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Author Topic: C1 and C2 DOUBTS HERE!!!!!  (Read 88605 times)

Offline Dibss

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Re: C1 DOUBTS HERE!!!!!
« Reply #375 on: January 09, 2011, 07:40:40 pm »
Quote from: 8T on January 09, 2011, 07:32:02 pm
Wait x is not zero it's -1 & + 1

Edited my post.

Quote
Q 9
Help ?

Trying now. (:
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Offline Dibss

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Re: C1 DOUBTS HERE!!!!!
« Reply #376 on: January 09, 2011, 07:50:57 pm »
Given that
f ?(x) = 6 ? 4x ? 3x2,
(a) find an expression for y in terms of x, (5)
(b) show that AB = k 7 , where k is an integer to be found.
---
a)To find y, integrate f'(x)
To find c use (0,0) since the curve passes though origin
? c = 0
y = 6x ? 2x^2 ? x^3

b) Find the roots of the eq or set y=0
x(6 ? 2x ? x2) = 0
Use quadratic formula to solve for x coord at A and B
A (?1 ? root7 , 0), B (?1 + root7 , 0)
to find AB the length, just subtract.
? AB = (?1 + root7 ) ? (?1 ? root7 ) = 2root7

(:
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Offline 8T

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Re: C1 DOUBTS HERE!!!!!
« Reply #377 on: January 09, 2011, 08:28:53 pm »
Yes I knew that one once I solved for x THANKSS :D
Now I'll read ur solving for 9 (:
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Offline 8T

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Re: C1 DOUBTS HERE!!!!!
« Reply #378 on: January 09, 2011, 08:38:07 pm »
Quote from: Dibss (Whisper) on January 09, 2011, 07:50:57 pm
Given that
f ?(x) = 6 ? 4x ? 3x2,
(a) find an expression for y in terms of x, (5)
(b) show that AB = k 7 , where k is an integer to be found.
---
a)To find y, integrate f'(x)
To find c use (0,0) since the curve passes though origin
? c = 0
y = 6x ? 2x^2 ? x^3

b) Find the roots of the eq or set y=0
x(6 ? 2x ? x2) = 0
Use quadratic formula to solve for x coord at A and B
A (?1 ? root7 , 0), B (?1 + root7 , 0)
to find AB the length, just subtract.
? AB = (?1 + root7 ) ? (?1 ? root7 ) = 2root7

(:
Thank YOU TAHNK YOU THANK YOU :D

OKAY now can I please have question paPer 2010 january (: & the mark schme for JUNE 2010
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Offline 8T

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Re: C1 DOUBTS HERE!!!!!
« Reply #379 on: January 09, 2011, 08:42:13 pm »
4 B
I know how I just want to make sure I got the final answer is possible to offer help (: ?
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Offline 8T

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Re: C1 DOUBTS HERE!!!!!
« Reply #380 on: January 09, 2011, 08:47:04 pm »
Do I wait or is no one trying it  ?
Oh & is anyone sending me jan 2010 question paer
marsk scheme june 2010 ?
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Offline astarmathsandphysics

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Re: C1 DOUBTS HERE!!!!!
« Reply #381 on: January 09, 2011, 10:43:43 pm »
4b answer below
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Offline The Golden Girl =D

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #382 on: February 11, 2011, 01:46:17 pm »
the following Questions ;

C2 book page 159 Qs 6] a) and 10]b)

Q6]a) find ,without using calculator -.- ,the values of :

sin theta ,cos theta ,given tan theta = 5/12 and theta is acute.

Q10]b)

in each of the following , eliminate theta to give an equation relating to X and Y;

x = sin theta , y= 2cos theta.

I'm having trouble in solving ALL parts of Q 6 and 10 :S
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elemis

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #383 on: February 11, 2011, 01:56:42 pm »
Gimme 2 miutes.
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Offline The Golden Girl =D

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #384 on: February 11, 2011, 01:59:07 pm »
okay.
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elemis

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #385 on: February 11, 2011, 02:00:46 pm »
Quote from: ~ The Golden Girl ~ on February 11, 2011, 01:46:17 pm
the following Questions ;

C2 book page 159 Qs 6] a) and 10]b)

Q6]a) find ,without using calculator -.- ,the values of :

sin theta ,cos theta ,given tan theta = 5/12 and theta is acute.

Q10]b)

in each of the following , eliminate theta to give an equation relating to X and Y;

x = sin theta , y= 2cos theta.

I'm having trouble in solving ALL parts of Q 6 and 10 :S


Question 6 A

Draw the triangle as shown with 5 and 12 representing sides of the triangle as per the tan ratio given.

Use opposite/hypotenuse to find the sine and adjacent/hypotenuse to find the cosine ratios in fractions.

As all are acute then all of the above angles are in the FIRST QUADRANT and hence ALL ARE POSITIVE.
« Last Edit: February 11, 2011, 02:20:24 pm by Ari Ben Canaan »
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elemis

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #386 on: February 11, 2011, 02:07:06 pm »
x=sin\theta <br /><br />y=2cos\theta

Using sin^2\theta + cos^2\theta = 1

We first divide the second equation by two to remove the 2 from 2cos\theta

Hence, \frac{y}{2}=cos\theta

Next, we square equation 1 and equation 2 to get :

x^2=sin^2\theta<br /><br />\frac{y^2}{4}=cos^2\theta

Adding the two gives :

x^2 + \frac{y^2}{4} = 1
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Offline The Golden Girl =D

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #387 on: February 11, 2011, 02:17:04 pm »
Quote from: Ari Ben Canaan on February 11, 2011, 02:00:46 pm
Question 6 A

Draw the triangle as shown with 5 and 12 representing sides of the triangle as per the tan ratio given.

Use opposite/hypotenuse to find the sine and adjacent/hypotenuse to find the cosine ratios in fractions.

As all are acute then all of the above angles are in the FIRST QUADRANT and hence ALL ARE POSITIVE.

I got sin theta = 5/7  and cos theta =12/7   o.O  answer is sin theta = 5/13   and cos theta 12/13 :S
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elemis

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #388 on: February 11, 2011, 02:20:54 pm »
Quote from: ~ The Golden Girl ~ on February 11, 2011, 02:17:04 pm
I got sin theta = 5/7  and cos theta =12/7   o.O  answer is sin theta = 5/13   and cos theta 12/13 :S

I forgot to attache the picture in my first post.

Check now.
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Offline The Golden Girl =D

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Re: C1 and C2 DOUBTS HERE!!!!!
« Reply #389 on: February 11, 2011, 03:31:55 pm »
Thanks mate =]
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