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Author Topic: Need HELP for CIE AS Maths P1  (Read 1303 times)

Offline The SMA

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Need HELP for CIE AS Maths P1
« on: April 19, 2010, 07:26:52 am »
Hi guys  :), i have a doubt on the O/N 2008 paper 1, question 9 (iii). How do you find the acute angles? Please help. Thanks in advance  ;D
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nid404

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Re: Need HELP for CIE AS Maths P1
« Reply #1 on: April 19, 2010, 07:57:51 am »
In 5 hrs...im on the phone
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Offline anthonychy

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Re: Need HELP for CIE AS Maths P1
« Reply #2 on: April 19, 2010, 08:31:51 am »
Hi there,the answer for your question is 19.4 degree.

Here's the solution,
Initially, you differenciate the equation then substitute x=0 and x=1 which is the two points P and Q.
Now find the angle using this two gradients. (use tan [gradient of each]) you should get 56.3 degree and 36.9 degree.
Finally, find the difference of the angles. Done.
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Offline The SMA

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Re: Need HELP for CIE AS Maths P1
« Reply #3 on: April 19, 2010, 09:35:56 am »
Quote from: anthonychy on April 19, 2010, 08:31:51 am
Hi there,the answer for your question is 19.4 degree.

Here's the solution,
Initially, you differenciate the equation then substitute x=0 and x=1 which is the two points P and Q.
Now find the angle using this two gradients. (use tan [gradient of each]) you should get 56.3 degree and 36.9 degree.
Finally, find the difference of the angles. Done.

Ahhhh..Finally i got the answers! Thank you very much anthonychy  :)
but did you mean use (tan inverse [dy/dx]) for x=0 and x=1 right :P?
i already got the answer though. thanks  ;D

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Offline The SMA

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Re: Need HELP for CIE AS Maths P1
« Reply #4 on: April 19, 2010, 02:17:38 pm »
hey guys ;),
i have more doubts from the O/N 2008 paper 1.
its Q5, how do we relate maximum/minimum values to the given function?
is it from graph or something?

and Q10 (ii), i don't know how to prove max. value of gf(x)=9.
Can someone help me step by step?
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Offline astarmathsandphysics

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Re: Need HELP for CIE AS Maths P1
« Reply #5 on: April 19, 2010, 03:39:37 pm »
Will answer when I get home
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Offline Saladin

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Re: Need HELP for CIE AS Maths P1
« Reply #6 on: April 19, 2010, 04:01:02 pm »
Quote from: The SMA on April 19, 2010, 02:17:38 pm
hey guys ;),
i have more doubts from the O/N 2008 paper 1.
its Q5, how do we relate maximum/minimum values to the given function?
is it from graph or something?

and Q10 (ii), i don't know how to prove max. value of gf(x)=9.
Can someone help me step by step?

For question 5:
(i)The maximum value of the curve will occur when it is a+b, and thus in this case the value is 10.
The minimum value will occur when it is a-b, and thus the value is -2. We have to remember that both a and b are positive constants. The actual value of x does not matter in this question.

(ii) 4-6cos(x)=0
cos(x)=\frac{2}{3}

I got, 48.2 and 311.8 degrees

(iii) Should look something like a hill, with to intersections with the x axis in the positive region or the first quadrant.

For question 10:

This is simple, gf(x)=-9x^2+30x-16

Thus simply differentiate it to get the point at which the value is the maximum,
\frac{d}{dx}=-18x+30
And the value that you get for \frac{d}{dx}=0 is \frac{5}{3}

If you put this value of x to the formula for gf(x), then you get 9.

Hope this helped.
« Last Edit: April 19, 2010, 04:16:26 pm by The Ultimate Dude 321 »
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Offline The SMA

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Re: Need HELP for CIE AS Maths P1
« Reply #7 on: April 21, 2010, 09:31:35 am »
Quote from: Mr. Mysterous on April 19, 2010, 04:01:02 pm
For question 5:
(i)The maximum value of the curve will occur when it is a+b, and thus in this case the value is 10.
The minimum value will occur when it is a-b, and thus the value is -2. We have to remember that both a and b are positive constants. The actual value of x does not matter in this question.

(ii) 4-6cos(x)=0
cos(x)=\frac{2}{3}

I got, 48.2 and 311.8 degrees

(iii) Should look something like a hill, with to intersections with the x axis in the positive region or the first quadrant.

For question 10:

This is simple, gf(x)=-9x^2+30x-16

Thus simply differentiate it to get the point at which the value is the maximum,
\frac{d}{dx}=-18x+30
And the value that you get for \frac{d}{dx}=0 is \frac{5}{3}

If you put this value of x to the formula for gf(x), then you get 9.

Hope this helped.
Thank you for the help mr. Mysterous,
 i finally understand and got the answer. all answers corrrrrecct  :)
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