# Help needed with C1 Questions!!!

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

This discussion is closed.

Can sombody please work out these Qs for me?!

6) Given that dy/dx=3 sqrt x –x^2, and that y=2/3 when x=1, find the value of y when x=4. (7 marks)

7) The first three terms of an arithmetic series are (12-p), 2p and (4p-5) respectively, where p is a constant.

a) Find the value of p (2 marks)

b) Show that the sixth term of the series is 50 (3 marks)

c) Find the sum of the first 15 terms of the series. (2 marks)

d) Find how many terms of the series have a value of less than 400 (3 marks)

8) f(x)=2x^2 +3x –2.

a) Solve the equation f(x)=0 (2 marks)

b) Sketch the curve with equation y=f(x), showing the coordinates of any points of intersection with the coordinate axes. (2 marks)

c) Find the coordinates of the points where the curve with equation y=f(1/2x) crosses the coordinate axes. (3 marks)

d) When the graph of y=f(x) is translated by 1 unit in the positive x-direction it maps onto the graph with equation y=ax^2 +bx +c, where a, b and c are constants. Find the values of a, b and c. (3 marks)

6) Given that dy/dx=3 sqrt x –x^2, and that y=2/3 when x=1, find the value of y when x=4. (7 marks)

7) The first three terms of an arithmetic series are (12-p), 2p and (4p-5) respectively, where p is a constant.

a) Find the value of p (2 marks)

b) Show that the sixth term of the series is 50 (3 marks)

c) Find the sum of the first 15 terms of the series. (2 marks)

d) Find how many terms of the series have a value of less than 400 (3 marks)

8) f(x)=2x^2 +3x –2.

a) Solve the equation f(x)=0 (2 marks)

b) Sketch the curve with equation y=f(x), showing the coordinates of any points of intersection with the coordinate axes. (2 marks)

c) Find the coordinates of the points where the curve with equation y=f(1/2x) crosses the coordinate axes. (3 marks)

d) When the graph of y=f(x) is translated by 1 unit in the positive x-direction it maps onto the graph with equation y=ax^2 +bx +c, where a, b and c are constants. Find the values of a, b and c. (3 marks)

0

Report

#2

6) Given that dy/dx=3 sqrt x –x^2, and that y=2/3 when x=1, find the value of y when x=4. (7 marks)

7) The first three terms of an arithmetic series are (12-p), 2p and (4p-5) respectively, where p is a constant.

For (b) You find the common difference using the work done in (a). Then use the formulae for the nth term.

For (c) The method is similar to (b), but with the sum formulae.

For (d) You find the first term with a value greater than 400. The terms are all those below that.

8) f(x)=2x^2 +3x –2.

a) Solve the equation f(x)=0 (2 marks)

a) Solve the equation f(x)=0 (2 marks)

b) Sketch the curve with equation y=f(x), showing the coordinates of any points of intersection with the coordinate axes. (2 marks)

c) Find the coordinates of the points where the curve with equation y=f(1/2x) crosses the coordinate axes. (3 marks)

d) When the graph of y=f(x) is translated by 1 unit in the positive x-direction it maps onto the graph with equation y=ax^2 +bx +c, where a, b and c are constants. Find the values of a, b and c. (3 marks)

0

Report

#3

6.

dy/dx = sqrt x - x^2

dy/dx = x^(1/2) - x^2

integrate (indefinite integral)

y = (2/3)x^(3/2) - (1/3)x^3 + C

to find the constant of integration C, substitute the known paid of values of (x, y) in:

2/3 = (2/3)*1^(3/2) - (1/3)*1^3 + C

2/3 = 2/3 - 1/3 + c

c = 1/3

Now sub this back into the equation:

y = (2/3)x^(3/2) - (1/3)x^3 + 1/3

when x=4,

y = (2/3)*4^(3/2) - (1/3)*4^3 + 1/3

y = 16/3 - 64/3 + 1/3

y = -47/3

check through this yourself becuase:

(a) I might have made a mistake

(b) you will probably learn the method better if you try it too

dy/dx = sqrt x - x^2

dy/dx = x^(1/2) - x^2

integrate (indefinite integral)

y = (2/3)x^(3/2) - (1/3)x^3 + C

to find the constant of integration C, substitute the known paid of values of (x, y) in:

2/3 = (2/3)*1^(3/2) - (1/3)*1^3 + C

2/3 = 2/3 - 1/3 + c

c = 1/3

Now sub this back into the equation:

y = (2/3)x^(3/2) - (1/3)x^3 + 1/3

when x=4,

y = (2/3)*4^(3/2) - (1/3)*4^3 + 1/3

y = 16/3 - 64/3 + 1/3

y = -47/3

check through this yourself becuase:

(a) I might have made a mistake

(b) you will probably learn the method better if you try it too

0

Report

#4

lol... I'm getting slow.

Oh well.

(x-1)

Oh well.

(Original post by

Replace x in f(x) by (x+1) and equate coefficients.

**Gaz031**)Replace x in f(x) by (x+1) and equate coefficients.

0

X

Page 1 of 1

Go to first unread

Skip to page:

new posts

Back

to top

to top