## Friday, 31 December 2010

### Core1 Worked solutions January 2007 paper

Link to worked solutions for Jan 2007 C1 paper:

Notes supporting worked solutions:

1. Just differentiate - bring power down and reduce power

2. (a) the form of the answer required gives away that 3 is one of the factors to split 108 into

(b) expand and simplify

3. f(x) + 3 is translate 3 up - to find point crossing x axis make equation equal to zero

4. substitute equation 1 into 2 and solve the quadratic to find solutions for x - then put these solutions into equation 1 to find solutions for y

5. no real roots means discriminant is less than zero - identify a,b,c from original equation, substitute and solve the inequality

6. (a) expand and simplify ensuring the required form is reached
(b) use the answer to part (a) as the starting point then change the root x to power of 1/2 and finally integrate - increase the power and divide by that power - don't forget the +C

7. (a) first change the last term of f '(x) to power of -2 then integrate to find f(x) don't forget the +C then use the given point to substitute in values for x and y to find C then finally restate f(x) with the value for C
(b) substitute x=2 into original equation for f '(x) to find gradient of tangent at that point then use either straight line equation method to find the equation - ensuring it is given in the required from (y=mx+c)

8. (a) just differentiate bring power down and reduce the power
(b) just substitute x=4 into the original equation for y
(c) substitute x=4 into dy/dx found in (a) to find gradient of tangent at P
then if gradient of tangent is -3 then gradient of normal is 1/3 (perpendicular) - then using m=1/3 and point P in either straight line equation method to find equation of normal
(c) find x coordinate of Q by substituting y=0 into the equation then use pythagoras to find length PQ

9. (a) method 1 is using formula given in formula sheet, method 2 is basically old GCSE style method
(b) n,a&d are known so just substitute into formula for Sn
(c) a&d are known and use n=k in formula for Sn and then make less than 1750 and simplify to reach required expression
(d) negative value for k is not possible to must be the positive value and k must be an integer as it represents a number of rows so 33 is the most rows possible

10. (a) obtain roots as shown - for (i) a basic cubic shape with double root at zero and single root at 2 - for (ii) a quadratic but negative coefficient of x squared term so upside down and roots at 0 and 6
(b) equate the two equations and solve - ensure the solutions make sense from the graph sketches

## Thursday, 30 December 2010

### Core1 Worked solutions May 2006 paper

Link to worked solutions for May 2006 C1 paper:

Notes supporting worked solutions:

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

2. factorise, identify roots, sketch graph, identify part of graph needed (in this case "outside" the roots then state inequality solution (in this case two separate inequalities

3. (a) basic quadratic translated 3 left - substitute x=0 to get crossing point on y-axis
(b) from (a) to (b) is translated "k" units up - so crosses y-axis "k" units higher than in (a)

4. (a) substitute a1 into the equation to find a2 and subtitute a2 into the equation to find a3
(b) question means find the sum of the first five terms - so first find a4 and a5 in the same way as a2 and a3 were found - then add the terms

5. (a) first change root x to power of 1/2 then differentiate - bring power down and reduce power
(b) expand and simplify the numerator then put into separate fraction terms and simplify each term - then differentiate

6. (a) just expand and simplify
(b) to rationalise the denominator multiply top and bottom by the expression in the denominator but with the opposite sign

7. Using the information that the 11th term is 9, an equation in terms of a and d can be set up. Using the information that the sum of the first 11 terms is 77, another equation in terms of a and d can be set up. These are then solved simultaneously

8. (a) equal roots means discriminant = 0, identify a,b,c and substitute then solve for p (the question states p>0 so p=4 is only solution)
(b) substitute p=4 from (a) into the original equation and solve - expecting equal roots solution as stated in the original problem

9. (a) expand and simplify then factorise common factor of x
(c) basic cubic shape with roots at 0,3,5

