Mathematics: Post your doubts here!

[Ashjay]

ohhhh .. I got it now .. such question met me before .. can I know why can't we divide the 3x (3)^1/2 by 2 and thats it ?? .. and thanks in advance for your help ^_^


and how did you get the angles 30 and 90 to give 120 !!

We cannot divide it by two because we are dealing with vectors here. If we add or subtract vectors we have to take into account their directions too, The forces T1 and T2 have different directions so we cannot just add or subtract their magnitudes, we have to construct a vector triangle. So to construct the vector triangle we have to join the head of one vector with the tail of the other, if we do this the angle between them = 90 + 30 = 120.

 

[Serenia]

View attachment 50504

Sorry, just saw this today!

Thanks for the help! It cleared some things up =)

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_41.pdf

Q5 can someone help me please whats with the signs ? .. I got the answers but i dont know what sign should I put greater than or less than !!

RoOkaYya G Ashjay

 

[Ashjay]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_41.pdf

Q5 can someone help me please whats with the signs ? .. I got the answers but i dont know what sign should I put greater than or less than !!

RoOkaYya G Ashjay

In the first part we know that the block does not move, this means that the friction is 'greater than' the applied force in the horizontal direction. So the sign for "U" is 'greater than'. In the second part the block is moving so the friction is "less than" the force in the horizontal direction, so the sign for "U" is also "less than". In these types of questions remember these rules:
1. If the block is stationary (not moving) "u" is "greater than"
2. If the block is moving "u" is "less than"

 

[kylee33579]

How to do question no 6??http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_31.pdf

 

[Andhikasm]

PLEASE help me in this statistics question...I would be grateful
A die is biased. The mean and variance of a random sample of 70 scores on this die are found to be
3.61 and 2.70 respectively. Calculate a 95% confidence interval for the population mean score.
I don't understand why the variance is done as such? = (70/69)*2.7

http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_71/
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_ms_71/

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_43.pdf

Q6

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_43.pdf
Q7

 

[SitiPutri]

PLEASE help me in this statistics question...I would be grateful
A die is biased. The mean and variance of a random sample of 70 scores on this die are found to be
3.61 and 2.70 respectively. Calculate a 95% confidence interval for the population mean score.
I don't understand why the variance is done as such? = (70/69)*2.7

http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_71/
http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_ms_71/

Because n is > 30 and you have to use the formula for unbiased estimate of population variance (s^2). Formula attached.

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_43.pdf
Q7

Let's first discuss what is going to happen here - the masses are connected by a piece of inextensible string - this means that the string will not stretch no matter how much tension is applied on it, and so each part of the string will move with the same speed and same acceleration (if the string could stretch, different parts would stretch at different rates - these parts would stretch along the direction of movement and accelerate with the rest of the string, so the force on these parts would be different from the forces on the other parts, and so the tensions in the string would not be the same at every point). The consequence of this is at both masses will move with the same velocity and acceleration as long as there is any tension in the rope.

Mass A will accelerate downwards; B would accelerate to the left with the same acceleration magnitude, until Mass A hits the ground. When Mass A hits the ground, we can assume that it will come to a stop, and therefore will not pull on the string anymore - therefore, the tension in the string will become 0 Newtons. However, there is no friction to oppose the motion of mass B, so it will keep moving with the velocity it has accelerated to until that moment, until it hits the pulley.

Right, now to the mathematics - let's first take Mass A, take downwards as the positive direction, and apply Newton's Second Law:

0.32 * g - T = 0.32 * a (Where T is the tension in the rope and a is the acceleration magnitude of the mass)

Let's now apply the same law to Mass B, taking the leftward direction to be positive:

T = 0.48 * a (T is the same tension we used in the previous equation, since the string exerts equal tension forces on either mass, and a is again the same
acceleration magnitude we used in the previous equation, since the string in inextensible and the acceleration of both masses has the same magnitude)

Substituting this value of T in the first equation,

0.32 * 10 - 0.48 * a = 0.32 * a
0.32 * 10 = 0.8 * a

So a = 4 ms^-2

Substituting this value of a in the second equation,

T 0.48 * 4 = 1.92 Newtons.

I'll post the rest of the solution as soon as I can, sorry for the delays!

Hope this helped!
Good Luck for all your exams!

