HSC 2016 Maths Marathon (archive) (1 Viewer)

leehuan · Apr 3, 2016

Re: HSC 2016 2U Marathon

$$Instructive question$:\\$Explain why $\frac{d}{dx}\int{f(x)\,dx}$ does not necessarily have to equal to $\int{\frac{d}{dx}f(x)\,dx}$

Nailgun · Apr 3, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$$Instructive question$:\\$Explain why $\frac{d}{dx}\int{f(x)\,dx}$ does not necessarily have to equal to $\int{\frac{d}{dx}f(x)\,dx}$

$$Take an arbitrary function of $x,$ $ f(x) \\ \\ \frac { d }{ dx } \int { f\left( x \right) dx } =\frac { d }{ dx } \left( F\left( x \right) +C \right) =f\left( x \right) \\ \\ \\ \int { \frac { d }{ dx } f\left( x \right) } dx=\int { f^{ ' }\left( x \right) } dx=f\left( x \right) +D\\ \\ \\ f\left( x \right) \neq f\left( x \right) +D,\quad unless\quad D=0 \\ \\ \therefore \quad \frac { d }{ dx } \int { f\left( x \right) dx } \neq \int { \frac { d }{ dx } f\left( x \right) } dx\ $ is true for some $ f(x)$

Drsoccerball · Apr 3, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$$Take an arbitrary function of $x,$ $ f(x) \\ \\ \frac { d }{ dx } \int { f\left( x \right) dx } =\frac { d }{ dx } \left( F\left( x \right) +C \right) =f\left( x \right) \\ \\ \\ \int { \frac { d }{ dx } f\left( x \right) } dx=\int { f^{ ' }\left( x \right) } dx=f\left( x \right) +D\\ \\ \\ f\left( x \right) \neq f\left( x \right) +D,\quad unless\quad D=0 \\ \\ \therefore \quad \frac { d }{ dx } \int { f\left( x \right) dx } \neq \int { \frac { d }{ dx } f\left( x \right) } dx\ $ is true for some $ f(x)$

When doing questions such as "prove that the statement isn't necessarily true," one random example is enough. It wasn't difficult in this question to prove the general case may not be true but for other questions it may be too hard so just find one example that doesn't work and that proves it isn't always true. Good work by the way keep it up! Also make fb and join the convo.

Nailgun · Apr 3, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
When doing questions such as "prove that the statement isn't necessarily true," one random example is enough. It wasn't difficult in this question to prove the general case may not be true but for other questions it may be too hard so just find one example that doesn't work and that proves it isn't always true. Good work by the way keep it up! Also make fb and join the convo.

What is this convo lel

Paradoxica · Apr 3, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
What is this convo lel

2016er chat and BOSer chat

porcupinetree · Apr 3, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
What is this convo lel

You're missing out tbh

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

$\noindent Let $f(x)=mx+b$, where $m,b$ are real constants.$

$\noindent (i) Show that $f(x+1)-f(x)= f^\prime (x)$.$

$\noindent (ii) Does the above result surprise you? If not, why not?$

Drsoccerball · Apr 5, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Let $f(x)=mx+b$, where $m,b$ are real constants.$

$\noindent (i) Show that $f(x+1)-f(x)= f^\prime (x)$.$

$\noindent (ii) Does the above result surprise you? If not, why not?$

MVT ?

Nailgun · Apr 5, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Let $f(x)=mx+b$, where $m,b$ are real constants.$

$\noindent (i) Show that $f(x+1)-f(x)= f^\prime (x)$.$

$\noindent (ii) Does the above result surprise you? If not, why not?$

$\noindent f(x)=mx+b,\quad \therefore \quad f'(x)=m\\ \\ i)\quad L.H.S=f(x+1)-f(x)\\ =m(x+1)+b-mx-b\\ =mx+m-mx\\ =m\\ L.H.S=R.H.S$

