Mathematics: Post your doubts here!

[princess787]

dear XPC Bot..... wenevr i post a doubt u post some blank comment... u alright? or got any problems? :/

 

[aleezay]

i got that much but in the ms the value of z is 1.253 for 6 and i am not getting that value of z for my equation

P(X<6)=0.105 Let (6-mean)/sd be a
P(Z<a)=0.105
a is negative (im not sure how you've been taught to determine whether a is -ve or +ve. What we do is that if both the sign is less than, i.e <, and the probability is less than 0.5, or if both are more than, then a is -ve)
1- Fi(a)= 0.105
Fi(a)= 0.895
a= -1.253

 

[yousef]

The identities you have to know is tanx = sinx/cosx
and sin^2x + cos^2x = 1
and from the above u can also see cos^2x = 1-sin^2x

Now this is very messy, i would suggest to write what you see in paper...since it isnt neat on pc. And check Steel Arm post below...it is nicely represented !

1-tan^2(x) / 1+tan^2(x)
1-[ sin^2(x) / cos^2(x)] / 1 + [sin^2(x) / cos^2(x)]
now multiply every term with cos^2(x)
cos^2(x) - sin^2(x) / cos^2(x) + sin^2(x)
cos^2(x) - sin^2(x) / 1
(1-sin^2(x)) - sin^2(x)
1-2sin^2(x)

can any one help me bro's ???
q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

 

[abcde]

can any one help me bro's ???
q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf

AoA! You could maximize the expression (2 sin^2 x - 3 cos^2 x) by maximizing the value with the positive coefficient and minimizing the value with the negative coefiicient. Max (sin^2 x) = 1 and min(cos^2 x) = 0. So, the max value of f(x) = 2(1) - 3(0) = 2.

To minimize the expression, minimize the value with the positive coefficient and maximize the value with the negative coefficient. Min(sin^2 x) = 0 and max(cos^2 x) = 1. So, the minimum would be = 2(0) - 3(1) = -3.

Tee hee!

 

[yousef]

AoA! You could maximize the expression (2 sin^2 x - 3 cos^2 x) by maximizing the value with the positive coefficient and minimizing the value with the negative coefiicient. Max (sin^2 x) = 1 and min(cos^2 x) = 0. So, the max value of f(x) = 2(1) - 3(0) = 2.

To minimize the expression, minimize the value with the positive coefficient and maximize the value with the negative coefficient. Min(sin^2 x) = 0 and max(cos^2 x) = 1. So, the minimum would be = 2(0) - 3(1) = -3.

Tee hee!

thanks dude ... thats not the way we got taugh in school ... so probably i wont understand it ::: any way thanks for helping ......... if there is other way .. it will be much better ..if not (thanks )
can u do no. 1

 

[salvatore]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Please help me solve qn no. 8 (i & ii).. an explanation will be appreciated.

Thanks!

 

[abcde]

thanks dude ... thats not the way we got taugh in school ... so probably i wont understand it ::: any way thanks for helping ......... if there is other way .. it will be much better ..if not (thanks )
can u do no. 1

No problem, dude. Even if it isn't the way you were taught, I hope it made some sense.
Q5. Use the identity sin^2 x = 1 - cos^2 x over here and you'll have it solved.
(P.S. Just a bit of advice: If you learn to adapt to new methods, things can work quite smoothly for you. )

 

[Just visiting]

Can anyone please help me in solving this question and if he/she would be extra clear when explaining
Thanks in advanc

 

[Rutzaba]

dear XPC Bot..... wenevr i post a doubt u post some blank comment... u alright? or got any problems? :/

xpc bot is a way of advertising
u cant post a blabk post at xpc

 

[19islandprincess96]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
Can someone help me?
Q10) i)
I made an equation in u but I don't understand how to integrate it.

