DFranklin
Badges: 18
Rep:
?
#1
Report Thread starter 6 years ago
#1
STEP I Solutions now complete. Well done everyone!

STEP I:
1: Solution by DFranklin
2: Solution by Tarquin Digby
3: Solution by DFranklin
4: Solution by DFranklin
5: Solution by DFranklin
6: Solution by Tarquin Digby
7: Solution by Farhan.Hanif93
8: Solution by Pascal678
9: Solution by JosepthML
10: Solution by JosephML
11: Solution by DFranklin
12: Solution by DFranklin
13: Solution by Smaug123
0
reply
DFranklin
Badges: 18
Rep:
?
#2
Report Thread starter 6 years ago
#2
STEP I, Q1

(i) 3 = 2^2 - 1^2, 5 = 3^2 - 2^2, 8 = 3^2 - 1^2, 12 = 4^2 -2^2, 16 = 5^2 - 3^2

(ii) If n = 2k+1 is odd, then n = (k+1)^2 - k^2

(iii) if n = pq (p < q), then if we let a = (q - p) / 2, b = (p+q)/2, then b^2-a^2 = pq. If n = 4k, take p = 2, q = 2k to get a = k-1, b = k+1.

(iv) Suppose 4k+2 = (a^2-b^2), then (a-b)(a+b) = 4k+2. Note that since a+b = (a-b) + 2b, and 2b is even, either both (a+b), (a-b) are even or both are odd. Since 4k+2 is even, they must both be even, and so 4k+2 must be divisible by 4, which is impossible.

(v) Using (iii) we find a = (q - p) /2, b = (p+q) /2 as one solution, and factorising n = 1.(pq) we get a second solution a = (pq - 1)/2, b = (pq+1) / 2. These are clearly distinct since pq > q - p. Note that a 3rd solution would provide a new factorization of pq which is not possible since p, q are prime.

(vi) 675 = 25 * 27 = 5^2 * 3^3. Every factorization 675 into two factors p, q will have both p, q odd. So (q-p)/2 and (p+q)/2 will always both be integers. So the question is how many factorizations there are.

Factorize 675 = ( 5^r 3^s)(5^{2-r}3^{3-s}); there are 3 choices for r and 4 for s givinig 12 possible factorizations. But this double counts (e.g. 1 x 675 and 675 x 1 are both counted), so we need to divide by 2 giving a final answer of 6.
2
reply
Pascal678
Badges: 0
Rep:
?
#3
Report 6 years ago
#3
Might as well do the only question I could get a full solution for

STEP I 2014 question 8

part i)
Spoiler:
Show

Let the gradient of L_{a} be m

where m=\dfrac{1-a-0}{0-a}=\dfrac{a-1}{a}

The line L_{a} is y-0=\dfrac{a-1}{a}(x-a)

i.e. y=\dfrac{a-1}{a}x-(a-1)

If L_b is defined similarly then we have the same sets of coordinates and gradient but this time in terms of b

The line L_{b} is y=\dfrac{b-1}{b}x-(b-1)

L_{a} and L_{b} intersect when:

\dfrac{a-1}{a}x-(a-1)=\dfrac{b-1}{b}x-(b-1)

\Rightarrow b(a-1)x-ab(a-1)=a(b-1)x-ab(b-1)

so x(b(a-1)-a(b-1))=ab(a-1-(b-1))

x(a-b)=ab(a-b)

a\not=b \therefore a-b\not=0

So x=ab Thus y=\dfrac{a-1}{a}(ab)-(a-1)=b(a-1)-(a-1)=(a-1)(b-1)

Thus the point of intersection is T(ab,(a-1)(b-1))



part ii)
Spoiler:
Show

y=(1-\sqrt{x})^{2}

If the point of intersection lies on the curve, x=ab when y=(a-1)(b-1)

LHS=(a-1)(b-1)

RHS=(1-\sqrt{ab})^{2}

Thus as b->a

LHS=(a-1)(a-1)=(a-1)^{2}

RHS=(1-\sqrt{a^{2}})^{2}=(1-a)^{2}=(-1)^{2}(a-1)^{2}=(a-1)^{2}

So the point of intersection T lies on the curve C as LHS=RHS at this point as b->a



part iii)
Spoiler:
Show

First we find the gradient of the curve C at any point on the curve in terms of x

