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Reply 1
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions
Reply 2
I Question 2

Part (i) 1+2xβˆ’x2>2x1+2x-x^2>\frac{2}{x}, xβ‰ 0x\neq 0.

First it might help to find the points of equality: multiplying by x and rearranging, we get x3βˆ’2x2βˆ’x+2=(xβˆ’2)(xβˆ’1)(x+1)=0x^3-2x^2-x+2 = (x-2)(x-1)(x+1)=0, xβ‰ 0x\neq 0. The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for βˆ’1<x<0-1 < x < 0 (the zero due to the asymptote) and 1<x<21<x<2.

Part (ii) 3x+10>2+x+4\sqrt{3x+10} > 2 + \sqrt{x+4}, xβ‰₯βˆ’103x \geq -\frac{10}{3}.

Because of the condition, squaring both sides still leaves an iff, giving: 3x+10>4+x+4+4x+4β€…β€ŠβŸΉβ€…β€Š2x+2>4x+4β€…β€ŠβŸΉβ€…β€Šx+1>2x+4 3x + 10 > 4 + x + 4 + 4\sqrt{x+4}\implies 2x + 2 > 4\sqrt{x+4} \implies x + 1 > 2\sqrt{x+4}. Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, x2+2x+1>4(x+4)β€…β€ŠβŸΉβ€…β€Šx2βˆ’2xβˆ’15>0x^2+2x+1 > 4(x + 4) \implies x^2 - 2x - 15 > 0, giving (x+3)(xβˆ’5)>0(x+3)(x-5)>0. From this, we can deduce that x>5x > 5 and x<βˆ’3 x<-3, but our new condition, x>βˆ’1x>-1 only gives us x>5x>5 as the final solution.
Reply 3
STEP II 2001, Question 6

Part One

(i)

(ii)

STEP II Question 2

Sketch not included, but pretty simple (y=0 for 0<x<N, and y=1 for N<x<2N)

i) [k/N] [k/N] for 1<k<2N1<k<2N can be split into 3 regions:
[k/N]=0 [k/N]=0 for 1<k<Nβˆ’1 1<k<N-1
[k/N]=1 [k/N]=1 for N<k<2Nβˆ’1 N<k<2N-1 and
[k/N]=2 [k/N]=2 for k=2N k=2N

So by splitting the summation into these three ranges we get

βˆ‘k=12N(βˆ’1)[k/2N]k=βˆ‘k=1Nβˆ’1kβˆ’βˆ‘k=N2Nβˆ’1k+2N \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N

=12(Nβˆ’1)(N)βˆ’(12(2Nβˆ’1)(2N)βˆ’12(Nβˆ’1)(N))+2N = \displaystyle\frac{1}{2}(N-1)(N) -(\displaystyle\frac{1}{2}(2N-1)(2N) -\displaystyle\frac{1}{2}(N-1)(N)) +2N

=(Nβˆ’1)(N)βˆ’(2Nβˆ’1)(N)+2N=(N-1)(N)-(2N-1)(N)+2N

=N2βˆ’Nβˆ’2N2+N+2N=N^2-N-2N^2+N+2N

βˆ‘k=12N(βˆ’1)[k/2N]k=2Nβˆ’N2\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k =2N-N^2 as required

ii) SN=βˆ‘k=12N(βˆ’1)[k/2N]2βˆ’kS_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k}

Splitting into 3 as before:

SN=βˆ‘k=1Nβˆ’12βˆ’kβˆ’βˆ‘k=N2Nβˆ’12βˆ’k+2βˆ’2NS_N=\displaystyle\sum_{k=1}^{N-1} 2^{-k} - \displaystyle\sum_{k=N}^{2N-1} 2^{-k} + 2^{-2N}

Using the formula for the sum of a geometric progression
Sn=a(1βˆ’rn)1βˆ’rS_n=\displaystyle\frac{a(1-r^n)}{1-r} we get

