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

[iluvme]

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

Hey Welcome to SF
We are here to help you
Feel free to post your doubts. And I am glad you understood

[Dibss]

+ Rep for your help, Iluvme

[iluvme]

+ Rep for your help, Iluvme

Thankyou Ma'am.

[Tohru Kyo Sohma]

doubt.. ..;
1. the curve C has equation y=Kx^2 +4/x, where K is a constant. the tangent to C at the point with x-coordinate 2 is parallel to the line with equation y=3x+8 . find the value of K and the equation of the tangent?
2. A curve has the equation y=6/1-2x . find an expression for dy/dx. hence find the equation of the normal to the curve at the point where x=2?
3. For the following function, find the stationary point and its nature; 4x+1/x ?
4. find the coordinates of the stationary point on the curve y=x^4 -2x^2 distinguishing between maxima and minima ?

sorry for asking so many questions but i dont seem to get the answers right.............an explanation would be helpful too!!

[elemis]

On it.

[elemis]

Question 1

First differentiate the given equation :

dy/dx = 2kx - 4x^-2

Since the tangent is parallel to the line y=3x+8 it is obvious that dy/dx at x=2 will have a value 3.

Hence, 4k - 1 = 3

k = 1

I'll let you find the equation of the tangent.... thats the easy part.

[elemis]

Question 4

dy/dx = 4x³ - 4x

thus, 4x³ - 4x =0   at the stationary points.

Solving gives x as 1, -1 or 0

Differentiating a second time : 12x² - 4

Substituting x for 1 gives 8. This is the minimum point.

Substituting x for -1 gives 8. This is also a minimum point.

Substituting x for 0 gives -4. This is maximum point.

[Tohru Kyo Sohma]

Question 1

First differentiate the given equation :

dy/dx = 2kx - 4x^-2

Since the tangent is parallel to the line y=3x+8 it is obvious that dy/dx at x=2 will have a value 3.

Hence, 4k - 1 = 3

k = 1

I'll let you find the equation of the tangent.... thats the easy part.

thank you ari ben!!!!

[elemis]

thank you ari ben!!!!

You're welcome.

Please dont call me Ari Ben, if you consider the meaning of my name you will realise it makes no sense calling me Ari Ben.

Ari - Lion in Hebrew

Ben - Son Of in Hebrew

Canaan - Israel in Hebrew

[Tohru Kyo Sohma]

You're welcome.

Please dont call me Ari Ben, if you consider the meaning of my name you will realise it makes no sense calling me Ari Ben.

Ari - Lion in Hebrew

Ben - Son Of in Hebrew

Canaan - Israel in Hebrew
oppss!!!...sorry ok i will not call you ari ben...........so............then can i call you ari ben canaan???

[elemis]

Yeah, call me Ari or Ari Ben Canaan.

[Tohru Kyo Sohma]

Yeah, call me Ari or Ari Ben Canaan.

ok.......i ll call you ari.........nice to meet you!!!

[Tohru Kyo Sohma]

i have a doubt;
 The curve with the equation y=k(x^3 -3x^2 -24x +c)
where k and c are non-zero constants, has two stationary points. show that the coordinates of one stationary point are (4, kc-80k ), and find, in a similar form the coordinates of the second stationary point. given that the mid-point of the line joining the two stationary points lies on the x-axis, show that c=26. given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?

[Dibss]

i have a doubt;
 The curve with the equation y=k(x^3 -3x^2 -24x +c)
where k and c are non-zero constants, has two stationary points. show that the coordinates of one stationary point are (4, kc-80k ), and find, in a similar form the coordinates of the second stationary point. given that the mid-point of the line joining the two stationary points lies on the x-axis, show that c=26. given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?

This question is pretty long. General steps you'll need to go through are listed below. Ask if you need help/elaboration of a particular step, K.
(:

find dy/dx
set it to 0
you'll get two values for x
substitute them into the original equation to get the y coordinate of that particular stationary point

Now that you the two points, find the equation of the line joining them
Find the midpoint of this line (which lies on the x-axis therefore y=0)

given further that the tangent to the curve at the point of which x=2 passes through the point (5,-140), find the exact value of k.?

Find dy/dx at x =2
That's the gradient of the tangent.
Proceed as need indicates... I haven't solved the qs so I don't know.

[Tohru Kyo Sohma]

This question is pretty long. General steps you'll need to go through are listed below. Ask if you need help/elaboration of a particular step, K.
(:

find dy/dx
set it to 0
you'll get two values for x
substitute them into the original equation to get the y coordinate of that particular stationary point

Now that you the two points, find the equation of the line joining them
Find the midpoint of this line (which lies on the x-axis therefore y=0)
Are you certain you've copied the qs correctly?

i cant get the correct answer.....................any further elaboration....!!!