10. (a) integrate - increase power and divide by that - don't forget +C - then use point to substitute values of x and y to find C - then restate equation with value for C
(b) just substitute x=2 into equation found in (a)
(c) substitute x=-2 into original equation of f '(x) to find gradient of tangent at that point then use either straight line equation method to find equation - finally stating in correct form with integer coefficients

11. (a) Find gradient from two points. The use either straight line equation method with either point to find the equation.
(b) perpendicular so gradient goes from 1/2 to -2 and point known so use either straight line equation method to find equation - then equate l1 and l2 for x coordinate of intersection (substitute into either equation for y coordinate
(c) pythagoras using two points to find distance
(d) pythagoras again to find length PQ and then sketch identifies triangle (l1 and l2 known to be perpendicular)

## Wednesday, 29 December 2010

### Core1 Worked solutions January 2006 paper

Link below to the worked solutions for January 2006 C1 paper.

Notes below to support the worked solutions.

1. Common factor of x factorised out first. Then factorise remaining quadratic into two brackets.

2. Substitute each term into the expression to get the next. In part (a) it is shown that all odd terms will be 1 and even terms will be 4, so 20th term will be 4.

3. (a) substituting x=3 into the equation shows that P lies on the line.

(b) gradient of -2 is just identified from the equation (the m in the form y=mx+c) - so perpendicular gradient is 1/2 - then use either straight line equation method to find the equation - finally leave in the correct form with all coefficients as integers

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

5. (a) split the 45 into a factor pair where one of the factors is a square number (so 9 and 5) then simplify
(b) to rationalise the denominator multiply top and bottom by the expression in the denominator but with the opposite sign - then simplify

6. (a) f(x+1) is translate one left - inside the bracket so horizontal and the opposite way to that the sign suggests
(b) 2f(x) is stretch vertically scale factor 2 - outside so vertical change
(c) f(1/2 x) is stretch horzontally scale factor 2 - inside so horiztonal and the opposite way to the coefficient suggests

7. (a) difference given so 12th birthday is just £200 more than 11th then add them up
(b) 18th birthday is 8th term so add 7 lots of the difference to the first term
(c) first term and last (8th) term are known so substitute into formula
(d) Sn,a,d are known so n is only unknown - substitute into formula and simplify

8. First separate terms of fraction, then simplify so that the expression can be integrated.
Integrate - increase power and divide by it and don't forget the +C
Use given point to substitute values for x and y to find C
Restate equation with value for C

9. second bracket is difference of two squares so factorise completely
(a) identify roots from completely factorised equation - root at x=1 is already shown on graph so P and Q are the others
(b) expand brackets and differentiate
(c) substitute x=-1 into equation for dy/dx to find gradient of tangent at that point then use either straight line equation method to find the equation
(d) make dy/dx = 1 then solve, one solution has already been used (when x=-1 so R is the other solution, find y coordinate by substituting x value in to original equation for y

10. (a) completing the square
(b) completing the square form shows translation left 1 and up 2 from basic x squared graph - substitute x=0 for point crossing y axis
(c) identify a,b,c from original equation then subsitute to find discriminant - less then zero so no real roots
(d) identify a,b,c and substitute into inequality and solve inequality

## Tuesday, 21 December 2010

### Core1 Worked solutions May 2005 paper

The link is to worked solutions of the May 2005 C1 paper and there are some additional explanations below.

1. (a) power of 1/3 is cube root
(b) negative power gives one over - the numerator of 2 in the power is squared and the denominator of 3 in the power is cube root

2. (a) first change over x squared to power of -2 then differentiate (bring power down and reduce power by one)
(b) integrate - increase power and divide by that and don't forget the +C

3. (a) completing the square
(b) method 1 is to follow on from part (a) using completing the square - method 2 is quadratic formula

4. (a) y=3f(x) is stretch s.f. 3 vertical
(b) y=f(x+2) is translation 2 left

5. make x the subject of first equation and substitute into second (could have made y the subject but this is more efficient)
then solve the quadratic to obtain two solutions for y
substitute the values of y into the equation with x as the subject

