Welling Maths Year 12: 2011

The Further Maths Support Programme run online revision sessions for A-Level Maths applications units and Further Maths units.

So the relevant session for Welling Year 12 Maths students is the one for S1.

Also for Further Maths students the ones for M1, D1 & FP1.

This is the e-mail I received from FMSP which explains what you need in terms of equipment and how to access the sessions:
https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZjA2YTZiNmItMmUzNC00NmUxLTk0ZTMtNzdkYjVkMmEwNGE2&hl=en

The S1 session is on Tuesday 10th May at 5pm run by Bob Francis. The link to access the session is:
http://tinyurl.com/63ur68q

The M1 session is on Tuesday 10th May at 4.30pm run by Krys Golabek. The link to access the session is:
http://tinyurl.com/6ewlrgm

The D1 session is on Monday 9th May at 4.30pm run by Martin Bamber. The link to access the session is:
http://tinyurl.com/6crqupc

The FP1 session is on Wed 15th June at 4.30pm run by Sharon Tripconey. The link to access the session is:
http://tinyurl.com/66f2d5b

Core2 Jan 2011 worked solutions

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWUwZTgyYWMtYTI0My00MDA3LTk3YzMtMjNlYTQ0NGMzODky&hl=en

Thursday, 28 April 2011

Core2 General revision 28/04/11

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMWU1ZGYyZWQtZTg3Yy00YTQyLWEzMWItZTJkZmNhOWE4NDNi&hl=en

Thursday, 7 April 2011

Core2 June 2010 Worked solutions

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWRjOTQxZGQtMzgzYS00MWQzLThmNWUtNzIwNWI5NjkzZDNi&hl=en

Core2 June 2010 Worked solutions

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWRjOTQxZGQtMzgzYS00MWQzLThmNWUtNzIwNWI5NjkzZDNi&hl=en

Thursday, 31 March 2011

Core2 Solving trigonometric equations 31/03/11

Went through homework questions

Particular focus on finding alternative solutions

Reminder of trigonometric identities

Exam questions worksheet

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDc3NWQ3ZDctZTA3Ni00OGU3LWIyMGEtYjRjZTVmOGYwYzg1&hl=en

Worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpNjk2MDBlNmEtYTc0Ni00MWFjLWE0OTYtODcyYTE0MTk1MGJh&hl=en

Thursday, 24 March 2011

Core2 Trapezium rule 24/03/11

Discovery of trapezium rule for approximating area under a curve
Using different numbers of intervals
Using the rule
Reminder of definite integral to find exact area
HOMEWORK SET see below

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMzlhNWZiM2MtNDkyYi00NDJkLWFjNTMtNDI4ZWIzNGE3NjNi&hl=en

Worksheet

https://docs.google.com/leaf?id=0B2J1T-insOmpMDRmMzBhOTctNTU2Ny00ZDRjLWJjMmItMWNlZDdiZTJjNzk4&hl=en

Homework - revising solving trigonometric equations

https://docs.google.com/leaf?id=0B2J1T-insOmpMTIxN2UyYTItMDdmOC00NzQ4LWFhNjgtNzYxOTA4NDhmY2Fm&hl=en

Tuesday, 22 March 2011

Core 2 Worked solutions

Core 1 Worked Solutions

Thursday, 17 March 2011

Core2 Area under a curve 17/03/11

Integrating reminder
Area under a curve
Proof that area under curve is found by definite integral
Finding definite integrals
Finding area under a curve
Finding area between two curves
Finding area between a line and a curve
HOMEWORK SET

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNDIxY2U2N2YtZWZlZC00ZmIyLWE0NTUtNTMzNWI5OTExODg2&hl=en

Homework sheet (to do questions 1 d f , 2 a b e , 3 d , 4 a , all of 5)

https://docs.google.com/leaf?id=0B2J1T-insOmpYzU3NzFmMDAtZGM0My00MjZmLWIyNjktYjY4NzAzYWNkNTI5&hl=en

