p$, where $p-\Delta p+7=\sqrt{\frac{4000}{Q+1}}$. So we have $\Delta p=\sqrt\frac{4000}{Q}-\sqrt{\frac{4000}{Q+1}}=\sqrt{\frac{4000}{Q}}\left(1-\sqrt{\frac{Q}{Q+1}}\right)$ $=(p+7)\left(1-\sqrt{\frac{1}{1+\frac{1}{Q}}}\right)$. We now ...


2

I think you meant to write $x$ in terms of $y$? $$y=2x^2+8x+13$$$$y-13=2x^2+8x$$$$\frac{y-13}{2}=x^2+4x=(x+2)^2-4$$$$\frac{y-13}{2}+4=(x+2)^2$$$$\pm\sqrt{\frac{y-13}{2}+4}=x+2$$$$x=\pm\sqrt{\frac{y-13}{2}+4}-2$$


2

If $x$ is a real number, then all three quantities must be positive and intersection of all three sets will be final value of $x$ i.e $2x^2-1>0$ & $\sin(x)>0$. So $x\in[\frac{1}{\sqrt{2}},\frac{\pi}{2}] \cup [2n\pi,2n\pi+\frac{\pi}{2}]$ and same for the negative values of $x$. Hope this is helpful


2

You indeed get a fourth degree equation, but $x$ only appears with even exponent: $$ y=\frac{\sqrt{1+2x^2}}{1+x^2} $$ means that $y>0$ and that $$ (1+x^2)^2y^2=1+2x^2 $$ Expanding and reordering gives $$ y^2x^4+2(y^2-1)x^2+y^2-1=0 $$ and the usual quadratic formula provides the value for $x^2$; it's common to advise setting $z=x^2$ and solving $y^2z^2+2(y^...


2

Starting from $n=0$, even-$n =2k$ terms are $(4^k - 1)/3$ and the subsequent odd-$n = 2k+1$ terms are $2\times (4^k- 1)/3$. So overall: \begin{align} \sum_{k=0}^{49}\left[(1+2)\frac{4^{k} - 1}{3}\right] + \frac{4^{50}-1}{3} &= \sum_{k=0}^{49}(4^{k} - 1) + \frac{4^{50}-1}{3} \\ & = \sum_{k=0}^{49}(4^{k}) + \frac{4^{50}-1}{3}-50 \\ &= \frac{4^{50}-...


2

Graph it using Desmos. $$\tan\left(\frac{\pi}{2}\text{floor}(x)\right)$$ Now, it's easy to see that this is $0$ when $\text{floor}(x)$ is even and undefined when $\text{floor}(x)$ is undefined, so we have a period of $2$.



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