HSC 2016 Maths Marathon (archive) (1 Viewer)

leehuan · Jan 24, 2016

Re: HSC 2016 2U Marathon

Meh, school exams (that aren't bought) only reflect the capability of the school.

If I was to give that question in the HSC I would at least say differentiate x^x (if not x^(2x)) first rather than let them figure out f(x) by themselves

But tbh, post-2000 I've never seen d/dx a^x where a≠e

Drsoccerball · Jan 24, 2016

Re: HSC 2016 2U Marathon

$New Question$

$$ Solve : $ \log_{\frac{1}{2}}(x^2-3x+1) > \log_{\frac{1}{2}}(6-x^2+4x)$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$New Question$

$$ Solve : $ \log_{\frac{1}{2}}(x^2-3x+1) > \log_{\frac{1}{2}}(6-x^2+4x)$

$\log _{ \frac { 1 }{ 2 } } (x^{ 2 }-3x+1)\times \frac { 1 }{ 2 } >\log _{ \frac { 1 }{ 2 } } (6-x^{ 2 }+4x)\times \frac { 1 }{ 2 } \\ x^{ 2 }-3x+1>6-x^{ 2 }+4x\\ 2x^{ 2 }-7x-5\quad >0\\ Roots\quad are\quad \frac { 7+\sqrt { 89 } }{ 4 } and\frac { 7-\sqrt { 89 } }{ 4 } \\ Hence\quad x<\frac { 7-\sqrt { 89 } }{ 4 } \quad and\quad x>\frac { 7+\sqrt { 89 } }{ 4 }$

Also I've tried that other question ( $\int { x^{ 2x }(1+\ln { x } ) } dx$ ) a billion times. Can someone show me how to do it lele
I even learned integration by parts because I thought I would be able to do it that way. No dice

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$\log _{ \frac { 1 }{ 2 } } (x^{ 2 }-3x+1)\times \frac { 1 }{ 2 } >\log _{ \frac { 1 }{ 2 } } (6-x^{ 2 }+4x)\times \frac { 1 }{ 2 } \\ x^{ 2 }-3x+1>6-x^{ 2 }+4x\\ 2x^{ 2 }-7x-3>0\\ Roots\quad are\quad \frac { 7+\sqrt { 73 } }{ 4 } and\frac { 7-\sqrt { 73 } }{ 4 } \\ Hence\quad x<\frac { 7-\sqrt { 73 } }{ 4 } \quad and\quad x>\frac { 7+\sqrt { 73 } }{ 4 }$

$\noindent Hey, why did you multiply both sides of the inequality by $\frac{1}{2}$ at the start? Also, Drsoccerball's question had a trick in it! Remember that $y=\log_a x$ is a \textit{decreasing} function if $a<1$. So you need to flip the inequality sign when removing the log. I.e. $\log _{\frac{1}{2}} x_1 < \log _{\frac{1}{2}} x_2 \Longleftrightarrow x_1 > x_2 $. (And also when solving the inequation, make sure that you restrict $x$ so that both $x^2 - 3x +1$ and $6-x^2 + 4x$ are strictly positive, since we only take logs of positive numbers in HSC.)$

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Also I've tried that other question ( $\int { x^{ 2x }(1+\ln { x } ) } dx$ ) a billion times. Can someone show me how to do it lele
I even learned integration by parts because I thought I would be able to do it that way. No dice

$\noindent To do this integral, we use the fact that $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right) = 2x^{2x} \left(\ln x + 1 \right)$ (in the HSC, they would probably ask you to show this as a sub-part before asking the integral; we'll show this below). So rearranging gives$

$\begin{align*}x^{2x} \left(\ln x + 1 \right) &= \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right) \\ \Rightarrow \int x^{2x} \left(\ln x + 1 \right) \text{ d}x &= \frac{1}{2}x^{2x}+C.\end{align*}$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

I was thinking that if log y = log f(x)
then y = f(x)?
Lol dw about the 1/2 I kind of got the log laws mixed up

What do you mean restrict them?

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

$\noindent Here's the proof that $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right)=2x^{2x}\left(\ln x +1 \right)$. In order to do the differentiation here, we need to make the base into something we can differentiate with, so we convert it to base e. So we have $x^{2x}=\mathrm{e}^{2x\ln x}$, since $x=\mathrm{e}^{\ln x}$ and using index laws. So we have$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(x^{2x}\right) &= \frac{\mathrm{d}}{\mathrm{d}x}\left( \mathrm{e}^{2x\ln x}\right) \quad (\text{chain rule}) \\ &= \mathrm{e}^{2x\ln x} \times \frac{\mathrm{d}}{ \mathrm{d}x} \left(2x \ln x \right)\\ &= 2\mathrm{e}^{2x\ln x} \times \frac{\mathrm{d}}{ \mathrm{d}x} \left(x \ln x \right) \quad (\text{bringing the 2 out the front}). \end{align*}$

$But$

$\begin{align*} \frac{\mathrm{d}}{ \mathrm{d}x} \left(x \ln x \right) &= 1\cdot \ln x + x \cdot \frac{1}{x} \quad (\text{product rule}) \\ &= \ln x + 1.\end{align*}$

$\noindent Hence$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(x^{2x}\right) &= 2\mathrm{e}^{2x\ln x} \times \left(\ln x + 1\right) \\ &= 2x^{2x}\left(\ln x +1 \right) \quad \Big{(}\text{as }\mathrm{e}^{2x\ln x} = x^{2x} \Big{)}.\end{align*}$

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
I was thinking that if log y = log f(x)
then y = f(x)?