Thursday, 10 March 2011

Core2 Special triangles and identities 10/03/11

Finding values of sin,cos,tan of 30,45,60 degrees
Proof of trig identities
Solving quadratic trig equations
Using trig identities to solve trig equations

Lesson flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMzIxOTVkYjUtNmJmYS00YWMwLTg3YTMtYjQ3MjkzNTE4ODk0&hl=en

HOMEWORK SET - revision of Core 1 Integration - some hints below

Homework sheet -

https://docs.google.com/leaf?id=0B2J1T-insOmpMzYyODQxYjktMWY4My00MTJhLWFkMWEtNGI1Njg3Y2E3OGMy&hl=en

Flipchart with a couple of homework questions completed -

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYjM2NTE1MDctMjM1Yi00OThlLWFiYjYtYmRiODQ0MDk0NmZi&hl=en

First part of lesson - Interim Assessment - copy below

https://docs.google.com/leaf?id=0B2J1T-insOmpMjc2ZGY0YjUtNWEzYS00YmRlLWIzMzktMjA2ZWY5YTI1ZWRi&hl=en

Thursday, 3 March 2011

Core2 Unit circles & more trigonometry 03/03/11

From unit circles to trigonometric graphs
Trig graphs by quadrant
Multiple angles for same trig value
Simple trigonometric equations
Further trigonometric equations
HOMEWORK SET

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZmUzNTcxZDUtZDZkMC00ODVlLWFmMWMtNWY4MTRkNDQxZGRk&hl=en

Trig graphs sheets:

https://docs.google.com/leaf?id=0B2J1T-insOmpZTEwZjZhYzctYTlmZC00NmE3LTgwN2MtMmQzYzQ4MjBlZmRl&hl=en

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpNDkyNWI3NmEtODIyMy00ZDIwLWI2MDYtZTJmNThmNjMyNWZk&hl=en

(set questions 1 j k l , 2 d g , 3 b d , 4 b e , 7 a)

Friday, 18 February 2011

Core2 Jan/Feb Exam question review 18/02/11

A selection of Core 2 exam questions on topics covered so far

Question sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpMDY5MDY4YzMtN2RhZC00NjZhLWIwZmEtMDMxZGUwMTQyYmNm&hl=en

Worked solutions:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOWYzMzJiNzctYTg0MS00MDNiLWIyMDgtMzNjOTZkNTgzNDYz&hl=en

Thursday, 17 February 2011

Core2 Triangles and circles 17/02/11

Using cosine rule, sine rule, arc length, sector area
Defining radian measure
Using radian measure for arc length and sector area
Solving problems involving circles and triangles using angles in radians
HOMEWORK SET - plotting accurate graphs of y=sin x, y=cos x and y=tan x (with x in degrees) on the provided A3 graph paper

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOTA3OTBmNjYtYzFjNy00YzZlLTg2ZTAtYjExZDUwMmE2ZTJk&hl=en

Worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpYTU1ZTg1ZWQtY2FkZi00MzUxLTgzZjEtMDY0YmQxYjc1Nzlj&hl=en

Friday, 11 February 2011

Core2 Trigonometry revision 11/02/11

Revision of GCSE trigonometry
Using sin,cos,tan to find angles and sides
Using sine rule and cosine rule
HOMEWORK SET

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNzJlOGY1MGMtMTEyYS00ZjhjLWJjODgtNzc2OTBmMjljY2Fk&hl=en

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpZjQzMjE1NDgtMDI5ZS00NTg4LWJmYWYtNTNkZGY0ZGZlZmY2&hl=en

Core2 Binomial series exam questions 11/02/11

Three past exam questions
Including using for approximation

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNDlmMTM2MDMtN2FjYi00ZTQ4LTgxMGItYzFlMmY3NGZkNzFl&hl=en

Worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpZWFlYzE4ZmItNTRkZS00MGJiLTg3NGUtN2ZmNWQwNjk3ZWRl&hl=en

