[CIE] All Pure Mathematics (P1, P2 and P3) doubts here !

[islu_jf]

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

[elemis]

For tan keep adding/subtracting 180

[islu_jf]

Okay implicit diff. 

same paper question number 5 part two?

[Vampire-Love4ever]

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.

[Arthur Bon Zavi]

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

[Dibss]

^ +Rep (:

[Arthur Bon Zavi]

^ +Rep (:

Anytime.

[Vampire-Love4ever]

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]

[acash09]

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

[Dibss]

Please post the question or attach the paper. Thanks

[acash09]

i have another question too, its attached below:

i need full explanation too.

thanks in advance

[iluvme]

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?

[Tohru Kyo Sohma]

i have a doubt
differentiate the following with respect to x;
a.
b. 1/3x-5
c. 5/root 5x-2

[iluvme]

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.

[Tohru Kyo Sohma]

Here you go.
Hope you understood.

thanks alot..........i'm new at SF....i heard that they help regarding subject doubts here........thanks alot