This discussion is now closed.

Check out other Related discussions

- GCSE November Series 2023 Exam Discussions Hub
- Edexcel A Level Further Mathematics Paper 4B (9FM0 4B) - 24th June 2024 [Exam Chat]
- A-level Exam Discussions 2024
- Edexcel IGCSE Higher tier Mathematics A Paper 1 1H (4MA1) - 8th November 2023
- formula sheet in gases 2025
- Edexcel A Level Further Mathematics Paper 4D (9FM0 4D) - 24th June 2024 [Exam Chat]
- C1 January 2002?
- Edexcel A Level Further Mathematics Paper 4A (9FM0 4A) - 24th June 2024 [Exam Chat]
- Edexcel GCSE Mathematics Paper 3 (1MA1 3) - 13th November 2023 [Exam Chat]
- GCSE Exam Discussions 2023
- Edexcel A Level Further Mathematics Paper 4C (9FM0 4C) - 24th June 2024 [Exam Chat]
- Edexcel Past Papers
- GCSE Mathematics Study Group 2024-25
- Edexcel GCSE Mathematics Paper 2 (1MA1 2) - 10th November 2023 [Exam Chat]
- Edexcel GCSE Foundation tier Maths Paper 3 3F (1MA1) - 14th June 2023 [Exam Chat]
- GCSE Exam Discussions 2024
- Edexcel GCSE Mathematics Paper 2 Foundation (1MA1 2F) - 3rd June 2024 [Exam Chat]
- Edexcel GCSE Mathematics Paper 1 (1MA1 1) - 8th November 2023 [Exam Chat]
- Want to try go for a paramedic apprenticeship, what should I do to ensure my success?
- GCSE Revision Resource

Just a thread for anyone doing C1 in January or any time for that matter. If you have any problems post em here to prevent spamming.

Here are the topics in C1.

1. Algebra and functions

2. Coordinate geometry in the (x, y) plane

3. Sequences and series

4. Differentiation

5. Integration

Here are the topics in C1.

1. Algebra and functions

2. Coordinate geometry in the (x, y) plane

3. Sequences and series

4. Differentiation

5. Integration

Scroll to see replies

Im doin C1 this yr but I dnt seem to have too much troubles... Other than applying differentiation to questions as I dnt know how 2 use the information to solve the question... Ive only done about half of the course so far.. so I cant comment on the rest.

Isnt ther more sections 2 the module btw??

Isnt ther more sections 2 the module btw??

Fade Into Black

Im doin C1 this yr but I dnt seem to have too much troubles... Other than applying differentiation to questions as I dnt know how 2 use the information to solve the question... Ive only done about half of the course so far.. so I cant comment on the rest.

Isnt ther more sections 2 the module btw??

Isnt ther more sections 2 the module btw??

Those are the main headings of the specification.

If you don't understand differentiation, ask me or widowmaker.

The principles of differentiation I can understand.. But its just the application of diff. lik when they ask you to find the co-ordinate when the curve passes through the x-axis... etc.. could you perhaps give some tips of how to approach them and each step you might take in your head.

Fade Into Black

The principles of differentiation I can understand.. But its just the application of diff. lik when they ask you to find the co-ordinate when the curve passes through the x-axis... etc.. could you perhaps give some tips of how to approach them and each step you might take in your head.

give an example question ..

(but basically remember dy/dx will give you the equation for the gradient)

integration= basically the area under curve. also needed for volumes of revolution.

or u wanna get f(x) if yuve been given f'x

or u wanna get f(x) if yuve been given f'x

Jack Schitt

give an example question ..

(but basically remember dy/dx will give you the equation for the gradient)

(but basically remember dy/dx will give you the equation for the gradient)

I get the grasp of all the dy/dx's and wat they allow me 2 find...

But its questions such as..

the curve of the equation, y=ax^2+bx+c passes through the point (1,2). The gradient of the curve is zero at the point (2,1).Find the values of a,b and c.