Thursday, 10 February 2011

Core2 Binomial series 10/02/11

Expanding expressions to larger powers
Expanding (1+x) to increasing powers
Expanding (a+b) to increasing powers
Pascal's triangle - linking to coefficients of expansions
Permutations - factorial notation - nCr notation
Link to coefficients of binomial expansions
Formulae book - binomial series
Two examples of using binomial series expansions

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMzgyOWE3NTItYmRmOS00ODI1LTgzYjItZWVkMzA5MWFhYmMw&hl=en

Friday, 4 February 2011

Core2 Geometric series exam questions 04/02/11

Past exam questions using geometric series
Including use of logs to solve problems
HOMEWORK SET
Lesson and homework sheets linked below

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMDVjMTZlMDQtYzNjNy00Mzc0LWE3N2ItM2Q3NjI1MjcyMzBm&hl=en

Worksheet from lesson:

https://docs.google.com/leaf?id=0B2J1T-insOmpMjdmNGZiNmUtZDkxNi00ZDNmLWIyYmQtMGVhMDMxMzczM2Ni&hl=en

Homework worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpYWVjNmU1MGQtZmE1NC00Y2Y0LWE3NzItM2IxZTI4OGNmZjlj&hl=en

Thursday, 3 February 2011

Core2 Geometric series 03/02/11

Describing sequences
Converging, diverging and oscillating sequences
Concept of geometric series - common ratio
General term and formula for sum of geometric series
Using these formulae
Summing converging series
The concept of sum to infinity
Finding terms, sums and sums to infinity

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTlhYjM3ZGItZWQ2MC00YTkyLWFhMTAtN2I1YWIyNzY3ZDk2&hl=en

Friday, 28 January 2011

Core2 Solving exponential equations 28/01/11

Reminder of solviong a simple exponential equation
Solving various different types of exponential equations including quadratics
Changing base of a log expression
HOMEWORK SET solving exponential equations

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYTM4MGQyNTMtZGRhYi00NTY1LWFjNzEtNzM1NWE1ZTVlYjVk&hl=en

Homework sheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpNjYzZjcxMmMtZjlmNi00ODc4LWFlN2MtMTMwMGJiZjcwZDRj&hl=en

Thursday, 27 January 2011

Core2 Exponentials and logarithms 27/01/11

Reminders of index form
Intro to exponential functions
Review of graph plotting homework
Intro to log as inverse function of exponential
Converting between exponetial and log forms
Discovering laws of logarithms
Using laws of logarithms to simplify
Brief look at starting to solve equations involving exponentials

Flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYzY0ZTMxMGYtMDlmZS00OWNjLWI4OTAtYWJlNTQ5MmY5OWUw&hl=en

Worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpMmZjZWI1ZGUtMjFkNC00ZGExLWJiOGEtOTgyYWY0ZjcyYmM2&hl=en

Friday, 21 January 2011

Core2 Circles exam questions 21/01/11

A few Core 2 exam questions involving circles
HOMEWORK SET - a few more Core 2 exam questions involving circles
HOMEWORK SET - investigate exponential graphs using MyMaths graph plotter

Lesson flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMjI5NzM3YmYtNTBhZi00MjVmLWI4YWItODJkOWYxZGU4Y2E0&hl=en

Lesson worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpMzAyNTcwNGYtZTMyMi00OTdjLWJkY2ItMDYyNmVkMWI0N2Qy&hl=en

Circles homework:

https://docs.google.com/leaf?id=0B2J1T-insOmpMzJkMzc0NDQtYmMxNi00NTE2LTlhNzItYWFjNWIzMWU0ZTA4&hl=en

Exponential graph plotting homework:

https://docs.google.com/leaf?id=0B2J1T-insOmpZDQzNTYzYWYtMmQ5Ny00ZGQ1LTgzNGYtYWEwNjNlZDgzMjgx&hl=en

Thursday, 20 January 2011

Core2 Equation of a circle 20/01/11

Reminder of the basic form of the equation of a circle
Using completing the square to rearrange an equation into circle form (including deciding whether an equation is a circle)
Solving problems involving circles, tangents and normals

