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STEP Maths I, II, III 1997 Solutions

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Step I Q3:

(i) a1=3,a2=277mod10a_1 = 3, a_2 = 27 \equiv 7 \mod 10. So a3733mod10a_3 \equiv 7^3 \equiv 3 \mod 10. So a4,a67mod10,a5,a73mod10a_4,a_6 \equiv 7 \mod 10, a_5,a_7 \equiv 3 \mod 10. So the last digit of a7a_7 is 3.

Unparseable latex formula:

\text{(ii) }a_2 = 27 > 10^1 \implies a_3 > (10^1)^3 = 10^3 \implies a_4 > 10^9 \implies a_5 > 10^27 \\[br]\implies a_6 > 10^{81} \implies a_7 > 10^{343} > 10^{100}

.

(iii) a7+12a7+1=12(2a7+1)+12a7+1=1/2+1212a7+1<1/2+10100.\displaystyle \frac{a_7+1}{2a_7+1} = \frac{1}{2}\frac{(2a_7+1)+1}{2a_7+1} = 1/2 + \frac{1}{2}\frac{1}{2a_7+1} < 1/2 + 10^{-100}.

Thus the fraction = 0.50 to 2d.p. (and rather a lot more!).

[Am I missing something? This seems too easy, and I expected to need to use (i) in answering (iii)]
Reply 21
STEP III

Q1.
(i)
(x^2 + 5x + 4)e^x = n=0\sum_{n=0}^{\infty} x^(n+2)/n! + 5 n=0\sum_{n=0}^{\infty} x^(n+1)/n! + 4 n=0\sum_{n=0}^{\infty} x^n/n!

= n=2\sum_{n=2}^{\infty} x^n/(n-2)! + 5 n=1\sum_{n=1}^{\infty} x^n/(n-1)! + 4 n=0\sum_{n=0}^{\infty} x^n/n!

= 4 + 9x + n=2\sum_{n=2}^{\infty} [(1/(n-1)! + 5/(n-1)! + 4/n!) x^n]

= 4 + 9x + n=2\sum_{n=2}^{\infty} [(n(n-1) + 5n + 4)/n! x^n]

= 4 + 9x + n=2\sum_{n=2}^{\infty} (n+2)^2/n! x^n

Now plug in x=1 to get:
10e = 4 + 3^2/1! + 4^2/2! + 5^2/3^2 + ..., as required.

(ii)
10e = n=0\sum_{n=0}^{\infty} (n+2)^2/n!

= n=0\sum_{n=0}^{\infty} (n^2 + 2n + 4)/n!

= n=0\sum_{n=0}^{\infty} (n+1)^2/n! + n=0\sum_{n=0}^{\infty} (2n+3)/n!

= [n=0\sum_{n=0}^{\infty} (n+1)^2/n!] + 3 + n=1\sum_{n=1}^{\infty} (2n+3)/n!

= [n=0\sum_{n=0}^{\infty} (n+1)^2/n!] + 3 + 2 n=1\sum_{n=1}^{\infty} 2/(n-1)! + 3 n=1\sum_{n=1}^{\infty} 1/n!

= [n=0\sum_{n=0}^{\infty} (n+1)^2/n!] + 3 + 2e + 3(e-1)

So
10e = [n=0\sum_{n=0}^{\infty} (n+1)^2/n!] + 3 + 5e - 3

=> 5e = n=0\sum_{n=0}^{\infty} (n+1)^2/n!, as required.

(iii)

n=0\sum_{n=0}^{\infty} (n+1)^3/n! = n=0\sum_{n=0}^{\infty} (n+1)(n+1)^2/n!

= n=0\sum_{n=0}^{\infty} n(n+1)^2/n! + n=0\sum_{n=0}^{\infty} (n+1)^2/n!

= 10e + 5e

= 15e
DFranklin
Inspirato: does it come out simpler if you write:

x+2x2+2x+2dx=x+1x2+2x+2+1x2+2x+2dx\displaystyle \int \frac{x+2}{x^2+2x+2} dx = \int \frac{x+1}{x^2+2x+2} + \frac{1}{x^2+2x+2}dx as the 1st bit is just 12ddxln(x2+2x+2)\frac{1}{2}\frac{d}{dx} ln(x^2+2x+2).

(And similarly for the other part).

I actually set essentially the same question as this for some students, once upon a while...


I often take the stupid route to the solution. Yes it is and i knew there would be a much simpler way to do than the way i did. Ill correct it later.