I can start the question but not finish it..

Fade Into Black

I get the grasp of all the dy/dx's and wat they allow me 2 find...

But its questions such as..

the curve of the equation, y=ax^2+bx+c passes through the point (1,2). The gradient of the curve is zero at the point (2,1).Find the values of a,b and c.

I can start the question but not finish it..

But its questions such as..

the curve of the equation, y=ax^2+bx+c passes through the point (1,2). The gradient of the curve is zero at the point (2,1).Find the values of a,b and c.

I can start the question but not finish it..

You have to find simultaneous equations.

dy/dx = 2ax + b

We know that dy/dx = 0 when x = 2

therefore, 0 = 2a*2 + b

0 = 4a + b (1)

(1,2) is a point on the curve. Substitute these values into f(x)

2 = a*1^2 + b*1 + c

2 = a + b + c (2)

Finally we also know that (2,1) is a point on the curve so substitute these values into f(x);

1 = a*2^2 + b*2 + c

1 = 4a + 2b + c (3)

Now solve the simultaneous equations.

(3) - (2) gives;

1-0 = (4a + 2b + c) - (a + b + c) which gives;

3a + b (4)

(1) - (4) gives a = 0--1 = 1

Now put this value of a into equation (1) to find b.

0 = 4a + b

0 = 4*1 + b

b = -4

Now put these values of a and b into equation (2) to find c.

2 = a + b + c

2 = 1 + (-4) + c

2 = -3 + c

c = 5

So a = 1, b = -4, c = 5

Hope that helps.

I understand evrythin therrr....!!! thanx... I neva knew it was that simple I must overcomplicate things in my head.....lol. But one query as to y u used x=1 is that a general thing you do or did you do that jst to get rid of the x... I think Im missing something

Thanx Chameleon

Thanx Chameleon

The Chameleon

You have to find simultaneous equations.

dy/dx = 2ax + b

We know that dy/dx = 0 when x = 2

therefore, 0 = 2a*2 + b

0 = 4a + b (1)

(1,2) is a point on the curve. Substitute these values into f(x)

2 = a*1^2 + b*1 + c

2 = a + b + c (2)

Finally we also know that (2,1) is a point on the curve so substitute these values into f(x);

1 = a*2^2 + b*2 + c

1 = 4a + 2b + c (3)

Now solve the simultaneous equations.

(3) - (2) gives;

1-0 = (4a + 2b + c) - (a + b + c) which gives;

3a + b (4)

(1) - (4) gives a = 0--1 = 1

Now put this value of a into equation (1) to find b.

0 = 4a + b

0 = 4*1 + b

b = -4

Now put these values of a and b into equation (2) to find c.

2 = a + b + c

2 = 1 + (-4) + c

2 = -3 + c

c = 5

So a = 1, b = -4, c = 5

Hope that helps.

dy/dx = 2ax + b

We know that dy/dx = 0 when x = 2

therefore, 0 = 2a*2 + b

0 = 4a + b (1)

(1,2) is a point on the curve. Substitute these values into f(x)

2 = a*1^2 + b*1 + c

2 = a + b + c (2)

Finally we also know that (2,1) is a point on the curve so substitute these values into f(x);

1 = a*2^2 + b*2 + c

1 = 4a + 2b + c (3)

Now solve the simultaneous equations.

(3) - (2) gives;

1-0 = (4a + 2b + c) - (a + b + c) which gives;

3a + b (4)

(1) - (4) gives a = 0--1 = 1

Now put this value of a into equation (1) to find b.

0 = 4a + b

0 = 4*1 + b

b = -4

Now put these values of a and b into equation (2) to find c.

2 = a + b + c

2 = 1 + (-4) + c

2 = -3 + c

c = 5

So a = 1, b = -4, c = 5

Hope that helps.

Yeh that clarifys evrythin... as was confused about one part but I think it was a mistake u made... either way... I understand fully now

THanx Chameleon once again

Fade Into Black

Yeh that clarifys evrythin... as was confused about one part but I think it was a mistake u made... either way... I understand fully now

THanx Chameleon once again

THanx Chameleon once again

That's okay.