Link to flipchart:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpMGMyYjVlZjktODI5NC00OTMxLTgxYzAtOTE5N2IxM2MxYjRh&hl=en

Link to worksheet:

https://docs.google.com/leaf?id=0B2J1T-insOmpMjFlZjk1NTEtNGJlYS00MjZjLTg0NTAtMzRlMDFlYTA0NjQ2&hl=en

Monday, 17 January 2011

Link to Further Maths blog

As requested this is the link to the blog with Further Maths resources:

http://wellingfurthermaths12.blogspot.com/

Friday, 14 January 2011

Core2 Rem/fact theorem exam qu & Intro to circles 14/01/11

A look at a few past C2 exam questions testing remainder & factor theorems and fully factorising cubic expressions:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYWE2NTYyNzEtMjM2NS00Y2RkLWIzMjUtMmIwZjQ3MmU1NDEy&hl=en

Homework set to complete the remaining questions:

https://docs.google.com/leaf?id=0B2J1T-insOmpZDc1YTkyMjAtOTcyNC00N2FmLTg3NGUtZjUzNjlhZmNkNjhm&hl=en

Revision of some key circle theorems

Introduction to finding equation of a circle with centre at the origin

Homework set to find equation of a circle with a different centre

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYWJmZjg3YzgtMWRhMi00NWYyLTg5MWMtY2Y1YTA3MjAzN2Jk&hl=en

Homework set to consolidate circle theorem knowledge:

https://docs.google.com/leaf?id=0B2J1T-insOmpYTVkNGM2N2YtNDliOS00YzAwLTgwYmUtNWIxYzQzYjk5ZjQy&hl=en

Thursday, 13 January 2011

Core2 Remainder and factor theorems 13/01/11

Dividing cubic/quartic expression by linear expressions
With and without remainders
Discover remainder and factor theorems
Use remainder and factor theorems

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTVlZGI1YjgtZGIwOS00ZmY4LTg2NWUtMjg4YWJkYWNjNjJi&hl=en

Thursday, 6 January 2011

Core1 Revision - sequences, series and integration 06/01/11

Some past C1 exam questions
Recurrence relations
Arithmetic series
Integration
Finding an equation using integration

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZWY1ZjFjZmYtMmIwNC00YjZhLWEyZGItMDU0NDdhM2IxMjU2&hl=en

Tuesday, 4 January 2011

Core1 Worked solutions May 2010 paper

Link to worked solutions for May 2010 C1 paper:

https://docs.google.com/leaf?id=0B2J1T-insOmpZWE2ZDU5YzItZjU2ZS00NWFlLTg1ODMtMGU4N2JhYTM3MDNi&hl=en

Further supporting notes for worked solutions:

1. both root 75 and root 27 can be simplified to multiples of root 3

2. integrate - increase the power an divide by that - don't forget the +C

3. (a) solve the linear inequality - the steps used here avoid the need to divide by a negative
(b) solve the quadratic inequality by identifying the roots, sketching the graph, identifying which part of the graph is required which is between the roots in this case so state the final solution as a double inequality
(c) show the solutions to (a) and (b) on a number line and state an inequality for the overlap

4. (a) completing the square method
(b) using the answer to (a) the curve is a basic quadratic translated 3 left and 2 up - the point of intersection on the y-axis is found by substituting x=0 into the equation
(c) from the original equation identify a,b,c substitute into the discriminant (expecting the answer to this to be negative as there are no real roots

5. (a) substitute a1 into the equation to find a2 and then substitute a2 into the equation to find a3
(b) substitute a3 into the equation to find a4 and then substitute a4 into the equation to find a5

6. (a) f(x+3) is inside the bracket so horizontal and opposite direction to sign so translate 3 left
(b) 2f(x) is outside the bracket so stretch vertical scale factor 2
(c) for the minimum to be a (3,0) then the graph would have been translated 5 up so outside the bracket giving f(x)+5 so a=5

