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I have received a question about q22(a) by DM, so am posting about it here in case others want to know. An indicator for an acid-base titration will itself be a weak acid or base, so take it as having HInd as its acidic form. It is therefore present in solution as an equilibrium: HInd (aq) +...

With this graph of \Re{(z)} \geqslant \text{Arg}(z), I wonder how many sketches noted that there should be an open circle at the origin and a closed circle at z=-\pi.

Yes and No. The velocity at impact with the ground is indeed -10\sqrt{69}\ \text{m s}^{-1} or 10\sqrt{69}\ \text{m s}^{-1} in a downwards direction. The particle is not slowing down, however, it is speeding up as it is accelerating in the negative direction and moving in the negative...

I feel old :(

Oops, I'm wrong... I read their working for the time as t = \cfrac{-70 \pm \sqrt{70^2 - 4(-5) \times 400}}{2(-5)} = \cfrac{-70 \pm 10\sqrt{129}}{-10} = 7 \mp \sqrt{129} but I have misread and it is actually t = \cfrac{-70 \pm \sqrt{70^2 - 4(-5) \times 100}}{2(-5)} = \cfrac{-70 \pm...

Is it just mean, or is the arithmetic in 26(b) wrong? isn't it -10\sqrt{129}?

Q25 is the exact sort of mistake that everyone should be checking for after each answer by asking yourself "is my answer reasonable?" As several people have noted, turning $8,000 into over $60,000 in 8 years at less than 1% interest pa = unreasonable. Just like the normal distribution / number...

I posted a version of that in the discussion in the MX2 forum... I wonder how markers will evaluate anything like it?

One will be released once the marking is complete and marks are announced. In the meantime, if you have specific questions, feel free to ask.

\begin{align*} P(\text{IQ}\ >\ \text{130}\ |\ \text{solve time} <\ \text{2 h}) &= \cfrac{P(\text{IQ}\ >\ \text{130 AND solve time}\ <\ \text{2 h})}{P(\text{solve time}\ <\ \text{2 h})} \\ &= \cfrac{0.025 \times 0.8}{\log_{10}{2}} \approx 0.066 \end{align*} ...using the z-score of +2 (above the...

It's D. It can't be A or B as the particle will not continue along the path AB. It isn't C as the acceleration will result in a decreasing speed, whereas D has an increasing speed, as required.

As did I, but I'd like to check it again

The answer given for q33(d) is incorrect, it is a conditional probability problem as the question asks for P(IQ > 130 given can solve the problem in under 2 h)

Actually, proving the more general result that, for positive integers a and b, and for all integers n > 1 a^n + b^n \neq (a + b)^n might have been easier.

The answer posted for Q23 is wrong, too. The derivative of P = 5000b^{-\frac{t}{10}} is \cfrac{dP}{dt} = -500b^{-\frac{t}{10}} (as the solutions state) only if b = e. The derivative is actually \cfrac{dP}{dt} = -500b^{-\frac{t}{10}}\ln{b}.

Yes, that's my impression, though it strikes me more as challenging than unreasonable.

The answer to q10 in this is wrong. It is B. When x=a, the line is at (a,\ am) and the y-coordinate must be between 0 and 1, so we have that am < 1\ \implies\ m < \cfrac{1}{a}.

You are correct

I wonder what they would make of a proof that invoked Fermat's Last Theorem and noted that this was proven by Andrew Wiles a few years ago, pointing at that if there are no solutions in positive integers a, b, and c for the statement a^n + b^n = c^n for any integer n > 2, and then pointing out...

Can't you just prove this by contradiction? \begin{align*} \text{Suppose that the theorem is false, and so that}\ 2^n + 3^n &= 5^n \\ 5^n - 2^n - 3^n &= 0 \\ \text{However, LHS}\ &= (2 + 3)^n - 2^n - 3^n \\ &= \binom{n}{0} 2^n + \binom{n}{1}2^{n-1} \times 3 + \binom{n}{2}2^{n-1} \times 3^2 +...