Specialist 1/2 Question Thread!

[S_R_K]

For these types of questions (only MCQ so you always have your CAS and don't need to show working out) if you are struggling to convert the question into an answer a not so eloquent solution would be to assume x = some random number (you can go to heaps of decimal places) and then substitute that same number for all 6 expressions (The question + 5 answers) then you can select the one that matches.

EDIT: the above is more of an exam technique when you can't do the question rather than something you should always rely on

A similar technique is to try substituting in some numbers, and if the two expressions give you a different result, they can't be the same in general. This helps you rule out some of the options.

[pm_me_ur_eggs]

how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

[kiwikoala]

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

I'd start by seeing which numbers between 13 and 19 can divide evenly into 1050.

[AlphaZero]

how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

Just going to expand on what Steven said. Ages have to be natural numbers (positive whole numbers). So, we are looking for three natural numbers whose product is another natural number (mainly, 1050), with two of them being between 13 to 19. So, we should look for the factors of 1050 that are between 13 and 19.

I'll start you off. Using long division, or some other method, we find that 1050 is not divisible by 13, and so none of the children can be 13 years old. Now try 14, and see if it gives you progress

[studyingg]

how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

is the answer 5, 14,15

[S_R_K]

The methods suggested by other people will work, but they are not efficient.

The best method is to use prime factorisation. The prime factorisation of 1050 is 2*3*5*5*7. Now just use these prime factors to find the ages of three children, two of which are between 13 and 19.

[RuiAce]

The methods suggested by other people will work, but they are not efficient.

The best method is to use prime factorisation. The prime factorisation of 1050 is 2*3*5*5*7. Now just use these prime factors to find the ages of three children, two of which are between 13 and 19.

I think, in this case they are more efficient. The efficiency would break down if the range of numbers were greater, but here you only had to guess and check a total of 7 numbers. Here, the prime factorisation is perhaps moderate in difficulty to compute, however having to pair things off can also take time. You could've started by doing, say, 2*3 instead of going to 3*5. (When I looked at those, I seemed to be drawn to the last 3 numbers first, making numbers exceeding 20 before I considered the whole bundle in the one go.)

On the other hand, your method would be more formal in my opinion here. It would be more efficient only after the amount of numbers you have to guess and check goes haywire, say, more than 15 of them.

[S_R_K]

Perhaps... I think the prime factorisation is pretty immediate, since 2*5 is clearly a factor, and there are very few ways to use the prime factors to get a product between 13 and 19...

Then again, once you find that 1050 is divisible by 14, the other two ages are pretty obvious.

[pm_me_ur_eggs]

Perhaps... I think the prime factorisation is pretty immediate, since 2*5 is clearly a factor, and there are very few ways to use the prime factors to get a product between 13 and 19...

Then again, once you find that 1050 is divisible by 14, the other two ages are pretty obvious.

sorry if this is obvious/assumed knowledge but what does this mean from the worked solutions?

7 × 2 = 14
5 × 3 = 15

does it mean that any of the factors multiplied together also equals another factor of the original number?

[S_R_K]

sorry if this is obvious/assumed knowledge but what does this mean from the worked solutions?

7 × 2 = 14
5 × 3 = 15

does it mean that any of the factors multiplied together also equals another factor of the original number?

Not quite. Notice that both 525 and 210 are factors of 1050, but 525 * 210 is not a factor of 1050.

What is true is that any product of the prime factors of x will also be a factor of x. So, for example, since 2, 5, and 7 are prime factors of 1050, 2*5*7 = 70 is also a factor of 1050.

[Scribe]

When do you use tildes (~) and 'hats' (^) with regards to vectors, mechanics, etc? Also, with mechanics, do you write 'i-j direction', 'direction of motion', 'DOM', or 'positive direction', etc?

Thanks 

[AlphaZero]

When do you use tildes (~) and 'hats' (^) with regards to vectors, mechanics, etc? Also, with mechanics, do you write 'direction of motion', 'DOM', or 'positive direction', etc?

Thanks 

Tildes are used to denote any vector. For example: \[\underset{\sim}{\text{a}}=3\underset{\sim}{\text{i}}+4\underset{\sim}{\text{j}}.\]
Hats are used to denote unit vectors. For example: \[\underset{\sim}{\hat{\text{a}}}=\frac{1}{\sqrt{3^2+4^2}}\left(3\underset{\sim}{\text{i}}+4\underset{\sim}{\text{j}}\right)=\frac{3}{5}\underset{\sim}{\text{i}}+\frac{4}{5}\underset{\sim}{\text{j}}\]
As for wording in mechanics, any of those phrases are fine as long as you are consistent. Although, I will say, using some are more appropriate than others, but that depends on the context of the problem

[Scribe]

Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks 

[RuiAce]

Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks 

Your answer is correct. I can see that you solved \(a^2 - 4(4)(3) > 0\) because Wolfram gives the same answer.

They're both valid in theory because they do mean the same thing. But I wouldn't know if one is preferred in the VCE over the other.

[Sine]

Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks 

Both are fine in terms of vce

Although I prefer the first one since they answer is very clear and that is the usually notation vcaa will use in questions