\dfrac{dy}{dx}=2(1-\sqrt{x})^{1}*-\dfrac{1}{2}x^{-\frac{1}{2}}=\dfrac{1-\sqrt{x}}{-\sqrt{x}}=\dfrac{\sqrt{x}-1}{\sqrt{x}}

Compare this to the gradient of L_{a} which \dfrac{a-1}{a} These are very similar and as b->a the gradient of C at the general point of intersection which lies on L_{a} is \dfrac{\sqrt{a^{2}}-1}{\sqrt{a^{2}}}=\dfrac{a-1}{a}

Furthermore the point of intersection lies on the curve and L_{a} and so satisfies for the curve for various values of x and y, thus every tangent to C is of the form L_{a} for some a

Not entirely sure about this part, feel free to correct me

0
reply
DFranklin
Badges: 18
Rep:
?
#4
Report Thread starter 6 years ago
#4
STEP I, Q4.

Choose coordinate axes so the short hand has position (0, a) and the short hand has position (b \sin \theta, b \cos \theta).
If x is the distance between the hands, then  x^2 = b^2 \sin^2 \theta + (a - b \cos \theta)^2. Write S for \sin \theta, C for \cos \theta, then we have
x^2 = b^2 S^2 + a^2 - 2ab C + b^2 C^2 = b^2 + a^2 - 2abC.

Differentiate w.r.t. theta: 2x \dfrac{dx}{d\theta} = 2abS, so \dfrac{dx}{d\theta} = \dfrac{abS}{x}. Differentiate again:

\dfrac{d^2x}{d\theta^2} =  ab \dfrac{xC - S \frac{dx}{d\theta}}{x^2} = ab\dfrac{xC - abS^2 / x} {x^2}

This = 0 when xC = ab S^2 /x = 0, or equivalently x^2 C = ab S^2 (and x not = 0).

So (a^2+b^2-2abC)C = ab(1-C^2), so (a^2+b^2)C - ab = abC^2, so abC^2 -(a^2+b^2)C + ab = 0.

So C = \dfrac{1}{2ab} ((a^2+b^2) \pm \sqrt{(a^2+b^2)^2 - 4a^2b^2)} = \dfrac{1}{2ab} ((a^2+b^2) \pm (a^2-b^2)) = a/b or b/a. Since b > a, and c <=1, b/a is impossible, so C = a / b.

So for a max/min rate of change, we must have x^2 = b^2 + a^2 - 2ab (a/b) = b^2 - a^2 as desired. (I'm not going to worry about whether it's a max or min, because the everything "flips" for theta negative, and so we get into worrying about whether "largest" means most positive, or "biggest magnitude" etc. It's not terribly well worded).

When b = 2a, this first occurs when C = 1/2, i.e. when \theta = 60^o.
After t minutes have passed, the minute hand has moved through 6t degrees (60 minutes = 360 degrees), and the hour hand through 0.5t degrres (60 minutes = 360/12 =30 degrees). So they are apart by 60 degrees after 60/(6-0.5) = 60/5.5 minutes. Since 11 * 5.5 = 55 + 5.5 = 60.5 minutes, 60/5.5 = 11 - (0.5/5.5) minutes, which is slightly less than 11 minutes.
0
reply
DFranklin
Badges: 18
Rep:
?
#5
Report Thread starter 6 years ago
#5
STEP I, Q5.

(i) Not going to do the sketch, but if y = (x+2a)^3 -27a^2x, then dy/dx = 3(x+2a)^2 - 27a^2. When x = a, y = dy/dx = 0, so y = (x-a)^2(x-b) for some value b. Comparing constant coefficients, -a^2b = 8a^3, so b = -8 and so y >=0 for x >= -8. Note that for x > 0, y = 0 iff x = a.