SN=2βˆ’1(1βˆ’(1/2)Nβˆ’1)1βˆ’2βˆ’1+2βˆ’N(1βˆ’(1/2)N)1βˆ’2βˆ’1+2βˆ’2NS_N=\displaystyle\frac{2^{-1}(1-(1/2)^{N-1})}{1-2^{-1}} +\displaystyle\frac{2^{-N}(1-(1/2)^{N})}{1-2^{-1}} + 2^{-2N}

SN=1βˆ’(1/2)Nβˆ’1+21βˆ’Nβˆ’21βˆ’2N+2βˆ’2NS_N=1-(1/2)^{N-1}+2^{1-N}-2^{1-2N}+2^{-2N}
Which could probably be made simpler, so now we can see asNβ†’βˆžN\to\infty then SNβ†’1S_N\to1
(edited 13 years ago)
Reply 5
I Question 2

f(x)=(xβˆ’p)(xβˆ’q)(xβˆ’r)f(x) = (x-p)(x-q)(x-r), p<q<rp<q<r. Well, we sketch the graph to show the two stationary points and crossing the x-axis at x=p x = p, x=qx=q, x=rx=r. We can conclude the two stationary points exist, so there must be a real x-value for them to be found.

fβ€²(x)=3x2+2x(p+q+r)+(pq+qr+rp)=0f'(x) = 3x^2+2x(p+q+r)+(pq+qr+rp) = 0.
Since the two stationary points are distinct and real, by b2βˆ’4ac>0b^2-4ac>0, we get (2(p+q+r))2βˆ’4β‹…3β‹…(pq+qr+rp)>0β€…β€ŠβŸΉβ€…β€Š(p+q+r)2>3(pq+qr+rp)(2(p+q+r))^2-4\cdot 3 \cdot (pq+qr+rp) > 0 \implies (p+q+r)^2 > 3(pq+qr+rp).

Considering (x2+gx+h)(xβˆ’k)(x^2+gx+h)(x-k), to have three roots, there must be a real solution to the quadratic, giving g2>4hg^2>4h. This implies that the two other roots are Ξ±=βˆ’gβˆ’g2βˆ’2h2\alpha = \frac{-g-\sqrt{g^2-2h}}{2} and Ξ²=βˆ’g+g2βˆ’2h2\beta = \frac{-g+\sqrt{g^2-2h}}{2}, and Ξ³=k\gamma = k. Letting p=Ξ±p = \alpha, q=Ξ²q = \beta, r=Ξ³r = \gamma in any order (since they can be changed around due to triple-symmetry ). Knowing that p+q+r=(kβˆ’g)p+q+r= (k-g) and pq+qr+rp=Ξ±Ξ²+Ξ±Ξ³+Ξ³Ξ²=h+Ξ³(Ξ±+Ξ²)=h+k(βˆ’g)=hβˆ’gkpq+qr+rp = \alpha\beta + \alpha\gamma + \gamma\beta= h +\gamma(\alpha+\beta) = h + k(-g) = h - gk. Substituting into our previous result, we obtain: (kβˆ’g)2>3(hβˆ’gk)β€…β€ŠβŸΉβ€…β€Š(gβˆ’k)2>3(hβˆ’gk)(k-g)^2 > 3(h-gk) \implies (g-k)^2>3(h-gk), as required. We can show that it is not necessary by letting g=1g=1, k=1k = 1, (1βˆ’1)2>3(hβˆ’1)β€…β€ŠβŸΉβ€…β€Šh>1β€…β€ŠβŸΉβ€…β€Š12>4hΒ isΒ notΒ true(1-1)^2>3(h-1)\implies h > 1 \implies 1^2 > 4h \text{ is not true}.
Reply 6
STEP I - Q13

Given that plate is broken, the probability that it is broken by the mathematician is 0.25.

Therefore, the number of plates, XX, broken by the mathematician can be represented by the following binomial distribution:

XX~B(n,0.25)B(n,0.25)

We need to find P(Xβ‰₯3)P(X \geq 3) when n=5.