Luckly im quite careful so i do usually get it out. Im actually quite suprised i could do this question to be honest. Its STEP II and im not exactly that good at maths not good enough to do STEP anyways.
Step I 1997, Q6:

x4(1x)4=x84x7+6x64x5+x4x^4(1-x)^4 = x^8-4x^7+6x^6-4x^5+x^4. The easiest way of doing the next bit would probably be long division, but that will be painful to layout in TeX. So, let's compare coefficients...

Coeff of x8:a6=1x^8: a_6 = 1
Coeff of x7:a5=4x^7: a_5 = -4
Coeff of x6:a6+a4=6    a4=5x^6: a_6+a_4 = 6 \implies a_4 = 5
Coeff of x5:a5+a3=4    a3=0x^5: a_5+a_3 = -4 \implies a_3 = 0
Coeff of x4:a4+a2=1    a2=4x^4: a_4+a_2 = 1 \implies a_2 = -4
Coeff of x3:a3+a1=0    a1=0x^3: a_3+a_1 = 0 \implies a_1 = 0
Coeff of x2:a2+a0=0    a0=4x^2: a_2+a_0 = 0 \implies a_0 = 4
Coeff of x1:a1=0x^1: a_1 = 0 (Which we knew it should be, but it never hurts to check).
Coeff of x0:a0+b=0    b=4x^0: a_0+b = 0 \implies b=-4

Using this in the integral we get:

01x4(1x)41+x2dx=01x64x5+5x44x2+44/(1+x2)dx\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx = \int_0^1 x^6-4x^5+5x^4-4x^2+4 - 4/(1+x^2) dx

RHS = 1/74/6+14/3+4π=22/7π1/7 - 4/6 + 1 - 4/3 + 4 - \pi = 22/7 - \pi as required.

01x4(1x)4dx=01x84x7+6x64x5+x4dx=1/91/2+6/74/6+1/5\int_0^1 x^4(1-x)^4 dx = \int_0^1 x^8-4x^7+6x^6-4x^5+x^4 dx \\ = 1/9 - 1/2 + 6/7 - 4/6 + 1/5
=(70315+540420+126)/630=1/630 = (70 - 315 + 540 - 420 + 126) / 630 = 1 / 630

But 1+x2>1 for x>01+x^2 > 1 \text{ for } x > 0 and so 0<01x4(1x)41+x2<01x4(1x)4=1/6300 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} < \int_0^1 x^4(1-x)^4 = 1 / 630.

So since we know π=22/7x4(1x)41+x2\pi = 22/7 - \frac{x^4(1-x)^4}{1+x^2} we deduce 22/7>π>22/71/63022/7 > \pi > 22/7 - 1/630.

Funnily enough, I also set a question very similar to this one, once!
Reply 24
STEP III question 3:

The roots of the polynomial zn1z^n - 1 are 1,ω,ω2,1, \omega, \omega^2, \ldots and hence the polynomial can be written.

(z1)(zω)(zω2)(zωn1)=zn1(z - 1)(z - \omega)(z-\omega^2)\cdots(z -\omega^{n-1}) = z^n - 1.

Dividing by z1z - 1, and converting the formula for the sum of a geometric sequence into and actual sum of a geometric sequence, we get

(zω)(zω2)(zωn1)=1+z+z2+z3++zn1(z - \omega)(z - \omega^2)\cdots(z-\omega^{n-1}) = 1 + z + z^2 + z^3 + \cdots + z^{n-1}

as required.

Next part: Embedding the problem in the Argand plane, we can let A1,A2,,AnA_1, A_2, \ldots, A_n be represented by the complex numbers r,rω,rω2,,rωn1r, r\omega, r\omega^2, \ldots, r\omega^{n-1}, respectively. the point O then corresponds to the complex number 0.

But then, let z=ωz = \omega in the above proved formula. Then, the LHS becomes zero, and the sum in the RHS is precisely the sum of the position vectors of A1,A2,,AnA_1, A_2, \ldots, A_n, which must therefore also be zero, as required.

Let the point OO have coordinates (0,0)(0, 0), the point A1A_1 have coordinates (r,0)(r, 0).

Then the coordinates of AkA_k for 1kn 1 \leq k \leq n are given by (rcos(kπ/n),rsin(kπ/n))(r\cos(k\pi/n), r\sin(k\pi/n)).

