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Mathematics: Post your doubts here!

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may u help me in any of above posts
View attachment 59922 help in II june 2013 p12
For this expansion you're only interested in finding the term in x^2. So to make calculations simple, find the terms in x^2, and x^0 only in the expansion of (2x -1/2x)^6 .
Then multiply out those two terms with the remaining expression (1 + x^2), considering only terms in x^2 after, and ignoring the term in x^4.

Hope you'll be able to solve the rest by yourself. :)
 
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View attachment 59922 help in II june 2013 p12
This can more easily be solved by finding a general term first,
General term T(r+1) = 6Cr (2x)^6-r (-1/2x)^r
collect only powers of x now: (2x^6-r) (1/2x^r)
sums up to x^6-r/x^r =x^6-2r
Now, since we want to find the coefficient of x ^2
x^2=x^6-2r
only collect the powers (as there is the same index x) : 6-2r=2
2r=4 and r=2
Now, we simply put that into the formula we just made above, for the general term, (we just found the value of r):
T(r+1) = 6Cr (2x)^6-r (-1/2x)^r
6C2 (2x)^6-2 (-1/2x)^2
15(16x^4)(1/4x^2) = 60x^2
co-efficient of x^2 is 60
 
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This you have to imagine in mind first or just sketch out so that you have an idea what the question means. when you draw out, you will see that the point (-1,3) and the point R if joined together will form a line exactly perpendicular to the line in which R is reflected.

Since we are given the equation of the line in which the reflection takes place: 3y+2x=33 ---(i)
we can easily find the gradient of this, by making y the subject: y=33-2x/3 , you might be able to see already from the equation that gradient of this line is -2/3
since the product of the gradients of perpendicular lines = -1
the gradient of perpendicular line, on which R or the point (-1,3) exists is: m ( -2/3)=-1, which becomes 3/2
Now we just got the gradient of the perpendicular line and we already know that it also passes through (-1,3) equation of that perpendicular line can be easily made:
y-3=3/2 (x+1)
y-3=(3x+3)/2
y=(3x+3)/2 +3 ---(ii)
Now solving the both perpendicular equations simultneously will give us the coordinates, which is the intersection point of both lines
It will get us (3,9)
but still we haven't got the reflected point, we are exactly halfway through that, which in other words is the mid-point of both points ( the (-1,3) and the R)
assume R(x,y)
use the mid-point formula (we have the mid-point (3,9) and the point (-1,3):
x-1/2 , y+3/2 = 3, 9
seperate them
x-1/2 =3 and y+3/2 =9
here we get the required coordinates x =7 and y=15
 
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No. of ways of arranging 3 cards without restriction (also found in iii) = 9C3 x 3!
= 504
( Two cards out of 3 are green and pink together) = 7C1 x 1C1 x 1C1 x 2P2 x 2P2
= 28
504-28= 476
ه
i didnt understand they said how many araangement of 3 cards in part III contain a pink card why u wrote ( Two cards out of 3 are green and pink together) they didnt mention anything about greeen in the question part v is clear now but my doubt is in IV
 
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Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive.
(i) Find the probability that at least 2 of the 5 integers are less than or equal to 4. [3]
Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the
number of these nintegers which are less than or equal to a certain integer k between 1 and 9 inclusive.
It is given that the mean of X is 96 and the variance of X is 32.
(ii) Find the values of n and k.

Some1 tell me how to get the value of k plz rest I know.
k will be equal 6
 
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Robert uses his calculator to generate 5 random integers between 1 and 9 inclusive.
(i) Find the probability that at least 2 of the 5 integers are less than or equal to 4. [3]
Robert now generates n random integers between 1 and 9 inclusive. The random variable X is the
number of these nintegers which are less than or equal to a certain integer k between 1 and 9 inclusive.
It is given that the mean of X is 96 and the variance of X is 32.
(ii) Find the values of n and k.

Some1 tell me how to get the value of k plz rest I know.
k will be equal 6
https://in.answers.yahoo.com/question/index?qid=20140529123109AAtHhJS
 
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Please anybody give full solution to this question
 

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