P(Xβ‰₯3)=P(X=3)+P(X=4)+P(X=5)P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)
=(53)(0.25)3(0.75)2+(54)(0.25)4(0.75)+(55)(0.25)5=\displaystyle \binom{5}{3}(0.25)^3(0.75)^2 + \displaystyle \binom{5}{4}(0.25)^4(0.75) + \displaystyle \binom{5}{5}(0.25)^5
=0.0879+0.0146+0.0010=0.0879+0.0146+0.0010
=0.1035=0.1035
Reply 7
I shall get going. :smile:

SimonM
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions


Meikleriggs only has solutions since 2004 now :frown:
II/11:

Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be v22g\dfrac{v^2}{2g}.

P will hit the ground when its vertical displacement is βˆ’v22g-\dfrac{v^2}{2g}, so one must solve the quadratic equation vetβˆ’12gt2=βˆ’v22gv_et - \dfrac{1}{2}gt^2 = -\dfrac{v^2}{2g} giving t=ve+ve2+v2gt = \dfrac{v_e + \sqrt{v_e^2 + v^2}}{g} (using the quadratic formula). The horizontal distance travelled in this time will be u(ve+ve2+v2)g\dfrac{u(v_e + \sqrt{v_e^2 + v^2})}{g}.

Adding the distances: R=uvg+u(ve+ve2+v2g⇒gRu=v+ve+ve2+v2R = \dfrac{uv}{g} + \dfrac{u(v_e + \sqrt{v_e^2 + v^2}}{g} \Rightarrow \dfrac{gR}{u} = v + v_e + \sqrt{v_e^2 + v^2}, as required.

If R > D then v+ve+ve2+v2>Dguβ‡’ve2+v2>Dguβˆ’(v+ve)β‡’ve2+v2>D2g2u2βˆ’2Dgu(v+ve)+v2+2vve+ve2v + v_e + \sqrt{v_e^2 + v^2} > \dfrac{Dg}{u} \Rightarrow \sqrt{v_e^2 + v^2} > \dfrac{Dg}{u} - (v + v_e) \Rightarrow v_e^2 + v^2 > \dfrac{D^2g^2}{u^2} - \dfrac{2Dg}{u}(v + v_e) + v^2 + 2vv_e + v_e^2

β‡’2Dgveuβˆ’2vve>D2g2u2βˆ’2Dgvuβ‡’ve(2Dguβˆ’2v)>Dgu(Dguβˆ’2v)β‡’ve>Dg2u(Dguβˆ’2vDguβˆ’v)\Rightarrow \dfrac{2Dgv_e}{u} - 2vv_e > \dfrac{D^2g^2}{u^2} - \dfrac{2Dgv}{u} \Rightarrow v_e \left( \dfrac{2Dg}{u} - 2v \right) > \dfrac{Dg}{u} \left( \dfrac{Dg}{u} - 2v \right) \Rightarrow v_e > \dfrac{Dg}{2u} \left( \dfrac{\frac{Dg}{u} - 2v}{\frac{Dg}{u} - v} \right)

β‡’ve>Dg2u(Dgβˆ’2vuDgβˆ’vu)\Rightarrow v_e > \dfrac{Dg}{2u} \left( \dfrac{Dg - 2vu}{Dg - vu} \right), as required.
Reply 9
III Question 1

y=ln⁑(x+x2+1)β€…β€ŠβŸΉβ€…β€Šdydx=1+xx2+1x+x2+1=\displaystyle y = \ln\left(x+\sqrt{x^2+1}\right) \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} =
=1x2+1x2+1+xx2+1+x=1x2+1, given x2+1+x≠0, which is clearly true\displaystyle = \frac{1}{\sqrt{x^2+1}}\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}} \text{, given } \sqrt{x^2+1}+x \neq 0 \text{, which is clearly true}

