VCE Specialist 3/4 Question Thread!

[Ionic Doc]

Mate, if someone helps you out, even if they're wrong, there's no need to be so rude and passive-aggressive. No one here owes you anything.It's actually convenient to use 30. You have 8 factors of 30, which means you can form four pairs of factors that multiply to give 30. Thus, the product of all factors is 30^{4[\sup]. Divide out 30 to give the correct answer.}

sorry for my uneeded comment i didn't really try to sound agressive, just more of a smart ass..but thats also rude.  and your solution is also correct but you dont need to divide out by 30 in the end (what my teacher said) idk if he/she is 100%correct

[Peas]

Hi,

According to a multiple choice question in the ATARNotes Specialist Maths vectors topic test 1, a = -8i + 6j and b = 4i - 3j are linearly dependent (as they are parallel).

However, I'm struggling to understand why a = 2i - j and b = -6i + 3j are not also linearly dependent. Are the two vectors not parallel or is it because not all parallel vectors are linearly dependent?

Thanks 

[RuiAce]

Hi,

According to a multiple choice question in the ATARNotes Specialist Maths vectors topic test 1, a = -8i + 6j and b = 4i - 3j are linearly dependent (as they are parallel).

However, I'm struggling to understand why a = 2i - j and b = -6i + 3j are not also linearly dependent. Are the two vectors not parallel or is it because not all parallel vectors are linearly dependent?

Thanks 

They are linearly dependent as you said.

Where did you encounter this error? Was it also in our topic tests? (If so can you state which test and question number?)

[Peas]

They are linearly dependent as you said.

Where did you encounter this error? Was it also in our topic tests? (If so can you state which test and question number?)

It was in the ATARNotes Specialist Maths topic II (vectors) test 1, question 4 of section A multiple choice.

[RuiAce]

It was in the ATARNotes Specialist Maths topic II (vectors) test 1, question 4 of section A multiple choice.

Reported. Thank you

[hawkrook]

I've spent ages trying to think about how to solve these kinds of problems with no success. Can someone help me out?

[undefined]

I've spent ages trying to think about how to solve these kinds of problems with no success. Can someone help me out?

Treat it like a composite function where the range of the inside function has to be a subset of the domain of the outside function. If it isn’t you need to restrict said range which changes the domain of said inside function. The (potentially restricted) domain of the inside function becomes the implied domain and range of the outside function becomes the implied range. For example to solve the domain and range of cos(2arcsin(x)),

dom(2arcsin(x))=[-1,1]
ran(2arcsin(x))=[-pi,pi]

dom(cos(x))=[0,pi]
ran(cos(x))=[-1,1]

but the range of 2arcsin(x) must fit into the domain of cos(x) therefore the range of 2arcsin(x) is restricted to [0,pi]. since the range is restricted the domain must also change to [0,1] (which becomes the final domain) and the range of the outside function becomes the final range so [-1,1]

Edit: usually for trig questions like these if you can't find the implied range but have the implied domain you can sub the domain back into the original function since they'll spit out the range since they're one to one but I would only do that if you really can't find the range.

[hawkrook]

Treat it like a composite function where the range of the inside function has to be a subset of the domain of the outside function. If it isn’t you need to restrict said range which changes the domain of said inside function. The (potentially restricted) domain of the inside function becomes the implied domain and range of the outside function becomes the implied range. For example to solve the domain and range of cos(2arcsin(x)),

dom(2arcsin(x))=[-1,1]
ran(2arcsin(x))=[-pi,pi]

dom(cos(x))=[0,pi]
ran(cos(x))=[-1,1]

but the range of 2arcsin(x) must fit into the domain of cos(x) therefore the range of 2arcsin(x) is restricted to [0,pi]. since the range is restricted the domain must also change to [0,1] (which becomes the final domain) and the range of the outside function becomes the final range so [-1,1]

Edit: usually for trig questions like these if you can't find the implied range but have the implied domain you can sub the domain back into the original function since they'll spit out the range since they're one to one but I would only do that if you really can't find the range.

Thank you so much! This was really clear and I understand it now.

[undefined]

How would you solve for the range of 3xarctan(x)? or is the only way to recognise that all inputs come out as positive?

[Unsplash]

How would you solve for the range of 3xarctan(x)? or is the only way to recognise that all inputs come out as positive?

Good question, almost identical to VCAA 2014 SM E1 Q7a where 4% of the state got the mark. So fair to say a challenging question if you don't think about it.

In my opinion, simplest way is to split up into two functions "3x" and "arctan(x)".

Then consider what happens when x<0, x=0 and x>0 for both functions.

[Ansaki]

hello guys! i need help with two questions
(1) Given that tan(a)=b,π<a<3π/2, find tan(3a) in terms of b
(2) given that the cross section of a road on a hill is represented by the equation
H=50+100tan^-1(x/20-5),0≤x≤400
Where x meters is the horizontal distance of the road from a railway crossing and H meters is the height of the hill above sea level. How many meters is the road above sea level at the point of the road's greatest gradient?

[AlphaZero]

hello guys! i need help with two questions
(1) Given that tan(a)=b,π<a<3π/2, find tan(3a) in terms of b
(2) given that the cross section of a road on a hill is represented by the equation
H=50+100tan^-1(x/20-5),0≤x≤400
Where x meters is the horizontal distance of the road from a railway crossing and H meters is the height of the hill above sea level. How many meters is the road above sea level at the point of the road's greatest gradient?