(ii) Want to maximize xy^2 subject to x+2y <= 3. By the above, we know that (x+2y)^3 >= 27xy^2, so 27xy^2 <= (x+2y)^3 <= 3^3 = 27. We also know we need x = y if we want (x+2y)^3 = 27xy^2. So it is only in this case that we have 27xy^2 = (x+2y)^3, and of course we only have (x+2y)^3 = 27 when x+2y = 3. So xy^2 is maximized then x = y = 1.

(iii) (p+q+r)^3 = (p + 2(q+r)/2) ^2 >= 27 p (q+r)^2 / 4. But (q+r)^2 / 4 - qr = (q^2 + 2qr +r^2 -4qr) / 4 =(p-q)^2 / 4 > = 0. So (q+r)^2 / 4 >= qr, so 27p(q+r)^2/4 >= 27pqr.

For equality in (q+r)^2 / 4 >= qr, we need q = r. For equality in (p + 2(q+r)/2) ^2 >= 27 p (q+r)^2 / 4 we need p = (q+r)/2. Hence we have equality iff p = q = r.
0
reply
DFranklin
Badges: 18
Rep:
?
#6
Report Thread starter 6 years ago
#6
STEP I, Q3.
(i) \int_0^b x^2\,dx  =\left( \int_0^b x \,dx \right)^2, so b^3/3 = (b^2/2)^2 = b^4 / 4. So b/4 = 1/3, so b = 4/3.

(ii) Now we have (b^3 / 3 - 1/3) = (b^2 / 2 - 1/2 )^2
So 4b^3 - 4= 3(b^2 - 1)^2
So 4(b-1)(b^2+b+1) = 3(b-1)^2(b+1)^2. Since b>1, can divide by b-1.
So 4(b^2+b+1) = 3(b-1)(b+1)^2
So 3(b^2-1)(b+1) -4b^2 -4b -4 = 0.
So 3b^3 +3b^2 -3b - 3 - 4b^2 -4b - 4 =0.
So 3b^3 -b^2 - 7b - 7 = 0 as desired.

The "show there's only one root and it lies between 2 and 3" is definitely easier by a sketch; I would draw 3b^3 and b^2+7b+7 on the same graph and it should be reasonably obvious.

(iii) In general we have  (b^3-a^3) / 3 = (b^2-a^2)^2 / 4
So 4 (b-a)(b^2+ab+a^2) = 3(b-a)^2(b+a)^2. We are given b>a, so b-a is non-zero, s we can divide by it:
4(b^2+ab+a^2) = 3(b-a)(b+a)^2. If p = (b+a), q = (b-a), then the RHS is immediately 3p^2q.
The LHS is  4b^2 + 4ab + 4a^2. But p^2 = a^2+b^2 +2ab, and q^2 = a^2+b^2-2ab. So 3p^2+q^2 = 4a^2+4ab+4b^2 as desired.

If 3p^2+q^2 = 3p^2q, then 3p^2q - 3p^2 = q^2, so 3p^2(q-1) = q^2, so [latexp^2 = q^2/(3(q-1))[/latex].

Since p > 0 we must have q-1 > 0 and so b-a > 1.
Note that in fact p >=q, so we must also have 3(q-1) <= 1, so q <= 4/3, and so b-a <= 4/3.
0
reply
BabyMaths
  • Study Helper
Badges: 0
Rep:
?
#7
Report 6 years ago
#7
(Original post by DFranklin)
Thought we might as well create a solution thread... I'll do the indexing here as people post solutions.
Where's the paper?
1
reply
Pascal678
Badges: 0
Rep:
?
#8
Report 6 years ago
#8
(Original post by BabyMaths)
Where's the paper?
Here https://www.dropbox.com/sh/yyl2eax4r...%20Q7%20Q8.JPG
2
reply
Tarquin Digby
Badges: 9
Rep:
?
#9
Report 6 years ago
#9
STEP 1 Q2

i)
Integrating by parts
\displaystyle \int \ln x \, \mathrm d x = x \ln x  - \int \, \mathrm d x = x \ln x - x + C

Now substitute x \mapsto 2-x and remember to divide by -1.

