STEP Maths I, II, III 1988 solutions Watch

Glutamic Acid
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Shaky try at I/3.

OP, PQ and QO will form a triangle. The triangle with the longest shortest side will have its vertices touching the edges, as otherwise the triangle can always be stretched to increase the shortest side.

Using the sine rule, the shortest side is opposite the sine of the shortest angle etc. and the sine of the shortest side has the least value etc. Therefore, we want to maximize the smallest value of A+B+C = 180, which will be when A=B=C=60 degrees. Ie, an equilateral triangle.

Therefore, the triangle will look something like the sketch I've made.

\cos 15 = \dfrac{1}{h} \implies h = \frac{1}{\cos 15}

\cos 15 = \cos(45-30) = \cos 45 \cos 30 + \sin 45 \sin 30
= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2}
 = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}.

\implies h = \dfrac{2\sqrt{2}}{\sqrt{3}+1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} = \dfrac{2\sqrt{6} - 2\sqrt{2}}{2} = \sqrt{6} - \sqrt{2}.
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squeezebox
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(Original post by Glutamic Acid)
II/3.

- Let the roots be k, 1/k. The sum = (k^2+1)/k and the product = 1. Therefore b = -\dfrac{k^2+1}{k} and d = 1.

- Since we know c = 1, the quadratic will be of the form k^2 + bk + 1 = 0
Also (1-k)^2 + b(1-k) + 1 = 0 will satisfy it. This gives  1-2k+k^2+b-bk+1=0 \implies k^2 + (-2-b)k + (2+b) = 0

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is x^2 - x + 1 = 0.

- Call the roots of the equation \alpha, 1 - \alpha, \dfrac{1}{\alpha}.

Therefore
\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\

(1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\

\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0.

Multiplying through last equation by alpha^3 and dividing by r, you get:
\alpha^3 + \dfrac{q}{r}\alpha^2 + \dfrac{p}{r}\alpha + \dfrac{1}{r} = 0.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.
1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\

\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\

\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0.
I'm not trying to pick holes in your solution, but you may want to consider the case (in the first part), where the equation has roots -1 and 1. Which would make b zero.
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Glutamic Acid
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(Original post by squeezebox)
I'm not trying to pick holes in your solution, but you may want to consider the case (in the first part), where the equation has roots -1 and 1. Which would make b zero.
No problem . Thanks for that, I think I've gone about the first part in the wrong way.
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squeezebox
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STEP III - Question 2

au_{n+2} + bu_{n+1} + cu_{n} = 0

If  u_{n} = A\alpha^{n} + B\beta^{n}, where \alpha and \beta satisfy ax^{2} + bx +c = 0 (*), then au_{n+2} + bu_{n+1} + cu_{n} = aA\alpha^{n+2} + aB\beta^{n+2} + bA\alpha^{n+1} + bB\beta^{n+1} + cA\alpha^{n} + cB\beta^{n} = A\alpha( a\alpha^{2} + b\alpha + c) + B\beta^{n}(a\beta^{2} + b\beta + c) = 0 . (as alpha and beta satisfy (*).)

So the difference equation is satisfied by u_{n} = A\alpha^{n} + B\beta^{n}, for all A,B.

We also need:

 u_{0} = A+B and  u_{1} = A\alpha + B\beta , and so:

 B = \frac{u_{1} - \alpha u_{0}}{ \beta - \alpha} and  A = \frac{u_{1} - \beta u_{0}}{\alpha - \beta}.

~~~~~~~~~

Define u'_{n} to be a possible solution, so that u'_{n} \neq u_{n}. And define also:  w_{n} = u_{n} - u'_{n}.

Suppose that w_{k-1} = w_{k}=0. We shall show that w_{k+1} =0

We have: u_{k-1} - u_'{k-1} = 0  and u_{k} - u_'{k} = 0 .

Now: au_{k+1} + bu_{k} + cu_{k} = 0 and au'_{k+1} + bu'_{k} + cu'_{k} = 0.

\Rightarrow aw_{k+1} + bw_{k} + cw_{k} = 0

\Rightarrow w_{k+1} = 0, as  w_{k} = w_{k-1} = 0.

Also, for the determined values of A and B, the fist two terms are u_{0} and u_{1}, and so there is no alternative u'_{0} and u'_{1}. Hence by induction,  u_{n} are the only solutions.

