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Old 22nd June 2013, 11:23 AM   #1 (permalink)
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A MATH HELP!

1. Given that x^1/4 + x^-1/4 = 3, find the value of x + 1/x

2. The points (5, -1) and (5,15) are on the circumference of a circle whose centre, C, lies above the x-axis. The line y = -3 is a tangent to the circle. Find

(i) the radius of the circle,
(ii) the two possible x-coordinates of C,
(iii) the two possible equations of the circle.

3. The figure shows part of the graphs of the curve y = e^2x+1 and y= 2cosx.
Find the (a) coordinates of A and (b) the shaded area.


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Last edited by snowmine; 22nd June 2013 at 11:23 AM.
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Old 22nd June 2013, 11:44 AM   #2 (permalink)
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Re: A MATH HELP!

Worked out the first one for you:

x^1/4 + x^-1/4 = 3 ------------(1)

Taking fourth power on both sides for (1):

we have x + 4x^(1/2) + 6 + 4x^(-1/2) + 1/x = 81

x +1/x + 4 [ x^(1/2) + 2 + x^(-1/2) ] = 75 ------------(2)

Squaring both sides for (1):

x^(1/2) + 2 + x^(-1/2) = 9

x^(1/2) + 2 + x^(-1/2) = 7 ----------------(3)

Substituting (3) into (2):

x +1/x + 4(7) = 75

Hence, x +1/x = 47 (shown)

Hope this helps. Peace.


Last edited by whitecorp; 22nd June 2013 at 11:44 AM.
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Old 22nd June 2013, 12:03 PM   #3 (permalink)
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Re: A MATH HELP!

Here's the solution for Q2:

If the centre of the circle is denoted by (a, b), recognize that the point of tangency for the line y=-3 with
the circle is actually ( a, -3) . (ie either the highest or the lowest point on the circle)

Equation of circle is given by (x-a)^2 +(y-b)^2 = r^2 ----------(1)

Now we know that the points (a ,-3) , (5, -1) and (5, 15) lie on the circle, substituting these into (1) gives

(-3-b)^2 = r^2 ====> (3+b)^2 =r^2 ----------(2)

(5-a)^2 +(-1-b)^2 = r^2 =======> (5-a)^2 +(1+b)^2 = r^2 ----------(3)

(5-a)^2 +(15-b)^2 = r^2 ----------(4)

(4)-(3): (15-b)^2 - (1+b)^2 =0

Expanding gives -32b + 224 =0, ie b=7

Substituting b=7 into (2) gives r=10

Hence radius of circle is 10 units. (shown)

Substituting b=7 and r=10 into (3) gives (5-a)^2 +64 = 100

(5-a)^2 = 36

5-a =6 or 5-a =-6

therefore, the two possible x-coordinates of the circle are a=-1 or a= 11 (shown)

Lastly, with all the information in place, the two possible equations of the circle are given by

(x+1)^2 +(y-7)^2 = 100 or (x-11)^2 +(y-7)^2 = 100 (shown)


Hope this helps. Peace.


Last edited by whitecorp; 22nd June 2013 at 12:05 PM.
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Old 22nd June 2013, 12:19 PM   #4 (permalink)
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Re: A MATH HELP!

1.Squaring both sides,
(x^(1/4)+x^(-1/4))˛
=x^(1/2)+2x^(1/4)x^(-1/4)+x^(-1/2)
=x^(1/2)+x^(-1/2)+2
=3˛=9 -> x^(1/2)+x^(-1/2)=7
Squaring both sides again,
x+2x^(1/2)x^(-1/2)+x^(-1)=49
x+x^(-1)+2=49
x+1/x=47

2.

(i) The easiest way I see to get the radius of the circle is to see that since y=-3 is 2 units below (5,-1), then the higher horizontal tangent will be 2 units above (5,15), which is y=17. So the radius is (17+3)/2=10 units.

(ii) The y-coordinate of the 2 centres is easily found by 17-10=7.
There are multiple ways I see to get the x-coordinate, the fastest that I can see is this:
Let the centre of C2 be (a,7).
See that a-5 is the horizontal distance between C2 and the vertical line cutting through (5,-1) and (5,15).
By Pythagoras theorem, (a-5)˛+((15+1)/2)˛=10˛ which gives a-5=6 and a=11.
The x-coordinate of C1 is 5-6=-1.
So the 2 centres are (-1,7) and (11,7)

(iii)Should be fairly straightforward.

3. I think you're missing some info, can't get A.

Edit: Would like to point out that my solution for Q2 lacks rigour, whitecorp's solution would be more 'complete'.

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Last edited by Betakuwe; 22nd June 2013 at 12:33 PM.
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Old 22nd June 2013, 10:15 PM   #5 (permalink)
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Re: A MATH HELP!

Thanks whitecorp and betakuwe for your help!

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