ii)
Image

Symmetry about the y axis due to x^2 term. Intersects x axis at \pm \sqrt 3 and \pm \sqrt 5. Intersects y axis at \ln 4.

iii)
\displaystyle \int_0^{\sqrt 3} \ln {|x^2 - 4|} \, \mathrm d x = \int_0^{\sqrt{3}} \ln (4 - x^2) = \int_0^{\sqrt{3}} \ln (2 - x) + \ln (2 + x) \, \mathrm d x \\ = 4 \ln (2 + \sqrt {3} ) - 2 \sqrt{3}

Note that - \ln(2 - \sqrt{3}) = \ln(2 + \sqrt{3}).

iv)
\displaystyle \int_0^2 |\ln|x^2-4|| \, \mathrm d x = \int_0^2 |\ln(4-x^2)| \, \mathrm d x \\ = \int_0^{\sqrt{3}} \ln(4-x^2) \, \mathrm d x - \int_{\sqrt{3}}^2 \ln(4-x^2) \, \mathrm d x = 4 - 4 \sqrt 3 - 8 \ln 2 + 8 \ln (2 + \sqrt 3 )

Now multiply by 2 to get 8 - 8 \sqrt 3 - 16 \ln 2 + 16 \ln (2 + \sqrt 3 )
2
reply
BabyMaths
  • Study Helper
Badges: 0
Rep:
?
#10
Report 6 years ago
#10
Thanks.
0
reply
Tarquin Digby
Badges: 9
Rep:
?
#11
Report 6 years ago
#11
STEP 1 Q6

i)
u_1 = \sin^2 2 \theta

u_2 = \sin^2 4 \theta

Suppose u_k = \sin^2 (2^k \theta) for some k.

Then u_{k + 1} = 4 \sin^2 (2^k \theta) (1 - \sin^2 (2^k \theta)) = 4 \sin^2 (2^k \theta) \cos^2 (2^k \theta) = \sin^2 (2^{k+1} \theta).

Hence u_n = \sin^2 (2^n \theta) for all n \ge 1 by induction.

ii)

Let \alpha = \dfrac 4 p and \beta = \dfrac {q-4} {2p}.

Substitute v_n = \alpha u_n + \beta.

Then we have:

\dfrac 4 p u_{n+1} + \dfrac {q-4} {2p} = -p \left(\dfrac 4 p u_n + \dfrac {q-4} {2p} \right)^2  + q \left( \dfrac 4 p u_n + \dfrac {q-4} {2p} \right) + r

Rearranging:

u_{n+1} = -4u_n^2 + 4u_n +  \frac 1 {16} \left( 4pr - 8 - 2q + q^2 \right)

We're given 4pr = 8 + 2q - q^2 , so

u_{n+1} = 4u_n (1 - u_n) as required.

If v_{n+1} = - v_n^2 + 2 v_n +2 then p=1, q=2 and r=2.

This satisfies 4pr = 8 + 2q - q^2.

Also, \alpha = 4 and \beta = -1.

Therefore u_n = \sin^2 (2^n \theta) where v_n = 4u_n - 1.

v_0 = 1, so u_0 = \dfrac 1 2 which implies \theta = \dfrac {\pi} 4.

Hence \displaystyle v_n = 4 \sin^2 \left(2^n \frac {\pi} 4\right) - 1 = 4 \sin^2 (2^{n-2} \pi) - 1.
1
reply
DFranklin
Badges: 18
Rep:
?
#12
Report Thread starter 6 years ago
#12
STEP I, Q12. Note that we can regard the dice as a 3 sided dice that takes values 1,2,3 with probability 1/2, 1/3, 1/6 respectively. Note also that E[|Y|^2] = E[Y^2].