Now we have

k=1nA1Ak2\displaystyle \sum_{k=1}^n |A_1A_k|^2

=k=1n((rcos(kπ/n)r)2+(rsin(kπ/n))2)\displaystyle = \sum_{k=1}^n \left((r\cos(k\pi/n) - r)^2 + (r\sin(k\pi/n))^2\right)

=k=1n(2r22r2cos(kπ/n))\displaystyle = \sum_{k=1}^n \left(2r^2 - 2r^2\cos(k\pi/n)\right)

But

k=1nrcos(kπ/n)=0\sum_{k=1}^n r\cos(k\pi/n) = 0

(This is basically the horizontal bit of the vector identity proved in the last part)

So

k=1nA1Ak2\displaystyle \sum_{k=1}^n |A_1A_k|^2

=k=1n2r2\displaystyle = \sum_{k=1}^n 2r^2

=2r2n\displaystyle = 2r^2n

as required.
Step I, 1997, Q7:

Fortunately, this question only asks us to find constants, it doesn't ask us to prove they are the only solution. If we guess that the answer will be in some sense symmetrical, this is about the shortest STEP question ever!

Consider what happens when P(x) = 1 (constant polynomial).

Then 11dt=a1+a2    a1+a2=2\int_{-1}^1 dt = a_1 + a_2 \implies a_1+a_2 = 2. Now I don't know about you, but at this point I'm thinking, surely we're going to have a1=a2=1a_1 = a_2 = 1. So let's assume that and hope for the best.

Consider P(x) = x. Then 11tdt=a1u1+a2u2    0=a1u1+a2u2\int_{-1}^1 t dt = a_1 u_1 + a_2 u_2 \implies 0 = a_1 u_1 + a_2 u_2. But we've assumed a1=a2=1a_1 = a_2 = 1 and so we find u1=u2u1 = - u2.

Consider P(x) = x^2. Then 11t2dt=a1u12+a2u22    2/3=a1u12+a2u22\int_{-1}^1 t^2 dt = a_1u_1^2 +a_2u_2^2 \implies 2/3 = a_1u_1^2 +a_2u_2^2

So if a1=a2=1a_1 = a_2 = 1 and u1=u2u_1 = -u_2, we get 2/3=2u122/3 = 2 u_1^2 and so we put u1=1/3,u2=1/3u_1 = -1/\sqrt{3}, u_2 = 1/ \sqrt{3}.

So we've found our coefficients, and we know they work for P(x) = 1, x, x^2.

Consider P(x) = x^3. The integral is odd so = 0, and our formula gives us u1^3+u2^3 = 0 as u1 = - u2.

So our formula works for 1,x,x^2,x^3 and so works for all cubic polynomials by linearity.
I'll do Paper II, Q4,Q8.
Step II, 1997 Q4:

Let the remainder when p(x) is divided by (x-a) be r, so p(x) = (x-a)q(x)+r for some polynomial q. Then p(a) = (a - a)q(a) + r = r.

(i) p(1)=3    r(1)=3.p(2)=1    r(2)=1.p(3)=5    r(3)=5p(1) = 3 \implies r(1) = 3. \quad p(2) = 1 \implies r(2) = 1. \quad p(3) = 5 \implies r(3) = 5. We know r(x) has degree < 3, so has the form Ax2+Bx+cAx^2+Bx+c
r(1)=3    A+B+C=3r(1) = 3 \implies A+B+C = 3 (Eq 1)
r(2)=1    4A+2B+C=1    3A+B=2r(2) = 1 \implies 4A+2B+C = 1 \implies 3A+B = -2 (Eq 2) (obtained by subtracting (Eq 1))
r(3)=5    9A+3B+C=5    5A+B=4r(3) = 5 \implies 9A+3B+C = 5 \implies 5A+B = 4 (Eq 3) (obtained by subtracting (Eq 2)).

Comparing (Eq 2) and (Eq 3), we see A = 3, so B = -11 and C = 11. So r(x) = 3x^2-11x+11.

(ii) If we didn't need a polynomial of degree (n+1), then p(x) = x would work, as p(a) = a for a = 0,...,n.
Analogously to (i), if we find a polynomial q of degree n+1 with q(a)=0 for a = 0,...,n; then q(x)+x is a polynomial of degree (n+1) with q(a) = a for a = 0,...,n.
Such a q(x) is x(x-1)(x-2)...(x-n). So our solution is x(x-1)(x-2)...(x-n)+x.
Reply 28
STEP I, question 5:

(Here, for not-getting-killed-by-LaTeX reasons, I'll use a,b,c,da, b, c, d to denote vectors a,b,c,d\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}. Furthermore, I won't write out the dot product operator symbol, so you should just assume abab means ab\mathbf a\cdot \mathbf b. And, I'll use the notation a2a^2 to mean aa\mathbf a \cdot \mathbf a.)