~~~~~~~~

Put v_{n} = (n+1)t_{n}.

Then you get: (n+1)(n+2)(n+3)[ 8t_{n+2} - 2t_{n+1} - t_{n}] = 0

\Rightarrow 8t_{n+2} - 2t_{n+1} - t_{n} = 0 .

Let the roots of this difference equation be p and q. So from the first part, p and q must satisfy:  8x^{2} -2x -1 = 0, whose roots are: \frac{1}{2} and \frac{-1}{4}.

So let p be 0.5, and so q = -0.25. Then  t_{n} = C(\frac{1}{2})^{n} + D(\frac{-1}{4})^{n}. Where C = \frac{t_{1} - q t_{0}}{ p -q} and D = \frac{t_{1} - p t_{0}}{ q-p}.

So t_{n} = \frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n} .

Hence: v_{n} = (n+1)[\frac{4}{3}(\frac{1}{2})^{n} -\frac{4}{3}(\frac{-1}{4})^{n}].
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squeezebox
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(Original post by Glutamic Acid)
No problem . Thanks for that, I think I've gone about the first part in the wrong way.
I supose a small improvement would be letting the roots be a and b, and then say: if b = k, then for the condition to be satisfied k^{-1} is also a root. So then either b = \pm 1 or a = k^{-1}.
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nota bene
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I/15

First of all this question looks more or less trivial, so I'm possibly missing something


In Fridge football each team scores two points for a goal and one point for a foul committed by the opposing team. In each game, for each team, the probability that the team scores n goals is \frac{3-|2-n|}{9} for 0\le n \le 4 and zero otherwise, while the number of fouls committed against it will with equal probability be one of the numbers from 0 to 9 inclusive. The number of goals and fouls for each team are mutually independent. What is the probability that in some particular game a particular team gains more than half its points from fouls?
P(0 goals)=1/9 => For more than half points any foul is allowed 1-P(no foul)=9/10
P(1 goal)=2/9 => For more than half points: P(3 fouls or more)=7/10
P(2)=3/9=1/3 => For more than half points: P(5 fouls or more)=5/10=1/2
P(3)=2/9 => For more than half points: p(7 fouls or more)=3/10
P(4)=1/9 => For more than half points: P(9 fouls)=1/10

So probability that a team in a game scores more than half its points from fouls committed against it is: \frac{1}{9}\frac{9}{10}+\frac{2}  {9}\frac{7}{10}+ \frac{1}{3}\frac{1}{2}+\frac{2}{  9}\frac{3}{10}+ \frac{1}{9}\frac{1}{10}= \frac{45}{90}=\frac{1}{2}

In response to criticisms that the game is boring and violent, the ruling body increases the number of penalty points awarded for a foul, in the hope that this will cause large numbers of fouls to be less probable. During the season following the rule change, 150 games are played and on 12 occasions (out of 300) a team committed 9 fouls. Is this good evidence for a change in the probability distribution of the number of fouls? Justify your answer.
\frac{12}{300}=\frac{1}{25} which is larger than 1/10, so there has been a change in the probability for 9 fouls, which suggests the strategy has worked, although not a very drastic decrease. However, only knowing data for 9 fouls creates a bit of insecurity as for the accuracy of our findings.
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brianeverit
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Here are some of my solutions for paper I
Numbers 4 6 and 8 attached.
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brianeverit
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Here are 10, 11,12,13,14, and 16 on Paper I
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SimonM
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Does anyone have this paper anywhere? (If so could they send it to me )
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Horizontal 8
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(Original post by SimonM)
Does anyone have this paper anywhere? (If so could they send it to me )

Check your PM inbox
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SimonM
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(Original post by Horizontal 8)
Check your PM inbox
Done
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SimonM
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STEP II, Question 2

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2a-3y = \dfrac{(z-x)^2}{y} and 2a-3z = \dfrac{(x-y)^2}{z}

The first two equations give us

2ay - 3y^2 = (z-x)^2 and
2az - 3z^2 = (x-y)^2

Subtracting yields

2a(y-z)-3(y^2-z^2)= (z-x)^2 - (x-y)^2

Which is

(y-z)(2a - 3(y+z)) = (z-y)(z+y-2x)

