Mathematics Question Thread

[Fahim486]

The other question is already addressed in the compilation.

So when you sub in x = 0, does the cos and sec squared disappear? because when i put cos(0) into my calculator, it gives me 1 so i thought cos3x = 3

[RuiAce]

So when you sub in x = 0, does the cos and sec squared disappear? because when i put cos(0) into my calculator, it gives me 1 so i thought cos3x = 3

[MisterNeo]

Hey guys, this may be easy, but how do you do part ii ?

[Opengangs]

Hey guys, this may be easy, but how do you do part ii ?

[RuiAce]

Hey guys, this may be easy, but how do you do part ii ?

To observe why, consider a few examples.
30 has 2 digits. Additionally, \(\log_10 30\approx 1.47\) and 1.47 rounded up gives you 2.
987 has 3 digits. Additionally, \(\log_10 987 \approx 2.99\) and 2.99 rounded up gives you 3.

The simple reason behind this is just the observation that log is a monotonic increasing function, and furthermore,
\( \log_{10} 10 = 1, \log_{10} 100 = 2, \log_{10} 1000 = 3, \log_{10} 10000 = 4 \dots\)
So one of infinitely many consequences is that if \(10 < x < 100 \), then \( 1\le \log_{10} x < 2\), so if you round up \( \log_{10}x\) where x is a two digit number that's not 10 itself, you get back to 2.

[Mounica]

Can someone pls help me with this question

[RuiAce]

Can someone pls help me with this question

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[Mounica]

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Thanks you

[Becky234]

Hey
 Can someone pls help me with this question

[Shadowxo]

Hey
 Can someone pls help me with this question

Which part are you stuck on?

[sidzeman]

I'm at a complete loss as how to do this question, even from part i

[RuiAce]

I'm at a complete loss as how to do this question, even from part i

Part i has already been addressed in the compilation.

Go back and think about the remaining parts again. They are standard applications of coordinate geometry, followed by a max/min problem (albeit with the wording overcomplicated). As a hint, for part ii, recall that BQ is a horizontal line. Also, Q is the point where the lines \( x \cos \theta + y\sin\theta = 1\) and \( y = 1\), which should now be all the information you need.

(You may want to anticipate a messy quotient rule for part iv.)

Please post your thought process if you need further help. Part i) is the hardest part of this question.

[_____]

Is integration of tanx a part of the 2 unit syllabus or is this trial just being hard? Wondering if it is because I haven't encountered it before (I know now to use sinx/cosx).



Seems straightforward, a + B = -B + B therefore b/a = 0. But why is (4k+1) = b here? Shouldn't it be (4k+1)x? Solution:



Thanks

[RuiAce]

(Image removed from quote.)

Is integration of tanx a part of the 2 unit syllabus or is this trial just being hard? Wondering if it is because I haven't encountered it before (I know now to use sinx/cosx).

(Image removed from quote.)

Seems straightforward, a + B = -B + B therefore b/a = 0. But why is (4k+1) = b here? Shouldn't it be (4k+1)x? Solution:

(Image removed from quote.)

Thanks

It really isn't if you ask me. As you've just seen, the trick was to use sin(x)/cos(x), but personally I find it unfair how a 2U student should be told to do that without any guidance whatsoever.
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[_____]

It really isn't if you ask me. As you've just seen, the trick was to use sin(x)/cos(x), but personally I find it unfair how a 2U student should be told to do that without any guidance whatsoever.
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Thanks.

For the second one, the equation is given as 2x^2 - (4k+1) + 2k^2 - 1 = 0. Shouldn't there be an x next to the -(4k+1) (so the original equation is 2x^2 - (4k+1)x + 2k^2 - 1 = 0) so that I know that -(4k+1) is b? If not, how do I know what's b and what's c? I should have explained this better earlier.