Mathematics: Post your doubts here!

[salvatore]

Try to list down the values in a table

x 0 1 2 3 4 5 6 7 8

y 13 3 -3 -5 -3 3 13
[Line of Symmetry]

Can you visualise it...?

10 (iv) it has an inverse because it is 1-1

10 (v) Let g(x) = y
we should take the completed form so that x is able to be made subject of formula

2 ( x - 3 ) ^ 2 - 5 = y

2 ( x - 3 ) ^ 2 = y + 5

( x - 3 ) ^ 2 = ( (y+5) / 2 )

( x - 3 ) = PlusMinus Square Root of ( (y+5) / 2 )

x = 3 PlusMinus Square Root of ( (y+5) / 2 )

Now we have to choose either the plus sign or the minus sign. Since the domain is greater than 4 we choose the "plus" sign.

Hence g inverse of x = 3 + ( ( x + 5 ) /2 ) ^ (1/2)

Thanks a lot bro.. ur awesome!
I think you forgot about no 11 (iii).. please explain if possible.
Sorry for bothering

 

[rockerzregmi]

guys can you solve this question ?
integration of (xe^2x) dx

 

[Buddila Wijeyesekera]

Cn anyone help me out ???? Write the following complex number in modulus argument form .
Question = 1+2i (also whr cn i find notes in complex number????)

 

[Binyamine]

Cn anyone help me out ???? Write the following complex number in modulus argument form .
Question = 1+2i (also whr cn i find notes in complex number????)

For the question; the real part is 1 while the imaginary part is 2.

The modulus; R = square root of [(1) ^2 + (2) ^ 2 ] = square root 5 = 5 ^ (1/2)
Its argument = tan inverse of ( 1/2 ) = & [ evaluate it, i am going to call it "&"]

In Trigonometric Form = R ( cos & + i sin & )
In Exponential Form = R e ^ ( i & )

 

[Binyamine]

Thanks a lot bro.. ur awesome!
I think you forgot about no 11 (iii).. please explain if possible.
Sorry for bothering

LOL, thanks for saying that i am awesome. hahaha but i want to become LEGEND...wait for it....LEGENDAAAAARYYYYYYYYYY. [ Dialogue of Barney Stinson ]

Well to answer 11 (iii), i presume that you have successfully tackled part (i) and part (ii), i.e. you already know the coordinates of A, B and the equation of the normal to the curve at C.

Now, the first step is to find coordinate of C, by solving simultaneously the equation of curve and the equation of the curve at C. Just find its x coordinate. Infact you will get two values of x, reject the value of x which is smaller than the x coordinate of point A.

Let us say that Area H = integration of the equation of normal to the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Let us say that Area K = integration of the equation of the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Try to visualise what the above area represents and you would come to the conclusion that

the area of the shaded region = Area H - Area K

Hope that it helps, if it does then make du'a for me and for all the oppressed muslims wherever they are especially the Syrian people and Afghan people.

 

[Binyamine]

guys can you solve this question ?
integration of (xe^2x) dx

We will have to solve by part. Consider Late, x is algebraic while e ^ ( 2*x) is exponential.

Hence; u = x ; v = ( e ^ ( 2* x ) ) / 2
du/dx = 1 ; dv/dx = e ^ ( 2*x)

integration of (xe^2x) dx = u * v - integration of v du
= x * ( e ^ ( 2* x ) ) / 2 - integration of 1 * ( e ^ ( 2* x ) ) / 2
= x * ( e ^ ( 2* x ) ) / 2 - ( e ^ ( 2* x ) ) / 4

 

[sweetiepie]

Q. In How Many Years a sum of Rs 3000 would amount Rs 6130.43 at 6% compounded quarterly ?
Q. Find the amount of Rs 250 invested at the end of each of 5 successive years at 6% interest compounded annually ?
plz help me as soon as possible with all steps

 

[19islandprincess96]

I need help with q 4 in math 9709/33/o/n/11. The link to the paper is: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
I don't get it at all.

 

[Binyamine]

I need help with q 4 in math 9709/33/o/n/11. The link to the paper is: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
I don't get it at all.

Here you are ::

 

[Rutzaba]

Here you are ::

Sorry to have left you to do it all alone stats just isnt my genre

 

[19islandprincess96]

Here you are ::

The video you posted is of 2012. My question was from 11.

