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secondary 3 | A Maths
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need help with all qns except for 12a
very confused with logarithms and binomal theorem so pls explain too :)
Date Posted: 3 years ago
Views: 232
If you need my explanations, let me know, provided that I can remember and I have time
yepp need explanation for all of the questions as i dont understand binomial theorem and i am confused for logarithms (especially for 12 b and c)
btw can you help me with the formulas (how do they work) and what i need to know for binomial theorem
Sorry, I was not free the entire of the day. Will look again some time today.
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Binomial theorem and multiplication
Date Posted:
3 years ago
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For this question, I didn't do the usual approach of "coefficient of 1/x is zero since the term is missing in the final expression".
This is because the question did not state "the first few terms in the expansion of...", but only stated "the expansion of...".
The problem is that if we were to rearrange our expression in descending powers of x from the start, the first term would be -ak^5 x. And because the starting expression at the end is given to be -3/16 (no term in x provided), then -ak^5 would be taken to be zero, but we are given that a and k are non-zero numbers, so the two data pieces are in conflict with each other (one says that -ak^5 must be zero while the other says that neither a nor k is 0).
So, we cannot say that the term -ak^5 x and the term in 1/x is zero. They could be well hidden within the expansion because of this, so we cannot make the assumption that the coefficient of 1/x would be zero.
It turns out to be the case though, but in any case I find this question very badly set. The setter did not put this clearly enough.
This is because the question did not state "the first few terms in the expansion of...", but only stated "the expansion of...".
The problem is that if we were to rearrange our expression in descending powers of x from the start, the first term would be -ak^5 x. And because the starting expression at the end is given to be -3/16 (no term in x provided), then -ak^5 would be taken to be zero, but we are given that a and k are non-zero numbers, so the two data pieces are in conflict with each other (one says that -ak^5 must be zero while the other says that neither a nor k is 0).
So, we cannot say that the term -ak^5 x and the term in 1/x is zero. They could be well hidden within the expansion because of this, so we cannot make the assumption that the coefficient of 1/x would be zero.
It turns out to be the case though, but in any case I find this question very badly set. The setter did not put this clearly enough.
Date Posted:
3 years ago
i still dont really understand the part starting from -ak^5x .....
i am very bad at binomial theorem as i find it extremely confusing and most of the time i dont really know what is going on in this topic...
i am very bad at binomial theorem as i find it extremely confusing and most of the time i dont really know what is going on in this topic...
The main objective of this topic is to expand large expressions without the need to do expansions manually.
In my expansions in the first few lines (before the part "since the constant term is -3/16..."), I expanded the expression for (4 - ax) (k + 1/x)^5 to the desired power of 1/x^2 and grouped the expressions according to the relevant power of x (so I will not have two separate terms in a given power of x later on).
The part "since the constant term is -3/16..." requires us to make comparisons with the relevant powers.
In reality, the question should have mentioned "given that the first two terms in the expansion of". With this, we can deduce that the reason why they skipped the term in 1/x at the end is because the term has zero coefficient.
This part onwards leading to the value of a and of k is basically equating and solving for a and k.
In my expansions in the first few lines (before the part "since the constant term is -3/16..."), I expanded the expression for (4 - ax) (k + 1/x)^5 to the desired power of 1/x^2 and grouped the expressions according to the relevant power of x (so I will not have two separate terms in a given power of x later on).
The part "since the constant term is -3/16..." requires us to make comparisons with the relevant powers.
In reality, the question should have mentioned "given that the first two terms in the expansion of". With this, we can deduce that the reason why they skipped the term in 1/x at the end is because the term has zero coefficient.
This part onwards leading to the value of a and of k is basically equating and solving for a and k.
i still dont really understand the by inspection... part tho
That portion is basically your cubic equation solving. You guess a factor out and then do long division thereafter.
root 1/2 and factor 2k-1 is by guessing or... as i dont really understand where it came from
sorry if i ask too much questions out of a sudden because i have add maths exam on monday...
sorry if i ask too much questions out of a sudden because i have add maths exam on monday...
That one is guessing. You can use your calculator to do the guessing.
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Q12b
Date Posted:
3 years ago
when to know whether to sub a letter to something?
like subbing u into 3^x
like subbing u into 3^x
is that change of base at the log part? im quite confused with change of base especially in this question as i was stucked there
can you explain to me change of base fornula? will really appreciate it :)
can you explain to me change of base fornula? will really appreciate it :)
The substitution helps us to look at the equation more easily, because the structure of the equation to be solved appears to be quadratic with a squared term existing. Using the substitution will allow the equation to be really quadratic.
The solving of the exponential part can be solved by taking logs on both sides. It’s not really a change of base as the equation to solve, 3^x = 5, is not in logarithmic form. I took log to base 10 on both sides of the equation in this case.
You can also do a change of base, however, by noting that writing 3^x = 5 in the equivalent logarithmic form gets us x = log3 5, which we can then use change of base to base 10 to get lg 5 / lg 3 which you will then put in the calculator.
