STEP I 2014 solutions

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).

Where did these come from?

I think this is too terse to be helpful, frankly. (Certainly you wouldn't get away with it in an exam).

I substituted it in, found alpha and beta by equating coefficients for u_n and u_n^2, then the constant simplifies to 0 by the given condition. It's just a lot of tedious messy algebra, nothing difficult. I'll expand it a bit.

Sorry, I just noticed the last bit of your answer to the question. Note that for n> 2 (or something like that), the angle you are taking the sin of is definitely a multiple of pi, and so the sin = 0. This seems a little unlikely to be the correct solution, but I haven't actually checked to be sure it isn't.

The sequence eventually becomes constant.

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.

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

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].

Didn't it ask us for the magnitude of the force exerted on the pulley by tension, i.e. 2T? for system I that is

"For the sum of two rounds to be > 25, we..."

How many marks do u think i would get if i did everything up to ( but not including)this point correctly. I ran out of time so couldnt complete the question

Yeah that's definitely one possibility. It's just that if I'd set the question I would have made the initial value of theta was something like pi/3 so that it never went constant.

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.

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 $\displaystyle v_n = 4 \sin^2 \left(2^n \frac {\pi} 4\right) - 1 = 4 \sin^2 (2^{n-2} \pi) - 1$.

I substituted it in, found alpha and beta by equating coefficients for u_n and u_n^2, then the constant simplifies to 0 by the given condition. It's just a lot of tedious messy algebra, nothing difficult. I'll expand it a bit.

Substituting $v_n=\alpha u_n+\beta \text{ into } u_{n+1}=4u_n(1-u_n)$ gives
$v_{n+1}=-p(\alpha v_n+\beta)^2 +q(\alpha v_n+\beta)+r$ so the equating the coeficients of $u_n^2$ gives $p\alpha^2=4$ so I still don't see how you got $p[br]\alpha=\frac{4}{p}$