$\noindent (ii) Do I lose marks if I say it's surprising lol.\\ \\ I believe it has to do with first principles of differentiation, and in this case, we have taken $h=1$. In a typical application of the first principles, one uses the limit of $\lim _{ h\rightarrow 0 }{ } $ as $h$ must be an infinitesimally small distance in order to determine the slope at a single point on the curve.\\ \\ However, in a linear curve following $f(x)=mx+b$ this is unnecessary, as the gradient is the same at every point on the line and so you can make $h$ any arbitrary length and it will still compute the gradient. It is this property of the linear curve which allows the use of the well-known formula $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } $ to calculate the derivative of the curve at any point.$

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$\noindent f(x)=mx+b,\quad \therefore \quad f'(x)=m\\ \\ i)\quad L.H.S=f(x+1)-f(x)\\ =m(x+1)+b-mx-b\\ =mx+m-mx\\ =m\\ L.H.S=R.H.S$

$\noindent (ii) Do I lose marks if I say it's surprising lol.\\ \\ I believe it has to do with first principles of differentiation, and in this case, we have taken $h=1$. In a typical application of the first principles, one uses the limit of $\lim _{ h\rightarrow 0 }{ } $ as $h$ must be an infinitesimally small distance in order to determine the slope at a single point on the curve.\\ \\ However, in a linear curve following $f(x)=mx+b$ this is unnecessary, as the gradient is the same at every point on the line and so you can make $h$ any arbitrary length and it will still compute the gradient. It is this property of the linear curve which allows the use of the well-known formula $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } $ to calculate the derivative of the curve at any point.$

Well done! Yeah, for the second one, the slope of the line is just [f(x+1) - f(x)]/[(x+1) - x] = f(x+1) - f(x), explaining why f'(x) comes out to be f(x+1) - f(x) (or we could've used any other place, like f(t+1) - f(t) for any real t, since the slope is constant).

Nailgun · Apr 5, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Well done! Yeah, for the second one, the slope of the line is just [f(x+1) - f(x)]/[(x+1) - x] = f(x+1) - f(x), explaining why f'(x) comes out to be f(x+1) - f(x).

Ohhh lol, I see

Is my explanation still valid though?

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Ohhh lol, I see

Is my explanation still valid though?

Pretty much. Like the main thing to say was that f'(x) can be calculated just using the gradient formula, which I think you said at the end.

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

Btw, expressions like f(x+1) - f(x) are known as finite forward differences (see here for more info: https://en.wikipedia.org/wiki/Finite_difference ). They can be used to approximate derivatives to help solve (numerically) differential equations.

From the series and sequences topic, you may remember that an arithmetic sequence has terms of the form f(n) = an + b. This "a" is f(n+1) - f(n) (the common difference), and is analogous to the derivative of the function mx+b. The sum of n terms of an arithmetic sequence is analogous to integrating a linear function from say 0 to x.

astroman · Apr 5, 2016

Re: HSC 2016 2U Marathon

ln(ln(x)) = 1
how to solve for x?

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

astroman said:
ln(ln(x)) = 1
how to solve for x?

Raise e to the power of both sides, then
ln(x) = e.

Hence x = e^e.

Nailgun · Apr 5, 2016

Re: HSC 2016 2U Marathon

astroman said:
ln(ln(x)) = 1
how to solve for x?

the way i would think about it is when a log = 1
that means the base is equal to the number

so e=ln(x)
and then x=e^e

but Integrands way is smarter lel

astroman · Apr 5, 2016

Re: HSC 2016 2U Marathon

solve for x:
e^ax = Ce^bx, where a ≠ b

leehuan · Apr 5, 2016

Re: HSC 2016 2U Marathon

If C<0 or C=0 then we have no solution

For C>0

e^(ax) = e^(lnC)*e^(bx)
ax = ln(C) + bx
If C<0 or C=0 then we have no solution

For C>0

e^(ax) = e^(lnC)*e^(bx)
ax = ln(C) + bx by equating indices
x = ln(C)/(a-b)
____________________
Alternatively

ax = ln(C) + ln(e^(bx)) using log laws
ax = ln(C) + bx
continue as above

astroman · Apr 5, 2016

Re: HSC 2016 2U Marathon

Differentiate
y = cos(a⁶+x⁶)

Paradoxica · Apr 5, 2016

Re: HSC 2016 2U Marathon

astroman said:
Differentiate
y = cos(a⁶+x⁶)

You haven't specified a variable.

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