The solving of the exponential part can be solved by taking logs on both sides. It’s not really a change of base as the equation to solve, 3^x = 5, is not in logarithmic form. I took log to base 10 on both sides of the equation in this case.
You can also do a change of base, however, by noting that writing 3^x = 5 in the equivalent logarithmic form gets us x = log3 5, which we can then use change of base to base 10 to get lg 5 / lg 3 which you will then put in the calculator.
for change of base formula do we just sub in any numbers as the base or theres a standard procedure to determine what number to use as the base
Depending on the question, some bases are more effective than others. Base 10 and e are for calculator use. Base 2, 3 etc are for “non calculator” working purposes.
bases 2,3,5 etc are usually used when the question ask : when log2 7 = 5, solve......?
“Solve” are mainly bases 10 and e.
If the question says
“given that log2 x = k, express log4 x in terms of k”
Then base 2 is more appropriate.
If the question says
“given that log2 x = k, express log4 x in terms of k”
Then base 2 is more appropriate.
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Q12c
Date Posted:
3 years ago
i dont really understand what is going on here
btw we are allowed to add and subtract logarithmic forms?
btw we are allowed to add and subtract logarithmic forms?
why isnt 1 changed to log16 16 tho
Basically the first step which I did here is to convert all the bases into base 2 for such equation solving. Bases 10 and e are more of solving exponential equations.
Logarithms can be added and subtracted. In fact, the laws of logarithms contain such additions and subtractions.
For example,
log2 5 + log2 8 = log2 40.
Even things like log2 x + log2 x = 2 log2 x make sense, because this is just like k + k = 2k.
You can actually change the “1” to log16 16, but I decided to keep the 1 at the end so that I can convert the logarithmic form to exponential form. I always leave a number on one side (in this case, the number 1) for the calculation.
5/12 log2 x = 1
From here, by leaving the 1 on the RHS, we are almost ready to solve such equations by converting the form.
For example,
log2 5 + log2 8 = log2 40.
Even things like log2 x + log2 x = 2 log2 x make sense, because this is just like k + k = 2k.
You can actually change the “1” to log16 16, but I decided to keep the 1 at the end so that I can convert the logarithmic form to exponential form. I always leave a number on one side (in this case, the number 1) for the calculation.
5/12 log2 x = 1
From here, by leaving the 1 on the RHS, we are almost ready to solve such equations by converting the form.
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Q13
Date Posted:
3 years ago
for the base = 4units, how do we know if 4units is from c1 to c2 for c1 to point B
how to know whether we need to find the length of the line or we can just use units
there was one question where i used units to solve but ended up with a wrong answer as it required us to find the actual length to continue solving the question
there was one question where i used units to solve but ended up with a wrong answer as it required us to find the actual length to continue solving the question
It’s from point B to one of the points labelled C. Those are just two possible positions of C for the question, but I included them in a single diagram.
For each case the base of the triangle will be 4 units long.
For each case the base of the triangle will be 4 units long.
As far as I can, I use units to solve the question. This is especially so when a point lies further down a line. Say AC is three times the length AB and these three points are on a line. In such cases, to find C, I will use unit comparisons to do measurements as compared to lengths which may complicate numbers with the decimals.
btw what is angle of inclination?
are you free to take at look at some questions today as i have quite a few questions that i need help with, but if you are not free today then its ok
are you free to take at look at some questions today as i have quite a few questions that i need help with, but if you are not free today then its ok
Angle of inclination, sometimes called the angle of elevation, is the angle at which something is raised from a horizontal line
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Coefficient of 1/x is zero
Date Posted:
3 years ago
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Use log base2 to solve all
Use triangle property and right angle triangle
|x-3|
Not easy questions but worth to practise and you will know more
Use triangle property and right angle triangle
|x-3|
Not easy questions but worth to practise and you will know more
Date Posted:
3 years ago
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Q10
The rest I look another time
The rest I look another time
Date Posted:
3 years ago
i still dont really understand part ii tho
btw for questions like part iii, if the constant term does not contain x, we have to see which term in the expansion combined together will give a term that does not contain x, if it contain x or x^2 we have to see which term combines to give a term with x and x^2?
btw for questions like part iii, if the constant term does not contain x, we have to see which term in the expansion combined together will give a term that does not contain x, if it contain x or x^2 we have to see which term combines to give a term with x and x^2?
For part ii, you need to realise that 0.999^8 is the same as (1 - 0.001)^8, since 1 - 0.001 = 0.999. This is because we are asked to use the results from the first part, and we must look at how part ii is related to part i. Here, the exponents are the same at 8, so to use part i result, we have to rewrite 0.999 as 1 - 0.001, or in this case, 1 - 0.003/3.
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"if it contain x or x^2 we have to see which term combines to give a term with x and x^2?"
Yes, you will have to compare the corresponding expressions.
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"if it contain x or x^2 we have to see which term combines to give a term with x and x^2?"
Yes, you will have to compare the corresponding expressions.
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