Since y \not = z

2a - 3y - 3z = -z-y + 2x \Rightarrow \boxed{x+y+z = a}

2a - 3y- 3x = 2z-x-y

So 2a(y-x) -3(y-x)(y+x) = (y-z)(2z-x-y)

So 2a(y-x)-3(y^2-x^2) = (z-x)^2 - (y-z)^2

Subtracting from 2ay - 3y^2 = (z-x)^2 gives

2ax - 3x^2 = (y-z)^2 \Rightarrow \boxed{2a - 3x = \frac{(y-z)^2}{x}}

No, take x = \frac{2}{3} a, y = z = \frac{1}{6} a we can easily verify this satisfies all equations.
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SimonM
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STEP II, Question 6

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(i) y = x-\tanh x \ge 0

y' = 1 - \sech^2 x = \tanh^2 x \ge 0

y(0) = 0 - 0 = 0

Therefore x > 0 \Rightarrow y > 0 as y is increasing

(ii) y = x \sinh x -2 \cosh x + 2 \ge 0

y' = \sinh x + x \cosh x - 2 \sinh x = x \cosh x - \sinh x = \cosh x \left ( x - \tanh x ) \ge 0 (by (i))

Therefore x > 0 \Rightarrow y > 0

(iii) y = 2x \cosh 2x - 3 \sinh 2x + 4x \ge 0

y' = \cosh u + u \sinh u - 3 \cosh u + 2 = u \sinh u - 2\cosh u + 2 \ge 0 (by (ii))

Consider \ln f(x) = \ln x +\frac{1}{3} \ln \cosh x - \ln \sinh x

Differentiating,
\displaystyle \frac{f'(x)}{f(x)} = \frac{1}{x} + \frac{1}{3} \tanh x - \cotanh x

Since f(x) \ge 0 if there are turning points that expression is zero

\displaystyle \frac{f'(x)}{f(x)} = \frac{1}{3 x \cosh x \sinh x} \left ( 3 \sinh x \cosh x + x \sinh^2 x - x 3\cosh ^2x \right ) =


\displaystyle \frac{1}{3 x \cosh x \sinh x} \left ( \frac{3}{2} \sinh 2x - 2x -x \cosh 2x \right ) < 0 (by (iii))

And we're done
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SimonM
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STEP II, Question 9

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Let G_1, G_2 \leq G. Given G_1 \cup G_2 = G we have |G| = |G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|

We know Lagrange's Theorem states that the order of a subgroup divides the order of a group. If, for the sake of contradiction, neither of G_1,G_2 is G then |G_1|, |G_2| \leq |G|/2

However, |G_1\cap G_2| \ge 1 since e \in G_1\cap G_2

Therefore |G_1| + |G_2| - |G_1 \cap G_2| \leq |G| - 1 < |G|. Contradiction.

For our example, consider the Klein four-group, an abelian group, order 4, where ever element has order two. That is H = \{1,a,b,ab\}, a^2 = b^2 = 1

Consider the subgroups H_1 = \{1,a\},H_2 = \{1, b\}, H_3 = \{1, ab\}.


Why aren't group theory questions in STEP any more :'(
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SimonM
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STEP III, Question 9

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(i) Proceed by contradiction. Suppose n \ndiv N

Therefore kn < N < (k+1)n \Rightarrow 0 < N-kn < n

But g^{N-kn} = g^N  (g^n)^{-k} = e. Which contradicts n's minimality. Therefore n|N

(ii) There is a trivial way of doing this using induction and associativity. However, we can say something stronger, which proves this result. That is \phi ( h) = ghg^{-1} (conjugation) is an Inner automorphism. (This isn't hard to verify) and \phi  (x)^m = \phi (x^m) for any homomorphism.

(iii) ghg^{-1} = h^2 \Rightarrow h^4 = (ghg^{-1})^2 = gh^2g^{-1} = g(ghg^{-1})g^{-1} = g^2hg^{-2} as required.

h^8 = (gh^2g^{-1})^2 = gh^4g^{-1}= (g^2hg^{-2})^2 = g^2h^2g^{-2} = g^3hg^{-3}
h^{16} = (gh^4g^{-1})^2 = gh^{8}g^{-1} = (g^3hg^{-3})^2 = g^3h^2g^{-3} = g^4hg^{-4}
h^{32} = (gh^8g^{-1})^2 = gh^{16}g^{-1} = (g^4hg^{-4})^2 = g^4h^2g^{-4} = g^5hg^{-5} = h

Therefore h^{31} = e. If the order of h is less than 31, then it divides 31. However, 31 is prime. Therefore it is the order of h.
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maltodextrin
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(Original post by Glutamic Acid)
II/3.

