Mathematics: Post your doubts here!

[Rizwan Javed]

Rizwan Javed Wow you're really at good at permutations and combinations. I find them rather difficult actually so any general tips that you could pass on?

The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!

 

[malk mohamed]

please i want formulas of AS math please

 

[Copy Cat]

The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!

Thanks!!!

If possible then can you please post the harder one(s) over here.Once again Thanks

 

[Farhan Ismail]

ln(2 + e^-x) = 2
2 + e^-x = e^2
e^-x = e^2 - 2
-xlne = e^2 - 2
-x = 5.39
x = -5.39

you skipped one step ,you forgot to take natural log of 5.39 right answer is -1.68
thanks alot tho

 

[Konstantino Nikolas]

you skipped one step ,you forgot to take natural log of 5.39 right answer is -1.68
thanks alot tho

Yes yes
No problem.

 

[Saad the Paki]

The only thing to remember here is that if there's some kind of arrangement involved, then permutations must used. But if you're to select something the you're to use the combinations. You can refer to this website for learning this topic: http://www.mathsisfun.com/combinatorics/combinations-permutations.html . If you want to master this topic, solve as many questions as you can. I also used to be bad at this topic, so I solved a lot of questions on this topic and those questions which i solved were much much harder than these. So now as i had solved the harder ones, I don't feel problem in these easier ones. So just practice, that's all that I can say. Best of luck!

Alright thanks!

 

[Raj21999]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_11.pdf

please help me with questions 8 (ii) , (iii) and (iv)

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf

please help me with questions 8 (ii) , (iii) and (iv)

(x + a)^2 + b
Here b = range
-a = domain

 

[Iris Dylan Lane]

Prove that the 13th term of an A.P. is zero, if five times it's fifth term is equal to 8 times its eighth term. Anyone?

 

[Rizwan Javed]

Prove that the 13th term of an A.P. is zero, if five times it's fifth term is equal to 8 times its eighth term. Anyone?

Let the first term of the A.P be T1.
d is the common difference.

The 5th term would then be: T1 + d(5-1) = T1 + 4d
The 8th term would be : T1 + d(8-1) = T1 + 7d

5(T1 + 4d ) = 8(T1 + 7d)

^ Solve it and get 'd' in terms of T1.

d = (-1/12) T1

The 13th term would be:

T1 + d(13-1)

substitute the value of 'd' we found in terms on T1.

T1 -1/12T1(13-1)
= T1 - T1
= 0 (shown)

 

[Iris Dylan Lane]

Let the first term of the A.P be T1.
d is the common difference.

The 5th term would then be: T1 + d(5-1) = T1 + 4d
The 8th term would be : T1 + d(8-1) = T1 + 7d

5(T1 + 4d ) = 8(T1 + 7d)

^ Solve it and get 'd' in terms of T1.

d = (-1/12) T1

The 13th term would be:

T1 + d(13-1)

substitute the value of 'd' we found in terms on T1.

T1 -1/12T1(13-1)
= T1 - T1
= 0 (shown)

Wow. Thanks... a lot.

 

[farhan141]

Q.1? My fourth ordinate is the same as the 3rd. How is the 4th 0.693?



This is MJ 32 2015

 

[qwertypoiu]

Q.1? My fourth ordinate is the same as the 3rd. How is the 4th 0.693?

View attachment 59633

This is MJ 32 2015

How come? I have just checked with my calculator:

f(x) = ln(1+sin(x))

f(0) = 0
f(pi/6) = 0.4055
f(2*pi/6) = 0.6238
f(3*pi/6) = 0.6931

 

[farhan141]

How come? I have just checked with my calculator:

f(x) = ln(1+sin(x))

f(0) = 0
f(pi/6) = 0.4055
f(2*pi/6) = 0.6238
f(3*pi/6) = 0.6931

Oh sorry i took the last ordinate as 2/3π

 

[farhan141]

Can anyone tell me why Tan (98.8) is equal to -6.459.. But when I do tan inverse (-6.459) then it gives me -81? Crazy..

 

[qwertypoiu]

Can anyone tell me why Tan (98.8) is equal to -6.459.. But when I do tan inverse (-6.459) then it gives me -81? Crazy..

The problem is that there are infinitely many values of x such that tan(x) = -6.459
Some examples are:
x = ..., -261.2, -81.2, 98.8, 278.8, 458.8, .....

Doing tan on all the above numbers will give you answer as -6.459. So poor calculator, which one should it tell you when you perform inverse tan on -6.459…?

We have therefore defined something known as the PRINCIPAL VALUES of the inverse trigonometric functions. These are ranges from which an inverse trig function should output an answer:

sin-1(x) always gives answer between -90 and +90
cos-1(x) always gives answer between 0 and +180
tan-1(x) always gives answer between -90 and +90

So this is why your calculator gives answer as -81.2, since it falls within the range of principal values.

 

[farhan141]

The problem is that there are infinitely many values of x such that tan(x) = -6.459
Some examples are:
x = ..., -261.2, -81.2, 98.8, 278.8, 458.8, .....

Doing tan on all the above numbers will give you answer as -6.459. So poor calculator, which one should it tell you when you perform inverse tan on -6.459…?

We have therefore defined something known as the PRINCIPAL VALUES of the inverse trigonometric functions. These are ranges from which an inverse trig function should output an answer:

sin-1(x) always gives answer between -90 and +90
cos-1(x) always gives answer between 0 and +180
tan-1(x) always gives answer between -90 and +90

So this is why your calculator gives answer as -81.2, since it falls within the range of principal values.

One question had two answers for tan x as it was quadratic and I was only getting one correct answer while marking scheme had both. What should we do in that situation?

 

[phanttasy]

Need help on this question ! I couldn't get the answer after many times of trials,

2010 O/N paper 33 question 8 ii) b)

 

[lara dalal]

Hello guys. I have a confusion when it comes to drawing argand diagrams, do we draw them as approximations or do we use a compass and protactor. Also do we sketch them on graph paper or on normal paper.
Thank you

 

[awesomaholic101]

Hello guys. I have a confusion when it comes to drawing argand diagrams, do we draw them as approximations or do we use a compass and protactor. Also do we sketch them on graph paper or on normal paper.
Thank you

Just sketches. No graph paper needed, no protractor, no compass (unless you can't draw a circle that doesn't look like an egg )