$\noindent Yes, this is true. But the statement ``if $\log _a A > \log_a B$ ($A,B>0$), then $A>B$'' is correct if and only if $a>1$. If $a<1$ (and of course $a>0$ since log bases are positive), then the correct statement is ``if $\log _a A > \log_a B$ ($A,B>0$), then $A<B$'' (in other words, the function is decreasing).$

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
What do you mean restrict them?

$\noindent Restrict them as in, as part of your solution to the inequation, you need to ensure that the quantities inside the logs originally will always be positive. Generally speaking, whenever you get inequations with logs, you will need to do this. So for Drsoccerball's question, you need to also solve $x^2-3x+1 >0$ and $6-x^2+4x>0$. Then in the end, the solution to the inequation will be the places where all these inequalities (i.e. the ones for restricting $x$ and the ones obtained by solving the given inequation) hold simultaneously. In other words, the intersection of all these solution sets.$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Here's the proof that $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2x}\right)=2x^{2x}\left(\ln x +1 \right)$. In order to do the differentiation here, we need to make the base into something we can differentiate with, so we convert it to base e. So we have $x^{2x}=\mathrm{e}^{2x\ln x}$, since $x=\mathrm{e}^{\ln x}$ and using index laws. So we have$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(x^{2x}\right) &= \frac{\mathrm{d}}{\mathrm{d}x}\left( \mathrm{e}^{2x\ln x}\right) \quad (\text{chain rule}) \\ &= \mathrm{e}^{2x\ln x} \times \frac{\mathrm{d}}{ \mathrm{d}x} \left(2x \ln x \right)\\ &= 2\mathrm{e}^{2x\ln x} \times \frac{\mathrm{d}}{ \mathrm{d}x} \left(x \ln x \right) \quad (\text{bringing the 2 out the front}). \end{align*}$

$But$

$\begin{align*} \frac{\mathrm{d}}{ \mathrm{d}x} \left(x \ln x \right) &= 1\cdot \ln x + x \cdot \frac{1}{x} \quad (\text{product rule}) \\ &= \ln x + 1.\end{align*}$

$\noindent Hence$

$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left(x^{2x}\right) &= 2\mathrm{e}^{2x\ln x} \times \left(\ln x + 1\right) \\ &= 2x^{2x}\left(\ln x +1 \right) \quad \Big{(}\text{as }\mathrm{e}^{2x\ln x} = x^{2x} \Big{)}.\end{align*}$

$\frac { d }{ dx } ({ e }^{ 2x\ln { x } })={ e }^{ 2x\ln { x } }\times \frac { { d } }{ { d }x } \left( 2x\ln { x } \right) \\ $ Is this a standard form? Like I thought that you multiply by the coefficent of x, not the derivative of the power? $$

But I get how to do it now so much appreciated. Fun fun

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$\frac { d }{ dx } ({ e }^{ 2x\ln { x } })={ e }^{ 2x\ln { x } }\times \frac { { d } }{ { d }x } \left( 2x\ln { x } \right) \\ $ Is this a standard form? Like I thought that you multiply by the coefficent of x, not the derivative of the power? $$

But I get how to do it now so much appreciated. Fun fun

$\noindent In general, we multiply by the derivative of the power. The reason why we so often just multiply by the coefficient of $x$ is that the exponential is in the form $\mathrm{e}^{kx}$ ($\mathrm{e}^{2x}$ for instance). In this case, the derivative of the power is just the coefficient of $x$, since the power is a linear function. But in general, we need to multiply by the derivative of the power, which may not be the thing being multiplied by $x$ (in fact, this would be the case if and only if the power is a linear function).$

$\noindent I.e. $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{f(x)}\right) = f^\prime (x) \mathrm{e}^{f(x)}$, where $f$ is a (differentiable) function of $x$. This fact follows from the chain rule. So for example, $\frac{\mathrm{d}}{\mathrm{d}x}\left( \mathrm{e}^{\sin x}\right) = \cos x \cdot \mathrm{e}^{\sin x}$.$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent you need to also solve $x^2-3x+1 >0$ and $6-x^2+4x>0$. Then in the end, the solution to the inequation will be the places where all these inequalities (i.e. the ones for restricting $x$ and the ones obtained by solving the given inequation) hold simultaneously. In other words, the intersection of all these solution sets.$

As in use the x>n that holds for the x>n inequalities and the x < n that holds for all the x < n inequalities and put them together?