6.(a) solve linear inequality (the method used avoids any divison by a negative which would require a change of sign)
- factorise
- identify roots
- sketch
- identify outside of roots
(c) combine solutions to (a) and (b) - number line clearly identifies the overlap (between 1/4 and 1/2 and greater than 3)

7. (a) expand the top, simplify, split into separate fractions, simplify
(b) integrate (increase power and divide by that) don't forget +C then use given value to find C and restate equation

8. (a) given gradient and point - use either straight line equation method
(b) the given info for l2 gives the equation as y = -2x
lines intersect - equate and solve for x and substitue into either equation to find y
(c) sketch of triangle identifies that one side is on y-axis so half x base x height is simple from coordinates

9. (a) show list of terms then backwards to show that 2Sn can be simplfies to n lots of (2a + (n-1)d)
(b) formally or informally use a=149 and d=-2
(c) a=149, d = -2 and Sn=5000 so only unknown is n
(d) just factorise and solve
(e) a=149 and d=-2 so after 75 terms the value of the term is negative which does not make sense for the given situation - loan repayments

10. (a) just subst. x=3 into the equation for y
(b) differentiate and substitute x=3 to find gradient of tangent to curve at P then use either straight line equation method given gradient of -7 and point (3,0)
(c) parallel lines have same gradient - make dy/dx equal to -7 and solve - the solution of x=3 gives the point P already known so the other solution if for point Q - substitute into original equation for y

## Saturday, 18 December 2010

### Core1 Worked solutions January 2005 paper

Link to worked solutions for January 2005 C1 paper and then further explanatory notes below.

1. (a) power of 1/2 is square root

(b) the minus in the power gives one over; the over two in power is square root; and the three is cubed

2. (i) basic rules of differentiation of differentiation - bring the power down and reduce the power - and for part (b) repeat to obtain second differential

(ii) first change the root to power 1/2 and the one over x squared to power of -2 then basic rules of integration - increase the power and divide by it - and don't forget the plus C - remember dividing by 3/2 is same as multiplying by 2/3

3. remember equal roots means discriminant is equal to zero then substitue values and solve for k

4. rearrange first equation (could make x the subject rather than y as shown) then substitute into second equation - solving quadratic obtains two solutions for x then substitute back into the equation which has y as its subject to find the solutions for y

5. (a) just substitute r=1,2,3 in turn into the expression

(b) common difference identified from part (a) results

(c) just substituting a=-3 and d=2 as identified in (a) and (b) into the formula for the sum to n terms (which is given on the formula sheet)

6. (a) reflection in x axis - because the minus is outside the transformation is in the vertical direction

(b) squash s.f.2 in x direction - the 2 is inside the bracket so in x direction and squash rather than stretch

7. (a) first rearrange to split the fraction and simplify then differentiate then substitute to find the value of the gradient of the tangent at the given point

(b) substitute given value of x into original equation for y to find y coordinate of P. Then use either straight line equation method with gradient of 3 and point (1,8).

(c) just substituting y=0 into equation found in (b)

8. (a) identify coordinates so that D is midpoint of AC

(b) calculate gradient of AC using difference in y over difference in x (being careful with negative signs) - then gradient of line l is opposite sign and fraction upside down as they are perpendicular - then use eitherstraight line equation method and leave in correct form

(c) substitute y=7 (because AB is the line y=7 and E is on AB) into equation found in (b) to find x coordinate

9. (a) substitute x = 1 into given dy/dx to find gradient of tangent at P - gradient of normal to P is minus one over that - then use either straight line method to find equation of normal at P

(b) integrate given dy/dx to find y - don't forget + C and then use point (1,3) to find C and then restate equation for y

(c) just showing that there are no solutions for the given equation for dy/dx to be negative (would need to be -2 to be parallel to y=1-2x)

10. (a) complete the square

(b) from y=x^2 just a translation right 3 and up 9 and substitute x=0 to get point on y-axis

(c) solve by completing the square (could also use quadratic formula)

## Thursday, 16 December 2010

### Core1Revision Using integration 15/12/10

A few questions looked at from this worksheet...

