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TMUA Preparation Thread 2017

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Original post by zeldor711
You don't have to release your results to unis and none of them require TMUA (just a bonus to have). STEP is the real monster :smile:

(I think 6.5 will be about 28/40 marks this year)


Good point, thanks for the reassurance. Maybe I did better than I think, guess I'll just have to wait to find out.

28/40 seems a bit high in my opinion, I'm thinking (well hoping really) it'll be a couple marks less because I believe 28 was a 9.0 last year. Then again, we did have longer and it was easier so you could definitely be right.
Original post by Eldronyx
Good point, thanks for the reassurance. Maybe I did better than I think, guess I'll just have to wait to find out.

28/40 seems a bit high in my opinion, I'm thinking (well hoping really) it'll be a couple marks less because I believe 28 was a 9.0 last year. Then again, we did have longer and it was easier so you could definitely be right.


Yeah, 28/40 was pretty arbitrary - just figured it because this year we've got another years worth of past papers, more time and I found it easier. Perhaps a more realistic estimate would be 26/40.
Original post by zeldor711
You don't have to release your results to unis and none of them require TMUA (just a bonus to have). STEP is the real monster :smile:

(I think 6.5 will be about 28/40 marks this year)


I suspect it will be less. I think grade 9.0 will jump from 28/40 to approx 35/40 and 1.0 will stay at 6/40 - the same as last year. So using these approximations I imagine it will be 25 or 26/40
Does 6.5 equate to receiving a reduced offer from Warwick/Durham btw? Or is it just a loose guideline
Original post by Doonesbury
To remind you - no discusion of specific questions please, until tomorrow.


Are we able to discuss now?
Reply 85
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Doones
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Original post by Integer123
Are we able to discuss now?


AFAIK yes :smile:
Original post by marsbar384
Does 6.5 equate to receiving a reduced offer from Warwick/Durham btw? Or is it just a loose guideline


Yeah 6.5 gets you a reduced offer from Warwick and Durham.
Original post by Doonesbury
AFAIK yes :smile:


Great, thanks :smile:
Unofficial mark scheme anyone? Or just any answers ppl remember
These are some of the answers that I got, and I've put my reasoning if I can remember it/the question. I've put the question numbers I remember in order, and then those I don't with a ? at the end. I could have got some of these answers wrong, so please feel free to correct me :smile:

Paper 1
Currently 13/20 questions

Question 1 (finding equation of curve)
Answer of: y=x3+x−2−3x−1+6y=x^3+x^{-2}-3x^{-1}+6
Method: dydx=3x2−2−3xx3\frac{dy}{dx}=3x^2-\frac{2-3x}{x^3}. Integrate and find constant cc since y(1)=5y(1)=5.

Question 2 (second derivative)
Answer of: f′′(1)=5f''(1)=5
Method: differentiate

Question 3 (area enclosed by lines)
Answer of: 161516\frac{1}{5}
Method: The line y=6−2xy=6-2x has the points (3, 0) and (0, 6) on it. The perpendicular line has gradient of 1/2 (I can't remember which lines it passes through). The area enclosed is 81/5=161581/5 =16\frac{1}{5}.

Question 5 (sequence (xn)(x_n))
Answer of: -5
Method: The sequence is periodic.

Question 6 (tangent to circle)
Answer of: 15 (I got 12 using this method, oops)
Method: Sketch the circle x2+y2=144x^2+y^2=144, i.e. with radius 12 and centre (0,0). Then since the tangent is perpendicular to the radius, we can construct a right-angled triangle composed of two smaller right-angled triangles. Letting y be the desired value, and use t so that t+16 is the length of the tangent (assuming it is a line segment), we get 202+y2=(t+16)220^2+y^2=(t+16)^2 and 122+t2=y212^2+t^2=y^2. Solving these gives y=15y=15.

Question 7 (arithmetic/geometric sequences)
Answer of: 95/8
Method: use the sequences

Question 19 (quadratic inequality)
Answer of: qc<x<pcqc<x<pc
Method: The quadratic formula gives the roots of the new quadratic to be those of the previous quadratic multiplied by c. Since c<0, we swap the inequality around.