7. first change to root x to power of 1/2, split the fraction into separate terms and simplify then differentiate - bring power down and reduce the power

8. (a) find the gradient of AB by using difference in y / difference in x then use this gradient and either of the points in either straight line equation method - leaving in correct form with all coefficient as integers
(b) length of AB is found by using pythagoras on the two pairs of coordinates
(c) use pythagoras with the length AC and the unknown and make equal to the answer to (b) and solve to find the unknown t
(d) sketch of the coordinates and triangle demonstrates how to find the area of the triangle

9. (a) just substituting values into the formula
(b) in a similar way substitue values into the sum formula then rearrange the answer to (a) so that it can be substituted into the sum formula and rearranged to the required equation
(c) solve the answer to (b) to find a and then substitute this value into the answer to (a) to find d

10. (a) (i) quadratic shape but upside down because of the -x in the bracket - roots at 0 and 4
(ii) cubic shape but upside down because of the -x in the bracket - a double roots at 0 and another root at 7
(b) x-coordinates of the points of intersection found by equating the two equations and rearranging
(c) one solution to the equation given in (b) is x=0 so interested in solutions to the quadratic inside the bracket - use either quadratic formula or completing the square to solve - the two solutions for the x-coordinates are both positive but either by substituting to find y or with reference to the graph, the lower value of x gives a positive value of y

11.(a) first change the second term to power of -1/2 then integrate to find f(x) - don't forget the +C - then substitute values from the given point to find the C and restate the equation for f(x) with the value for C included
(b) substitute x=4 into the original equation for dy/dx to find the gradient of the tangent then use this gradient and the given point in either straight line equation method to find the equation of the tangent - leaving in the correct form with coefficients as integers

Core1 Worked solutions January 2010 paper

Link to worked solutions for January 2010 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZDFmOGVhODgtMDk2NC00ZGQ0LTg5MDgtYTg3ZDRlYjcxOGQ5&hl=en

Further supporting notes for worked solutions:

1. differentiate - bring the power down and reduce the power

2. (a) just expand and simplify
(b) to rationalise the denominator multiply the top and bottom by the expression in the denominator with the opposite sign - use the answer to part (a) for the numerator

3. (a) find the gradient of l1 by rearranging the equation to the form y=mx+c and indentifying th m
(b) gradient of l1 is -3/5 so gradient of perpendicular l2 is 5/3 - then use this gradient and the given point with either straight line equation method to find the equation of l2

4. - first change root x to power of 1/2 and simplify the second term
- then to find y need to integrate dy/dx - increase the power and divide by that - don't forget the +C
- then use the given point to substitue values of x and y to find the C
- then finally retstate the equation for y with the value for C included

5. first rearrange equation 1 to make y the subject and substitute into equation 2 and solve to find solutions for x - substitute these solutions into the rearranged equation to find the solutions for y

6. (a) - expand and simplify the top
- break into separate terms
- simplify
- then differentiate - bring the power down and reduce the power
(b) - substitute x=2 into dy/dx to find the gradient of the tangent at that point
- substitue x=2 into the original equation for y to find the y coordinate of the point
- then use this gradient and point with either straight line equation method to find the equation of the tangent

7. from the given information identify u1,u2,u3 and then a & d
(a) Year 10 is 9 steps of the common difference on from Year 1
(b) just substitute values for a,n,d into the sum formula
(c) subsitute values for n& d and the unknown A and make equal to double the previous total - solve for A

8. (a) f(x)+2 outside the bracket so translation up 2 - don't forget the equation of the asymptote
(b) 4f(x) outside the bracket so stretch 4 vertically
(c) f(x+1) inside the bracket so horizontal and opposite direction to sign - so translate 1 left

9. (a) common factor of x first then the remaining quadratic is a difference of two squares
(b) identify roots from answer to (a) and basic cubic shape
(c) separately substitue x=-1 and x=3 into the equation to find the y-coordinates of A and B then find the gradient of AB by difference in y over difference in x then use this gradient and one of the points A or B in either straight line equation method to find the equation of the line
(d) length AB is found using pythagoras on the two pairs of coordinates for A and B