Case when n=0n = 0:
(x2+1)(βˆ’x(x2+1)βˆ’3/2)+x(x2+1)βˆ’1/2=x(x2+1)βˆ’1/2(βˆ’1+1)=0(x^2+1)(-x(x^2+1)^{-3/2}) + x(x^2+1)^{-1/2} = x(x^2+1)^{-1/2}(-1+1) = 0.
Differentiating the given relation, we get:
(x2+1)y(n+3)+(2+(2n+1))xy(n+2)+(2n+1+n2)y(n+1)=0=(x^2+1)y^{(n+3)}+(2+(2n+1))xy^{(n+2)} + (2n+1+n^2)y^{(n+1)} = 0 =
=(x2+1)y((n+1)+2)+(2(n+1)+1)xy(n+2)+(n+1)2y(n+1)=0 = (x^2+1)y^{((n+1)+2)} + (2(n+1)+1)xy^{(n+2)} + (n+1)^2y^{(n+1)} = 0.
Since it is true for n = 0, and it is also true for n + 1, by mathematical induction we conclude that it must also be true for n = 1, n = 2, and so forth for all positive values of n.

Substituting x=0x = 0 into the relation, we get y(n+2)(0)=βˆ’n2y(n)(0)y^{(n+2)}(0) = -n^2y^{(n)}(0). Thus y(0)(0)=0β€…β€ŠβŸΉβ€…β€Šy(2k)(0)=0y^{(0)}(0) = 0\implies y^{(2k)}(0) = 0, all even powers are zero. y(1)(0)=1β€…β€ŠβŸΉβ€…β€Šy(3)(0)=βˆ’1β€…β€ŠβŸΉβ€…β€Šy(5)(0)=9β€…β€ŠβŸΉβ€…β€Šy(7)(0)=βˆ’225y^{(1)}(0) = 1\implies y^{(3)}(0) = -1 \implies y^{(5)}(0) = 9 \implies y^{(7)}(0) = -225.

Maclaurin series with terms of even powers equal to zero is y(x)β‰ˆxy(1)(0)+13!x3y(3)(0)+15!x5y(5)(0)+17!x7y(7)(0)y(x) \approx xy^{(1)}(0) + \frac{1}{3!}x^3y^{(3)}(0) + \frac{1}{5!}x^5y^{(5)}(0) + \frac{1}{7!}x^7y^{(7)}(0), so y(x)β‰ˆx+βˆ’13!x2+95!x5+βˆ’2257!x7=xβˆ’16x3+340x5βˆ’5112x7y(x)\approx x + \frac{-1}{3!}x^2 + \frac{9}{5!}x^5+\frac{-225}{7!}x^7 = x-\frac{1}{6}x^3 + \frac{3}{40}x^5 - \frac{5}{112}x^7.
Mark13
STEP I - Q13

Given that plate is broken, the probability that it is broken by the mathematician is 0.25.

Therefore, the number of plates, XX, broken by the mathematician can be represented by the following binomial distribution:It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.
Reply 11
Question 4, STEP I 2001

Spoiler

II/5:

dy/dx = 0 when x = 0, so there's a stationary point. d2ydx2=eβˆ’x2(2x4βˆ’5x2+1)\dfrac{\text{d}^2y}{\text{d}x^2} = e^{-x^2}(2x^4 - 5x^2 + 1), which equals 1 at x = 0, so it's a minimum point. Consider ∫x(1βˆ’x2)eβˆ’x2 dx\displaystyle \int x(1 - x^2)e^{-x^2} \, \text{d}x, letting u = x^2 gives du = 2x dx so it becomes
∫12eβˆ’uβˆ’12ueβˆ’u du\displaystyle \int \frac{1}{2}e^{-u} - \frac{1}{2}ue^{-u} \, \text{d}u. From inspection, y=12ueβˆ’uβ‡’y=12x2eβˆ’x2y = \frac{1}{2}ue^{-u} \Rightarrow y = \frac{1}{2}x^2e^{-x^2}. No constant as the curve passes through the origin. At x = 1, y = 12eβˆ’1\frac{1}{2}e^{-1} and d2ydx2=βˆ’2eβˆ’1\dfrac{\text{d}^2y}{\text{d}x^2} = -2e^{-1}; negative, so a maximum.