Question 1

Hint:  \(\tan(3a)=\tan(2a+a)\)

Solution

\begin{align*}\tan(3a)&=\tan(2a+a)\\
&=\frac{\tan(2a)+\tan(a)}{1-\tan(2a)\tan(a)}\\
&=\frac{\frac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\frac{2\tan(a)}{1-\tan^2(a)}\!\times\tan(a)}\\
&=\frac{\frac{2b}{1-b^2}+b}{1-\frac{2b^2}{1-b^2}}\\
&=\frac{2b+b-b^3}{1-b^2-2b^2}\\
&=\frac{3b-b^3}{1-3b^2} \end{align*}

Question 2

Hint:  consider the graph of  \(y=\arctan(x)\).  Where does the point of steepest ascent occur?

Solution

The graph of  \(y=\arctan(x)\)  has a point of inflection (and has the greatest gradient) at \((0,\,0)\).

Thus, applying the relevant transformations the graph of \[H(x)=50+100\arctan\!\left(\frac{x}{20}-5\right)\] will have a point of inflection at \[\big(20(0+5),\ 100\!\times\!0+50\big)=(100,\;50).\] That is, the road's steepest gradient occurs at a height of  \(H=50\ \text{m}.\)

[DBA-144]

In regards to the above question, could someone explain how arctan has a point of inflection, even though we cannot solve its derivative for zero? What am I missing?

And in the same way, how come the point of greatest gradient is at x=0 if the gradient is zero at this point, but is positive elsewhere? My main point of confusion is that if the derivative of arctan is 1/1+x^2, where the derivative is always greater than zero, how come we have a point of inflection?

Wait ok nvm the bit about the point of greatest gradient makes sense but the stationary point bit still doesn't. Could someone please give a formal explanation of each please?

My reasoning for why the point of greatest gradient is at x=0 is that the reciprocal of x^2 +1 has a maximum value at x=0 and hence maximum gradient is at this point. Yet, the reciprocal of the above function never equals zero. Lol i am ao lost.

[AlphaZero]

...

I'll be writing a concise guide on inflection points soon, but I'll give you a brief overview.

Definitions (these are the two useful definitions for Specialist Maths):

1.  A differentiable function  \(f:D\to\mathbb{R}\)  has a point of inflection at  \((a,\,f(a))\),  where  \(a\in D\),  if and only if the first derivative, \(f'\), has an isolated extremum at \(x=a\).

2.  A \(k\) times (\(k\geq2\)) differentiable function  \(f:D\to\mathbb{R}\)  has a point of inflection at  \((a,\,f(a))\),  where  \(a\in D\),  if and only if the concavity (curvature) of \(f\) changes sign at \(x=a\).

In regards to the above question, could someone explain how arctan has a point of inflection, even though we cannot solve its derivative for zero? What am I missing?

A point of inflection does not simultaneously also have to be a stationary point. (If it is, we call it a "stationary point of inflection", but otherwise, it is a "non-stationary point of inflection).

And in the same way, how come the point of greatest gradient is at x=0 if the gradient is zero at this point, but is positive elsewhere? My main point of confusion is that if the derivative of arctan is 1/1+x^2, where the derivative is always greater than zero, how come we have a point of inflection?

Same as above. Points of inflection do not have to have a 0 gradient. The fact that \[\dfrac{d}{dx}\Big[\arctan(x)\Big]=\dfrac1{1+x^2}>0\ \ \ \forall x\in\mathbb{R}\] bears no affect on whether the graph of  \(y=\arctan(x)\)  has a point of inflection or not.

The graph of  \(g(x)=\arctan(x)\)  has a point of inflection at  \((0,\,0)\) because the graph of the first derivative, \(g'\), has a local maximum at \(x=0\).  Or, using the second definition,  the graph of \(g\) changes concavity (from concave up to concave down) at \(x=0\).

My reasoning for why the point of greatest gradient is at x=0 is that the reciprocal of x^2 +1 has a maximum value at x=0 and hence maximum gradient is at this point.

You are correct.  The fact that  \(g'\)  is maximum at  \(x=0\)  makes this the point of steepest ascent as well as a point of inflection.

[lzxnl]

I'll be writing a concise guide on inflection points soon, but I'll give you a brief overview.

Definitions (the two following definitions are equivalent):

1.  A differentiable function  \(f:D\to\mathbb{R}\)  has a point of inflection at  \((a,\,f(a))\),  where  \(a\in D\),  if and only if the first derivative, \(f'\), has an isolated extremum at \(x=a\).

2.  A \(k\) times (\(k\geq2\)) differentiable function  \(f:D\to\mathbb{R}\)  has a point of inflection at  \((a,\,f(a))\),  where  \(a\in D\),  if and only if the concavity (curvature) of \(f\) changes sign at \(x=a\).
A point of inflection does not simultaneously also have to be a stationary point. (If it is, we call it a "stationary point of inflection", but otherwise, it is a "non-stationary point of inflection).
Same as above. Points of inflection do not have to have a 0 gradient. The fact that \[\dfrac{d}{dx}\Big[\arctan(x)\Big]=\dfrac1{1+x^2}>0\ \ \ \forall x\in\mathbb{R}\] bears no affect on whether the graph of  \(y=\arctan(x)\)  has a point of inflection or not.

The graph of  \(g(x)=\arctan(x)\)  has a point of inflection at  \((0,\,0)\) because the graph of the first derivative, \(g'\), has a local maximum at \(x=0\).  Or, using the second definition,  the graph of \(g\) changes concavity (from concave up to concave down) at \(x=0\).
You are correct.  The fact that  \(g'\)  is maximum at  \(x=0\)  makes this the point of steepest ascent as well as a point of inflection.

Going to be a bit picky, but the two definitions that you mentioned are not identical for the sole reaaon that the conditions are different. Your second definition requires the existence of more derivatives than the first one, and thus is a weaker/less general definition.