Mathematics: Post your doubts here!

[ashcull14]

Show tan(3x+x) = 2tan (60-x) in the form of tan^2x+ 6 (3)^1/2 tanx-5 =0
2) hence solve the equation for 0<x <180

 

[randomcod]

Does anyone have the M/J 15 Paper 32 and 72 mark schemes or grade thresholds?

 

[cool Hassan]

If f(x)= (36x^2) -(2x^4)
Find the interval for which f(x) is a decreasing function.

^Can anyone please solve it, along with explanation.

its simple cuz u need to get the derivative first then put it less than 0
then solve that inequality to get the critical values of xand then u can randomly put the values of x into the inequality.. if inequality satisfies then u get the correct range
which can either be b/w the critical values or it can be above the greater critical value and lesser than the smaller critical value

 

[cool Hassan]

If f(x)= (36x^2) -(2x^4)
Find the interval for which f(x) is a decreasing function.

^Can anyone please solve it, along with explanation.

dy/dx=72x-8x^3
putting dy/dx=0
72x-8x^3=0
multiply both sides with -1
8x^3-72x=o
take x common
x(8x^2-72)=0
u get the critical values as
x=0
x=+3
x=-3
u get the range of the values of x for which f(x) is decresing as
x>3
It can be totally wrong though

 

[Copy Cat]

f(x)= (6+7x)/(2-x)(1+x^2)= A/(2-x) + Bx+C/(1+x^2)
Cross Multiply: 6+7x = A(1+x^2) + Bx+C(2-x)
Let x=2: 6+7(2) = A(1+4) A = 4
Coefficient of x^2: Ax^2 - Bx^2= 0x^2
A=B , B=4
Coefficient of x: Bx - Cx = 7x
4x - Cx = 7x
-Cx = 3x
C= -3

How did you got that Bx+C?

 

[eliyeap]

How did you got that Bx+C?

Its the formula that my teac gave. You can alternatively break it into A/2-x + B/1+x + C/1+x^2 and solve it but its kind of complicated and takes more effort, so feel free to use this formula. The Bx+C is only for quadratic factors, like here you have 1 term that's x^2.

 

[eureka!]

Assalamu `Alaykum.
I was wondering if we need to know how to derive laws, so in P3 do they ask questions like, prove that log a - log b = log (a/b)?

 

[iSean97]

How would you explain it for (iii) ?

 

[Turki AbdulAziz]

I need topic wise past papers for Pure 3 and Statistics 1.
I kept looking everywhere but I couldnt find anything.
I hope someone could help!!

 

[Rizwan Javed]

I need topic wise past papers for Pure 3 and Statistics 1.
I kept looking everywhere but I couldnt find anything.
I hope someone could help!!

cool Hassan <-- Ask this guy

 

[RadzMau]

Can anyone please help me out with this question?

 

[eureka!]

Can anyone please help me out with this question?

7a.
3(a+bi) + 2i(a-bi) = 17+8i
Expand:
3a+3bi+2ai+2b = 17+8i
Equate imaginary numbers on both sides and real numbers on both sides:
(3b+2a)i = 8i
3b+2a=8 and 2b+3a=17
Solve simultaneously:
(Multiply eq.1 with 2 and eq.2 with 3 or vice-versa.)
6b+4a=16
6b+9a=51
a=7, b=-2
So w=7-2i.

Don't know about 7b, sorry about that!

 

[manya]

can someone help me with part b

 

[manya]

and part b of this aswell

 

[#SponsoredWalk]

Having problem with Permutations and Combinations also in probability chapter 4 I never get this stuff into my head, rest is a piece of cake. Any help?
Notes or sth like how to tackle specific type of questions...

 

[Rectified Spirit]

Hello friends can anyone help me to understand no. ( i i i ) in this particular question. I am facing difficulty in understanding what happens when someone refuses. This question is from S1 Permutation and combination

A committee of 5 people is to be chosen from 6 men and 4 women. In how many ways can this be done
(i) if there must be 3 men and 2 women on the committee,
( i i ) if there must be more men than women on the committee,
( i i i ) if there must be 3 men and 2 women, and one particular woman refuses to be on the committee
with one particular man?

 

[Rizwan Javed]

Hello friends can anyone help me to understand no. ( i i i ) in this particular question. I am facing difficulty in understanding what happens when someone refuses. This question is from S1 Permutation and combination

A committee of 5 people is to be chosen from 6 men and 4 women. In how many ways can this be done
(i) if there must be 3 men and 2 women on the committee,
( i i ) if there must be more men than women on the committee,
( i i i ) if there must be 3 men and 2 women, and one particular woman refuses to be on the committee
with one particular man?

First, find the no. of ways in which none of them is included (i.e. neither the woman who refuses, nor the man is included)
No. of ways in which neither of the two particular man and woman are included: 5C3*3C2 = 30

Then find the no. of ways in which the man is included but the woman is not: 6C3*3C2 = 60

Hence total no. of ways in which the committee can be selected is : 30+60 = 90 <-- so here goes your answer

 

[Rectified Spirit]

Thanks Rizwan Zaved, it was of great help.

 

[GriM ReapeR]

Assalamualaykum everyone, I have a formula doubt. This was written in my statistics notebook without any context, someone please explain.
thank you

 

[Ziki629]