Question 20 (angle inequality)
Answer of: 60<θ<12060<\theta<120
Method: use the cosine rule

Question ? (hexagon inside circle)
Answer of: 36336\sqrt{3}.
Method: Completing the square gives the circle a radius of 434\sqrt{3} (the centre is irrelevant). We may split the hexagon up into six isosceles triangles with angles π/3\pi/3 and sides 434\sqrt{3} arranged in a manner for the area of each to be Δ=12(43)2×sin⁡π3=63\Delta = \frac{1}{2}(4\sqrt{3})^2\times \sin\frac{\pi}{3}=6\sqrt{3}. Hence the hexagon has area 36336\sqrt{3}.

Question ? (binomial expansion, find a+b)
Answer of: 5
Method: Use the binomial expansion formula and find a+b

Question ? (area A)
Answer of: ∫02[f(x+2)+1]  dx=A+2\int^{2}_0\left[f(x+2)+1\right]\;\mathrm{d}x=A+2.
Method: We have A=∫24f(x)  dxA=\int^{4}_2 f(x)\;\mathrm{d}x and f(x+2) is a translation 2 units in the negative x-direction. Hence the area between 0 and 2 of this curve is A and the result follows.

Question ? (trig inequality)
Answer of: 0⩽x⩽π60\leqslant x \leqslant \frac{\pi}{6}, π2⩽x⩽5π6\frac{\pi}{2} \leqslant x \leqslant \frac{5\pi}{6}.
Method: For (1−2sin⁡x)cos⁡x⩾0(1-2\sin x)\cos x \geqslant 0 we require both factors to be positive or negative. Sketching the graphs reveals the answer.

Question ? (trapezium rule)
Answer of: B (I think it was under-under-over)
Method: Consider the curves and their shapes.

Question ? (greatest value of M)
Answer of: 2/3
Method: differentiate and complete the square. Since we are taking the reciprocal, we have a maximum

Paper 2
Currently 13/20 questions

Question 1 (differentiation)
Answer of: ?
Method: differentiate

Question 2 (area of PQRS)
Answer of: 40
Method: Sketch the rectangle and find the area using Pythagoras' theorem to help with lengths.

Question 16 (counterexample to even number function)
Answer of: C
Method: A and B did not satisfy the if part and so could not be counterexamples, and D was an actual example.

Question 17 (stapled set)
Answer of: F
Method: Use the definition of a stapled set.

Question 18 (solution to equation with logarithms)
Answer of: Incorrect method, error on step I
Method: They used the incorrect result that a<b⇒log⁡1/2a<log⁡1/2ba<b \Rightarrow \log_{1/2}a<\log_{1/2}b since log⁡1/2x\log_{1/2}x is a decreasing function.

Question 19 (sufficient condition for one real root)
Answer of: ∣a∣>4|a|>4
Method: We want to consider a sufficient condition on a such that x3−3x2+a=0x^3-3x^2+a=0 has exactly one real root. First sketching y=x3−3x2y=x^3-3x^2 gives a repeated root at the origin and a minimum point (differentiating) at (2,-4). Considering translations in the vector (0, a), the necessary condition on a is that a<0 or a>4. Hence ∣a∣>4|a|>4 is sufficient.

Question 20 (passwords)
Answer of: three attempts.
Method: I can't remember the exact question, but I got that c was always in position 4, and that positions 2,3 were occupied by b and d, while 1,5 were occupied by a and e (I might have got the letters wrong but they are unimportant). We have in total up to 2*2=4 possible passwords to try. We are looking for the maximum number of trials before a deduction may be made. Trying out the pairs (a,e) and (b,d) in either order with the c in its position, we can assume this is wrong. We then swap one of the pairs around, and then assume that this password is wrong. Hence, we have tried both the correct and incorrect ordering of this pair and obtained an incorrect password, that is the other pair is in the wrong ordering, and we must swap it around. Finally, we trial this correct pair with the previous pair we were referring to in either order to deduce the password on the third attempt.

Question ? (incorrect trig inequality)
Answer of: sin⁡2x<cos⁡2x<tan⁡x\sin 2x<\cos 2x<\tan x
Method: Sketch y=tan⁡xy=\tan x on the sketch provided and test which are correct to find the incorrect one.