- Both k and 1/k will satisfy the equation.
x^2 + bk + c = 0 \\

\dfrac{1}{k^2} + \dfrac{b}{k} + c = 0 \implies 1 + bk + ck^2 = 0 \implies k^2 + \dfrac{b}{c}k + \dfrac{1}{c} = 0.
So 1/c = c hence c^2 = 1 and c = +-1.
And b/c = b so b = cb. When c = 1, b can be any real value. When c = -1, b = 0.

- Since we know c = 1, the quadratic will be of the form k^2 + bk + 1 = 0, as k^2 - 1 = 0 does not satisfy the conditions (because -1 is a root, but 1--1 = 2 isn't a root).
Also (1-k)^2 + b(1-k) + 1 = 0 will satisfy it. This gives  1-2k+k^2+b-bk+1=0 \implies k^2 + (-2-b)k + (2+b) = 0

Comparing coefficients, -2-b = b ==> b = -1 This also satisfies 2 - 1 =1. So the quadratic is x^2 - x + 1 = 0.

- Call the roots of the equation \alpha, 1 - \alpha, \dfrac{1}{\alpha}.

Therefore
\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\

(1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\

\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0.

Multiplying through last equation by alpha^3 and dividing by r, you get:
\alpha^3 + \dfrac{q}{r}\alpha^2 + \dfrac{p}{r}\alpha + \dfrac{1}{r} = 0.

Comparing coefficients, 1/r = 1 so r = 1.
q/r = p p/r = q so p = q.

Replacing q by p and expanding the second equation we get.
1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\

\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\

\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0

Comparing coefficients, -3-p = p ==> 2p = -3 and p = -3/2. Since q = p, q = -3/2.
So the cubic equation is x^3 - \dfrac{3}{2}x^2 - \dfrac{3}{2}x + 1 = 0.
I'm probably missing something really obvious here but I'm confused as to how c can equal -1 in the first part. I understand how you've come to that conclusion but if c = -1 and b = 0 then the equation is x^2 - 1 = 0 with roots 1 and -1. But -1 isn't 1/1 so doesn't that mean the conditions aren't satisfied?
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Glutamic Acid
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(Original post by maltodextrin)
I'm probably missing something really obvious here but I'm confused as to how c can equal -1 in the first part. I understand how you've come to that conclusion but if c = -1 and b = 0 then the equation is x^2 - 1 = 0 with roots 1 and -1. But -1 isn't 1/1 so doesn't that mean the conditions aren't satisfied?
That's true, and on reflection my method of doing it was... odd. It's clear the product of the roots = c = k*k^(-1) = 1. I must have been on crack. Thanks.
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maltodextrin
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(Original post by Glutamic Acid)
That's true, and on reflection my method of doing it was... odd. It's clear the product of the roots = c = k*k^(-1) = 1. I must have been on crack. Thanks.
Ahh thanks that's a relief, I thought i must be wrong. To be fair the method you used seems to be the only way of solving the second part of the question so I can see why you'd use it
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cheena
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(Original post by nota bene)
STEP I Q1

h(x)=\frac{\log x}{x}
h'(x)=-\frac{\log x}{x^2}+\frac{1}{x^2}=\frac{1}{x  ^2}(1-\log x)
For max/min set h'(x)=0 i.e. 0=\frac{1}{x^2}(1-\log x) \Rightarrow x=e
To find the nature of the stationary point, consider e.g. h'(e-0.2)=positive and h'(e+0.2)=negative hence a (global) maximum. For sketching purposes might be worth noting that \displaystyle\lim_{x\to0^+}h(x)=-\infty and \displastyle\lim_{x\to\infty}h(x  )=0. Graph see attached (Mathematica because of lack of scanner:p:).

Solving n^m=m^n is equivalent of solving \log n^m=\log m^n\Leftrightarrow mlog n=n\log m \Leftrightarrow \frac{\log n}{n}=\frac{\log m}{m} (last step valid as neither m nor n can be 0).
m=n is the trivial solution.
Consider the line f=c (i.e. a constant). For 0< c< \frac{1}{e} there are two solutions. There's one solution when c=\frac{1}{e} or c\le0. And for c>\frac{1}{e} there are no solutions.

