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$\noindent The function $\sin^{-1} x + \cos^{-1} x$ is continuous because it is the sum of two continuous functions.$

Re: MATH1131 help thread $\noindent That's just another notation for vectors. The top number refers to the first component, and the bottom number refers to the second component.$

Re: MATH1131 help thread $\noindent $\mathbb{R}$ refers to the set of all real numbers (think real number line). $\mathbb{R}^2$ refers to the set of all ordered pairs of real numbers (think of the Cartesian plane).$

$\noindent It means that the definition of a function being injective is that $f(x) = f(y)$ implies $x=y$. (Don't worry about the $\Leftarrow$ part, that's obvious by definition of any function, not just injective ones. That just says if we sub. in the same value to the function, the output is...

$\noindent For 1, let's assume we have defined the domain and codomain as $\mathbb{R}$. It is injective as it is monotonic, as you proved. It is also surjective, because given any $y$ in the codomain ($\mathbb{R}$), there exists $x\in \mathbb{R}$ such that $f(x)=y$ (since this is a cubic...

$\noindent Another way for part (b) (though I considered intervals differently to them, I find this way easier). Let $g(x) \equiv 1 + x + x^2 + x^3 + x^4$. Clearly $g(x) >0$ for all $x\geq 0$. Now consider $x<0$. Note that we can use the G.P. sum formula to write $g(x) \equiv \frac{x^5...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We can put the equality part on either branch and it'll still be correct, since if we sub. in $x =2$ into either branch's equation, we get the correct $y$-value of $-1$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The answer is $y = |x-2| -1$. To write it as a piecewise function, it'd be $y = \begin{cases} -x+1, \quad x\leq 2 \\ x -3, \quad x>2 \end{cases} .$

$\noindent That's true iff $a$ and $b$ have the same sign (obviously neither can be $0$).$

$\noindent Let $y = x^3 \ln (x+1)$. Then using the product rule, the derivative is$ $$\begin{align*} y^\prime &= 3x^2 \ln (x+1) + x^3\cdot \frac{1}{x+1} \\ &= 3x^2 \ln (x+1) +\frac{x^3}{x+1}.\end{align*}$$

$\noindent We are given the roots are $r$ (real) and $a\pm bi$, where $a$ and $b$ are integers. By the product of roots, we have $r\left(a^2 + b^2\right) = -2702$. Since $b^2 = 3a^2 - 3$ and $r=-2a$, we have$ $$\begin{align*}-2a\left(a^2 + 3a^2 - 3\right) &= -2702 \\ \Longleftrightarrow...

$\noindent (c) Let the roots be $r$ (real) and $a\pm bi$ ($a,b$ real). By sum of roots being 0, we have $2a + r = 0 \Rightarrow r = -2a$. Now the pairwise product of roots is easily found: $r(a+bi) + r(a-bi) + (a+bi)(a-bi) =- 3$. The LHS simplifies to: $LHS= ra + ra + a^2 + b^2 = 2ra + a^2 + b^2...

$\noindent It's a telescoping series (which means that if you write out the terms you'll see most of them cancel). The answer is $\frac{1}{20^2} -\frac{1}{(2-1)^2} = \frac{1}{400} -1 = -\frac{399}{400}$.$

Ah OK, cheers.

Depends on your definition of 'good', as this is a subjective term.

$\noindent Another way to write the domain is $D = \left \{x \in \mathbb{R}: -\frac{7\pi}{6}+2k\pi \leq x \leq \frac{\pi}{6}+2k\pi, \, k\in \mathbb{Z}\right \}$ (just to avoid writing something like $+2k\pi$ next to an interval, which I'm not sure whether or not is a standard notation).$

$\noindent One way to express the domain is simply $D = \left \{ x \in \mathbb{R} : \sin x \leq \frac{1}{2} \right \}$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent I think it's to do with some calculators computing cube roots as logarithms, which are undefined at negative values, so these calculators only end up working with positive values of $x$ for these power functions. For...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We have $S_n = \sum_{k=1}^{n} \frac{k}{2^{k-1}}$, for $n=1,2,3,\ldots$. Hence $2S_n = \sum_{k=1}^{n} \frac{2k}{2^{k-2}}= \sum_{k=1}^{n} \frac{k}{2^{k-2}}$. So we have$ $$\begin{align*}2S_n - S_n &=...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Was it part of a larger question where n had to be positive (e.g. because it was the number of some objects or something)?