The other stationary point is (βˆ’1,12eβˆ’1(-1, \dfrac{1}{2}e^{-1} as it's an even function (which is to be expected as the derivative is odd).

For C_2: d2ydx2=eβˆ’x3(3x5βˆ’3x3βˆ’3x2+1)\dfrac{\text{d}^2y}{\text{d}x^2} = e^{-x^3}(3x^5 - 3x^3 - 3x^2 + 1). At x = 0, this equals 1; so minimum. At x = 1, this equals βˆ’2eβˆ’1-2e^{-1}, negative, so maximum.

The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that k>12eβˆ’1k > \frac{1}{2}e^{-1}. This relies on showing that eβˆ’x3>eβˆ’x2⇔ex2>ex3⇔x2<x3⇔x2(xβˆ’1)<0e^{-x^3} > e^{-x^2} \Leftrightarrow e^{x^2} > e^{x^3} \Leftrightarrow x^2 < x^3 \Leftrightarrow x^2 (x - 1) < 0 for 0 < x < 1, which can be shown with a very simple sketch. As e^(x^2) and e^(x^3) < 1 for 0 < x < 1, when taking logarithms the inequality is reversed.
Reply 13
STEP III 2001, QUESTION 3

Consider the equation
x2βˆ’bx+c=0x^2 - bx + c = 0
where b and c are real numbers.

(i) Show that the roots of the equation are real and positive if and only if b>0b > 0 and b2β‰₯4c>0b^2 \geq 4c > 0, and sketch the region of the b-c plane in which these conditions hold.

Spoiler



(ii) Sketch the region of the b-c plane in which the roots are real and less than 1 in magnitude.

Spoiler



(Apologies for any dodgy copypasting, notepad is a bitch to work with sometimes.)
DFranklin
It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.


Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
Reply 15
STEP I 2001, Question 5

Spoiler



Which happens to be true. (For more info, look into differentiation under the integral sign)
Reply 16
Aurel-Aqua
I Question 2

Part (i) 1+2xβˆ’x2>2x1+2x-x^2>\frac{2}{x}, xβ‰ 0x\neq 0.

First it might help to find the points of equality: multiplying by x and rearranging, we get x3βˆ’2x2βˆ’x+2=(xβˆ’2)(xβˆ’1)(x+1)=0x^3-2x^2-x+2 = (x-2)(x-1)(x+1)=0, xβ‰ 0x\neq 0. The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for βˆ’1<x<0-1 < x < 0 (the zero due to the asymptote) and 1<x<21<x<2.

Part (ii) 3x+10>2+x+4\sqrt{3x+10} > 2 + \sqrt{x+4}, xβ‰₯βˆ’103x \geq -\frac{10}{3}.

Because of the condition, squaring both sides still leaves an iff, giving: 3x+10>4+x+4+4x+4β€…β€ŠβŸΉβ€…β€Š2x+2>4x+4β€…β€ŠβŸΉβ€…β€Šx+1>2x+4 3x + 10 > 4 + x + 4 + 4\sqrt{x+4}\implies 2x + 2 > 4\sqrt{x+4} \implies x + 1 > 2\sqrt{x+4}. Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, x2+2x+1>4(x+4)β€…β€ŠβŸΉβ€…β€Šx2βˆ’2xβˆ’15>0x^2+2x+1 > 4(x + 4) \implies x^2 - 2x - 15 > 0, giving (x+3)(xβˆ’5)>0(x+3)(x-5)>0. From this, we can deduce that x>5x > 5 and x<βˆ’3 x<-3, but our new condition, x>βˆ’1x>-1 only gives us x>5x>5 as the final solution.


You're genius you know, I've got 98% in my C2 - but I was clueless in this question - you're really a maths genius :smile:
Glutamic Acid
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.
Yeah, I'd actually expect Mark13's method to give the right answer, I just don't think it's justified. I was a little surprised to get a different answer doing it by hand, and I probably have made a mistake - that's why I said I'd need to check.
Can someone do 2001 III Q8?
Reply 19
STEP I, Question 14

Spoiler



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