STEP maths I, II, III 1990 solutions Watch

brianeverit
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#181
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(Original post by *bobo*)
STEP I
10) i)total resistive forces= 50+2(v-3)= 2(22 +v) where v is pos. in direction of turning point from starting point.
using P=Fv, where F=resistive forces due to no acceleration
300=2v(22+v)
2v^2 + 44v- 300=0
v=0.25 (in forward journey), v=1.25(backward journey) to 2dp
using v=s/t:
total time for race= 5000/0.25 + 5000/1.25
= 24000 seconds (supposed to answer to nearest second but don't have a calc on me)


ii) total energy expended using P=W/t
W=300x24000=7200000

x(1)t(1)+x(2)t(2)= 7200000

x(1)= v(1)(50+2(v(1)-3))= v(1)(44 + 2v)
t(1)= 5000/x(1)

x(2)= v(2)(50+2(v(2) +3))=v(2)(56+2v(2))
t(2)=5000/v(2)

=> 5000(44+2v(1))+5000(56+2v(2))=72 00000

=> 100+ 2v(1) + 2v(2)= 1440
=> v(1) +v(2)=670 so is independant of x(1) and x(2) values.
I don't think it would take over 6 hours to run 10000 metres.
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cxs
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#182
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the last part can be derived naturally by written k as a function of p,and we find the range when p<0
(Original post by SimonM)
STEP II Question 7

f'(t)+f(t)-kf(t-1)=0

If there is a solution of the form f(t) = Ae^{pt} then

Ape^{pt} + Ae^{pt} -kAe^{p(t-1)} = 0

That is pe^{pt} +e^{pt} = ke^{p(t-1}}

Or \boxed{p+1 = ke^{-p}} as required.

If k &gt; 0 there will be one solution. (I'm not going to stick my graphs up, (no scanner)) (However, visualise the graphs y=x+1 and y = ke^{-x}

If our two graphs are tangential, there will be one solution (this is also the critical point between two solutions and no solutions for k&lt;0

So \frac{d}{dx} (x+1) = 1 and \frac{d}{dx} (ke^{-x}) = -ke^{-x}

That gives us 1 = -ke^{-x} \Rightarrow x = \ln (-k)

Substituting back in gives us

\ln (-k)+1 = ke^{-\ln (-k)} \Rightarrow \ln(-k)+1 = -1 \Rightarrow \boxed{k = -e^{-2}}

So we have

0 solutions if k &lt; -e^{-2}
1 solution if k = -e^{-2} or k &gt;0
2 solutions if 0&gt;k&gt;-e^{-2}

The final question asks us to find where the solutions are negative.

Consider the critical values when x = 0 is a solution. This gives us k = 1

When k&lt;0, ke^{-x} &lt; 0 so for this to equal x+1, we must have x&lt;-1, which gives a negative p

When k&gt;0, if ke^{-x}&lt;x+1, consider at x=0, then the solution will be negative, that is k&lt;1

So this gives negative p for \boxed{k&lt;1}
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cxs
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#183
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EXM, why in the first part, m is neglected?
I think the better way is to suppose ka-c/ka-b= ka-c^2/ka-b^2=ka-c^3/ka-b^3
thus we can have one variable, and by substitute finally we have (a-b)(a-c)=0 and that is an contradiction.
(Original post by Simba)
III/2:

i) OA = m(a^3 i + a^2 j + ak)

BC = b^3 i + b^2 j + bk + n[(c^3 - b^3)i + (c^2 - b^2)j + (c - b)k]

For an intersection:

b^3 + n(c^3 - b^3) = a^3. --- (1)
b^2 + n(c^2 - b^2) = a^2. --- (2)
b + n(c - b) = a. --- (3)

Sub (3) in (2):

b^2 + n(c^2 - b^2) = [b + n(c - b)]^2 = b^2 + 2bn(c - b) + n^2.(c - b)^2.

c^2 - b^2 = 2b(c - b) + n(c - b)^2.

(c + b)(c - b) - 2b(c - b) - n(c - b)^2 = 0.

(c - b)[c + b - 2b - n(c - b)] = 0.

(c - b)[c - b - n(c - b)] = 0.

(c - b)(c - b)(1 - n) = 0.

But c =/= b, so n = 1 (if they intersect).

Sub in (3):

b + c - b = a.

c = a.

But c =/= a, so the lines don't intersect.

QED.

ii) cos(AOB) = a.b/|a||b| = [(ab)^3 + (ab)^2 + ab]/root(a^6 + a^4 + a^2).root(b^6 + b^4 + b^2)

= [(ab)^3 + (ab)^2 + ab]/ab.root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)]

= [(ab)^2 + ab + 1]/root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)].

