Mathematics: Post your doubts here!

[Muhammad Talha]

Assalamo Alaikum...
I wanna ask a question about Paper 6 that is S1...
Can anybody explain me about the continuity corrections....that where they would be applied....b/c in some ques they are applied bt in the other they r not....
Plz reply asap.....

 

[s1294]

jxt start with 6c1 always regardless of itx saying thats what i used to do and aftr getting some terms take ur required one whether ascending or decending

Thanks that clears my confusion

 

[leadingguy]

Assalamo Alaikum...
I wanna ask a question about Paper 6 that is S1...
Can anybody explain me about the continuity corrections....that where they would be applied....b/c in some ques they are applied bt in the other they r not....
Plz reply asap.....

I am not sure As I havenot studied it deeply but I can say that itx only applied when the value of meu and varience is big like it could be (50,4) this value is not readable by table so we convert it into standard one by continuity correction .

 

[s1294]

when you are given f ' (x) as 2x-6 and that the range of function f(x)>= -4 and you have to find the value of x for which f(x) has a stationary value, do you differentiate the f ' (x) agn so that you get f '' (x)= 2 ... then would 2 be the answer?
and to find the original function from f ' (x) do you integrate it to get x^2 - 6x + C?

 

[umarashraf]

Assalamo Alaikum...
I wanna ask a question about Paper 6 that is S1...
Can anybody explain me about the continuity corrections....that where they would be applied....b/c in some ques they are applied bt in the other they r not....
Plz reply asap.....

mate you wish to solve a bionomial probability question with the help of standard normal distribution, then the continuity correction is applied... i can explain its reason as well if you are really interested

 

[umarashraf]

when you are given f ' (x) as 2x-6 and that the range of function f(x)>= -4 and you have to find the value of x for which f(x) has a stationary value, do you differentiate the f ' (x) agn so that you get f '' (x)= 2 ... then would 2 be the answer?
and to find the original function from f ' (x) do you integrate it to get x^2 - 6x + C?

i dont think so").. i think just replace x in (f ' ) relation... f " will only indicate the idea of the nature of stationary point... either maximum or minimum...

 

[s1294]

i dont think so").. i think just replace x in (f ' ) relation... f " will only indicate the idea of the nature of stationary point... either maximum or minimum...

i thought f ' (x) gave us the gradient of the tangent line not the stationary point?

 

[leadingguy]

when you are given f ' (x) as 2x-6 and that the range of function f(x)>= -4 and you have to find the value of x for which f(x) has a stationary value, do you differentiate the f ' (x) agn so that you get f '' (x)= 2 ... then would 2 be the answer?
and to find the original function from f ' (x) do you integrate it to get x^2 - 6x + C?

to find original funcyion of f'x u should integrate and to find stationary point jxt put f'x to zero and find value of x no need for double diff.

 

[s1294]

i dont think so").. i think just replace x in (f ' ) relation... f " will only indicate the idea of the nature of stationary point... either maximum or minimum...

if i do that i get 2x-6=0, 2x=6, x=3 so 3 is the x value which will give a stationary value and what about the integral i asked about if you want to change f '(x) back into f (x) how would you have done it- i got x^2-6x +c as the answer

 

[s1294]

to find original funcyion of f'x u should integrate and to find stationary point jxt put f'x to zero and find value of x no need for double diff.

so after integrating it i get x^2 - 6x + C is that correct?

 

[leadingguy]

Need help with this as well (please draw and show as well)
i) sketch on a single diagram the graphs of y-cos2(theta) and y=1/2 for 0<= (theta)=<2(pi)

ii) write down the number of roots of the equation 2cos2(theta) - 1=0 in the interval 0<= (theta)=< 2(pi)

iii) deduce the number of roots of the equation 2cos2(theta) - 1 =0 in the interval 10(pi)<= (theta) =< 20(pi)

i really don't get the theta and pi stuff, pls explain

I cannot provide U with the diagram at the moment but can give U an idea.

draw a line of y = 1/2 which is parallel to x axis and will pass through y axis at y =1/2.

and then use the eq. y = cos2(theta put values of theta in it and mark the points drawing a sketch

2pi = 360
pi = 180
pi/2 =90
0pi = 0 the range is from o to 2pi means from o to 360

u will get y= cos2(theta as
y = cos2(o = 1
y = cos2(90 = cos(180 = -1
y= cos2(180 = cos(360 = 1
y = cos2(270 = cos(540 = -1
y = cos2(360 cos (720 = 1

plot these points and form a graph


now look at the intersection of these lines (y=1/2 and y =cos2(theta ) there are four intersections known as there roots

 

[umarashraf]

so after integrating it i get x^2 - 6x + C is that correct?

100% correct mate...

 

[umarashraf]

so after integrating it i get x^2 - 6x + C is that correct?

double derivative will only tell about the nature of ALREADY OBTAINED STATIONARY POINT....

 

[s1294]

100% correct mate...

thnak you- working late night, with u all helping give me courage, thnx

 

[s1294]

double derivative will only tell about the nature of ALREADY OBTAINED STATIONARY POINT....

ohh okay- thanks i get it now

 

[umarashraf]

thnak you- working late night, with u all helping give me courage, thnx

thanks bro...")

 

[leadingguy]

so after integrating it i get x^2 - 6x + C is that correct?

yes itx totaly correct.
if nothing else is given then this is ur ans
but if co-ordinates are given then put the x and y coordinates and solve for C (constant)

 

[s1294]

I cannot provide U with the diagram at the moment but can give U an idea.

draw a line of y = 1/2 which is parallel to x axis and will pass through y axis at y =1/2.

and then use the eq. y = cos2(theta put values of theta in it and mark the points drawing a sketch

2pi = 360
pi = 180
pi/2 =90
0pi = 0 the range is from o to 2pi means from o to 360

u will get y= cos2(theta as
y = cos2(o = 1
y = cos2(90 = cos(180 = -1
y= cos2(180 = cos(360 = 1
y = cos2(270 = cos(540 = -1
y = cos2(360 cos (720 = 1

plot these points and form a graph


now look at the intersection of these lines (y=1/2 and y =cos2(theta ) there are four intersections known as there roots

so the roots are the four roots shown on the diagram- which is unclear because i am not using a graphing paper- btw changing the scale from radians to degrees is allowed in exams?
and the roots i asked for was for 2cos(2 theta) - 1= 0 not cos(2 theta) and y= 1/2
after which i have to deduce the number of roots for 2cos(2 theta) - 1= 0 for an interval of 10 pi<= (theta)<= 20pi

please check n let me know

 

[s1294]

yes itx totaly correct.
if nothing else is given then this is ur ans
but if co-ordinates are given then put the x and y coordinates and solve for C (constant)

right finallyyyy getting the hang of it- thank you

 

[leadingguy]

so the roots are the four roots shown on the diagram- which is unclear because i am not using a graphing paper- btw changing the scale from radians to degrees is allowed in exams?
and the roots i asked for was for 2cos(2 theta) - 1= 0 not cos(2 theta) and y= 1/2
after which i have to deduce the number of roots for 2cos(2 theta) - 1= 0 for an interval of 10 pi<= (theta)<= 20pi

please check n let me know

change 2cos2(theta -1 = o by substituting y= cos2(theta
i.e
2y - 1 = o Y is substituted for cos2(theta now it will be y = 1/2

U can see frm graph that at how many positions is this line of y =1/2 cutting the graph (at four)