STEP 2018 Solutions Watch

123Master321
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Heres 6: Zacken
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\displaystyle \sum_{r=1}^n{2\sin\theta \sin(2r-1)\theta}

\displaystyle =\sum_{r=1}^n\cos(2r-2)\theta-cos(2r \theta)

\displaystyle =\cos(0)-cos(2n \theta)

\displaystyle =1-cos(2n \theta)

i)

\displaystyle A_n = \sum_{r=0}^{n-1}{\sin(\frac{\pi}{n}(r+1/2)) * \frac{\pi}{n}}

\displaystyle 2\sin(\frac{ \pi}{2n})A_n=\frac{\pi}{n} *\sum_{r=0}^{n-1}{2\sin(\frac{(2r+1)\pi}{2n})}\  sin(\frac{\pi}{2n})

\displaystyle 2\sin(\frac{ \pi}{2n})A_n=\frac{\pi}{n} *\sum_{r=0}^{n-1}{\cos(\frac{r\pi}{n})-cos(\frac{(r+1)\pi}{n}})

\displaystyle 2\sin(\frac{ \pi}{2n})A_n=\frac{\pi}{n}(2)

\displaystyle \sin(\frac{ \pi}{2n})A_n=\frac{\pi}{n}

ii)

\displaystyle B_n=\frac{\pi}{2n}(y_0+y_n+2(y_1  +y_2+...+y_{n-1}))

Note \displaystyle y_k = \sin\frac{k\pi}{n}

\displaystyle B_n=\frac{ \pi}{2n}(2\sum_{r=0}^n{\sin( \frac{k \pi}{n}}))

\displaystyle 2\sin(\frac{ \pi}{2n})B_n=\frac{ \pi}{n}(\sum_{r=0}^n{2\sin(\frac  {2k \pi}{2n}})\sin(\frac{ \pi}{2n}))

\displaystyle 2\sin(\frac{\pi}{2n})B_n=\frac{ \pi}{n}\sum_{r=1}^{n-1}{\cos\frac{(2k-1) \pi}{2n}-\cos{\frac{(2k+1) \pi}{2n}}}

\displaystyle 2\sin(\frac{\pi}{2n})B_n=\frac{ \pi}{n}(cos(\frac{ \pi}{n})-cos(\frac{(2n-1) \pi}{2n}))

\displaystyle 2\sin(\frac{\pi}{2n})B_n=\frac{ \pi}{n}(2cos(\frac{ \pi}{2n}))

\displaystyle \sin(\frac{\pi}{2n})B_n=\frac{ \pi}{n}(cos(\frac{ \pi}{2n}))

iii)
\displaystyle 0.5(A_n+B_n)=\frac{\frac{ \pi}{2n}}{sin\frac{ \pi}{2n}}(1+cos\frac{ \pi}{2n})

\displaystyle 0.5(A_n+B_n)=\frac{\frac{ \pi}{2n}}{sin\frac{ \pi}{2n}}(2cos^2\frac{ \pi}{4n})

\displaystyle 0.5(A_n+B_n)=\frac{\frac{ \pi}{2n}}{sin\frac{ \pi}{4n}}(cos\frac{ \pi}{4n})=B_{2n}

iv)

\displaystyle A_nB_{2n}=\frac{\frac{ \pi}{n}}{sin(\frac{ \pi}{2n})}\frac{\frac{ \pi}{2n}}{sin\frac{ \pi}{4n}}(cos\frac{ \pi}{4n})

\displaystyle A_nB_{2n}=(\frac{ \pi}{2n})^2 * \frac{1}{sin^2(\frac{ \pi}{4n})}=A^2_{2n}







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123Master321
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(Original post by Zacken)
Note that we have \displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x} so that we can see that f,f' = 0 when \ln^2 x = 1.

