VCE Methods Question Thread!

[secretweapon]

Attached. Great to have in your bound reference

THANKS!!
Would you by any chance know why my cas sometimes gives my wrong answers?
like i typed ∫(-cos(x)-sin(x),x,3*π/4,π)
into my cas and it gave me -0.82, then i reset my cas and it gave me the correct answer of 0.41. Is there a way to fix this problem?

[S200]

Double check your settings.
You could be in radians or degrees when you don't need to be...

[MAGGOT]

THANKS!!
Would you by any chance know why my cas sometimes gives my wrong answers?
like i typed ∫(-cos(x)-sin(x),x,3*π/4,π)
into my cas and it gave me -0.82, then i reset my cas and it gave me the correct answer of 0.41. Is there a way to fix this problem?

What are your document settings?
Settings > document settings?

[secretweapon]

I was in the same setting before and after resetting my cas, in other words both the original incorrect answer and the correct answer after i reseted my cas were both in the same setting. What else could be the problem with my cas randomly giving me wrong answers?
Also, if i said that the answer to a question was 48/5, does it have to be simplified to 9.6 to get full marks or is 48/5 accepted?

[MAGGOT]

I would rather use improper fractions and both answers should be accepted. However, always re read the whole question and highlight what form the examiners want it to be.

[secretweapon]

I would rather use improper fractions and both answers should be accepted. However, always re read the whole question and highlight what form the examiners want it to be.

If it asks for "exact  values" then are both 48/5 and 9.6 considered to be exact values?

[S200]

If it asks for "exact  values" then are both 48/5 and 9.6 considered to be exact values?

So the improper fraction is certainly an exact value.
But i am not sure that integers like 9.6 are exact values.

[MAGGOT]

If it asks for "exact  values" then are both 48/5 and 9.6 considered to be exact values?

Agree with s, 48/5 is an exact value but not 9.6.
Like 4/3 is an exact value but 1.33 is not an exact value

[Bri MT]

Decimals can be exact values too. Eg. 3/2 = 1.5.  Both are exact values

Basically you can't round decimals otherwise it isn't exact.
Eg. Pi = 3.14 isn't exact because it's a rounded version of 3.1415.... etc

1/3 = 0.3 isn't exact because it's a rounded version of 0.333.... etc


Not the most technical description but hopefully it helps you understand

[S200]

Oh, ok. So if they are finite, and short enough to be written without rounding, that is exact?

[Sine]

Oh, ok. So if they are finite, and short enough to be written without rounding, that is exact?

you always want to be giving the exact value (unless number of decimal places are required).

As miniturtle has already illustrated numbers with decimals can also be exact values. E.g. 1/4 = 0.25
So 1/4 and 0.25 are exact values for the same thing.
For numbers which don't "end" after a certain number of expressable decimal places (in terms of non zero digits)e.g. 1/7  = 0.142857143.... for this you couldn't just say 1/7 and 0.14 are the same since those decimals actual continue as non-zero digits.

[xdmemeguy]

Need some help with this question

Let P(x) = (x+2)(x-5)Q(x)+ax+b

where Q(x) is a polynomial and a and b are real numbers.

The polynomial P(x) has a factor of (x+2).

When P(x) is divided by (x-5) the remainder is 14

Find the values of a and b

[VanillaRice]

Need some help with this question

Let P(x) = (x+2)(x-5)Q(x)+ax+b

where Q(x) is a polynomial and a and b are real numbers.

The polynomial P(x) has a factor of (x+2).

When P(x) is divided by (x-5) the remainder is 14

Find the values of a and b

Which part are you stuck at? You have two pieces of information here - what do they tell you? Hint: consider the factor and remainder theorems. Can you use these to form two equations to solve for a and b?

Hope that helps Post if you're still stuck.

[xdmemeguy]

What does q(x) become though? does it just stay there? Or do I sub -2 and 5 into x?

[VanillaRice]

What does q(x) become though? does it just stay there? Or do I sub -2 and 5 into x?

If you substitute either -2 or 5 into x, Q(x) will disappear This is fine, since we only care about the values of a and b.