Let a, b, c, d be the position vectors of points A, B, C, D, respectively. Then position vectors of P, Q, R, S are (a + b)/2, (b + c)/2, (c + d)/2, (d + a)/2, respectively.

To prove:
AB2BC2+CD2DA2=2QS22PR2|AB|^2 - |BC|^2 + |CD|^2 - |DA|^2 = 2|QS|^2 - 2|PR|^2

We shall prove the equivalent form

2AB2+2CD24QS2=2BC2+2DA24PR22|AB|^2 + 2|CD|^2 - 4|QS|^2 = 2|BC|^2 + 2|DA|^2 - 4|PR|^2.

The LHS can be written in vector form as

2(ab)2+2(cd)2(b+cad)22(a-b)^2+2(c-d)^2 - (b+c-a-d)^2

=2(ab)2+2(cd)2(ba)2(cd)22(ba)(cd)= 2(a-b)^2 + 2(c-d)^2 - (b-a)^2 - (c-d)^2 - 2(b-a)(c-d)

=a22ab+b2+c22cd+d22bc2da+2ac+2bd= a^2 - 2ab + b^2 + c^2 - 2cd + d^2 - 2bc - 2da + 2ac + 2bd

=a2+b2+c2+d2+2ac+2bd2ab2bc2cd2da=a^2 + b^2 + c^2 + d^2 + 2ac + 2bd - 2ab - 2bc - 2cd - 2da

The RHS is

2(bc)2+2(ad)2(a+bcd)22(b-c)^2 + 2(a-d)^2 -(a+b-c-d)^2

=2(bc)2+2(ad)2(ad)2(bc)22(ad)(bc)= 2(b-c)^2 + 2(a-d)^2 - (a-d)^2 - (b-c)^2 - 2(a-d)(b-c)

=b22bc+c2+a22ad+d22ab2dc+2ac+2bd= b^2 - 2bc + c^2 + a^2 - 2ad + d^2 - 2ab - 2dc + 2ac + 2bd

=a2+b2+c2+d2+2ac+2bd2ab2bc2cd2da= a^2 + b^2 + c^2 + d^2 + 2ac + 2bd - 2ab - 2bc - 2cd - 2da

and so LHS = RHS, and we have proven

AB2BC2+CD2DA2=2QS22PR2=2QS22PR2|AB|^2 - |BC|^2 + |CD|^2 - |DA|^2 = 2|QS|^2 - 2|PR|^2 = 2|QS|^2 - 2|PR|^2.

Now, since QS2PR2|QS|^2 - |PR|^2 only depends on the lengths of the rods, and the rods are rigid, the expression is going to stay constans as the rods move.

Similarly, we shall prove

2ACBDcosθ=AB2+CD2BC2DA22|AC||BD|\cos\theta = |AB|^2 + |CD|^2 - |BC|^2 - |DA|^2

Written in vectors, this is

2(ac)(bd)=(bc)2+(da)2(ab)2(cd)22(a - c)(b - d) = (b-c)^2 + (d-a)^2 - (a-b)^2 - (c-d)^2

LHS:

2(ac)(bd)=2ab+2cd2bc2ad2(a-c)(b-d) = 2ab + 2cd - 2bc - 2ad

RHS:

(bc)2+(da)2(ab)2(cd)2(b-c)^2 + (d-a)^2 - (a-b)^2 - (c-d)^2

=b22bc+c2+d22ad+a2a2+2abb2c2+2cdd2 = b^2 - 2bc +c^2 + d^2 - 2ad + a^2 - a^2 + 2ab - b^2 - c^2 + 2cd - d^2

=2ab+2cd2bc2ad = 2ab + 2cd - 2bc - 2ad

and so LHS = RHS, and

2ACBDcosθ=AB2+CD2BC2DA22|AC||BD|\cos\theta = |AB|^2 + |CD|^2 - |BC|^2 - |DA|^2

Since ACBDcosθ|AC||BD|\cos\theta now only depends on the lengths of the rods, it will remain constant. QED.

(Actually, there are two angles θ\theta formed by AC and BD and depending on which one you choose, you'll end up with different values of ACBDcosθ|AC||BD|\cos\theta (one of the values is minus the other). But as long as you stick with the same angle, it's going to be the same scalar product, and so the value remains constant.)
Step II, 1997 Q8:

I'm not sure what the "standard explanation" is here. I think Siklos says you just need to draw a couple of vague wiggles and say the integral is the area under the curve. More formally, if f,g are cts, consider the integral of f-g, apply the fundamental theorem of calculus and the mean value theorem to show that ab(fg)(x)dx=(ba)[f(c)g(c)]\int_a^b (f-g)(x) dx = (b-a)[f(c)-g(c)] for some c(a,b)c \in (a,b) and so is greater than 0.