 

[Sadia Jahan Lisa]

I have statitics 1 in my mathematics. Is this table is required for Statistics 1 or it is required for statistics 2?

 

[Sadia Jahan Lisa]

Is this part of the normal distribution table is required for Statistics 1?If yes,is there anyone who can teach me when to look and how to look is part of the normal distribution table?

 

[AnujaK]

Solved AL Math papers! https://sites.google.com/site/thefinallevel3/ There's only math as of now, more papers coming up soon though. If you want the link to the AS math papers site, just comment here and I'll put it up.

 

[SitiPutri]

Solved AL Math papers! https://sites.google.com/site/thefinallevel3/ There's only math as of now, more papers coming up soon though. If you want the link to the AS math papers site, just comment here and I'll put it up.

Hi. Would've been better if you could compile all the images in one pdf file for each exam. It'd be much easier, instead of downloading the images one by one. Thanks for making the site, btw.

 

[lxelle]

Question 7ii) please!! http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w02_qp_3.pdf

 

[lxelle]

sagar65265

 

[iqbal]

Q4

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_3.pdf

Q7 (iii)

 

[iqbal]

Please somebody help me out with this question....Q4

 

[sagar65265]

Question 7ii) please!! http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w02_qp_3.pdf

(This solution is for 7(ii), not for 7(i), 7(iii) or 7(iv), so in case you'd like a discussion on those too, do post here)

From solving part (i), we know that α satisfies the equation (99/100)x = sin(x). We can rearrange this equation by subtracting sin(x) from both sides to give us a zero equality, namely

(99/100)x - sin(x) = 0

Therefore, the root of this equation will satisfy this equality, and any values just above or below it shouldn't (unless they themselves are roots).

Also, note that this function is continuous - there are no points where this function is invalid, so at all points the equation should be valid, and since the graph drawn from this would be a smooth curve, it is continuous at all points - you can confirm this by checking that the gradient of this curve is a valid function - differentiating this equation, we get

f'(x) = (99/100) - cos(x)

(Contrast this with the f(x) = |x| graph - the graph takes a sudden turn at x=0, and so the gradient at x=0 cannot be verified by looking at the graph. However, with the above curve, the gradient at all points is verifiable, so it is smooth).

This smoothness is an important point - for there to be a root, the curve must touch or cross the x-axis at some point on the other, at which point the value of f(x) will be equal to zero (what we wanted from the solution). Since the curve is continuous, it has to go from above the x-axis to below the x-axis (or vice-versa, from below the x-axis to above the x-axis), cross the x-axis in the process, giving us a root.

Therefore, the one-line solution for this is to prove that the curve has opposite signs at each end of the range in which there is a root. In other words, we have to prove that the sign of f(0.1) is the opposite of the sign of f(0.5) - this means that the curve will cross the x-axis between these values of x, and thus give us a root somewhere in between.

(Working in radians,) we substitute the value x = 0.1 radians in f(x) and find that

f(0.1) = (99/100) * 0.1 - sin(0.1 radians) = -8.3 * 10^-4

which is clearly a negative value - when x = 0.1 radians, the function holds a negative value, i.e. (99/100)x is slightly less than sin(x). Now let's try x = 0.5 radians:

f(0.5) (99/100) * 0.5 - sin(0.5 radians) = + 0.479

which has the opposite sign, i.e. f(0.5) is positive, and (99/100)x is slightly greater than sin(x). Therefore, since we are looking for the value of x for which
(99/100)x = sin(x) (and for which (99/100)x - sin(x) = 0), we must find it somewhere between 0.1 and 0.5 - therefore, a root of this equation exists between x = 0.1 and x = 0.5.

There is an interesting case that i've not mentioned here, and that is if the curve just touches the x-axis and continues with the same sign after doing so - take the equation of (x-2)2 = 0 , which only has one root at x = 2, and is positive on both sides of the root (and always even otherwise, since it is a square). In this case, this root check will not be possible, but I doubt such situations will ever come up in your exams.

One thing in these situations is that as long as the equation is continuous, the gradient of the function at the root point will be equal to zero. In our above example, if we differentiate (x-2)2=0 , we get f'(x) = 2(x-2) , which is indeed zero at x=2, the point at which we find our root.

Hope this helped!
Good Luck for all your exams!