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$\noindent Here's a `lazy' way of doing it.$ $\noindent \textbf{Claim.} Let $b\in \mathbb{R}$ be fixed. Then $\sin^2 x \cos^2 b - \cos^2 x \sin^2 b = \sin^2 x - \sin^2 b$ for all real $x$.$ $\noindent \textbf{Proof.} Note that the LHS and RHS (viewed as functions of $x$) are continuous and...

$\noindent Some of these are pretty tedious computationally. I'll provide some hints/starters on the process for now.$ $\noindent 106. (Btw, for this theorem, which is known as the mean value theorem for integrals, $g$ actually needs to be a function of fixed sign, rather than just non-zero.)...

Setting x = 0 and solving for t would give the time when x = 0 (of course), not in general the time when the ball started. (The time of starting is just t = 0 of course. In this Q., the ball starts at x = 2e.)

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The identity is equivalent to $\sum _{k=1}^{n-2} \binom{n-1}{k}=2^{n-1}-2$ (just dividing the given identity to-prove through by $n$). Now, from this,$ $$\begin{align*} LHS &= \sum _{k=0}^{n-1}\binom{n-1}{k}-...

I'll denote the integral of f over the interval by I. By the hint, we have m <= f(t) <= M for all t in [a,b]. Integrating the inequality from a to b, m(b - a) <= I <= M(b-a). Therefore, m <= (1/(b-a))*I <= M (dividing through by (b-a)). So (1/(b-a))*I is between m and M, so by the...

Re: HSC 2016 Complex Numbers Marathon Yes. Luckily it's pretty quick to prove. Angle OU1W is 90 degrees so by converse of 'angle in a semicircle', OW is the diameter.

$\noindent I am assuming it is $\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$. $\Big{(}$If writing without \LaTeX, it is best to put brackets around the numerators and denominators, i.e. $\left(\sin \theta \right)/\left(1+\cos \theta\right)$, etc.$\Big{)}$.$ $\noindent...

$\noindent I think you proved the results for small $n$. Assume the result holds for some particular odd integer $(n-1)$ and even integer $n \geq 2$. We will show this implies the result for $n+1$ (odd) and $n+2$ (even).$

$\noindent Here is one way to do it.$ $\noindent Since $P$ is an odd polynomial of degree 3, it only contains odd-power terms (you may like to prove this as an exercise, it is true for general degreed polynomials too, apart from the trivial case of the zero polynomial). So $P(x) = ax^3 + bx$...

Re: MX2 2016 Integration Marathon Mental substitution of u = 2x.

We do go outside them. I was just saying I think it'd have been a better question if they made it bound by the y-values too.

We don't need a calculator. $\noindent Define $f:\left[a,b\right] \to \mathbb{R}$, where $a=23,b=27$, and $f(t) = t^{\frac{1}{3}}$ for $t\in \left[a,b\right]$. Then $f$ is continuous on the closed interval $\left[a,b\right]$ and differentiable on the corresponding open interval, so by the Mean...

Well in fact the only commutativity used was P I = I P. If I is treated as the identity matrix (standard practice) and P some compatible matrix, then this would have been valid. :p (Well I guess I also used commutativity for the Ohm's Law bit.)

No, because the former has no mention of y. It is simply an inequation in x, which has a solution: -2 ≤ x ≤ 2. If we wanted to sketch this on the number line, we'd sketch -2 ≤ x ≤ 2 on the number line, i.e. we'd shade in the part of the number line between -2 and 2 and include the endpoints. But...

It doesn't mean that. I was just saying that would have made a better Q. If that was what was being meant, it would have been written as -√(4-x^2) ≤ y ≤ -|x|+2 .

= R I P I P = V P I P (from Ohm's Law, V = IR) = V I P P (by commutativity of multiplication) = VIP2 = Very Important Person2.

$\noindent Hint for the first one: we can use $f(t) = t^{\frac{1}{3}}$, defined for $t\in \left[a,b\right]$, where $a=23$ and $b=27$.$ $\noindent Hint for the second one: an arctan function would be used, with the domain having $1$ being involved of course, as well as $\frac{9}{8}$.$

$\noindent Well note that $\sqrt{4-x^2}$ is only real for $x$ between $-2$ and $2$ (otherwise the thing under the square root (called the radicand) is negative). So we should only consider $x$ in the range $-2\leq x \leq 2$. Now it is actually clear that for any $x$ in this interval, $-|x|+2...

$\noindent Intensity $I$ (power per unit area) is proportional to the the Power of the light $P$ and proportional to the inverse square of the distance to the light source $d$. That is, $I \propto P$ and $I \propto \frac{1}{d^2}$.$ (See: https://en.wikipedia.org/wiki/Inverse-square_law#Example...

A very common phrase on BOS.