Mathematics: Post your doubts here!

[Sarosh Jameel]

PLZ help with part (I) !!

 

[Anum96]

PLZ help with part (I) !!

Make a triangle ABO. Make a line in between. This line will bisect angle AOB

Label the end of this point D

Find AD using cos.

cos(x) = AD/r

AD= rcos(x)

So AB will twice of AD

Which is 2rcosx

AB =AC =2rcosx


Now make another triangle ABN; N being the second corner of shaded region on the line OA.

You have AB which is equal to AN(AN is the radius of the circle with center A)

So Find the length of arc BN

Which will be r(theta)

r =2rcosx

theta =2x

r(theta) = 2rcosx*2x =4xrcosx


Now the question says...

perimeter of shaded region = 1/2(circumference of circle

AB +AC +BC =1/2(2*pi*r)

2rcosx+ 2rcosx+ 4xrcosx = pi*r

All r's cancel out

Now you have 4cosx +4xcosx =pi

Take cos common

cosx(4+ 4x) = pi

x= cos^-1 [(pi/(4 +4x)] Shown

 

[The Sarcastic Retard]

Anyone taking P7 ?

Me.

 

[The Sarcastic Retard]

I got two values of x. 1 is pi/6 and other is 5pi/6. I also found that pi/6 is maximum point, but i cant find 5pi/6 to be minimum point. Help required.



(iii) No idea. Help needed.

Thanks.

 

[The Sarcastic Retard]

As there is no one active in chemistry thread, I want someone to help me in this question(A2 - Electrochemistry chapter).
https://www.xtremepapers.com/commun...st-your-doubts-here.9859/page-748#post-948570
Thanks. ^_^

 

[Anum96]

View attachment 58135
I got two values of x. 1 is pi/6 and other is 5pi/6. I also found that pi/6 is maximum point, but i cant find 5pi/6 to be minimum point. Help required.


View attachment 58136
(iii) No idea. Help needed.

Thanks.

For a second, I doubted my eyesight. Very clear pictures I must say.
Lol, Wait up!

 

[Anum96]

Equation of the tangent
y = mx + c
c will be 0 since the tangent passes through origin.
gradient(Differentiate the equation of curve)
You will get e^x(3 -x)

Plug in the values.
y = mx + c [c = o & y = (4-x)e^x]
(4 - x)e^x = e^x(3 - x)x + 0
(4 - x)e^x = e^x(3 - x)x
e^x cancels out.
4-x = (3-x)x
4-x = 3x - x^2
4 - x - 3x + x^2 = 0
x^2 - 4x + 4 = 0
x(x-2) - 2(x-2) = 0
x or p is 2 Ans.

 

[The Sarcastic Retard]

Equation of the tangent
y = mx + c
c will be 0 since the tangent passes through origin.
gradient(Differentiate the equation of curve)
You will get e^x(3 -x)

Plug in the values.
y = mx + c [c = o & y = (4-x)e^x]
(4 - x)e^x = e^x(3 - x)x + 0
(4 - x)e^x = e^x(3 - x)x
e^x cancels out.
4-x = (3-x)x
4-x = 3x - x^2
4 - x - 3x + x^2 = 0
x^2 - 4x + 4 = 0
x(x-2) - 2(x-2) = 0
x or p is 2 Ans.

Thanks. Can u help me in other question as well.

 

[The Sarcastic Retard]

For a second, I doubted my eyesight. Very clear pictures I must say.
Lol, Wait up!

Ah. You are totally correct, they were clearest pictures I ever saw in my life. Thanks to GCE guide. (Y)

 

[Anum96]

Thanks. Can u help me in other question as well.

The other qs is so boring. :/
Lol, Wait

 

[Anum96]

Ah. You are totally correct, they were clearest pictures I ever saw in my life. Thanks to GCE guide. (Y)

Alright! Blame the innocent guide.

 

[The Sarcastic Retard]

The other qs is so boring. :/
Lol, Wait

Aha!! Waiting ^_^

 

[The Sarcastic Retard]

Alright! Blame the innocent guide.

Blame is on the correct thing They are innocent, but so naaughty.

 

[Anum96]

Aha!! Waiting ^_^

Look. You got the right answers and yes pi/6 is maximum. That means your working for derivative and double derivative is correct.
Second point is 5pi/6
Double derivative is -2cos(x)
Plug in the value. -2cos(5pi/6) [5pi/6 is 150]
Absolute value of cos 150 is -undrt3/2
Therefore -2(-undrt3/2) will give a positive value which comes out to be the minimum value of undrt3

 

[The Sarcastic Retard]

Look. You got the right answers and yes pi/6 is maximum. That means your working for derivative and double derivative is correct.
Second point is 5pi/6
Double derivative is -2cos(x)
Plug in the value. -2cos(5pi/6) [5pi/6 is 150]
Absolute value of cos 150 is -undrt3/2
Therefore -2(-undrt3/2) will give a positive value which comes out to be the minimum value of undrt3

I dont get

 

[Anum96]

I dont get

Use the same method you applied for first point. Plug in the value of x
If you dont remember the absolute value use your calculator. Pi is 180. So 5(180)/6 is 150
cos 150 is negative. And negative*negative gives out positive, And positive values are always minimum

 

[The Sarcastic Retard]

Use the same method you applied for first point. Plug in the value of x
If you dont remember the absolute value use your calculator. Pi is 180. So 5(180)/6 is 150
cos 150 is negative. And negative*negative gives out positive, And positive values are always minimum

ok thanks.

 

[Anum96]

ok thanks.

Welcome. Lol.

 

[The Sarcastic Retard]

Welcome. Lol.

can u help me in chem?

 

[Anum96]

can u help me in chem?

I don't take chemistry