 

[19islandprincess96]

Please help me with question 10 from the paper: http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf

 

[salvatore]

LOL, thanks for saying that i am awesome. hahaha but i want to become LEGEND...wait for it....LEGENDAAAAARYYYYYYYYYY. [ Dialogue of Barney Stinson ]

Well to answer 11 (iii), i presume that you have successfully tackled part (i) and part (ii), i.e. you already know the coordinates of A, B and the equation of the normal to the curve at C.

Now, the first step is to find coordinate of C, by solving simultaneously the equation of curve and the equation of the curve at C. Just find its x coordinate. Infact you will get two values of x, reject the value of x which is smaller than the x coordinate of point A.

Let us say that Area H = integration of the equation of normal to the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Let us say that Area K = integration of the equation of the curve with lower limit the x ordinate of coordinate C and upper limit the x ordinate of coordinate B

Try to visualise what the above area represents and you would come to the conclusion that

the area of the shaded region = Area H - Area K

Hope that it helps, if it does then make du'a for me and for all the oppressed muslims wherever they are especially the Syrian people and Afghan people.

Lool.. dw, u'll become a legend
The part about Area K is correct.. but I think you have made an error in Area H.
According to the marking scheme, you are supposed to find the area of a trapezium and subtract it from area K. Now where did the trapezium come from??

And yeah.. may god help the oppressed Muslims around the world!

 

[Binyamine]

Lool.. dw, u'll become a legend
The part about Area K is correct.. but I think you have made an error in Area H.
According to the marking scheme, you are supposed to find the area of a trapezium and subtract it from area K. Now where did the trapezium come from??

And yeah.. may god help the oppressed Muslims around the world!

My friend, analyse well, put your hand on the line which is normal to the curve at C and go down from the value of x of point A and value of X for point C. You see the area H that i talked of is infact that area of trapezium that is mentioned in the marking scheme.

 

[VampGirl]

Plz help me with q:8 (ii) of oct/nov/05 p1

Thnx

 

[salvatore]

My friend, analyse well, put your hand on the line which is normal to the curve at C and go down from the value of x of point A and value of X for point C. You see the area H that i talked of is infact that area of trapezium that is mentioned in the marking scheme.

Oh yeah.. hadn't figured that out earlier.. that was dumb of me.
Thanks for your help mate!

 

[salvatore]

Plz help me with q:8 (ii) of oct/nov/05 p1

Thnx

Inverse:

(i) Make equate the function to y
-----> y = (2x − 3)^3 − 8

(ii) Make x the subject
-----> y + 8 = (2x - 3)^3
(y + 8)^1/3 = (2x - 3)^1/3.... ^1/3 refers to cubic root
(y + 8)^1/3 = 2x - 3
(y + 8)^1/3 + 3 = 2x
x = ( (y + 8)^1/3 + 3 ) / 2

(iii) The switch the y with x.. this becomes the inverse
---> f−1(x) = ( (x + 8)^1/3 + 3 ) / 2

Domain:

The range of the function becomes the domain of the inverse.. and vice versa
Hence, use the domain of the function to find the range i.e 2 ≤ x ≤ 4. Substitute the values in the expression:

Lower limit:
y = (2 (2) − 3)3 − 8
= -7

Upper limit:
y = (2 (4) − 3)3 − 8
=117

The domain becomes -7 ≤ x ≤ 117 ​

 

[VampGirl]

Inverse:

(i) Make equate the function to y
-----> y = (2x − 3)^3 − 8

(ii) Make x the subject
-----> y + 8 = (2x - 3)^3
(y + 8)^1/3 = (2x - 3)^1/3.... ^1/3 refers to cubic root
(y + 8)^1/3 = 2x - 3
(y + 8)^1/3 + 3 = 2x
x = ( (y + 8)^1/3 + 3 ) / 2

(iii) The switch the y with x.. this becomes the inverse
---> f−1(x) = ( (x + 8)^1/3 + 3 ) / 2

Domain:

The range of the function becomes the domain of the inverse.. and vice versa
Hence, use the domain of the function to find the range i.e 2 ≤ x ≤ 4. Substitute the values in the expression:

Lower limit:
y = (2 (2) − 3)3 − 8
= -7

Upper limit:
y = (2 (4) − 3)3 − 8
=117

The domain becomes -7 ≤ x ≤ 117 ​

Thank u

 

[kronix6]

Need help in this question plz.

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


Find the exact values of tanQ

 

[Rutzaba]

Need help in this question plz.

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


Find the exact values of tanQ

are u sure this question is ryt ? :S