Mathematics: Post your doubts here!

[The Sarcastic Retard]

-b/2a??

Vertex y value, as it will be the lowest the range will be from there to the highest value of domain substituted in f(x).

 

[awesomaholic101]

Vertex y value, as it will be the lowest the range will be from there to the highest value of domain substituted in f(x).

Ok thanks

 

[The Sarcastic Retard]

good luck

 

[lara dalal]

Hi guys I need help in solving q9 part i. Thankyou in advance

 

[Rectified Spirit]

Hi can anyone help me to understand no 2, I didn't get how to find out summations of x squared I need to understand it ASAP as I have exam the day after tomorrow pls someone help.....

 

[Brianlee97]

Hi can anyone help me to understand no 2, I didn't get how to find out summations of x squared

I need to understand it ASAP as I have exam the day after tomorrow pls someone help.....

Here is the solution for this question. By the way ,I am also sitting for the same paper. Good luck

 

[nehaoscar]

Please help!

 

[Rectified Spirit]

Thanks man was stuck with this question for a long time. Good to know that you are also giving the same exam, wish you good luck!!!!

 

[Brianlee97]

Thanks man was stuck with this question for a long time. Good to know that you are also giving the same exam, wish you good luck!!!!

Wish you luck also friend !

 

[qwertypoiu]

Hi guys I need help in solving q9 part i. Thankyou in advance

d/dx((1-x)/(1+x)) = -2 / (1+x)^2
Let u = (1-x)/(1+x)
d/dx(√u) = u' / 2√u = [ -2 / (1+x)^2 ] / [ 2 * √((1-x)/(1+x)) ]
= -1 / [ (1+x)^2 * √((1-x)/(1+x)) ]

Now the gradient of the normal is the negative reciprocal of the gradient of tangent:

(1+x)^2 * √((1-x)/(1+x))
= (1+x)^2 * (1-x)^1/2 * (1+x)^-1/2
= (1+x)^3/2 * (1-x)^1/2
= (1+x)^1 * (1+x)^1/2 * (1-x)^1/2
=(1+x)^1 * [(1+x)(1-x)]^1/2
= (1+x) * (1-x^2)^1/2

 

[Rectified Spirit]

Hello friends I do have S 1 exam tomorrow would anyone pls share their experiences, what should I do go through over now for last time preparation........

 

[Asif1223]

Plz help me with this problem.. Its urgent plz!!!

 

[Asif1223]

Plz guyz anyone solve this problem

 

[musiclover gurl]

Help please?
Nov 2013 P 32

 

[Brianlee97]

[/QUOTE]

View attachment 57230 Plz guyz anyone solve this problem

Here is the solution. Hopefully it is right.
[Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime.
][Maimonides]

 

[nehaoscar]

Please help!

 

[qwertypoiu]

Help please?
Nov 2013 P 32
View attachment 57231

2 * |3^x - 1| = 3^x

Square both sides to get rid of modulus (cuz its ugly)

4*(3^2x - 2* 3^x + 1) = 3^2x
Expand bracket:
4•3^2x - 8•3^x + 4 = 3^2x
Move terms:
3•3^2x - 8•3^x + 4 = 0
Make u=3^x :
3u^2 - 8u + 4 = 0
Factorise:
3u^2 - 6u - 2u + 4 = 0
3u(u-2) -2(u-2) = 0
(3u-2)(u-2) = 0
u=2, u=2/3
3^x = 2, 3^x = 2/3
x = lg2/lg3 , x=lg(2/3) / lg3
sorry don't have a calculator

 

[Mahreen Mustafa]

Any suggestion for tomorrow's statistics exam?

 

[Asif1223]

Yeah

Here is the solution. Hopefully it is right.
[Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime.
][Maimonides]
View attachment 57232[/QUOTE] Thanks bro for the solution .. but in 7 (ii) it doesnot match with the mark scheme answer

 

[Hala Ezzat]

J12 P6 Q1 , how to solve this?? Please tell me if there is any online website that can help me out with this type of question. My teacher never taught me this