Author Topic: [CIE] All Pure Mathematics (P1, P2 and P3) doubts here !  (Read 80237 times)

Offline islu_jf

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #60 on: December 24, 2010, 08:47:04 pm »
Hey its Oct/Nov 2001 P2

Question number one!.. i know how to do the whole thing, but i always get confused what to do with the result,
for ex.
if i got tanx=-1 limits are 0<x<360 (both equal sign but don't know how to write it)
x=-45
so then what should i do?
-45(-+)180 or 180(-+)-45?
and for 360/?

and if u can put the sinx part as well.. ??
plz help

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #61 on: December 25, 2010, 04:08:52 am »
For tan keep adding/subtracting 180

Offline islu_jf

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #62 on: December 25, 2010, 01:45:46 pm »
Okay implicit diff.  ???

same paper question number 5 part two?

Offline Vampire-Love4ever

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #63 on: December 27, 2010, 01:12:09 pm »
Q1- The function 3-2x-x^2 = a-(x+b)^2, find the value of a and b and hence find it's maximum value.

Q2- Express x^2+4x in the form a-(x+a)^2, stating values of a and b.

Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.
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Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #64 on: December 27, 2010, 03:00:50 pm »
Q1- The function 3-2x-x^2 = a-(x+b)^2, find the value of a and b and hence find it's maximum value.

Q2- Express x^2+4x in the form a-(x+b)^2, stating values of a and b.

Q3- Express 2x^2-6x in the form a(x-b)^2+c stating the numerical values of a, b and c.

I do all these by a simple method, and if you don't understand, use the formula given in the book :
ax2 + bx + c = a ( x2 + b    )
                             (         ---x )
                             (          a    )


x2 + bx + c = (x2 + bx) + c = {(x + 1/2 b)2 - 1/4 b2} + c

(1)

 -x2 - 2x - 3
-(x2 + 2x - 3)
-(x2 + 2x + 1 - 4)
-(x + 1)2 + 4           ---------> a = 4; b = -1; maximum value = 4.

(2)

x2 + 4x
(x2 + 4x)
(x2 + 4x + 4 - 4)
(x + 2)2 - 4             ----------> a= -4; b= -2

(3)

2x2 - 6x
2 (x2 - 3x)
2 (x2 - 3x + 2.25 - 2.25)
2 (x - 1.5)2 - 4.5                ------------> a= 2; b= 1.5; c= -4.5
« Last Edit: December 27, 2010, 03:09:58 pm by Dhirubhai Ambani »

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Offline Dibss

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #65 on: December 27, 2010, 07:38:52 pm »
^ +Rep (:

Offline Arthur Bon Zavi

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #66 on: December 28, 2010, 04:01:07 am »

Continuous efforts matter more than the outcome.
- NU

Offline Vampire-Love4ever

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #67 on: December 28, 2010, 05:56:20 pm »
I do all these by a simple method, and if you don't understand, use the formula given in the book :
ax2 + bx + c = a ( x2 + b    )
                             (         ---x )
                             (          a    )


x2 + bx + c = (x2 + bx) + c = {(x + 1/2 b)2 - 1/4 b2} + c

(1)

 -x2 - 2x - 3
-(x2 + 2x - 3)
-(x2 + 2x + 1 - 4)
-(x + 1)2 + 4           ---------> a = 4; b = -1; maximum value = 4.

(2)

x2 + 4x
(x2 + 4x)
(x2 + 4x + 4 - 4)
(x + 2)2 - 4             ----------> a= -4; b= -2

(3)

2x2 - 6x
2 (x2 - 3x)
2 (x2 - 3x + 2.25 - 2.25)
2 (x - 1.5)2 - 4.5                ------------> a= 2; b= 1.5; c= -4.5



Thanks!:) [rep]
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Offline acash09

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #68 on: December 28, 2010, 07:14:39 pm »
I have a problem here:

Question 4 --> P1 M/J 2009 (CIE AS-LEVEL) (9709/01/M/J/09)

I need explanation on how to exactly solve this question. Help is very much appreciated.
My friend told me its about amplitude and wavelength etc.

Thanks in advance.
Acash09
« Last Edit: December 29, 2010, 04:35:50 pm by acash09 »
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Offline Dibss

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #69 on: December 28, 2010, 07:18:37 pm »
Please post the question or attach the paper. Thanks :)

Offline acash09

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #70 on: December 29, 2010, 04:38:52 pm »
i have another question too, its attached below:

i need full explanation too.

thanks in advance
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Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #71 on: December 29, 2010, 07:15:48 pm »
i have another question too, its attached below:

i need full explanation too.

thanks in advance

y=12/(x2+3)
y=12(x2+3)-1

dy/dx= -12(x2+3)-2(2x)
       =-24x/(x2+3)2

ii) Equation of the normal, First find the gradient at P on the curve, then using m1*m2=-1 Find the gradient of the normal.

Thus,
  dy/dx at P x=1
     -24(1)/(12+3)2
      -24/16
      -3/2

Using m1*m2=-1 to find the gradient of normal,
   gradient of normal =2/3

Using the formula,
   (y-y1)=m(x-x1)
y-3=2/3(x-1)
y=2x/3 + 7/3

As for the third part could you confirm if the answer is -0.018?
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Offline Tohru Kyo Sohma

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #72 on: December 31, 2010, 03:26:49 pm »
i have a doubt
differentiate the following with respect to x;
a. \sqrt{4x-1}
b. 1/3x-5
c. 5/root 5x-2
« Last Edit: April 04, 2016, 09:17:36 pm by astarmathsandphysics »

Offline iluvme

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #73 on: December 31, 2010, 03:53:42 pm »
i have a doubt
differentiate the following with respect to x;
a. root 4x-1
b. 1/3x-5
c. 5/root 5x-2

Here you go.
Hope you understood. :)
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Offline Tohru Kyo Sohma

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Re: All Pure Mathematics DOUBTS HERE!!!
« Reply #74 on: December 31, 2010, 04:30:51 pm »
Here you go.
Hope you understood. :)
thanks alot..........i'm new at SF....i heard that they help regarding subject doubts here........thanks alot