So E[X^2] = \frac{1}{2}(\frac{1}{2} (k-1)^2 + \frac{1}{3} (2k-1)^2 + \frac{1}{6}(3k-1)^2) + \frac{1}{2} (\frac{1}{2} (k-1)^2 + \frac{1}{3}(k-2)^2 + \frac{1}{6}(k-3)^2)

=\frac{1}{12} (3(k-1)^2 + 2(2k-1)^2 + (3k-1)^2 + 3(k-1)^2 + 2(k-2)^2 + (k-3)^2)

=\frac{1}{12} (6(k-1)^2 + 2(2k-1)^2 + (3k-1)^2 + 2(k-2)^2 + (k-3)^2)

=\frac{1}{12}(6k^2 - 12k+6 + 8k^2 - 8k+2 + 9k^2 - 6k + 1 + 2k^2 -8k + 8 + k^2 - 6k + 9)

=\frac{1}{12}(6+8+9+2+1)k^2 - (12+8+6+8+6)k + (6+2+1+8+9))

=\frac{1}{12}(26k^2 - 40k + 26) = \frac{1}{6}{13k^2 - 20k + 13}

= \frac{1}{6}(13k^2 - 26k + 13 + 6k)= k + \frac{13}{6}(k-1)^2 as desired.

If k is an integer and E[X^2] is also an integer, then 6 must divide (k-1)^2, and so 6 must divide (k-1) (since both 2 and 3 must divide it). So if k is a single digit integer, we must have k = 7.

PDF of X: The "head" values for X are 6, 13 and 20 with probabilities 1/4, 1/6, 1/12 respectively.
The "tail" values for X are 6, 5, 4 with probabilities 1/4, 1/6, 1/12 respectively.
So combining these we finally have
P(X = 4) = 1/12, P(X=5) = 1/6, P(X=6) = 1/2, P(X=13) = 1/6, P(X=20) = 1/12.

For the sum of two rounds to = 25, we need either X=5, then X=20 or vice versa, with probability 2 x (1/6) x (1/12) = 1/36.
For the sum of two rounds to be > 25, we need either (6, 20), (13, 20), (20, 20), (13, 13) or (20, 6) or (20, 13).
The probability of this happening is 2 x ((1/2) x (1/12) + (1/6) x (1/12)) + (1/6)^2 + (1/12)^2 = 2 x (1/24 + 1/72) + 1/36 + 1/144 = (2 x (6+2) + 4 + 1) / 144 = 21 /144 = 7/48.

The expected return for the gambler is 7w/48+1/36 (ignoring the loss of the original stake).

The expected profit for the casino is 1 - 7w/48 - 1/36 = (144 - 21 w - 4) = (140-21w). So we require 21w < 140. If w is an integer, the greatest possible value is w = 6.

[Alternative calc for last bit: p(X>=13) is 1/6+1/12 = 1/4. So p(both tries >=13) = 1/16. If both aren't >=13, the only other possibility for getting over 25 is to get (6,20) or (20,6). This has probability 2 x (1/2) x (1/12) = 1/12. 1/16+1/12 = 7/48].
0
reply
DFranklin
Badges: 18
Rep:
?
#13
Report Thread starter 6 years ago
#13
(Original post by Tarquin Digby)
Let \alpha = \dfrac 4 p and \beta = \dfrac {q-4} {2p}.
Where did these come from?

Substituting v_n = \alpha u_n + \beta and simplifying we obtain (*).
I think this is too terse to be helpful, frankly. (Certainly you wouldn't get away with it in an exam).
0
reply
JosephML
Badges: 2
Rep:
?
#14
Report 6 years ago
#14
Question 9:

Resolving in the standard way we get:

 y = U \sin \theta t - \frac{1}{2} g t^2

 x =U \cos \theta t - \frac{1}{2}kgt^2

So we get T_H when \frac{dy}{dt} = 0, thus:

\displaystyle T_H = \frac{U \sin \theta}{g}

And so by symmetry we get \displaystyle T_L = \frac{2U \sin \theta}{g}

Now we define \displaystyle T = \frac{U \cos \theta}{kg}. So T is the point at which the x velocity is zero. For the sketches we consider the ratios:

\displaystyle \frac{T_H}{T} = k \tan \theta

\displaystyle \frac{T_L}{T} = 2k \tan \theta

So when k \tan \theta &lt; \frac{1}{2} we have the particle reaching its maximum height and hitting the ground without the x velocity changing direction.

When  \frac{1}{2} &lt; k \tan \theta &lt; 1 we have the particle reaching its maximum height then moving back towards  O then hitting the ground.