If p > q and x>=1 then xpxqx^p \ge x^q. So 1xtpdt1xtqdt\int_1^x t^p dt \ge \int_1^x t^q dt.
So xp1pxq1q\frac{x^p-1}{p} \ge \frac{x^q-1}{q} as desired.

If p > q and 0 <= x <1 then xpxqx^p \le x^q. So x1tpdtx1tqdt\int_x^1 t^p dt \le \int_x^1 t^q dt and so
1xpp1xqq\frac{1-x^p}{p} \le \frac{1-x^q}{q}. Multiplying by -1 confirms the earlier inequality for x >=0.

Doing the same trick on the inequality we've just found gives us
0x1p(xp1)dx0x1q(xq1)dx\int_0^x \frac{1}{p} (x^p - 1) dx \ge \int_0^x \frac{1}{q} (x^q - 1) dx and so

1p(xp+1p+1x)1q(xq+1q+1x)\frac{1}{p}\left(\frac{x^{p+1}} {p+1} - x \right) \ge \frac{1}{q}\left(\frac{x^{q+1}}{q+1} - x \right).

If x > 0, dividing by x gives the desired result, while if x=0 the result is obviously true.
Step II, Q7:

I'm not going to call this a solution, as I don't know how to do sketches here.

But y2=x2(a2x2)=a44(x2a22)2y^2 = x^2(a^2-x^2) = \frac{a^4}{4} - (x^2-\frac{a^2}{2})^2
and so the maximum value of y2y^2 is a4/4a^4/4, attained when x2=a2/2x^2 = a^2/2. This gives minimum, maximum values for y of ±a2/2\pm a^2/2.
Observe also y = 0 when x=0, x=a, and I think you've all you need for a sketch...
Reply 31
STEP I Q2

i
f(x)=arctanx+arctan(1x1+x)\displaystyle f(x) = arctanx + arctan\left(\frac{1-x}{1+x}\right)

f(x)=11+x2+11+(1x1+x)2×(1+x)(1x)(1+x)2\displaystyle f'(x) = \frac{1}{1+x^2} + \frac{1}{1+\left(\frac{1-x}{1+x}\right)^2} \times \frac{-(1+x)-(1-x)}{(1+x)^2}

f(x)=11+x22(1+x)2+(1x)2\displaystyle f'(x) = \frac{1}{1+x^2} - \frac{2}{(1+x)^2 + (1-x)^2}

f(x)=11+x222+2x2=0\displaystyle f'(x) = \frac{1}{1+x^2} - \frac{2}{2 + 2x^2} = 0

f(x)=C\therefore f(x) = C

f(1)=arctan1+arctan0=π4=Cf(1) = arctan1 + arctan0 = \frac{\pi}{4} = C

f(x)=π4\therefore f(x) = \frac{\pi}{4}

Hmm on second thought should this be (8n+1)pi/4?

ii
x=ysiny2x = ysiny^2

dxdy=siny2+y×2ycosy2=siny2+2y2cosy2\displaystyle\frac{dx}{dy} = siny^2 + y \times 2ycosy^2 = siny^2 + 2y^2 cosy^2

siny2=xy\displaystyle siny^2 = \frac{x}{y}

y2x2=y2y2sin2y2=y2(1sin2y2)=y2cos2y2y^2 - x^2 = y^2 - y^2sin^2y^2 = y^2(1-sin^2y^2) = y^2cos^2y^2

y2x2=ycosy2\sqrt{y^2 - x^2} = ycosy^2

dxdy=xy+2yy2x2=x+2y2y2x2y\displaystyle\frac{dx}{dy} = \frac{x}{y} + 2y\sqrt{y^2 - x^2} = \frac{x + 2y^2\sqrt{y^2-x^2}}{y}

dydx=yx+2y2y2x2\displaystyle\frac{dy}{dx} = \frac{y}{x + 2y^2\sqrt{y^2-x^2}}
I am currently attempting STEP I Q 13

I will post my reasoning soon, but would like someone to check it as I very well may have made mistakes...

Edit: OK here we go, it is probably a bit messy...

Probability for Mr's to plant the traps: Mr1 = 1/10, Mr2 = 2/10, Mr3 = 3/10, Mr4 = 4/10

The probabilities for the vicious attack canary can be disregarded as we get to know that Mr Blond hear it isn't such a thing in the room...