10. (a) completing the square method
(b) no real roots means discriminant is less than zero - identify a,b,c from original equation and substitute into discriminant (or alternatively identify discriminant from answer to (a)) - then solve the inequality
(c) substitue k=1 into the answer to (a) and identify the two translations from basic quadratic (left 2 and up 10) - y intercept found by substituting x=0

Core1 Worked solutions June 2009 paper

Link to worked solutions for June 2009 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOTgyYWZkOGMtODcyNi00MWE3LTkxNDQtMTgwNTBiZTIwNmMy&hl=en

Further supporting notes for worked solutions:

1. (a) just the term inside the bracket squared
(b) expand and simplify

2. express the 32 and the root 2 as powers of 2 and then multiply together (so add the indices together)

3. first change the second term to power of -2
(a) differentiate - bring the power down and reduce the power
(b) integrate - increase the power and divide by that - don't forget the +C

4. (a) solve the linear inequality - the steps used avoid any dividing by a negative number
(b) solve the quadratic inequality by factorising, identifying roots, sketching the graph, identifying which part of the graph is required - in this case between the roots so finally state the solution as a double inequality
(c) show the solutions to (a) and (b) on a number line and give the overlap as an inequality

5. (a) 1960 is the 10th year and 1990 the 40th year so the difference between them is 30 steps of the common difference - the answer is negative because the values are decreasing
(b) to find the first term need to go back 9 steps from 10th term
(c) last term is known (from year 1990) so either sum formula can be used

6. equal roots means discriminant is equal to zero - identify a,b,c then substitute into the discriminant an solve for p (the questions states that p is non-zero)

7. (a) substitute a1 into the equation to find a2
(b) substitute a2 into the equation to find a3
(c) the question means the sum of the first four terms = 43 so substitute a3 into the equation to find a4 then add the terms, equate to 43 and then solve for k

8. (a) find the gradient of AB using the given points as this is -5/2 the gradient of line l is 2/5 as they are perpendicular - then use this gradient and the coordinates of A with either straight line equation method to find the equation of line l - giving in correct form with all coefficients as integers
(b) method shown is to rearrange the equation into the form y=mx+c to find the y-intercept - could also have substituted x=0 into the equation to find y
(c) OC is the base of the triangle and the x coordinate of B is the height of the triangle

9. (a) expand and simplify the top, split into separate terms, make root x power of 1/2 and simplify each term
(b) differentiate the answer to (a) - bring the power down and decrease the power
(c) substitute 9 into the answer to (b)

10. (a) common factor of x then factorise the remaining quadratic
(b) basic cubic shape and from (a) double root at 3 and another root at 0
(c) comparing the original equation to this one all the x's have become x-2 so it is a translation 2 right

11. (a) substitute x=2 into the equation so get y=7 to show P is on the curve
(b) differentiate and subsitute x=2 to find the gradient of the tangent at P and then use either straight line equation method to find the equation of the tangent
(c) gradient of tangent at P is 3 to gradient of tangent at Q is -1/3 as they are perpendicular - make the equation for dy/dx equal to -1/3 and solve to find the x coordinate of P - from the quadratic use either quadratic formula or completing the square method

Sunday, 2 January 2011

Core1 Worked solutions January 2009 paper

Link to the worked solutions for the Jan 2009 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOGZkNWU0YzItZDk4My00ODMxLWEyYjEtMzNlZTU5OTBiZTQ1&hl=en

Supporting notes for the worked solutions:

1. (a) power of 1/3 is cube root
(b) negative power is one over, and 2/3 is cube root and squared

2. integrate - increase power and divide by that - don't forget +C

3. just expand and simplify

4. integate f '(x) to find f(x) - increase the power and divide by that - don't forget the +C - then substitute the values from the given point to find C - and finally restate the equation for f(x) with the value for C