When k \tan \theta &gt; 1 we have the particle moving back towards the origin before reaching its maximum height.

Finally when k \tan \theta = 1 we have the particle stationary when t = T = T_H. Also we have:

k y = U \cos \theta - \frac{1}{2} k g t^2 = x

so the trajectory is a straight line.

The last part might be a little unclear, if people can't understand then I'll rewrite it.
0
reply
Smaug123
  • Study Helper
Badges: 15
Rep:
?
#15
Report 6 years ago
#15
(Original post by DFranklin)
Thought we might as well create a solution thread... I'll do the indexing here as people post solutions.
Q13:
Finding the PDF:
Spoiler:
Show

Our PDF looks like a triangle with height g(c) and base b-a. Thus (since the area of the triangle is 1, by "probability density function") we have \frac{b-a}{2}g(c) = 1 - for ease, write this as g(c) = \frac{2}{b-a}. Also g(a) = 0.

Write g(x) = \alpha + \beta x. Then we have \alpha + \beta c = \frac{2}{b-a}, and \alpha + \beta a = 0. Eliminate \alpha: \beta c - \beta a = \frac{2}{b-a}.

Then \beta = \frac{2}{(b-a)(c-a)}, and \alpha = -\beta a, so g(x) = \beta (-a + x) = \frac{2(x-a)}{(b-a)(c-a)} as required.

Now, the h expression will be similar by symmetry. To get h(b) = 0 and a negative slope, the numerator will become 2(b-x). We still need h(c) = g(c) = \frac{2}{b-a}; hence h(x) = \frac{2(b-x)}{(b-c)(b-a)} by inspection.


Finding the mean:
Spoiler:
Show

The mean of the distribution is the integral of x against the PDF. That is \int_a^c g(x) x \text{d}x + \int_c^b h(x) x \text{d}x. We'll continue to write g(x) = \alpha + \beta x for the moment, and h(x) = \gamma + \delta x. We get that the mean is \displaystyle \int_a^c \alpha x + \beta x^2 \text{d}x + \int_c^b \gamma x + \delta x^2 \text{d}x = [\frac{\alpha}{2} x^2 + \frac{\beta}{3} x^3]_a^c + [\frac{\gamma}{2} x^2 + \frac{\delta}{3} x^3]_c^b. Expanding these gives \frac{\alpha c^2}{2} + \frac{\beta c^3}{3} - \frac{\alpha a^2}{2} - \frac{\beta a^3}{3} + \frac{\gamma b^2}{2} + \frac{\delta b^3}{3} - \frac{\gamma c^2}{2} - \frac{\delta c^3}{3}.

Now we have \alpha = -\beta a, \gamma = -\delta b by inspection, so we can tidy the expression up a bit:
\displaystyle \frac{- \beta a c^2}{2} + \frac{\beta c^3}{3} - \frac{-\beta a^3}{2} - \frac{\beta a^3}{3} + \frac{- \delta b^3}{2} + \frac{\delta b^3}{3} - \frac{-\delta b c^2}{2} - \frac{\delta c^3}{3} = - \frac{\beta a c^2}{2} + \frac{\beta c^3}{3} + \frac{\beta a^3}{6} - \frac{\delta b^3}{6} + \frac{\delta b c^2}{2} - \frac{\delta c^3}{3}.

Factorising: this is \beta (-\frac{a c^2}{2} + \frac{c^3}{3} + \frac{a^3}{6}) + \delta (-\frac{b^3}{6} + \frac{b c^2}{2} - \frac{c^3}{3}).


We'll consider the \beta term first. The bracket factorises as (c-a)(\frac{c^2}{3} - \frac{a c}{6} - \frac{a^2}{6}), so the \beta term is \displaystyle \frac{2}{b-a} (\frac{c^2}{3} - \frac{a c}{6} - \frac{a^2}{6}) (after cancelling the c-a from top and bottom).

Now the \gamma term is -\gamma (c-b)(\frac{c^2}{3} - \frac{b c}{6} - \frac{b^2}{6}), so we get -\frac{2}{b-a} (\frac{c^2}{3} - \frac{b c}{6} - \frac{b^2}{6}) for that term.