Probability for anaconda being at present: Mr1 = 2/7, Mr2 = 3/7, Mr3 = 4/7, Mr4 = 5/7

Probability for bomb being at present: Mr1 = 1/7, Mr2 = 2/7, Mr3 = 3/7, Mr4 = 4/7


Probability for placements of bombs: Mr1 = (1/10)(1/7), Mr2 = (2/10)(2/7), Mr3 = (3/10)(3/7), Mr4 = (4/10)(4/7)
Total probability of there being a bomb 1+22+32+4270=37 \frac{1+2^2+3^2+4^2}{70} = \frac{3}{7}

Probability for placements of anacondas: Mr1 = (1/10)(2/7), Mr2 = (2/10)(3/7), Mr3 = (3/10)(4/7), Mr4 = (4/10)(5/7)
Total probability of there being an anaconda 2+6+12+2070=47 \frac{2+6+12+20}{70}=\frac{4}{7}

Now Mr Blond has two choices, turn on the light or let it be turned off...
Probability of ending up dead by chosing:

Turning it on: (4/7)(1/3) by anaconda, (3/7)(1/2) by bomb, Total: 1742\frac{17}{42}

Leaving it turned off: (4/7)(1/2) by anaconda, (3/7)0 by bomb, Total: 1242\frac{12}{42}

This means that the probability of dying is higher if Mr Blond turns the light on, hence he should leave it off.

Edit2: small correction...
Reply 33
STEP III Question 9:

Let F1,F2F_1, F_2 be the tensions in the two springs AB, AC, and let their natural lengths be l1,l2l_1, l_2 and their stretched lengths be L1,L2L_1, L_2

Taking moments about the midpoint of BC, we get

F1sinB=F2sinCF_1\sin \angle B = F_2\sin \angle C

By the law of sines:

F1L2=F2L1F_1 L_2 = F_2 L_1

F1L1=F2L2\displaystyle \frac{F_1}{L_1} = \frac{F_2}{L_2}

Let

λ=F1L1=F2L2\displaystyle \lambda = \frac{F_1}{L_1} = \frac{F_2}{L_2}.

Now we have

l1l2=L1F1/κL2F1/κ=κL1F1κL2F2=κL1λL1κL2λL2=L1L2\displaystyle \frac{l_1}{l_2} = \frac{L_1 - F_1/\kappa}{L_2 - F_1 / \kappa} = \frac{\kappa L_1 - F_1}{\kappa L_2 - F_2} = \frac{\kappa L_1 - \lambda L_1}{\kappa L_2 - \lambda L_2} = \frac{L_1}{L_2}

as required.

(This seemed like a very easy question, considering it's on paper III. Well, so there might be hope after all...)
I'll do III Q2, Q5. I can see how to do 6, but it feels a bit of a grind, if no-one else wants to do it I'll have a go later.

Edit: whoops - didn't notice the sketch at the end of Q2. As I'm not going to do any sketches, maybe someone else should do it...
Reply 35
STEP I question 11

Change in kinetic energy = work done on the particle:

Δ(v22)=s1s2(kv2g)ds\displaystyle \Delta\left(\frac{v^2}{2}\right) = \int_{s_1}^{s_2} \left(- kv^2 - g\right)ds

Differentiating:

ddsv2=2kv22g\displaystyle \frac{d}{ds}v^2 = -2kv^2 - 2g

ddsv2+2kv2=2g\displaystyle \frac{d}{ds}v^2 + 2kv^2 = -2g

One particular solution to the equation is v2=g/kv^2 = -g/k. The corresponding homogeneous equation

ddsv2+2kv2=0\displaystyle \frac{d}{ds}v^2 + 2kv^2 = 0

has solutions

v2=Ce2ks\displaystyle v^2 = Ce^{-2ks}

Thus the general solution is

v2=Ce2ksgk\displaystyle v^2 = Ce^{-2ks} - \frac{g}{k}

At s=0s=0 we have v2=u2v^2 = u^2:

Unparseable latex formula:

\dislaystyle u^2 = C - \frac{g}{k}



C=u2+gk\displaystyle C = u^2 + \frac{g}{k}

Thus

v2=(u2+gk)e2ksgk\displaystyle v^2 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} - \frac{g}{k}

v2=u2e2ks+gk(e2ks1)\displaystyle v^2 = u^2e^{-2ks} + \frac{g}{k}\left(e^{-2ks} - 1\right)

as required.