5. (a) f(x+3) - inside the bracket means x direction and opposite way - so translate 3 left
(b) f(-x) - inside the bracket means x direction so horizontal change so reflection in y-axis

6. (a) change root x to power of 1/2 then split into separate terms and simplify
(b) differentiate the answer to (a) - bring the power down and decrease the power

7. 2 different real solutions means discriminant is greater than zero
(a) identify a,b,c then substitute into the discriminant and simplify
(b) solve the quadratic inequality - factorise, identify roots, sketch graph and identify required part of graph which in this case is outside of the roots so the solution is two separate inequalities

8. (a) to find the y coordinate of point P just substitute x=1 into the equation
(b) (i) is a cubic with negative coefficient of the cubed term (because of the -x in the second bracket) with a double root at x=-1 and another root at x=2
(ii) is a basic reciprocal shape
(c) the two equations equated finds points of intersection - so the two points of intersection on the diagram means there are two solutions

9. (a) one equation in terms and a & d can be written from each given term
(b) solve the simultaneous equations from (a)
(c) n is the only unknown - can substitute Sn, a & d into the formula - and then rearrange to the required equation
(d) solve the equation by factorising - n must be >0

10. (a) gradient and point given for l1 so just use either straight line equation method to find the equation for l1
(b) subsitute x=-2 into the answer from (a) and show y=7
(c) length AB found by using pythagoras in the two given pairs of coordinates
(d) subsitute x=p into the equation to find the y-coordinate then find an expression for the length by using pythagoras - make it equal to 5 and rearrange

11. (a)
-find y coordinate of point P by substituting x=2 into the equation
-differentiate and substitute x=2 to find gradient of tangent at P
-use gradient and point in either straight line equation method to find equation of tangent
(b) if gradient of tangent is -2 then gradient of normal is 1/2 as they are perpendicular - then use point and gradient in either straight line equation method to find equation of normal
(c) substitute y=0 into the two equations found in (a) and (b) to find the x coordinates of points A and B - diagram shows the triangle and how the area is calculated

Saturday, 1 January 2011

Core1 Worked solutions June 2008 paper

Link to worked solutions for June 2008 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpOWQ1OTUwM2MtMTVlOS00MGI4LWJhNDMtYjZhZmVhMTczNGIz&hl=en

Further supporting notes for the worked solutions:

1. integrate - increase the power and divide by that - don't forget the +C

2. first factorise the common factor of x then the remaining quadratic is a difference of two squares

3. (a) f(x) + 3 is translate 3 up (the +3 is outside the bracket)
(b) f(2x) is squash s.f.2 horizontally (the multiply by 2 is inside the bracket)

4. (a) differentiate - bring the power down and reduce the power
(b) make the answer to part (a) equal to 15 then solve

5. (a) substitute x1 into the equation to find x2
(b) substitute x2 into the equation to find x3
(c) make the answer to (b) equal to 7 and solve

6. (a) separately substitute values of x=0 and y=0 into each equation to find points crossing axes
(b) equate the two equations and solve to find x coordinates of the intersection points then substitute each x coordinate into either equation to find the y coordinates

7. first identify a & d from the given information
(a) three steps from the first term to the fourth term
(b) method 1 is using the formula from the formula book and method 2 is basically an old gcse type method
(c) substitue the values for a & d into the formula for Sn and simplify
(d) most efficient method is to use the answer to (b), make it equal to 43 and solve
(e) use the answer to part (c) rather than starting from scratch

8. no real roots means discriminant is less than zero
(a) identify a,b,c and substitute into discriminant
(b) solve the quadratic inequality by factorising, identifying roots, sketching and identifying part of the graph which in this case is between the roots so solutions is a double inequality

9. (a) differentiate - bring power down and reduce power
(b) find an expression for the gradient of the tangent at A by substituting x=-1/2 into the equation for dy/dx - the gradient of the given straight line equation is found by rearranging into the form y=mx+c - find k by equating these
(c) substitute x=-1/2 and the value of k into the original equation