 

[19islandprincess96]

No problem, dude. Even if it isn't the way you were taught, I hope it made some sense.
Q5. Use the identity sin^2 x = 1 - cos^2 x over here and you'll have it solved.
(P.S. Just a bit of advice: If you learn to adapt to new methods, things can work quite smoothly for you. )

I am jumping into you conversation out of nowhere but I had the same problem... My question is, how'd you know that max (sin^2 x) = 1 and min(cos^2 x) = 0?

 

[19islandprincess96]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Please help me solve qn no. 8 (i & ii).. an explanation will be appreciated.

Thanks!

f(x) = a+bcos 2x
f(0)=−1
=> a + bcos 2(0) = -1

Solve cos 2(0) in radian mode.
=> a + b = -1


and f(π/2) = 7
=> a + bcos 2(π/2) = 7
=> a - b =7

Solve them as simultaneous equations.

a + b = -1
a - b =7

a will be 3 and b -4.


Part ii)

y would be 0 when the curve intersects the x axis.
Putting the values of a and b found in i, the equation becomes
=> 3 - 4cos2x

And because y=0

3 - 4cos2x = o

I guess u can solve the rest.
Hope I helped!

 

[TERMINATOR]

A PROBABILITY QUESTION

Helen has some black tiles, some white tiles and some grey tiles. She places a single row of
8 tiles above her washbasin. Each tile she places is equally likely to be black, white or grey.
Find the probability that there are no tiles of the same colour next to each other

The ans is 128/2187

Any help will be highly appreciated.

 

[abcde]

I am jumping into you conversation out of nowhere but I had the same problem... My question is, how'd you know that max (sin^2 x) = 1 and min(cos^2 x) = 0?

The range of both the sine and cosine functions is -1 to 1. The range of their squares would be from 0 to 1 [ (-1)^2 = 1 ].

 

[Sandhya Mahat]

Plz help me with these questions guys..
1) Solve: 2^(2x-1) < 3^(3x-2)
2) Solve: (3* Mod(x)) / (x-1) < 2 , where Mod() means modulus or absolute value.
I would be very grateful.

 

[princess787]

A PROBABILITY QUESTION

Helen has some black tiles, some white tiles and some grey tiles. She places a single row of
8 tiles above her washbasin. Each tile she places is equally likely to be black, white or grey.
Find the probability that there are no tiles of the same colour next to each other

The ans is 128/2187

Any help will be highly appreciated.

yo terminator! so c now ther r 8 tiles! so 1st calculate the probability of getting d same color tiles! tht wld be (1/3)^8.... now its 1/3 cause therz an equal probability of al three colors... so (1/3)^8 and multiplied by 3 cause ther r 3 different colors! so u get 1/6561 into 3 = 3/6561 which reduces to 1/2187.... now the question asks tht no same color tile shld b aside so u subtract tht frm 1.... 1-1/2187.... which gives 2187-1/2187 giving 2186/2187....... thou i dunno hw they get 127.... well thts til where i cld reach..

 

[ahmed abdulla]

can any one help me bro's ???
q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf

i need also the same help >>> can any one help us please ?
WAITING

 

[Dug]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
Can someone help me?
Q10) i)
I made an equation in u but I don't understand how to integrate it.

I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
I = ⌡[tanⁿ x(tan² x + 1)] dx
u = tan x
dx = du/sec² x

I = ⌡[tanⁿ x(tan² x + 1)] du/sec² x
I = ⌡uⁿ du
I = uⁿ⁺¹/(n + 1)

u = tan x
tan(π/4) = 1
tan(0) = 0

I = 1/(n + 1)

 

[Farahhassan]

I need help with S2 please :'(
Hypothesis testing !!

 

[ahmed abdulla]

I = ⌡(tanⁿ⁺² x + tanⁿ x) dx
I = ⌡[tanⁿ x(tan² x + 1)] dx
u = tan x
dx = du/sec² x

I = ⌡[tanⁿ x(tan² x + 1)] du/sec² x
I = ⌡uⁿ du
I = uⁿ⁺¹/(n + 1)

u = tan x
tan(π/4) = 1
tan(0) = 0

I = 1/(n + 1)

q no. 1 q5 (ii)


http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_11.pdf can u help bro