Let u = \log t, then for x \in (0,1) we have

\displaystyle 

\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any x > 0 with x \neq 1, we have F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.
Would I lose marks if I defined my F(x) piecewise so it had ln(lnx)) for x>1 and ln(-lnx) for 0<x<1. Also how many marks would i lose if in my graph i had the curve shaping downards as x--> infinity. It still tends to infinity but the gradient is decreasing.
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DFranklin
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(Original post by gnsbrg)
STEP I Q5

b)
S(0)=2017 implies

abcd=16=2^4

Taking into account  a&lt;b&lt;c&lt;d , we find that there are 4 set of integers (-2,-1,2,4),(-1,1,2,8),(-4,-2,1,2),(-8,-2,-1,1)

This last bit is the one I'm unsure about (btw sorry for the bad latex first time ever)
I'm not following your logic; if each quarter of values is supposed to correspond to a,b,c,d then surely they need to multiply to 16?
But -1 x 1 X 2 X 8 = -16, for example.
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Zacken
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#24
(Original post by 123Master321)
Would I lose marks if I defined my F(x) piecewise so it had ln(lnx)) for x>1 and ln(-lnx) for 0<x<1. Also how many marks would i lose if in my graph i had the curve shaping downards as x--> infinity. It still tends to infinity but the gradient is decreasing.
Piecewise F(x) is fine (I was just too lazy)

you’d lose a mark or two at most for the second slip up
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JeremyC
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Question 7 is an absolute monster of algebra.

Spoiler:
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7(i)
Make the substitution and wade through a page of algebra. The equation transforms into:

(p^3+pq^2 -2p^2 q)z^3 + (q^3 + p^2 q - 2 p q^2) = 0

(Note that p and q are distinct - if they were equal then this would just be 0 = 0)

Superdjman1 kindly pointed out that this nasty expression has a factor of (b-a)^2. Since a and b are distinct we can divide through by this factor to leave us with the much more pleasant expression pz^3+q = 0

7(ii)
This requires solving the simultaneous equations
c = pq
d = pq(p+q)

Solving them leads to a quadratic formula (either cp^2 -dp + c^2 = 0 or the same equation with p replaced by q, depending on which variable you eliminate).

Solving with the quadratic equation, you find that:
p, q = \frac{d \pm \sqrt{d^2-4c^3}}{2c}
The condition d^2 &gt; 4c^3 follows from the requirement that p and q are real and distinct.

7(iii)
We now apply our results from parts (ii) and (i). We see that we need to pick c = -2, d = -2 (noting that the condition in (ii) is satisfied).

Using our solution to the quadratic equation in (ii) we find that p, q = -1, 2 (or the other way round).

Using our result in (i), we find that substituting
x = \frac{-z+2}{z+1}
into the equation leads to -z^3+2 = 0, with solution z = \sqrt[3]{2}. Therefore, our answer is
 x = \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1}

7(iv)
We cannot use the result from (i), as p and q are now equal. Using the remainder theorem we see that (x-p) is a factor of the equation x^3-3p^2x+2p^3=0, and we can then factorise the equation as (x-p)^2(x+2p) = 0, from which the roots immediately follow.

In the case d^2 = 4c^3 we get repeated roots from the quadratic equation we found in (ii). Therefore, p = q = \frac{d}{2c}, and so we can use the first part of (iv) to see that the roots are now x = \frac{d}{2c}, -\frac{d}{c}
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Electric-man7
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(Original post by Zacken)
Note that we have \displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x} so that we can see that f,f' = 0 when \ln^2 x = 1.

Let u = \log t, then for x \in (0,1) we have

\displaystyle 

\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any x &gt; 0 with x \neq 1, we have F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.

How many marks for the graph? I messed it up entirely
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Electric-man7
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(Original post by Notnek)
STEP I Q8 :

(i)

Spoiler:
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To show s^3 + c^3 is constant we can show that the derivative of the expression is 0.

(s^3+c^3)' = 3s^2s' + 3c^2c' = -3c's' + 3s'c' = 0

So we have shown that s^3+c^3 = k where k is constant.