Now if m\not=n then \frac{n}{m}=k (k\not=1).

\frac{\log n}{n}=\frac{\log \frac{n}{k}}{\frac{n}{k}}
\log n=\frac{nk(\log n-\log k)}{n}=k(\log n - \log k)
\log n=\frac{k}{k-1}\log k
Thus n=k^{\frac{k}{k-1}} and m=k^{\frac{                  1}{k-1}}.

(What am I missing here, this looks trivial for a STEP question...)

----

STEP I Q2

g(x)=f(x)+\frac{1}{f(x)} (f(x)\not=0)
g'(x)=f'(x)-\frac{f'(x)}{(f(x))^2}=f'(x)(1-\frac{1}{(f(x))^2})
g''(x)=f''(x)(1-\frac{1}{(f(x))^2})+f'(x)(\frac{  2f'(x)}{(f(x))^3})=f''(x)(1-\frac{1}{(f(x))^2})+\frac{2(f'(x  ))^2}{(f(x))^3}

Let f(x)=4+\cos(2x)+2\sin(x)
f'(x)=-2\sin(2x)+2\cos(x)=2\cos(x)(-2\sin(x)+1)
Now WLOG g'(x)=0\Leftrightarrow f'(x)=0 \text{or} (f(x))^2=1

f'(x)=0 \Leftrightarrow 2\cos(x)(1-2\sin(x))=0 Has solutions cos(x)=0 and sin(x)=1/2
i.e. x=\frac{\pi}{2},\frac{3\pi}{2},\  frac{\pi}{6},\frac{5\pi}{6}

f(x)=1\Leftrightarrow 1=4+1-2\sin^2(x)+2\sin(x) \Leftrightarrow 0=2(2-u^2+u) where u=\sin(x)
By quadratic formula the solutions are u=\frac{1\pm\sqrt{1+4\times2}}{2  } i.e. u=-1, 2
\sin(x)=2 has no real solutions and \sin(x)=-1 has the root x=\frac{3\pi}{2} (which we have already found).

f(x)=-1\Leftrightarrow -1=4+1-2\sin^2(x)+2\sin(x)\Leftrightarr  ow 0=2(3-u^2+u) where u=\sin(x)
Solutions to this are u=\frac{1\pm\sqrt{1+4\times3}}{2  } and as \frac{1\pm\sqrt{1+4\times3}}{2}>  1 sin(x)=u will have no real solutions.

f'(\frac{\pi}{2}-\frac{\pi}{16})<0, f'(\frac{\pi}{2}+\frac{\pi}{16})  >0 so a (local)min f(\frac{\pi}{2})=4-1+2=5 \Rightarrow g(\frac{\pi}{2})=5+\frac{1}{5}
f'(\frac{3\pi}{2}-\frac{\pi}{16})<0, f'(\frac{3\pi}{2}+\frac{\pi}{16}  )>0 so a (global) min f(\frac{3\pi}{2})=4-1-2=1 \Rightarrow g(\frac{3\pi}{2})=2
f'(\frac{\pi}{6}+\frac{\pi}{18})  <0, f'(\frac{\pi}{6}-\frac{\pi}{18})>0 so a (global) max f(\frac{\pi}{6})=4+\frac{\sqrt{2  }}{2}+1=5+\frac{\sqrt{2}}{2} \Rightarrow g(\frac{\pi}{6})=5+\frac{\sqrt{2  }}{2}+\frac{1}{5+ \frac{\sqrt{2}}{2}}
f'(\frac{5\pi}{6}+\frac{\pi}{18}  )<0,f'(\frac{5\pi}{6}-\frac{\pi}{18})>0 so a (local) max f(\frac{5\pi}{6})=4+\frac{1}{2}+  1=\frac{11}{2} \Rightarrow g(\frac{5\pi}{6})=\frac{11}{2}+\  frac{2}{11}

could someone explain the reasoning behind the last bit to question 2 (the nature of the points)

i also dont get the last partof question 1, dont we just see from the graph the only possible values that one of m or n must be is 2 (ignoring m=n) because that is the only integer between e and 1, and we just use trial and error for the other value???

thanks:cool:
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cheena
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