- Integrating and using given point to find equation of curve

## Monday, 13 December 2010

### Core 1 Worked solutions January 2008 paper

The link below is to a set of worked solutions for the January 2008 C1 paper.

There are then some notes below offering further explanations for each question.

**There is one small error in the woked solutions - see notes for 5(b).

1. Remember the general rule for integrating - increase the power and divide by that number - and don't forget the + C for the final mark

2. (a) power of 1/4 is fourth root : 2 x 2 x 2 x 2 = 16
(b) the power is applied to both the 16 and the x separately - for the x part the 12 and 3/4 are multiplied together and then simplified

3. First remember the general principle of multiplying top and bottom by the expression in the denominator but with the opposite sign. Then be careful multiplying out two pairs of double brackets. And then take care simplifying.

4. (a) First find the gradient from the two given points - be careful with negatives.
Then use one of two methods substituting the gradient and one of the points into either y-y1 = m(x-x1) or y = mx + c. Finally leave equations in correct form - the questions asks for integer coefficients.
(b) Pythagoras to find the distance between two given points. And simplify the surd.

5. (a) First split into separate terms. Then the first term gives x to the power of 1/2 divided by x to the power of 1 - so 1/2 subtract 1 gives minus 1/2. The second term is just power of -1.
(b) Remember the basic rule for differentiation - bring down the power and reduce the power.
**** ERROR in solutions - the second term should be minus (because it was power of -1/2 being brought down)

6. (a) stretch in y direction (vertically) scale factor 2 - so points on x axis don't change and maximum has y value doubled
(b) reflection in y-axis (horizontal direction - note minus is inside the bracket)
(c) needs translation two units left so a = 2 (remember f(x+2) is two left and f(x-2) is two right)

7. (a) subsitute in value of x1
(b) substitue in value of x2 and show all steps
(c) solve the quadratic - question gives p not equal to 0
(d) show pattern of alternating odd terms and even terms

8. (a) First make rearrange to make equations = 0
Then find a,b,c. And know that discriminate will be negative as no real roots. Then substitute values of a,b,c. Show all steps of rearranging.
(b) Factorise. Find roots. Sketch graph. Highlight "inside" part. Show inequality.

9. (a) Restate f ' (x) with powers of x. Remember basic integation rules. Remember dividing by 3/2 is same as multiplying by 2/3. Remember + C. Then use point given to find C. And finally restate whole equation including the C.
(b) From the original given equations for f ' (x) subst. x=4 to find gradient of tangent at P. Then find gradient of normal (minus one over gradient for tangent). Then use one of the two methods for finding equation with gradient found and point given.

10.(a) Basic shape from knowing it is cubic. Double root at x=1 so sits on axis there. Another root at x=-3 and subst. x=0 to get y=3 for crossing y-axis
(b) Just expand brackets
(c) Differentiate. Make equal to 3. Solve.

11. (a) Two possible approaches - informal or formal (equation for formal method given in formula book) - Remember 25th term is 24 steps after 1st term.
(b) Again formal or informal. To get from the 25th to the 21st is 4 steps. Also could have worked forwards from the 1st term - 20 steps to get to the 21st term.
(c) Lots of different methods - important point is identifying that largest total will be before terms become negative. Also realise that the total for 20 terms and 21 terms will be the same because the 21st term was zero.

## Thursday, 9 December 2010

### Core1 Integration 09/12/10

• Reminder of differentiation
• The opposite of differentiation
• Defined integration and notation
• Consolidating integration
• Using integration to find an equation of a curve

http://www.scribd.com/doc/44968497/Core1-Integration-091210