HSC 2015 MX1 Marathon (archive) (5 Viewers)

VBN2470 · Feb 4, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$y=\cos ^{-1}\left ( \frac{1-x^2}{1+x^2} \right ), \text{ for }-1\leq \frac{1-x^2}{1+x^2} \leq 1$

$\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1-x^2}{1+x^2} \right )=\frac{-2x(1+x^2)-(1-x^2)\cdot2x}{(1+x^2)^2}$

$=\frac{-4x}{(1+x^2)^2}$ .

$\therefore y^\prime = \frac{-\left ( \frac{-4x}{(1+x^2)^2} \right )}{\sqrt{1-\left ( \frac{1-x^2}{1+x^2} \right )^2}}\text{ for}-1< \frac{1-x^2}{1+x^2} < 1$ (using the chain rule)

$=\frac{4x}{(1+x^2)^2 \sqrt{\frac{(1+x^2)^2 - (1-x^2)^2}{(1+x^2)^2}}}$

$=\frac{4x}{(1+x^2)\sqrt{4x^2}}, \text{ as } (1+x^2)^2-(1-x^2)^2 =(1+x^2 + 1-x^2)(1+x^2 -1 + x^2)=2\times 2x^2 = 4x^2$

$=\frac{2x}{|x|(1+x^2)}.$

Alternatively, you could try substituting $$x=\tan\frac{\theta}{2}$$ and use the half-angle formula to make the differentiation easier to do. Should yield the same answer though

LostInHSC · Feb 4, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$y=\cos ^{-1}\left ( \frac{1-x^2}{1+x^2} \right ), \text{ for }-1\leq \frac{1-x^2}{1+x^2} \leq 1$

$\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1-x^2}{1+x^2} \right )=\frac{-2x(1+x^2)-(1-x^2)\cdot2x}{(1+x^2)^2}$

$=\frac{-4x}{(1+x^2)^2}$ .

$\therefore y^\prime = \frac{-\left ( \frac{-4x}{(1+x^2)^2} \right )}{\sqrt{1-\left ( \frac{1-x^2}{1+x^2} \right )^2}}\text{ for}-1< \frac{1-x^2}{1+x^2} < 1$ (using the chain rule)

$=\frac{4x}{(1+x^2)^2 \sqrt{\frac{(1+x^2)^2 - (1-x^2)^2}{(1+x^2)^2}}}$

$=\frac{4x}{(1+x^2)\sqrt{4x^2}}, \text{ as } (1+x^2)^2-(1-x^2)^2 =(1+x^2 + 1-x^2)(1+x^2 -1 + x^2)=2\times 2x^2 = 4x^2$

$=\frac{2x}{|x|(1+x^2)}.$

VBN2470 said:
Alternatively, you could try substituting $$x=\tan\frac{\theta}{2}$$ and use the half-angle formula to make the differentiation easier to do. Should yield the same answer though

The given answer is:

$\frac{2}{1+x^{2}} \,\,\,\, $if$\,\,\,\, x>0, \frac{-2}{1+x^{2}}\,\,\,\, $if$\,\,\,\, x<0,\,\,\, $undefined at$ \,\,\, x=0$

FrankXie · Feb 4, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION

$$ For what values of $ \theta, |(3\sin\theta-\cos\theta)+k(\cos\theta-2\sin\theta)|\leq1 $ is true for all real $ -1\leq k\leq1 ?$

InteGrand · Feb 5, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
The given answer is:

$\frac{2}{1+x^{2}} \,\,\,\, $if$\,\,\,\, x>0, \frac{-2}{1+x^{2}}\,\,\,\, $if$\,\,\,\, x<0,\,\,\, $undefined at$ \,\,\, x=0$

Yes, that's equivalent to the answer I posted.

LostInHSC · Feb 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Yes, that's equivalent to the answer I posted.

How do you work with the domain and know it's undefined? I solved to only get $\frac{2}{1+x^{2}}$

InteGrand · Feb 5, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
How do you work with the domain and know it's undefined? I solved to only get $\frac{2}{1+x^{2}}$

Well, I said the derivative was defined for $-1<\frac{1-x^2}{1+x^2}<1$ , and x can't be 0 in this domain (otherwise the fraction would EQUAL 1, which we can't have), so the derivative is undefined at x = 0.

In general, $y = \cos ^{-1} f(x)$ is defined for $-1\leq f(x)\leq +1$ , and the derivative $y^\prime$ is defined for $-1< f(x) < +1$ .

InteGrand · Feb 5, 2015

Re: HSC 2015 3U Marathon

Also, my solution has $\frac{x}{|x|}$ in it. This is equal to 1 when x is positive, and -1 when x is negative, so it agrees with your book's/source's answer.