Then plugging in the initial conditions:

s(0) + c(0) = k \Rightarrow 0 + 1 = k \Rightarrow k = 1

thus proving that

\displaystyle s(x)^3 + c(x)^3 = 1





(ii)

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Using the product rule:

\displaystyle \left(sc)' = s'c + c's = c^2 c - s^2 s = c^3-s^3 = c^3-(1-c^3) = 2c(x)^3-1

Using the quotient rule:

\displaystyle \left(\frac{s}{c}\right)' = \frac{cs' - sc'}{c^2} = \frac{c^3+s^3}{c^2} = \frac{1}{c(x)^2}





(iii)

Spoiler:
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\displaystyle \int s^2 \ \mathrm{d}x = -\int c' \ \mathrm{d}x= -c(x) + \text{constant}

\displaystyle \begin{aligned}\int s^5 \ \mathrm{d}x&= \int (1-c^3)s^2 \ \mathrm{d}x \\ &= \int s^2 - c^3s^2\ \mathrm{d}x \\ &= \int -c' + c^3c' \ \mathrm{d}x \\ &= -c(x) + \frac{1}{4}c(x)^4 + \text{constant}\end{aligned}






(iv)

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\displaystyle u = s \Rightarrow u' = s'

\displaystyle \begin{aligned} \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \ \mathrm{d}u &= \int \frac{1}{\left(1-s^3\right)^{\frac{2}{3}}} \cdot s' \ \mathrm{d}x \\ &= \int   \frac{1}{c^2} \cdot c^2 \ \mathrm{d}x \\ &= \int 1 \ \mathrm{d}x \\ &= x +   \text{constant} \\& = s^{-1}(u) + \text{constant}\end{aligned}






(v)

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Using the same substitution as above:

\displaystyle \int \frac{1}{\left(1-u^3\right)^{\frac{4}{3}}} \ \mathrm{d}u = \int \frac{1}{c^4} \cdot s' \ \mathrm{d}x = \int \frac{c^2}{c^4} \ \mathrm{d}x = \int \frac{1}{c^2} \ \mathrm{d}x

Then using the second result in (ii):

\displaystyle \begin{aligned} \int \frac{1}{c^2} \ \mathrm{d}x &= \frac{s}{c} + \text{constant} \\ &= \frac{u}{\left(1-u^3\right)^\frac{1}{3}} + \text{constant}\end{aligned}


Again using the same substitution:

\displaystyle \int \left(1-u^3\right)^{\frac{1}{3}} \ \mathrm{d}u = \int c\cdot s' \ \mathrm{d}x = \int c^3 \ \mathrm{d}x

Rearranging the first result in (ii) gives

\displaystyle \begin{aligned}\int c^3 \ \mathrm{d}x &= \frac{1}{2}\left(sc + \int 1 \ \mathrm{d}x \right) \\ &= \frac{1}{2}\left( u(1-u^3)^\frac{1}{3} + x\right) + \text{constant} \\ &= \frac{1}{2}u(1-u^3)^\frac{1}{3} + \frac{1}{2}s^{-1}(u) + \text{constant}\end{aligned}



(Original post by Rohan77642)
How much do you (and other people) think I would get for the question if I did the parts highlighted in the solution?
How many marks would I lose if I forgot the 1/2 in the very final part of question 8?
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123Master321
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(Original post by Electric-man7)
How many marks for the graph? I messed it up entirely
I would have thought about 4 or 5 but then again, this is a complete guess
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JeremyC
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Question 1.

Spoiler:
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Finding P & Q.
Set the equations equal to each other:
a^2x =x(b-x)^2
Expand out and factorise x to be left with the quadratic equation x^2-2bx+(b^2-a^2).

Solving with the quadratic equation, we find that x = b \pm a (with the minus value corresponding to the x-coordinate of P, and the plus value corresponding to the x-coordinate of Q).

The sketch has already been provided by igot99problems here.

Finding the tangent line at P.
We differentiate the curve y=x(b-x)^2 to find \frac{dy}{dx} = 3x^2-4bx+b^2.

We substitute in the x-coordinate at P (x = b - a), and deal with some algebra, to find that the gradient of the curve at P is  a(3a-2b).

We also substitute this x-coordinate into the equation for y (y = x(b-x)^2), and deal with the algebra, to find that the y-coordinate at P is -a^3+a^2b.

We substitute our results into the equation of a straight line through (x_1, y_1) = (b-a, -a^3+a^2b) with gradient m=a(3a-2b), namely
y-y_1 = m(x-x_1), and do some algebra to find that
y = a(3a-2b)x + 2a(b-a)^2
as required.

