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secondary 4 | A Maths
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Kathy
Kathy

secondary 4 chevron_right A Maths chevron_right Singapore

Thank you

Date Posted: 4 years ago
Views: 247
Eric Nicholas K
Eric Nicholas K
4 years ago
Later

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Eric Nicholas K
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5997 answers (Tutor Details)
1st
Q43
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Q44
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Q45

This question very very tricky because not only we have to find the range of values for which f’(v) is positive, we also must make sure that the square root of v3 - v2 must exist (in other words, we must solve for v3 - v2 >= 0).

Therefore, we get v >= 1 as the range of values.
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Eric Nicholas K
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Q46
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Q47

Kathy, you must not forget how to do binomial after not touching it for quite some time.
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Eric Nicholas K
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Q48, still halfway in (ii)

End of session. I am not free now and will continue at a much later time. You can post the next one first.
Eric Nicholas K
Eric Nicholas K
4 years ago
Out now and will only continue at a much later time
Kathy
Kathy
4 years ago
This part u not yet done?
Eric Nicholas K
Eric Nicholas K
4 years ago
I return to do later; forgot to bring that piece of paper out
Eric Nicholas K
Eric Nicholas K
4 years ago
My workings are on that piece of paper and I forgot to take it out of the scanner to continue writing
Kathy
Kathy
4 years ago
Don’t forget this qns. Thanks
Eric Nicholas K
Eric Nicholas K
4 years ago
I do the other three first. Now going back to Q48.
Kathy
Kathy
4 years ago
I finish this, I start new chapter
Eric Nicholas K
Eric Nicholas K
4 years ago
Currently stuck at that question, I trying to figure out how to get sq 3 / 2 - 3, I obtained a different answer altogether
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Q48, corrected

I have worked out the last part, and I believe that the final expression is incorrect. In fact, for x < negative sqrt 3, dy/dx can go from 0 to negative infinity in value. I will upload a graph of dy/dx in a moment to show you.
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Eric Nicholas K
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Kathy, here is the proof which shows that dy/dx should be greater than, instead of less than, 0.5 sqrt 3 - 3.
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Q48 proving part, updated for clarity
Eric Nicholas K
Eric Nicholas K
4 years ago
The given question is incorrect.