= 3/root(a^4.b^4 + a^4.b^2 + a^4 + a^2.b^4 + a^2.b^2 + a^2 + b^4 + b^2 + 1)

= 3/root(3 + a^2 + a^4 + b^2 + a^2 + b^4 + b^2)

= 3/root(a^4 + b^4 + 2a^2 + 2b^2 + 3)

Since a and b are variable, we can make a (or b) as large as we like, so the denominator can approach infinity. So cos(AOB) can be made arbitrarily close to 0. Values of a and b very close to 1 (but not equal to 1 since a and b are distinct) give cos(AOB) arbitrarily close to 1. It is not possible for a sum of squares (+3) to be negative, so cos(AOB) > 0, outruling pi/2 < AOB < pi.

Hence 0 < cos(AOB) < 1, and so 0 < AOB < pi/2.

QED.
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cxs
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#184
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we can continue the method of the complex numbers in part 3, quite easy
let z=cosθ+isinθ, and let w=secθ
LHS=Re(1+zw+z^2w^2+...)=Re (1-(zw)^n)/1-zw, which we can calculated easily
(Original post by SimonM)
Question 4

\begin{array} {l}

\displaystyle \cos \alpha + \cos ( \alpha + 2 \beta ) + \ldots + \cos ( \alpha + 2 n \beta ) \\

\displaystyle = \rm Re \, \left ( e^{i\alpha} + e^{i(\alpha+2\beta)} + \ldots + e^{i(\alpha+2n\beta)} \right ) \\

\displaystyle = \rm Re \,  \left ( e^{i\alpha} ( 1+e^{i2\beta}+\ldots+e^{i2n\beta  }) \right ) \\

\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (\frac{e^{i2(n+1)\beta} - 1}{e^{i2\beta}-1} \right ) \right ) \\

\displaystyle = \rm Re \, \left ( e^{i\alpha} \frac{e^{i(n+1)\beta}}{e^{i\beta  }} \frac{e^{i(n+1)\beta}-e^{-i(n+1)\beta}}{e^{i\beta}-e^{-i\beta}} \right ) \\

\displaystyle = \rm Re \, \left ( e^{i (\alpha + n\beta)} \frac{\sin(n+1)\beta}{\sin \beta} \right ) \\

\displaystyle = \frac{\sin (n+1)\beta \cos (\alpha + n \beta)}{\sin \beta} \end{array}

\begin{array}{l}

\displaystyle \cos \alpha + \binom{n}{1} \cos ( \alpha + 2\beta ) + \ldots + \binom{n}{r} \cos ( \alpha + 2r \beta ) + \ldots + \cos ( \alpha + 2n \beta ) \\

\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (1+e^{i2\beta} \right )^n \right ) \\

\displaystyle = \rm Re \, \left ( e^{i \alpha} \left ( 2 \cos \beta e^{i \beta} \right )^n \right ) \\

\displaystyle = \rm Re \, \left ( e^{i( \alpha + n \beta )} 2^n cos ^n \beta \right ) \\

\displaystyle = 2^n \cos^n \beta \cos ( \alpha + n \beta) 

\end{array}

For our final one, we proceed by induction.

Theorem

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Proof (by Induction)

Base Case

When n=0

LHS = 1
\displaystyle RHS = \frac{\sin \theta \times 1}{\sin \theta} = 1 = LHS as required

Inductive Hypothesis

Assume that, for all k\leq n,

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos k \theta \sec ^k \theta = \frac{\sin (k+1)\theta \sec^k \theta}{\sin \theta}

Inductive Step

If \displaystyle 1+\cos \theta \sec \theta + \ldots + \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Then by adding \cos (n+1)\theta \sec^{n+1} \theta to each side we get

\begin{array}{l} \displaystyle 1+\cos \theta \sec \theta + \ldots  + \cos n \theta \sec ^n \theta +\cos (n+1)\theta \sec^{n+1} \theta  \\ 

\displaystyle = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}  + \cos (n+1)\theta \sec^{n+1} \theta \\

=  \displaystyle \frac{\sin(n+1)\theta \sec^n \theta + \sin \theta \cos (n+1) \theta \sec^{n+1}}{\sin \theta} \\

=  \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \left ( \sin(n+1)\theta \cos \theta + \sin \theta \cos (n+1) \theta \right ) \\

=  \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \sin(n+2)  \end{array}

Therefore, if our inductive hypothesis is true for n, it is true for n+1. Since it is true for 0, by the principle of mathematical induction, it is true for all nonnegative integers
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cxs
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#185
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i find it pretty tricky to do the last part of Ⅱ13
we just need“the length of smaller 2sides of the triangle is larger the largest” to get what we want
and it‘s the key of this problem,rather than the angle---i think
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