Any other problems, post them here.

TheWolf

Can someone go through all the transformations and what it does to the x or y value. ie af(x), f(ax)...etc

f(x+a) => translation of -a in x-direction.

f(x-a) => translation of +a in the x-direction.

af(x) => stretch of a in the y-direction (multiply y-coordinates by a, x-coordinates remain the same)

f(ax) => stretch of 1/a in the x-direction (multiply x-coordinates by 1/a, y-coordinates remain the same)

y = a/x => stretch of +a in the y-direction (multiply y-coordinates by a, x-coordinates remain the same).

E.g. 3/x is a stetch of factor 3 of 1/x in the y direction. x-coordinates remain the same, y-coordinates remain the same.

y=f(-x) => reflection of original graph in the y-axis. Multiplying x-coordinates by -1. E.g. (3,0) becomes (-3,0)

y = -f(x) => reflection of original graph in the x-axis.

By the way, I can guarantee that a graph question will come up in C1. The fact that they've devoted a whole chapter on the topic in the heinemann book, means that the examiners are obviously keen on graphs.

Hey the chameleon,

thanks for this thread

I'm having trouble with sequence questions!!!

8) A salesman is paid comission of 10 pounds per week for each life insurance policy that he has sold. Each week he sells one new policy so that he is paid 10 pounds in the first week, 20 pounds in the second week, 30 pounds in the third week and so on.

a) Find his total commission in the first 52 weeks

b) In the second year the commission increases to 11 pounds per week on new policies sold, although it remains at 10 per week for policies sold in the first year. He continues to sell one policy per week. Show that he is paid 542 pounds in the second week of his second year.

c) Find the total commission paid to him in the second year.

10)a) Find the sum of the integers which are divisible by 3 and lie between 1 and 400

b) Hence, or otherwise find the sum of the integers, from 1 to 400 inclusive, which are not divisible by 3.

11) A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find, for this series:

a) the common difference

b) the first term

13) Prospectors are drilling for oil. The cost of drilling to a depth of 50m is 500 pounds. To drill a further 50m costs 640 pounds and, hence, the total cost of drilling a depth of 100 m is 1140 pounds. Each subsequent extra depth of 50m costs 140 pounds more to drill than the previous 50 m.

a) show that the cost of drilling to a depth of 500 m is 11300 pounds

b) The total sum of money available for drilling is 76000 pounds. Find, to the nearest 50 m, the greatest depth that can be drilled.

Thank you ever so much.

thanks for this thread

I'm having trouble with sequence questions!!!

8) A salesman is paid comission of 10 pounds per week for each life insurance policy that he has sold. Each week he sells one new policy so that he is paid 10 pounds in the first week, 20 pounds in the second week, 30 pounds in the third week and so on.

a) Find his total commission in the first 52 weeks

b) In the second year the commission increases to 11 pounds per week on new policies sold, although it remains at 10 per week for policies sold in the first year. He continues to sell one policy per week. Show that he is paid 542 pounds in the second week of his second year.

c) Find the total commission paid to him in the second year.

10)a) Find the sum of the integers which are divisible by 3 and lie between 1 and 400

b) Hence, or otherwise find the sum of the integers, from 1 to 400 inclusive, which are not divisible by 3.

11) A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find, for this series:

a) the common difference

b) the first term

13) Prospectors are drilling for oil. The cost of drilling to a depth of 50m is 500 pounds. To drill a further 50m costs 640 pounds and, hence, the total cost of drilling a depth of 100 m is 1140 pounds. Each subsequent extra depth of 50m costs 140 pounds more to drill than the previous 50 m.

a) show that the cost of drilling to a depth of 500 m is 11300 pounds

b) The total sum of money available for drilling is 76000 pounds. Find, to the nearest 50 m, the greatest depth that can be drilled.