$If\quad { x }^{ 2 }-3x+1>0\\ Then\quad \frac { 3+\sqrt { 13 } }{ 2 } >x\quad and\quad \frac { 3+\sqrt { 13 } }{ 2 } <x\\ If\quad 6-{ x }^{ 2 }+4x>0\\ Then\quad 2-\sqrt { 10 } <x<2+\sqrt { 10 } \\ \\ We\quad already\quad have\quad \frac { 7-\sqrt { 89 } }{ 4 } <x<\frac { 7+\sqrt { 89 } }{ 4 } \\ Hence\quad intersection\quad will\quad be\quad \frac { 3+\sqrt { 13 } }{ 2 } >x\quad and\quad \frac { 3+\sqrt { 13 } }{ 2 } <x$

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
As in use the x>n that holds for the x>n inequalities and the x < n that holds for all the x < n inequalities and put them together?

$If\quad { x }^{ 2 }-3x+1>0\\ Then\quad \frac { 3+\sqrt { 13 } }{ 2 } >x\quad and\quad \frac { 3+\sqrt { 13 } }{ 2 } <x\\ If\quad 6-{ x }^{ 2 }+4x>0\\ Then\quad 2-\sqrt { 10 } <x<2+\sqrt { 10 } \\ \\ We\quad already\quad have\quad \frac { 7-\sqrt { 89 } }{ 4 } <x<\frac { 7+\sqrt { 89 } }{ 4 } \\ Hence\quad intersection\quad will\quad be\quad \frac { 3+\sqrt { 13 } }{ 2 } >x\quad and\quad \frac { 3+\sqrt { 13 } }{ 2 } <x$

Those are not the answers.

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

$Please past graduates do not help in this next problem. I don't want to waste an amazing question.$

$$ Simplify : $ \frac{\log_{\frac{a}{b}}x \log_{\frac{a}{c}}{x}}{\log_a ^2 x} +\frac{\log_{\frac{b}{c}}x \log_{\frac{b}{a}}{x}}{\log_b^2 x} +\frac{\log_{\frac{c}{a}}x \log_{\frac{c}{b}}{x}}{\log_c ^2 x}$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Those are not the answers.

Okay I redid the question, and drew diagrams - I think I know where I went wrong.
*crossing fingers*
Is the answer
$\frac { 7-\sqrt { 89 } }{ 4 } <x<\frac { 3-\sqrt { 5 } }{ 2 } \quad or\quad \frac { 3+\sqrt { 5 } }{ 2 } <x<\frac { 7+\sqrt { 89 } }{ 4 }$

InteGrand · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Okay I redid the question, and drew diagrams - I think I know where I went wrong.
*crossing fingers*
Is the answer
$\frac { 7-\sqrt { 89 } }{ 4 } <x<\frac { 3-\sqrt { 5 } }{ 2 } \quad and\quad \frac { 3+\sqrt { 5 } }{ 2 } <x<\frac { 7+\sqrt { 89 } }{ 4 }$

(btw, we should actually say "or" rather than "and", since "and" means they are simultaneously true.)

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

I changed it lelel

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Okay I redid the question, and drew diagrams - I think I know where I went wrong.
*crossing fingers*
Is the answer
$\frac { 7-\sqrt { 89 } }{ 4 } <x<\frac { 3-\sqrt { 5 } }{ 2 } \quad or\quad \frac { 3+\sqrt { 5 } }{ 2 } <x<\frac { 7+\sqrt { 89 } }{ 4 }$

Ill post my answer hold up.

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$New Question$

$$ Solve : $ \log_{\frac{1}{2}}(x^2-3x+1) > \log_{\frac{1}{2}}(6-x^2+4x)$

$Firstly it is important to note that the graph's inside the $ \log $ must be positive.$

$x^2 -3x +2 >0 $ Equality occurs at $ x=1,2$

$6-x^2+4x>0 $ Equality occurs at $ x= -1.16, 5.16$

$$ Getting rid of the $ \log 's $ we get : $ 2x^2-7x-4 > 0 $ Equality occurs at $ x=4, \frac{-1}{2}$

$$ Using a number line testing all points in the original inequality we get the values :$

$\frac{-1}{2} < x < 1 or 2<x<4$

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$Firstly it is important to note that the graph's inside the $ \log $ must be positive.$

$x^2 -3x +2 >0 $ Equality occurs at $ x=1,2$

$6-x^2+4x>0 $ Equality occurs at $ x= -1.16, 5.16$

$$ Getting rid of the $ \log 's $ we get : $ 2x^2-7x-4 > 0 $ Equality occurs at $ x=4, \frac{-1}{2}$

$$ Using a number line testing all points in the original inequality we get the values :$

$\frac{-1}{2} < x < 1 or 2<x<4$

Erm not too jump in early
$$Wasn't the LHS $ x^2 -3x +1 $ not $ x^2 -3x + 2$

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