Mathematics: Post your doubts here!

[Rutzaba]

:

srry. i hate functions and complex numbers and this range and domain thingy :/

 

[VampGirl]

srry. i hate functions and complex numbers and this range and domain thingy :/

Even I hate functions

 

[izzahzainab]

Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)

Q10)
c) The answer to part two is f(x)= (x - 1.5)^2 - 2.25
we know that it will be graphed as a parabola that opens upwards (because x has a positive coefficient).
So we need too find the minimum value of f(x) ie -2.25. You can visualize that all the values of f(x) are greater than or equal to -2.25 since it is the minimum point of the curve.

iv) Since the graph is a parabola, every "y" value has 2 corresponding x values, ie the graph is not a 1:1 function. Only 1:1 functions have an inverse so this graph doesn't.

v) let 'radical x' be represented by z.
so x = z^2
z^2 - 3z - 10 = 0
solve the quadratic equation and you'll get z=-2 and z= 5
replace z with 'radical x'. Since anything squared is always positive, -2 is discarded. Now 'radical x' equals 5. square both sides. Final answer: x= 25

 

[Rutzaba]

Q10)
c) The answer to part two is f(x)= (x - 1.5)^2 - 2.25
we know that it will be graphed as a parabola that opens upwards (because x has a positive coefficient).
So we need too find the minimum value of f(x) ie -2.25. You can visualize that all the values of f(x) are greater than or equal to -2.25 since it is the minimum point of the curve.

iv) Since the graph is a parabola, every "y" value has 2 corresponding x values, ie the graph is not a 1:1 function. Only 1:1 functions have an inverse so this graph doesn't.

v) let 'radical x' be represented by z.
so x = z^2
z^2 - 3z - 10 = 0
solve the quadratic equation and you'll get z=-2 and z= 5
replace z with 'radical x'. Since anything squared is always positive, -2 is discarded. Now 'radical x' equals 5. square both sides. Final answer: x= 25

and izzah saved the day!!!!

ps. i love ur name

 

[izzahzainab]

and izzah saved the day!!!!

ps. i love ur name

lol I guess I had to , and thanks!

ps. I love your avatars. They leave me smiling and "aww'ing"

 

[Yousif Mukkhtar]

Got a question guys, please help:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3)

 

[VampGirl]

Q10)
c) The answer to part two is f(x)= (x - 1.5)^2 - 2.25
we know that it will be graphed as a parabola that opens upwards (because x has a positive coefficient).
So we need too find the minimum value of f(x) ie -2.25. You can visualize that all the values of f(x) are greater than or equal to -2.25 since it is the minimum point of the curve.

iv) Since the graph is a parabola, every "y" value has 2 corresponding x values, ie the graph is not a 1:1 function. Only 1:1 functions have an inverse so this graph doesn't.

v) let 'radical x' be represented by z.
so x = z^2
z^2 - 3z - 10 = 0
solve the quadratic equation and you'll get z=-2 and z= 5
replace z with 'radical x'. Since anything squared is always positive, -2 is discarded. Now 'radical x' equals 5. square both sides. Final answer: x= 25

Thank u
Nd cn u plz help me with q11 as well

 

[Rutzaba]

Got a question guys, please help:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3)

Yar frst u take lcm of this equation

That will give you xb + ay = ab

Then at p where on x axis y = 0 if u put that value of y into abv equ ul get that a=x

then at Q y axis x wud be zero... put that into eq abv ul get y=b

P (a,0) and Q ( 0, b)

 

[Rutzaba]

Yar frst u take lcm of this equation

That will give you xb + ay = ab

Then at p where on x axis y = 0 if u put that value of y into abv equ ul get that a=x

then at Q y axis x wud be zero... put that into eq abv ul get y=b

P (a,0) and Q ( 0, b)

then ul get 1 equation by putting it in the distance formula... and the othr equation by putting it in the gradient formula...

solve two eqs simultaneously and ud get the answer insha Allah

 

[Binyamine]

arghh functions!!! id confuse you more if id reply... this is a job for Binyamine

Lol.

I am sorry, i have been out today. Alhamdulillah "Izzah" was there.