Question ? (sufficient condition for area of zero)
Answer of: f(2+x)=−f(2−x)f(2+x)=-f(2-x)
Method: ∫13f(x)  dx=0\int^{3}_1 f(x)\;\mathrm{d}x=0 is that the 'area' (effectively) to be zero. If the curve is odd about x=2, then this integral will be zero. This means that f(4−x)=−f(x)f(4-x)=-f(x). Replace x with 2-x so that f(2+x)=−f(2−x)f(2+x)=-f(2-x).
Alternatively, note that ∫13f(x)  dx=0\int^{3}_1f(x)\;\mathrm{d}x=0 so that ∫−11f(x+2)  dx=0\int^{1}_{-1}f(x+2)\;\mathrm{d}x=0 (translation argument). So we have that f(x+2)f(x+2) is an odd function as sufficient condition, i.e. f(x+2)=−f(−x+2)f(x+2)=-f(-x+2).

Question ? (graph transformation)
Answer of: Whichever graph was translated to the right and down, I believe it was (F)
Method: The given sketch is y=x2−2bx+c=(x−b)2+c−b2y=x^2-2bx+c=(x-b)^2+c-b^2. We are considering y=x2−2Bx+c=(x−B)2+c−B2y=x^2-2Bx+c=(x-B)^2+c-B^2. So since B>b>0B>b>0 (b>0 by the first sketch), we have B2>b2B^2>b^2 too and we are looking for the graph translated to the right and down.

Question ? (functions)
Answer of: f(1000)−g(1000)=2f(1000)-g(1000)=2.
Method: We have f(x+1)={3f(x)+1,      f(x) is odd,12f(x),      f(x) is even\displaystyle f(x+1)=\begin{cases}3f(x)+1,\;\;\;f(x)\text{ is odd},\\\frac{1}{2}f(x),\;\;\;f(x)\text{ is even}\end{cases}, and f(1)=5. This gives us that for x=1,2,3,4,5,6, f(x)=5,16,8,4,2,1. But then note that f(7)=4 and we will continually oscillate 4,2,1... and so on so that f(7+3n)=4 and f(1000)=4. Using a similar method with g(x), we get g(1000)=2 for the desired result.

Question ? (smallest value)
Answer of: (2sin⁡π4)2\left(2\sin\frac{\pi}{4}\right)^2
Method: This is equal to 2, and all the others were larger than 2.

Question ? (91 = 7*13 counterexample)
Answer of: ?
Method: ?
(edited 6 years ago)
Nice - worked a lot of these out and made educated guesses to the others - thanks this has made me feel a lot better about getting my score!
Anyone any predictions on what a 4.5 will be to get a reduced offer from Lancaster?
Original post by Integer123
x


I wish I remembered all my answers xD. For instance, I have no idea if I put 2 or -2 for the functions question in paper 2.
Can anyone remember any other questions? I've got 26 so far
Original post by Integer123


Answer of: (2sin⁡π4)2\left(2\sin\frac{\pi}{4}\right)^2
Method: This is equal to 2, and all the others were larger than 2.: ?


I remember this as asking which is the largest value to which I put log base 2 of 7
(edited 6 years ago)
Original post by marsbar384
One question was which number is larger. I remember I put log base 2 of 7 for that one. I *think* it was an earlier q for paper 2


Oh, I thought it was find the smallest value so I put the (2sin(pi/4))^2...?
Original post by Integer123
Oh, I thought it was find the smallest value so I put the (2sin(pi/4))^2...?


Was it? Ah maybe. Can anyone else remember?
(edited 6 years ago)
Original post by marsbar384
Was it? Ah maybe. Can anyone else remember?


Haha I always get these ones the wrong way round! I'll put on both solutions until someone can confirm it :smile:
Original post by Integer123
Haha I always get these ones the wrong way round! I'll put on both solutions until someone can confirm it :smile:


It was definitely find the smallest - I had time to thoroughly check my answer to this one!!
Original post by Roquebrune2014
It was definitely find the smallest - I had time to thoroughly check my answer to this one!!


Ok, and did you get (2sin(pi/4))^2? :smile:

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