Thank you ever so much.

littlemisshala

Hey the chameleon,

thanks for this thread

I'm having trouble with sequence questions!!!

8) A salesman is paid comission of 10 pounds per week for each life insurance policy that he has sold. Each week he sells one new policy so that he is paid 10 pounds in the first week, 20 pounds in the second week, 30 pounds in the third week and so on.

a) Find his total commission in the first 52 weeks

b) In the second year the commission increases to 11 pounds per week on new policies sold, although it remains at 10 per week for policies sold in the first year. He continues to sell one policy per week. Show that he is paid 542 pounds in the second week of his second year.

c) Find the total commission paid to him in the second year.

10)a) Find the sum of the integers which are divisible by 3 and lie between 1 and 400

b) Hence, or otherwise find the sum of the integers, from 1 to 400 inclusive, which are not divisible by 3.

thanks for this thread

I'm having trouble with sequence questions!!!

8) A salesman is paid comission of 10 pounds per week for each life insurance policy that he has sold. Each week he sells one new policy so that he is paid 10 pounds in the first week, 20 pounds in the second week, 30 pounds in the third week and so on.

a) Find his total commission in the first 52 weeks

b) In the second year the commission increases to 11 pounds per week on new policies sold, although it remains at 10 per week for policies sold in the first year. He continues to sell one policy per week. Show that he is paid 542 pounds in the second week of his second year.

c) Find the total commission paid to him in the second year.

10)a) Find the sum of the integers which are divisible by 3 and lie between 1 and 400

b) Hence, or otherwise find the sum of the integers, from 1 to 400 inclusive, which are not divisible by 3.

formula to remember

sum of AP = (n/2)(2a+(n-1)d

a=first term = 10

d=common difference = 10

n= 52 (weeks)

so sum = 26*(20+510)

for the second year, you know that he earned 520 in the last week of the previous year, so for year 2

week 1 = 520+11

week 2 = 520+11+11

Q10 same formula, work out n. number of terms is 400/3 = 133

a=3

d=3

apply the formula!

second part: work out sum of all integers to 400, subtract result of part a.

you can use the same formula for this initially, or a reduced form of it, which is used to sum the first n natural numbers

(n/2)(n+1) [which]

=> 200*401

I need a drink + some lunch - might do some more later!

Aitch

Oh yes and you also need to know the proof for the sum of arithmetic series.

Sn = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + .... + (a + (n-1)d)

where a is the first term and d is a common difference.

Write this formula back to front.

Sn = a + (n-1)d + (a + (n-2)d) + .... + (a + d) + a

Adding both series gives;

2Sn = n(a + a + (n-1)d)

Sn = 0.5n(2a + (n-1)d)

Sn = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + .... + (a + (n-1)d)

where a is the first term and d is a common difference.

Write this formula back to front.

Sn = a + (n-1)d + (a + (n-2)d) + .... + (a + d) + a

Adding both series gives;

2Sn = n(a + a + (n-1)d)

Sn = 0.5n(2a + (n-1)d)

littlemisshala

11) A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find, for this series:

a) the common difference

b) the first term

13) Prospectors are drilling for oil. The cost of drilling to a depth of 50m is 500 pounds. To drill a further 50m costs 640 pounds and, hence, the total cost of drilling a depth of 100 m is 1140 pounds. Each subsequent extra depth of 50m costs 140 pounds more to drill than the previous 50 m.

a) show that the cost of drilling to a depth of 500 m is 11300 pounds

b) The total sum of money available for drilling is 76000 pounds. Find, to the nearest 50 m, the greatest depth that can be drilled.

Thank you ever so much.