Calculating the area of S
Here we need to integrate the area under the curve from 0 to the x-coordinate of P, minus the area under the line from 0 to the x-coordinate of P. Mathematically, we need to calculate:
\int_0^{b-a} x(b-x)^2 - a^2 x dx
Expanding the bracket and integrating, we get
[\frac{x^2b^2}{2}-\frac{2bx^3}{3}+\frac{x^4}{4}-\frac{a^2x^2}{2}]_0^{b-a}

Factorising x^2 out to make the algebra more bearable, and substituting in the limits we eventually find that the area of S is \frac{1}{12}(b-a)^3(b+3a) as required.

Showing the inequality
We use the fact that the area of a triangle is half its base times its height.

The height of triangle OPR equals the y-intercept of the tangent line, namely 2a(b-a)^2. The width of the triangle equals the x-coordinate of P, namely b-a. We therefore find that the area of the triangle, T, equals a(b-a)^3.

For the final inequality, we note that
S = \frac{1}{12} (b-a)^3(b+3a)
= \frac{3a}{12} (b-a)^3 + \frac{b}{12}(b-a)^3
= \frac{a}{4}(b-a)^3 + \frac{b}{12}(b-a)^3
= \frac{T}{4} + \frac{b}{12}(b-a)^3

Now, since b &gt; a we have \frac{b}{12}(b-a)^3 &gt; \frac{a}{12}(b-a)^3

Substituting this into the previous result, we find that
 S &= \frac{T}{4} + \frac{b}{12}(b-a)^3
&gt; \frac{T}{4} + \frac{a}{12}(b-a)^3
=  \frac{T}{4} + \frac{T}{12}
= \frac{T}{3}

as required.

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Orwellian49
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Could anyone enlighten me with their solutions for Q13? When I answered it on Tuesday I felt that it went alright, but now I think I may have missed out some parts in my working.

On a side note, I only managed to complete 3 & a quarter questions, which seemed pretty little compared to others. Not sure if I could meet the grades needed for a 2. ((
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esrever
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Problem 6


 = \sum_{i = 1}^n 2\sin(\theta)\sin ((2i-1)\theta )
 = \sum_{i = 1}^n \cos((2i-2)\theta \right)) - \cos(2i\theta)
 = \sum_{i = 0}^{n-1} \cos(2i\theta \right)) - \sum_{i = 1}^n \cos((2i)\theta \right))
 = \cos(0) + \sum_{i = 1}^{n-1} \cos(2i\theta \right)) - \cos(2n\theta) - \sum_{i = 1}^{n-1} \cos((2i)\theta \right))
 = 1 - \cos(2n\theta)

(i)
ith strip is between \frac{(i-1)\pi}{n} and \frac{i\pi}{n}. So midpoint is \frac{(2i-1)\pi}{2n}. So

A_n = \frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{(2i-1)\pi}{2n})

Set \frac{\pi}{2n} = \theta (n is a constant).

A_n = \frac{\pi}{n} \sum_{i=1}^{n} \sin((2i-1)\theta)
2A_n sin(\theta) = \frac{\pi}{n} \sum_{i=1}^{n} 2\sin(\theta)\sin((2i-1)\theta)
2A_n sin(\theta) = \frac{\pi}{n}(1 - \cos(2n\theta))
2A_n sin(\frac{\pi}{2n}) = \frac{\pi}{n}(1 - \cos(\pi))
2A_n sin(\frac{\pi}{2n}) = \frac{\pi}{n}

(ii) is similar but I'll type it later
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gnsbrg
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(Original post by DFranklin)
I'm not following your logic; if each quarter of values is supposed to correspond to a,b,c,d then surely they need to multiply to 16?
But -1 x 1 X 2 X 8 = -16, for example.
Absolutely, such a silly mistake (I think in the exam I did it right but I typed this so late at night)
Thanks for checking
(a,b,c,d)=(-2,-1,2,4),(-4,-2,1,2)(-4,-1,1,4)(-2,-1,1,8)
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DFranklin
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(Original post by gnsbrg)
Absolutely, such a silly mistake (I think in the exam I did it right but I typed this so late at night)
Thanks for checking
(a,b,c,d)=(-2,-1,2,4),(-4,-2,1,2)(-4,-1,1,4)(-2,-1,1,8)
What about (-8,-1,1,2)?
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gnsbrg
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(Original post by DFranklin)
What about (-8,-1,1,2)?
Yeah it works; I forgot it but I hope they don’t take out many marks (I think there aren’t more possibilities??)
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DFranklin
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(Original post by gnsbrg)
Yeah it works; I forgot it but I hope they don’t take out many marks (I think there aren’t more possibilities??)
I think it would only cost you a mark (and that's the only one you missed).
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Zacken
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(Original post by Electric-man7)
How many marks for the graph? I messed it up entirely
I'd say about 6 marks (in Tripos terms, not doing it would just about cost you the alpha).
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Student1256
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(Original post by JeremyC)
Question 1.