10. (a) find the length QR by using pythagoras on the two pairs of coordinates
(b) find the gradient of l1 from the two given points - found to be -1/2 so gradient of l2 is 2 as it is perpendicular to l1 - use this gradient and coordinates of Q in either straight line equation method to find an equation for l2
(c) P is the y intercept of l2 so coordinates identified from y=mx+c answer to part (b)
(d) diagram shows the triangle - PQ and QR are perpendicular as l1 and l2 are perpendicular so lengths PQ and QR needed to find area of triangle - QR was found in (a) & PQ now found in same way then half x base x height for area of triangle

11.(a) expand the numerator and then split into separate terms and simplify
(b) integrate from the answer to part (a) - don't forget the +C - then use the coordinates from the given point to find C - then restate the equation for f(x) with the value found for C

Core1 Worked solutions May 2007 paper

Link to the worked solutions forthe May 2007 C1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpYWE2ZmE5MWQtMzkwMC00ODMzLTk2OGMtM2VjYzhhYTM0NTYx&hl=en

Further supporting notes for worked solutions:

1. Just expand and simplify

2. (a) cube root of 8 is 2 then 2 the power of 4
(b) dividing means subtract the indices (4/3 - 1 = 1/3)

3. first change root x to power of 1/2
(a) differentiate - bring power down and reduce the power
(b) differentiate again teh answer to (a) to get the second differential
(c) integrate the original equation - increase the power and divide by that - don't forget the +C

4. identify a & d from given information
(a) 200th term is 199 steps from the first term so add 199 lots of the difference
(b) part (a) found the last term so formula used in method 1 can be used or alternatively that used in method 2 (both formulae are in the formulae book)

5. (a) as shown the graph is translated 2 units left - don't forget the show the asymptote with a dashed line
(b) state equations for both asymptotes (state y=0 rather than y-axis as the question specifically asks for equations)

6. (a) substitute equation 1 into 2 and simplify
(b) first solve the quadratic formed in (a) to find solutions for x and then substitute these values into equation 1 to find solutions for y (two possible methods for solving the quadratic - completing the square or using the quadratic formula - which must be recalled

7. different real roots means discriminant must be greater than zero
(a) identify a,b,c from original equation, substitute into discriminant, simplify to reach the required inequality
(b) to solve the quadratic inequality, first factorise, then identify roots, then sketch, then identify the correct part of the graph (in this case outside of the roots) and finally state solutions (in this case two separate inequalities)

8. (a) substitue a1 into the equation to find a2
(b) substitute a2 into the equation to find a3
(c) (i) question means find the sum of the first four terms, so first find a4 by substituting a3 into the equation, then add the terms
(ii) show that the expression is divisible by 10 by factorising the common factor of 10

9. (a) integrate - increase the power and divide by that - don't forget the +C - then use the given point to substitue x and y to find C - then restate the equation for f(x) with the value for C
(b) first factorise the common factor x and then factorise the remaining quadratic
(c) basic cubic shape with roots as found by solving the answer to (b) equal to zero

10. (a) find the y coordinates of P and Q by substituting the x value into the given equation then use pythagoras on the two pairs of coordinates to find the length PQ
(b) to find the gradients of the tangents at P and Q need to first differentiate and then substitue the values of the x coordinates of P and Q - these gradients being equal shows the tangents are parallel
(c) the normal at P is perpendiculat to the tangent so as gradient of tangent is -13 the gradient of the normal is 1/13 - use this gradient and the coordinates of P in either straight line equation method to find the equation of the normal - leave in the correct form with all coefficients as integers (key step here is to multiply through by 13)

11. (a) simplest way to find the gradient of l2 is to rearrange into the form y=mx+c
(b) x coordinate of the point of intersection is found by equating the equations of the two lines and then substitute the value into either equation to find the y coordinate
(c) find x coordinates of A and B by making each equation equal to 1 (as y=1) then the sketch shows the points ABP to show the calculation required to find the area of the triangle

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