11) A polygon has 10 sides. The lengths of the sides, starting with the smallest, form an arithmetic series. The perimeter of the polygon is 675cm and the length of the longest side is twice that of the shortest side. Find, for this series:

a) the common difference

b) the first term

13) Prospectors are drilling for oil. The cost of drilling to a depth of 50m is 500 pounds. To drill a further 50m costs 640 pounds and, hence, the total cost of drilling a depth of 100 m is 1140 pounds. Each subsequent extra depth of 50m costs 140 pounds more to drill than the previous 50 m.

a) show that the cost of drilling to a depth of 500 m is 11300 pounds

b) The total sum of money available for drilling is 76000 pounds. Find, to the nearest 50 m, the greatest depth that can be drilled.

Thank you ever so much.

OK Lunch over, claret still going, so Q11...

a and d to find

we know

n=10

sum=675

term(10) = a+(n-1)d = a+9d

this is twice a, so 2a=a+9d =>a=9d (1)

sum 675=(n/2)(2a+(n-1)d) =5(2a+9d) =10a+45d

=>135=2a+9d => 2a=135-9d (2)

put (1) in (2) and get 18d=135-9d

=> d=5

put this into (1) to get a = 45

[Checks in Excel OK]

q13a

50m = 1 unit

so for 500m, n=10

a=500

d=140

sum [same formula again] = 5(1000+(9*140))

part b

sum = 76000=(n/2)(1000+(n-1)140)

if you work this through, you end up with, I think, (claret-induced doubt creeping in...)

7n^2+43n-7600 = 0

apply the quadratic formula to get 30 (or thereabouts)

[remember this is *50m units]

If you have time with qs like this, check the sum for n=29, n=31.

Remember you have only £76K to spend (stop dreaming) so keep below the limit.

[I'll put up a post later about checking in XL.]

Aitch

[E&O blamed on quite puckish 2000 Bordeaux]

...but don't try this in your P1 exam.

littlemisshala and anyone else interested in the above sequences/series:

In the above posts I used XL to check the series sums.

Easy to do if you know replication and Autosum.

XLS extension appears not to be allowed as TSR attachments, so see

http://www.gwynne4.freeserve.co.uk/download/checkseries.xls

Aitch

littlemisshala and anyone else interested in the above sequences/series:

In the above posts I used XL to check the series sums.

Easy to do if you know replication and Autosum.

XLS extension appears not to be allowed as TSR attachments, so see

http://www.gwynne4.freeserve.co.uk/download/checkseries.xls

Aitch

- GCSE November Series 2023 Exam Discussions Hub
- Edexcel A Level Further Mathematics Paper 4B (9FM0 4B) - 24th June 2024 [Exam Chat]
- A-level Exam Discussions 2024
- Edexcel IGCSE Higher tier Mathematics A Paper 1 1H (4MA1) - 8th November 2023
- formula sheet in gases 2025
- Edexcel A Level Further Mathematics Paper 4D (9FM0 4D) - 24th June 2024 [Exam Chat]
- C1 January 2002?
- Edexcel A Level Further Mathematics Paper 4A (9FM0 4A) - 24th June 2024 [Exam Chat]
- Edexcel GCSE Mathematics Paper 3 (1MA1 3) - 13th November 2023 [Exam Chat]
- GCSE Exam Discussions 2023
- Edexcel A Level Further Mathematics Paper 4C (9FM0 4C) - 24th June 2024 [Exam Chat]
- Edexcel Past Papers
- GCSE Mathematics Study Group 2024-25
- Edexcel GCSE Mathematics Paper 2 (1MA1 2) - 10th November 2023 [Exam Chat]
- Edexcel GCSE Foundation tier Maths Paper 3 3F (1MA1) - 14th June 2023 [Exam Chat]
- GCSE Exam Discussions 2024
- Edexcel GCSE Mathematics Paper 2 Foundation (1MA1 2F) - 3rd June 2024 [Exam Chat]
- Edexcel GCSE Mathematics Paper 1 (1MA1 1) - 8th November 2023 [Exam Chat]
- Want to try go for a paramedic apprenticeship, what should I do to ensure my success?
- GCSE Revision Resource

Latest

Trending

Last reply 1 month ago

How do l find the min & max radius of a circle on an argand diagramMaths

2

4