Adding them up gives \frac{1}{3 (b-a)} (-a^2 + b^2 - a c + b c) = \frac{1}{3} (b+a+c) as required.


Finding the median:
Spoiler:
Show

For the median: there will be two cases, the first when the median is in the g(x) part of the PDF, and the second when the median is in the h(x) part. There is an obvious check: when c = \frac{a+b}{2}, then the median will be c.

Firstly, let's suppose the median lies leftwards of c. Then we have \int_a^m g(x) \text{d}x = \frac{1}{2}, for m the median. The left-hand side is \alpha (m-a) + \frac{\beta}{2} (m^2-a^2) = (m-a)(\alpha + \frac{\beta}{2} (m+a)). \alpha = -\beta a so this in turn is (m-a) \beta (-a + \frac{m+a}{2}) = (m-a) \beta \frac{m-a}{2} = \frac{1}{2}. Solving for m yields (m-a)^2 = \beta^{-1} = \frac{(b-a)(c-a)}{2}, so m = a + \sqrt{\frac{(b-a)(c-a)}{2}}. (We pick the root that gives us m&gt;a.)

Now let's suppose the median lies rightwards of c. We then have \int_m^b h(x) \text{d}x = \frac{1}{2}, for m the median. The left-hand side is \gamma (b-m) + \frac{\delta}{2} (b^2 - m^2) = -b \delta (b-m) + \frac{\delta}{2} (b+m)(b-m) = (b-m) \delta (-b+\frac{b+m}{2} = \frac{1}{2}. Hence (b-m)^2 = -\delta^{-1} and so m = b - \sqrt{-\frac{1}{\delta}}, picking the root that gives us m &lt; b. This becomes m = b - \sqrt{\frac{(b-c)(b-a)}{2}}. (Notice that our check-case of c = \frac{a+b}{2} works for both of these.)

Finally we need the conditions under which the median lies leftwards or rightwards of c. By symmetry it is clear that the median is leftwards of c iff the slope of g is less than the negative-slope of x, so that the plot of the PDF is "more shallow" on the left-hand side. That is the only way that the left-hand side of the diagram can contain more than half the area. This happens precisely when |\beta| &lt; |\delta|, or when \frac{2}{(b-a)(c-a)} &lt; \frac{2}{(b-c)(b-a)}, or when (b-c)(b-a) &lt; (b-a)(c-a), or when b-c &lt; c-a, or when b+a-2c &lt; 0.
0
reply
DFranklin
Badges: 18
Rep:
?
#16
Report Thread starter 6 years ago
#16
(Original post by JosephML)
Question 9:

Resolving in the standard way we get:

 y = U \sin \theta t - \frac{1}{2} g t^2

 x =U \cos \theta t - \frac{1}{2}kgt^2
Finally when k \tan \theta = 1 we have the particle stationary when t = T = T_H.
Note that if k \tan \theta = 1, then k = \cos \theta / \sin \theta, and then ky = U \cos \theta t - \frac{1}{2}kgt^2 = x. So the path is actually a straight line.

(I don't know that they expected this, but to be honest I think they probably did).
1
reply
31541
Badges: 2
Rep:
?
#17
Report 6 years ago
#17
Guys - in Q9, I correctly found T_H and T_L and identified what T meant (x vel becoming zero) - however, I misinterpreted the graphs (probably as I was in a rush) and forgot that x would eventually change direction. So I drew just a sudden drop (don't ask..). I also understood what the first 2 inequalities meant, and explained (i.e. x vel becoming zero before landing) - but the whole idea I'm getting at here is that I didn't spot that it would change direction. How many marks am I looking at for this?

Also - how many marks would you estimate the introductory part to Q13 to be worth?