 

[Binyamine]

Got a question guys, please help:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3)

Rutzaba showed you the way.

( x / a ) + ( y / b ) = 1
Use the first information; to find P and Q.

On x-axis, y=0

x = a. Hence Coordinate P ( a , 0 )

On y-axis, x = 0

y = b. Hence Coordinate Q ( 0, b)

Next nformation; Length PQ is (45^ (1/2) )

manipulating the data;

a^ 2 + b ^ 2 = 45..............................First Equation

Next Information; Gradient = -1/2

b / (-a) = -1/2
a = 2b .............................................Second Equation

Solving the First and Second Equation Simultaneously would yield the following answer

b = 3 ; a = 6

 

[Rutzaba]

Rutzaba showed you the way.

( x / a ) + ( y / b ) = 1
Use the first information; to find P and Q.

On x-axis, y=0

x = a. Hence Coordinate P ( a , 0 )

On y-axis, x = 0

y = b. Hence Coordinate Q ( 0, b)

Next nformation; Length PQ is (45^ (1/2) )

manipulating the data;

a^ 2 + b ^ 2 = 45..............................First Equation

Next Information; Gradient = -1/2

b / (-a) = -1/2
a = 2b .............................................Second Equation

Solving the First and Second Equation Simultaneously would yield the following answer

b = 3 ; a = 6

may i be darned if i eva get that neat and clean !

 

[Binyamine]

may i be darned if i eva get that neat and clean !

Lol, it is not that neat and clean. You are the girl and it is thought that girl do a cleaner and more neat work.

 

[Rutzaba]

Lol, it is not that neat and clean. You are the girl and it is thought that girl do a cleaner and more neat work.

well not really. i got the dirtiest wrk inda wrld

 

[kronix6]

are u sure this question is ryt ? :S

Yes its correct, my teacher gave it to me as a home assignment.

 

[izzahzainab]

Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)

Q11)
ii) replace f ' (x) with y : y = 6 (2x + 3)
2x+3 = 6/y
2x= 6/y - 3
x = (6/y - 3 ) / 2
replace x with f ' x and y with x ; so f ' (x) = (6/x - 3) / 2

Now to find the domain you need to visualize the graph of f ' (x). You can infer that x has a small range. Now, where does the graph of f ' (x) cut the x axis?
0 = (6/x - 3) /2
x= 2

and where does the graph meet y axis? You can assume that the graph meets y axis.
y = (6/x - 3x/x) / 2 (I have taken the LCM of the terms in the bracket)
y = (6/0 - 0/0) Note: Anything divided by zero is infinity. So the graph starts at (0 , infinity) and ends at ( 2, 0)

0<x<(or equal to) 2

Note that the x is greater than 0 , not greater than or equal to. It is because for any value of x greater than 0, the graph exists but on x=0 , y becomes infinity. I hope its kind of clear

 

[izzahzainab]

Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)

Q11)
iv)
f(x) = 6/(2x + 3)
g(x) = x/2
fg (x) = 6 /(2*x/2 + 3 ) You have to replace g(x) ie x/2 in f(x)
so fg(x) = 6 / (x + 3)
The value of fg(x) is 3/2 so,
3/2 = 6 / (x + 3)
12 = 3x + 9 (cross multiplied)
x= 1

 

[Rutzaba]

Need help in this question plz.

given that sinQ - cosQ /sinQ + cosQ= 6sinQ/cosQ


Find the exact values of tanQ

Yar mujh se nhi horha :/

 

[Minato112]

Need help in this question plz.

given that sinQ - cosQ /sinQ + cosQ= 6sinQ/cosQ


Find the exact values of tanQ

6 sin^2Q + 6 sinQ cosQ = sinQ cosQ - cos^2Q
6 sin^2Q + cos^2Q = -5 sinQ cos Q

(Divide by cos^2Q everywhere) : 6 tan^2Q + 1 = -5 tan Q

6 tan^2Q + 5 tanQ + 1 = 0
(3 tanQ + 1) (2 tanQ + 1) = 0
tan Q = (-1/3) or tan Q = (-1/2)

Hope It Helps

 

[Yousif Mukkhtar]

Thanks @Ruztaba and Binyamine

You guys are live savers.