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Mathematics: Post your doubts here!

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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
 
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upload_2015-12-26_8-49-32.png
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.


upload_2015-12-26_8-53-54.png
(iii) No idea. Help needed.

Thanks. :)
 
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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.
 
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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. :)
 
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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
 
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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 :(
 
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I dont get :(
:eek:
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
 
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:eek:
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.
 
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