Mathematics: Post your doubts here!

[Iffat]

can sum1 plz help with the foll qs
2007 may-jun:
q5
q8q11(i),(ii),(iii)

 

[badrobot14]

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

11 (ii).... well b is the x value where max point of curve occurs.... for max point.. all you do is take derivative of curve... and set it equal to zero.... you solve for x.. and that gives u x value at max pt...

 

[Rutzaba]

the awkward mo wen u solve ur own question before sumone else does

 

[Silent Hunter]

Aoa anybody got some good notes according to the syllabus (M1) ? need them urgent?

JazakAllah

 

[PhyZac]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.

Can anyone solve this, please?

 

[inquisitiveness]

Factorise 24x^2+34x-45 (2 marks)

 

[Dug]

Assalamu Alikum Wa Rahmatullahi Wa Barakatoho.

Can anyone solve this, please?

Walaikum AsSalam Wa Rahmatullahi Wa Barakatoho!!

-> Separate variables: 1/(x + 1) dx = cos2Θ/sin2Θ dΘ
-> Integrate: ln|x + 1| = (1/2)ln|sin2Θ| + c
-> Put values: ln|1| = (1/2)ln(1/2) + c
c = (-1/2)ln(1/2)
-> Simplify: ln|x + 1| = (1/2)ln|sin2Θ| - (1/2)ln(1/2)
ln|x + 1| = (1/2)ln|(sin2Θ)/(1/2)|
ln|x + 1| = ln|2sin2Θ|^1/2
x + 1 = √(2sin2Θ)
x = √(2sin2Θ) - 1

 

[PhyZac]

Walaikum AsSalam Wa Rahmatullahi Wa Barakatoho!!

-> Separate variables: 1/(x + 1) dx = cos2Θ/sin2Θ dΘ
-> Integrate: ln|x + 1| = (1/2)ln|sin2Θ| + c
-> Put values: ln|1| = (1/2)ln(1/2) + c
c = (-1/2)ln(1/2)
-> Simplify: ln|x + 1| = (1/2)ln|sin2Θ| - (1/2)ln(1/2)
ln|x + 1| = (1/2)ln|(sin2Θ)/(1/2)|
ln|x + 1| = ln|2sin2Θ|^1/2
x + 1 = √(2sin2Θ)
x = √(2sin2Θ) - 1

Thank you so much sir! Jazaka Allah khairan, May Allah give you success in this life and hereafter....Alhamdulilah, i got it now..!

 

[Dug]

Factorise 24x^2+34x-45 (2 marks)

24x^2+34x-45
= 24x^2 + 54x - 20x - 45
= 6x(4x + 9) - 5(4x + 9)
= (6x - 5)(4x + 9)

 

[Kumkum]

cos^2θ=(1+cos2θ)/2
sin^2θ=(1-cos2θ)/2
cos^2(2θ)={1+cos2(2θ)}/2=(1+c0s4θ)/2
LHS=sin^2θcos^2θ
={(1-cos2θ)/2}{(1+cos2θ)/2}
=(1-cos^2(2θ))/4
={1-(cos2θ)^2}/4
={1-(1+c0s4θ)/2}/4
=(1-4cosθ)/8=RHS
Hope this helps . Ask if u dnt understand

thank u so much....I understood

 

[inquisitiveness]

thanks alot!Really appreciate your help!!Is there anyway you found it the two numbers are 54 and 20?or do you just keep guessing?

 

[Dug]

thanks alot!Really appreciate your help!!Is there anyway you found it the two numbers are 54 and 20?or do you just keep guessing?

I did it with my calculator (fx-991es). You can solve for the roots and then work your way backwards. It's very helpful !!

 

[Iffat]

can sum1 plz help with the foll qs
2007 may-jun:
q5
q8q11(i),(ii),(iii)

can sum1 plz reply asap!

 

[Dug]

The detailed solution of:-
View attachment 22107

x̄ = Σx/n = 645/150 = 4.3

Expand:
Σ(x-x̄)^2 = Σx^2 - 2Σxx̄ + Σx̄^2

Put values, Σx^2 = 8287.5 ; Σx = 645 ; x̄ = 4.3
Σ(x-x̄)^2 = 8287.5 - 2(645)(4.3) + (150)(4.3)^2
Σ(x-x̄)^2 = 5514

 

[josephsai]

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

Q7i) first of all u should know that p is the x coordinate
and q is the y coordinate so substitute y=11-x^2 into other eqn & arrive at
11-x^2=5-x
rearrange them
x^2 - x - 6 =0
the reason why we equate them is bcuz they intersect @ a pt
so solve this quadratic eqn &
x=3 & x= -2
we take x=3 cause it's on the de positive axis
so thats our p
to find q put the value of p which is x=3 into
y=5-x
y=5-3
=2
meaning q is 2 and p is 3
simple inform me if it helped

 

[josephsai]

for

Assalamualikum
Can someone please help me with question 7(ii), 8(i) & 11(ii).
Thanks a bunch in advance.

SRY i thought u meant Q7i) btw for ii) i got f inverse to be-x^2 +10x -14 am nt really sure
Q8i)d^2y/dx^2= 9(3x-4)^1/2 -6
Q11(ii) dy/dx=2x(x-2) + (x-2)^2 using product rule
expand
2x^2 -4x + x^2 -4x + 4
simplify
3x^2 - 8x + 4=0
x=2/3 & x=2
therefore b=2/3

 

[tanmaydube]

Summer 11 paper 32
Question 4 i)
Please help.

Thank you!

 

[littlecloud11]

Summer 11 paper 32
Question 4 i)
Please help.

Thank you!

OCT is a right-angled triangle with ∠OCT being 90 degree.
So, tan x = CT/r
CT =r tanx
Area of triangle = 1/2 * base* height =1/2 *r * rtanx
Area of shaded region = 1/2 r^2 tanx- 1/2 r^2 x
Area of semi-circle = 1/2* r^2* π
Area of semi-circle= area of shaded region
1/2*r^2* π = 1/2 * r^2* (tanx -x)
After canceling from both sides you get
π + x= tanx

 

[Alice123]

Summer 11 paper 32
Question 4 i)
Please help.

Thank you!

If u dnt understand feel free to ask

 

[Alice123]

Aoa anybody got some good notes according to the syllabus (M1) ? need them urgent?

JazakAllah

http://www.examsolutions.net/maths-revision/syllabuses/CIE/period-1/M1/module.php
hope this helps