Luukas.2's latest activity

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    Luukas.2 replied to the thread 4u integration.
    They were standard integrals on the old syllabus, but that's going back a way - see page 16 of the 2001 paper accessible from...
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    Luukas.2 replied to the thread ionisation energies q.
    Since element B has the highest first IE amongst the four consecutive elements, it must be a noble gas. Thus: element A is a halogen...
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    You can also find the boundary values by solving \begin{align*} |3x - 2| &= |x + 4| \\ \\ \textbf{Case 1:} \qquad 3x - 2 = x + 4 \qquad...
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    Luukas.2 replied to the thread Rate of Change Question.
    Define a height axis, y, along the central axis of the cylinder, taking y = 0 at the top and placing the light source at y = h above the...
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    Right-angle triangle trigonometry should allow you to show that \text{LHS} = \sin{\frac{\theta}{2}} + \frac{1}{2}\cos{\frac{\theta}{2}}...
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    The cover is a modified version of that used by the Independent Schools papers, as is the table on page 2 for MCQ papers... but it's not...
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    Luukas.2 replied to the thread Writing vs Typing notes.
    Some excellent points. From a cognition perspective, there is a significant difference depending on how you are with writing on a...
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    Luukas.2 replied to the thread ???.
    I think the observation implicit in these questions / comments is important... why is this worth 4 marks? Is it just a mistake, or an...
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    Luukas.2 replied to the thread Integration.
    The problem can be approached as an area between two curves, just relative to the y-axis. The curves that bound the region are x=e^{2 -...
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    \text{Since} \qquad \tan^{-1}{x} + \tan^{-1}{\left(\frac{1}{x}\right)} = \frac{\pi}{2} \qquad \text{for all $x > 0$} \text{And} \qquad...
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    Luukas.2 replied to the thread Integration.
    Since the area is symmetric about the x-axis, its area can be found by doubling the integral of the function above the x-axis...
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    Luukas.2 replied to the thread is the answer wrong.
    Definitely positively certainly B and not D.
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    Question 1: \text{$y = \sin^{-1}{x}$ takes ratios between $-1 \le x \le 1$ and outputs angles between $-\frac{\pi}{2} \le y \le...
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    Luukas.2 replied to the thread perms and combs question.
    Sure... \begin{align*} \text{Going from Line 1:} \qquad \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ \text{to Line...
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    Luukas.2 replied to the thread perms and combs question.
    Your algebra is correct, you got to \begin{align*} \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ 4!(n - 5)(n - 4) -...
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