The maximum height is when v=0v = 0:

0=(u2+gk)e2ksgk\displaystyle 0 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} - \frac{g}{k}

gk=(u2+gk)e2ks\displaystyle \frac{g}{k} = \left(u^2 + \frac{g}{k}\right)e^{-2ks}

e2ks=u2+g/kg/k\displaystyle e^{2ks} = \frac{u^2 + g/k}{g/k}

e2ks=u2kg+1\displaystyle e^{2ks} = u^2\frac{k}{g} + 1

s=ln(u2k/g+1)2k\displaystyle s = \frac{\ln \left(u^2k/g + 1\right)}{2k}.

On the downward path, gravity is increasing the particle's kinetic energy instead, and we get the differential equation

ddsv2=2g2kv2\displaystyle \frac{d}{ds}v^2 = 2g - 2kv^2

which clearly has a solution that's different from the given equation. To see this, the general solution is

Ce2ks+gk\displaystyle Ce^{-2ks} + \frac{g}{k}

instead. We have that v=0v=0 at

s=ln(u2k/g+1)2k\displaystyle s = \frac{\ln \left(u^2k/g + 1\right)}{2k}.

and so

0=C1u2k/g+1+gk\displaystyle 0 = C\frac{1}{u^2k/g + 1} + \frac{g}{k}

Unparseable latex formula:

\displaystyle C = -\frac{g/k}\left(u^2k/g + 1\right)



C=u2+gk\displaystyle C = u^2 + \frac{g}{k}

and

v2=(u2+gk)e2ks+gk\displaystyle v^2 = \left(u^2 + \frac{g}{k}\right)e^{-2ks} + \frac{g}{k}

v2=u2e2ks+gk(e2ks+1)\displaystyle v^2 = u^2e^{-2ks} + \frac{g}{k}\left(e^{-2ks} + 1\right)

thus the difference from when the particle was heading upwards is only that there is a plus sign instead of a minus signe before the '1'. But nevertheless it's different and so the equation (*) does not hold on the downward path.
Reply 36
Question 2, STEP III 1997

f(t)=lnttf(t)=1lntt2f(t)=0 at t=ee,e1 is a maximum point \\f(t)=\frac{\ln t}{t}\\f'(t)=\frac{1-\ln t}{t^2}\\ f'(t)=0 \text{ at } t=e\\ \therefore e,e^{-1} \text{ is a maximum point }

f(t)0, as tf(t)\rightarrow0, \text{ as } t\rightarrow\infty
Also, note that f(t) is negative for 0<t<1 and tends to -\infty as t tends to zero (Althought strictly speaking, f(t) is undefined at t=0). f(1)=0 and f(t) positive for t>1. Use these information to sketch your graph, remember to show distinctly that the x-axis is an asymptote for large t. Then from your graph, it can be seen that 2 values of t correspond to a positive value of f(t).

xy=yxylnx=xlnylnxx=lnyy\\x^{y}=y^{x}\\ y\ln x=x\ln y\\ \frac{\ln x}{x}=\frac{\ln y}{y}

i)There's a single (positive) value of y which satisfy xy=yxx^{y}=y^{x} for a given positive value of x within 0<x10<x\leq1 (And x=y) ii)There're 2 values of y which satisfy xy=yxx^{y}=y^{x} for a given positive value of x for 1<x<1<x<\infty (and one of the 2 values of y always is equivalent to x)
iii)At x=e, y=e, so there's only one value of y for x=e

Putting together these information, you should be able to sketch your final graph. Your curve must show:

i)A straight line from (not through) the origin with unit gradient. (This correspond to the set of solutions for which x=y for xy=yxx^{y}=y^{x}
ii)A curve with x=1 as asymptote for large y and y=1 as asymptote for large x. This curve intersects the straight line at x=e. This curve looks similar to the shape of y=1/x

EDIT:I'll type out solution to question 6 (STEP III), indeed, it's a rather unexciting question, hope i will survive typing it out.
Reply 37
STEP I question 14

F(x) and T(x) are the probability (density) of the maximum height being x, and the cost if the maximum heigh becomes x, respectively. The expected cost, say, C, is then given by

C=0f(x)T(x)dx\displaystyle C = \int_0^\infty f(x)T(x)dx

C=0yexydx+yex(y+r+s(xy))dx\displaystyle C = \int_0^y e^{-x}ydx + \int_y^\infty e^{-x}(y + r + s(x-y))dx

C=0yexdx+yex(r+sxsy)dx\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r + sx - sy)dx

C=0yexdx+yex(rsy)dx+ysxexdx\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r - sy)dx + \int_y^\infty sxe^{-x}dx