FrankXie · Feb 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Also, my solution has $\frac{x}{|x|}$ in it. This is equal to 1 when x is positive, and -1 when x is negative, so it agrees with your book's/source's answer.

yes quite so. and a comment, the function x/|x| (which usually called sign function or Heaviside function) has no definition and is discontinuous at x=1, this justfifies that y' does not exist at x=0.
plus, guys, don't forget my last question, still there unanswered

Ambility · Feb 16, 2015

Re: HSC 2015 3U Marathon

I only just started my 3U Maths preliminary course, so I can't yet answer many of the questions here, but I do have one for you guys if you want it. It was a challenge question on a class test today. The answer surprised me.

Simplify:

$\left(1+\frac{45}{x-8} - \frac{26}{x-6}\right)\left(3-\frac{65}{x+7}+\frac{8}{x-2}\right)$

FrankXie · Feb 16, 2015

Re: HSC 2015 3U Marathon

Ambility said:
I only just started my 3U Maths preliminary course, so I can't yet answer many of the questions here, but I do have one for you guys if you want it. It was a challenge question on a class test today. The answer surprised me.

Simplify:

$\left(1+\frac{45}{x-8} - \frac{26}{x-6}\right)\left(3-\frac{65}{x+7}+\frac{8}{x-2}\right)$

answer is 3.

Ambility · Feb 16, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
answer is 3.

Correct.

Sy123 · Feb 22, 2015

Re: HSC 2015 3U Marathon

$\\ $Prove through mathematical induction that$ \ n! \geq 2^n \ $for integers$ \ n \geq 4$

Drsoccerball · Feb 22, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
$\\ $Prove through mathematical induction that$ \ n! \geq 2^n \ $for integers$ \ n \geq 4$

$n!-2^n >=0$

$Prove $ n=4 $ is true$

$LHS=8, 8>0 \therefore n=4 $ is true $$

$Assume $ n=k $ is true$

$k!-2^k >0$

$prove $ n=k+1 $ is true$

$(k+1)! -2^{k+1}>0$

$(k+1)(k! -2^k) +k 2^k +2^k >0$

$(k+1)(k! -2^k) +2^k (k+1) >0$

$\therefore n=k+1 $ is true as $ k>4 $ and $ k! -2^k >0\therefore $ never negative $$

$$ Hence if n=k is true then n=k+1 is true. However, n=4 is true $ \therefore $ by induction $ n!-2^n>0 $ is true $ \therefore n!-2^n >0$

braintic · Feb 23, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$n!-2^n >=0$

$Prove $ n=4 $ is true$

$LHS=8, 8>0 \therefore n=4 $ is true $$

$Assume $ n=k $ is true$

$k!-2^k >0$

$prove $ n=k+1 $ is true$

$(k+1)! -2^{k+1}>0$

$(k+1)(k! -2^k) +k 2^k +2^k >0$

$(k+1)(k! -2^k) +2^k (k+1) >0$

$\therefore n=k+1 $ is true as $ k>4 $ and $ k! -2^k >0\therefore $ never negative $$

$$ Hence if n=k is true then n=k+1 is true. However, n=4 is true $ \therefore $ by induction $ n!-2^n>0 $ is true $ \therefore n!-2^n >0$

How about:

Assuming k! > 2^k (and k>=4)

(k+1)! = (k+1).k! > 2.2^k = 2^(k+1)

[BTW, "n=k+1 is true" doesn't mean too much. Not that the examiners are going to care, but it should be "true for n=k+1 (when true for n=k)" ]

Sy123 · Feb 23, 2015

Re: HSC 2015 3U Marathon

$\\ $Prove using first principles that$ \ \frac{d}{dx} \sin x = \cos x$

integral95 · Feb 23, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
$\\ $Prove using first principles that$ \ \frac{d}{dx} \sin x = \cos x$

$\frac{d}{dx}sinx = \lim_{h \to 0}\frac{sin(x+h)-sinx}{h}$

Expand and simply and use

$\ \ \ \lim_{x \to 0}\frac{sinx}{x} = 1\\ \\ \\ \lim_{x \to0}\frac{cosx-1}{x}=0$

To get the answer

Drsoccerball · Feb 23, 2015

Re: HSC 2015 3U Marathon

A group consists of 9 people. These 9 people are to be seated around a circular table. The group of 9 people consists of 4 students 3 teachers and 2 parents. How many possible arrangements are there if all the students sit together, all the teachers sit together and none of the students sit next to the teachers?

Sy123 · Feb 23, 2015

Re: HSC 2015 3U Marathon

integral95 said:
$\\ \lim_{x \to0}\frac{cosx-1}{x}=0$

This would require proof

Drsoccerball · Feb 23, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
This would require proof

$$ as $ x \to 0 ,cos x \to 1 \therefore 1-1=0$

iStudent · Feb 23, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
This would require proof

L'hopital's?

Drsoccerball said:
$$ as $ x \to 0 ,cos x \to 1 \therefore 1-1=0$

That's kind of wrong, since your denominator "x" approaches 0 too.

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