IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: The SMA on April 19, 2010, 07:26:52 am

Title: Need HELP for CIE AS Maths P1
Post by: The SMA 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
Title: Re: Need HELP for CIE AS Maths P1
Post by: nid404 on April 19, 2010, 07:57:51 am: In 5 hrs...im on the phone
Title: Re: Need HELP for CIE AS Maths P1
Post by: 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.
Title: Re: Need HELP for CIE AS Maths P1
Post by: The SMA 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
Title: Re: Need HELP for CIE AS Maths P1
Post by: 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?
Title: Re: Need HELP for CIE AS Maths P1
Post by: astarmathsandphysics on April 19, 2010, 03:39:37 pm: Will answer when I get home
Title: Re: Need HELP for CIE AS Maths P1
Post by: Saladin 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.
Title: Re: Need HELP for CIE AS Maths P1
Post by: The SMA 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 :)