C=0yexdx+yex(rsy)dx+[sxex]y+ysexdx\displaystyle C = \int_0^\infty ye^{-x}dx + \int_y^\infty e^{-x}(r - sy)dx + \left[-sxe^{-x}\right]_y^\infty + \int_y^\infty se^{-x}dx

C=[yex]0+[(rsy)ex]y+[sxex]y+[sex]y\displaystyle C = \left[-ye^{-x}\right]_0^\infty + \left[-(r-sy)e^{-x}\right]_y^\infty + \left[-sxe^{-x}\right]_y^\infty + \left[-se^{-x}\right]_y^\infty

(Note that limxxex=0\lim_{x\rightarrow\infty} xe^{-x} = 0, since exe^x grows faster than xx)

C=y+(rsy)ey+syey+sey\displaystyle C = y + (r-sy)e^{-y} + sye^{-y} + se^{-y}

C=y+(r+s)ey\displaystyle C = y + (r+s)e^{-y}

Differentiating:

C=1(r+s)ey\displaystyle C' = 1 - (r+s)e^{-y}

which is strictly increasing. Setting C=0C' = 0

1=(r+s)ey\displaystyle 1 = (r + s)e^{-y}

ey=(r+s)\displaystyle e^y = (r + s)

we get the minimum at

Unparseable latex formula:

\displaytyle y = \ln (r + s)



as required.

If (r+s)<1(r + s) < 1, the minimum will still be at y=ln(r+s)y = \ln (r + s), only that since ln(r+s)<0\ln (r + s) < 0, the probability that this happens anyway is zero. If y < 0, they won't actually gain anything by lowering the prepared height (it is sensible to assume that they don't get any money back if they prepare for a negative height); the only thing that will happen is that if the actual water height is larger than zero, they will lose out more, because the (X-y) part of the cost is larger. So instead, they should aim at having y = 0. If y becomes larger than that, the derivative of the expected cost,

C=1(r+s)eyC' = 1 - (r + s)e^{-y}

will always be positive for y>=0. The expected cost is thus increasing for y >= 0, and so the lowest cost they can get is at y = 0. Thus the authorities shouldn't prepare at all when r + s < 1.
STEP II Question 11

At the ground, vtcosα=a+bvt\cos\alpha = a+b, Hvtsinα12gt2=0H-vt\sin\alpha - \frac{1}{2}gt^2=0. Substitute in t from horizontal equation into vertical one giving:
H(a+b)tanαg(1+b)22v2cos2α=0H-(a+b)\tan\alpha - \frac{g(1+b)^2}{2v^2\cos^2\alpha} = 0
v2=g(a+b)2sec2α2[H(a+b)tanα]v^2 = \frac{g(a+b)^2 \sec^2\alpha}{2[H-(a+b)\tan\alpha]}
v2=g(a+b)2(1+tan2α2[h(a+b)tanα]v^2 = \frac{g(a+b)^2(1+\tan^2\alpha}{2[h-(a+b)\tan\alpha]}
QED

At the net, vtcosα=avt\cos\alpha=a and Hvtsinα12gt2=0H-vt\sin\alpha - \frac{1}{2}gt^2=0. Again, subsititute in t from the horizontal equation, giving:
Hatanαga2sec2α2v2=hH-a\tan\alpha - \frac{ga^2\sec^2\alpha}{2v^2}=h
Substitute in v^2 from the first part, giving:
Hatanαa2[h(a+b)tanα](a+b)2=hH-a\tan\alpha - \frac{a^2[h-(a+b)\tan\alpha]}{(a+b)^2}=h
A little bit of rearranging to get tan(alpha) as the subject gives:
tanα=2a+ba(a+b)Ha+babh\tan\alpha = \frac{2a+b}{a(a+b)}H -\frac{a+b}{ab}h
QED

v^2 is always positive so H>(a+b)tanαH>(a+b)\tan\alpha (from the denominator of the first part). Substitute tan(alpha) from the previous section to get:
H<h(a+b)2b(ab)H<\frac{h(a+b)^2}{b(a-b)}

tan(alpha) is always positive too (the question says that the ball is hit downward, so 0<alpha<90). So:
H2a+ba(a+b)>a+babhH\frac{2a+b}{a(a+b)}>\frac{a+b}{ab}h
H>(a+b)2b(2a+b)hH>\frac{(a+b)^2}{b(2a+b)}h
I've done 5+6 from Paper III, but am finding it hard to get the time to type them up (baby daughter to take care of gives lots of time for pencil/paper work, but not so much for computer work...)

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