# SOLUTIONS - Terry Lee (1 Viewer)

#### seanieg89

##### Well-Known Member
is he wrong?
For that last part of the last question yeah, thats all I looked at.

The integer closest to root(n) doesn't have to be the same as the integer closest to root(n)+1/2. Eg n=2.

#### jyu

##### Member
For that last part of the last question yeah, thats all I looked at.

The integer closest to root(n) doesn't have to be the same as the integer closest to root(n)+1/2. Eg n=2.
But it was clearly stated that 4n+1 is not a perfect square.

#### seanieg89

##### Well-Known Member
But it was clearly stated that 4n+1 is not a perfect square.
Take n=5 then, same problem. My point was that the last step is not valid in general, and the square condition does not change this.

#### seanieg89

##### Well-Known Member
He has fixed it now though.

#### RealiseNothing

##### what is that?It is Cowpea
Last two lines:

P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}+1/2$ so P(k) is greatest when k is the integer closest to $\bg_white \sqrt{n}$

Nice try terry.
hahahhahaha

lost it at "nice try terry".

#### No no no

##### New Member
He has just added this footnote: If n = 2, k < sqrt 2 + 0.5 = 1.92, but not more than 1.92, so k = 1, i.e. closest to sqrt 2.
I think he was right before. What do you think, seanieg89?

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#### seanieg89

##### Well-Known Member
Indeed the footnote and the current solution are correct. My objection was the two lines quoted in my original post, P(k) is NOT necessarily maximised at the closest integer to root(n)+1/2, nor would this necessarily imply that P(k) is maximised at the integer closest to root(n).

The key is that P(k) is maximised at the floor of root(n)+1/2.

Also his 'solution' to 15biii) is pretty weird. How can the proposition that P(z) has real zeroes depend on z? k is the parameter that determines the nature of the polynomial. And if z is meant to be the postulated real root of the polynomial rather than just a complex variable, then how can we say z is not equal to 1 or -1 (the step he uses to deduce that k=+-1? In fact any real roots MUST be + or - 1!)

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#### Fus Ro Dah

##### Member
The footnote is 'correct', but for totally the wrong reason. The only reason why it is 'correct' in this case is because we are considering three functions that have a uniformly equidistant from each other. This distance happens to be less than 1 and therefore the conclusion is made.

No no no (or should I say Terry Lee as it is quite obvious it is you): Proving an upper bound is not sufficient to definitely conclude that the square root of n is the closest integer. Why do you think there exists the Squeeze Theorem? Why can't they just prove one direction of the inequality and since it converges to something, that is therefore the solution? Take a look at the 2010 HSC versus the 2002 HSC. Why can't we just suddenly conclude, from the 2002 HSC, that the limit is pi^2/6? What is it about the 2010 HSC that makes it more valid? Why do you think at the very end of the paper, they specifically stated that you can assume the other direction of the inequality?

It is very clear that both directions are required in order for it to be 'apt to declare' that P(k) is maximised, and I don't know why you are trying to justify your claim here without, at the very least, saying that the difference is a finite value less than 1.

#### seanieg89

##### Well-Known Member
Not trying to defend incomplete reasoning here, but if we ignore the footnote then the solution now is pretty much enough. He proves both of the required inequalities and goes from there to saying that this implies k is the closest integer to sqrt(n). This is certainly not a fallacious argument (unlike his solution to 15biii ), the reasoning is just not as clear as it should be.

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