Thanks!
0
reply
DFranklin
Badges: 18
Rep:
?
#18
Report Thread starter 6 years ago
#18
(Original post by CD315)
Guys - in Q9, I correctly found T_H and T_L and identified what T meant (x vel becoming zero) - however, I misinterpreted the graphs (probably as I was in a rush) and forgot that x would eventually change direction. So I drew just a sudden drop (don't ask..). I also understood what the first 2 inequalities meant, and explained (i.e. x vel becoming zero before landing) - but the whole idea I'm getting at here is that I didn't spot that it would change direction. How many marks am I looking at for this?
I'm not sure on this one, to be honest. There's very little content in the question, which makes me think the sketches were probably a significant amount of the marks. But on the other hand, unless you mess up, the sketches aren't really much work either. If I had to guess I'd think you're probably looking at between 5 and 12 marks for the graphs + explanation. With about 8 being my best absolute guess.

Also - how many marks would you estimate the introductory part to Q13 to be worth?
Just the "show that"? 3-4 marks at most.
0
reply
31541
Badges: 2
Rep:
?
#19
Report 6 years ago
#19
(Original post by DFranklin)
I'm not sure on this one, to be honest. There's very little content in the question, which makes me think the sketches were probably a significant amount of the marks. But on the other hand, unless you mess up, the sketches aren't really much work either. If I had to guess I'd think you're probably looking at between 5 and 12 marks for the graphs + explanation. With about 8 being my best absolute guess.
Yes, there's surprisingly little content in it, so I was confused as to how the marks would have been distributed. I'll take 8!

Just the "show that"? 3-4 marks at most.[/QUOTE]

Yeah, the show that + finding a similar expression for h(x).
0
reply
DFranklin
Badges: 18
Rep:
?
#20
Report Thread starter 6 years ago
#20
STEP I, Q11. (Interesting question, this!).

Disclaimer: I honestly can't remember the last pulley question I did, which means it's probably 20+ years ago. So methods etc. may not be standard.

For the first part, let T be the tension in the string. The acceleration of M is g - T/M, and the accel of m is g - T/m. For the string not to stretch, these must be equal.
So g - T/M = T/m - g, so Mmg - Tm = TM -Mmg, so 2Mmg = T(M+m), so T = 2Mmg/(M+m).
Then the acceleration of M = g - 2mg/(M+m) = g(M-m)/(M+m) as required.
The force on the pulley is 2T = 4Mmg/(M+m).

For the 2nd part. Label the M1, M2 pulley P, and let the tension it its string be T. Let the tension in the other string be S. For legibility, write A for the acceleration a2.

Again, the acceleration of M is g - T/M and the acceleration of m is g-T/m. But relative to the pulley, the accelerations are (g+A) - T/M and (g+A) -T/m. So (replacing g in the earlier calculation with g+A) we end up with a
new value for the tension T = 2(g+A)\dfrac{m_1m_2}{m_1+m_2} = 2(g+A)\mu.

Now, since the pulley is weightless, the net force on it must be zero, so we have 2T = S, so S = 4(g+A)\mu. Now since the acceleration of M is A, we know the net force on M is MA.

So we must have Mg - S = MA. So  Mg - 4(g+A)\mu = MA. So A(M+4\mu) = g(M-4\mu) so A = g\dfrac{M-4\mu}{M+4\mu} as required.

Now suppose g\dfrac{M-4\mu}{M+4\mu} = g \dfrac{M-m}{M+m}.
This is true iff \dfrac{M+4\mu - 8\mu}{M+4\mu} = \dfrac{M+m-2m}{M+m} so 1-8\mu/(M+4\mu) = 1 - 2m/(M+m)
iff 4\mu/(M+4\mu) = m/(M+m)
iff 4\mu(M+m) = m(M+4\mu). So g\dfrac{M-4\mu}{M+4\mu} = g \dfrac{M-m}{M+m} iff 4\mu M + 4\mu m = mM + 4 \mu M iff 4\mu M = mM, iff [/latex]4 \mu = m[/latex],

But 4\dfrac{m_1m_2}{m_1+m_2} = (m_1+m_2) iff 4m_1m_2 = (m_1+m_2)^2 iff (m_1-m_2)^2 = 0 iff m_1 = m_2.
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should the school day be extended to help students catch up?

Yes (78)
29.21%
No (189)
70.79%

Watched Threads

View All
Latest
My Feed