Spoiler:
Show


Finding P & Q.
Set the equations equal to each other:
a^2x =x(b-x)^2
Expand out and factorise x to be left with the quadratic equation x^2-2bx+(b^2-a^2).

Solving with the quadratic equation, we find that x = b \pm a (with the minus value corresponding to the x-coordinate of P, and the plus value corresponding to the x-coordinate of Q).

The sketch has already been provided by igot99problems here.

Finding the tangent line at P.
We differentiate the curve y=x(b-x)^2 to find \frac{dy}{dx} = 3x^2-4bx+b^2.

We substitute in the x-coordinate at P (x = b - a), and deal with some algebra, to find that the gradient of the curve at P is  a(3a-2b).

We also substitute this x-coordinate into the equation for y (y = x(b-x)^2), and deal with the algebra, to find that the y-coordinate at P is -a^3+a^2b.

We substitute our results into the equation of a straight line through (x_1, y_1) = (b-a, -a^3+a^2b) with gradient m=a(3a-2b), namely
y-y_1 = m(x-x_1), and do some algebra to find that
y = a(3a-2b)x + 2a(b-a)^2
as required.

Calculating the area of S
Here we need to integrate the area under the curve from 0 to the x-coordinate of P, minus the area under the line from 0 to the x-coordinate of P. Mathematically, we need to calculate:
\int_0^{b-a} x(b-x)^2 - a^2 x dx
Expanding the bracket and integrating, we get
[\frac{x^2b^2}{2}-\frac{2bx^3}{3}+\frac{x^4}{4}-\frac{a^2x^2}{2}]_0^{b-a}

Factorising x^2 out to make the algebra more bearable, and substituting in the limits we eventually find that the area of S is \frac{1}{12}(b-a)^3(b+3a) as required.

Showing the inequality
We use the fact that the area of a triangle is half its base times its height.

The height of triangle OPR equals the y-intercept of the tangent line, namely 2a(b-a)^2. The width of the triangle equals the x-coordinate of P, namely b-a. We therefore find that the area of the triangle, T, equals a(b-a)^3.

For the final inequality, we note that
S = \frac{1}{12} (b-a)^3(b+3a)
= \frac{3a}{12} (b-a)^3 + \frac{b}{12}(b-a)^3
= \frac{a}{4}(b-a)^3 + \frac{b}{12}(b-a)^3
= \frac{T}{4} + \frac{b}{12}(b-a)^3

Now, since b &gt; a we have \frac{b}{12}(b-a)^3 &gt; \frac{a}{12}(b-a)^3

Substituting this into the previous result, we find that
 S &= \frac{T}{4} + \frac{b}{12}(b-a)^3
&gt; \frac{T}{4} + \frac{a}{12}(b-a)^3
=  \frac{T}{4} + \frac{T}{12}
= \frac{T}{3}

as required.

Can anyone estimate how many marks I’ll get if I did everything up until the point you actually show S in the form they require? I even showed the integral and subtracted the triangle you get from the line OP. I just couldnt get it in the required form so I left the question from there.
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Zacken
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(Original post by Student1256)
Can anyone estimate how many marks I’ll get if I did everything up until the point you actually show S in the form they require? I even showed the integral and subtracted the triangle you get from the line OP. I just couldnt get it in the required form so I left the question from there.
12-ish?
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Aditya33
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Here is my attempt at the third question:
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Student1256
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(Original post by Zacken)
12-ish?
Wow thank you, pleasantly surprised with that.
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