c1 revision workbook

C1 Revision Workbook

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Glyn Technology School 1

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C1 Algebra

1. (a) Express √108 in the form a√3, where a is an integer.(1)

(b) Express (2 – √3)2 in the form b + c√3, where b and c are integers to be found.(3)

(Total 4 marks)

2. Simplify

3235

,

giving your answer in the form a+b√3, where a and b are integers.(Total 4 marks)

3. (a) Write down the value of 31

125 .(1)

(b) Find the value of 32

125 .

(2)(Total 3 marks)

4. Write

√(75) – √(27)

in the form k √x, where k and x are integers.(Total 2 marks)

5. (a) Find the value of 34

8 .

(2)

(b) Simplify .3

15 34

xx

(2)(Total 4 marks)

6. (a) Write down the value of 41

16 .

(1)

(b) Simplify 43

12 )16( x .

(2)(Total 3 marks)

7. Given that 32 √2 = 2a, find the value of a.(Total 3 marks)

8. Factorise completely

x3 – 9x.(Total 3 marks)


9. Solve the simultaneous equations

y – 3x + 2 = 0

y2 – x – 6x2 = 0(Total 7 marks)

10. Solve the simultaneous equations

y = x – 2,

y2 + x2 = 10.(Total 7 marks)

11. Find the set of values of x for which

(a) 4x – 3 > 7 – x(2)

(b) 2x2 – 5x – 12 < 0(4)

(c) both 4x – 3 > 7 – x and 2x2 – 5x – 12 < 0(1)

(Total 7 marks)

12. The equation 2x2 – 3x – (k + 1) = 0, where k is a constant, has no real roots.

Find the set of possible values of k.(Total 4 marks)

13. (a) Show that x2 + 6x + 11 can be written as

(x + p)2 + q

where p and q are integers to be found.(2)

(b) In the space at the top of page 7, sketch the curve with equation y = x2 + 6x + 11, showing clearly any intersections with the coordinate axes.

(2)

(c) Find the value of the discriminant of x2 + 6x + 11(2)

(Total 6 marks)

14. The equation x2 + kx + (k + 3) = 0, where k is a constant, has different real roots.

(a) Show that k 2 – 4k – 12 > 0.(2)

(b) Find the set of possible values of k.(4)

(Total 6 marks)


15. The equation 0)5(42 kxkx , where k is a constant, has 2 different real solutions for x.

(a) Show that k satisfies

.0452 kk(3)

(b) Hence find the set of possible values of k.(4)

(Total 7 marks) 16. f(x) = x2 + 4kx + (3 + 11k), where k is a constant.

(a) Express f(x) in the form (x + p)2 + q, where p and q are constants to be found in terms of k.(3)

Given that the equation f(x) = 0 has no real roots,

(b) find the set of possible values of k.(4)

Given that k = 1,

(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graph crosses a coordinate axis.

(3)(Total 10 marks)

Topic Marks Marks you scored1. January 2007 Indices & surds 4 marks2. January 2008 Indices & surds 4 marks3. January 2009 Indices & surds 3 marks4. June2010 Indices & surds 2 marks5. June 2007 Indices & surds 4 marks6. January 2008 Indices & surds 3 marks7. June 2009 Indices & surds 3 marks8. June 2008 Algebra 3 marks9. January 2010 Simultaneous equations 7 marks10. January 2007 Simultaneous equations 7 marks11. June 2009 Inequalities 7 marks12. January 2007 Inequalities 4 marks13. June 2007 Quadratics 6 marks14. June 2010 Inequalities 6 marks15. January 2009 Quadratics 7 marks16. January 2010 Quadratics 10 marks


C1 Coordinate geometry

1. The point A (–6, 4) and the point B (8, –3) lie on the line L.

(a) Find an equation for L in the form ax + by + c = 0, where a, b and c are integers.(4)

(b) Find the distance AB, giving your answer in the form k√5, where k is an integer.(3)

(Total 7 marks)

2. The line l1 has equation 3x + 5y – 2 = 0

(a) Find the gradient of l1.(2)

The line l2 is perpendicular to l1 and passes through the point (3, 1).

(b) Find the equation of l2 in the form y = mx + c, where m and c are constants.(3)

(Total 5 marks)

3. The line l1 has equation y = 3x + 2 and the line l2 has equation 3x + 2 y – 8 = 0.

(a) Find the gradient of the line l2.(2)

The point of intersection of l1 and l2 is P.

(b) Find the coordinates of P.(3)

The lines l1 and l2 cross the line y= 1 at the points A and B respectively.

(c) Find the area of triangle ABP.(4)

(Total 9 marks)

4. (a) Find an equation of the line joining A (7, 4) and B (2, 0), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(3)

(b) Find the length of AB, leaving your answer in surd form.(2)

The point C has coordinates (2, t), where t > 0, and AC = AB.

(c) Find the value of t.(1)

(d) Find the area of triangle ABC.(2)

(Total 8 marks)

5. The line l1 passes through the point A (2, 5) and has gradient 21

.

(a) Find an equation of l1, giving your answer in the form y = mx + c.(3)

The point B has coordinates (–2, 7).


(b) Show that B lies on l1.(1)

(c) Find the length of AB, giving your answer in the form 5k , where k is an integer.(3)

The point C lies on l1 and has x-coordinate equal to p.

The length of AC is 5 units.

(d) Show that p satisfies

.01642 pp(4)

(Total 11 marks)

6. The curve C has equation x

y 3 and the line l has equation y = 2x + 5.

(a) On the axes below, sketch the graphs of C and l, indicating clearly the coordinates of any intersections with the axes.

(3)

(b) Find the coordinates of the points of intersection of C and l.

x

y

O

(6)(Total 9 marks)

Question Topic MarksMarks you scored

1. January 2008 Coordinate Geometry 7 marks

2. January 2010 Straight lines 5 marks

3. June 2007 Straight lines 9 marks


5. January 2009 Coordinate Geometry 11 marks

6. June 2008 Coordinate Geometry 9 marks


C1 differentiation and Integration

1. Given that

y = 4x3 – 1 + 21

2x , x > 0,

find xy

dd

.

(Total 4 marks)

2. Given that y = x4 + ,331

x find .dd

xy

(Total 3 marks)

3. f(x) = 3x + x3, x > 0.

(a) Differentiate to find f′ (x).(2)

Given that f′ (x) = 15,

(b) find the value of x.(3)

(Total 5 marks)

4. ,)4–3(

)f(2

xx

x x > 0

(a) Show that ,9)f( 21

21

–BAxxx where A and B are constants to be found.

(3)

(b) Find f´(x).(3)

(c) Evaluate f´(9).(2)

(Total 8 marks)

5. The curve C has equation

y = x3 – 2x2 – x + 9, x > 0

The point P has coordinates (2, 7).

(a) Show that P lies on C.(1)

(b) Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

(5)

The point Q also lies on C.

Given that the tangent to C at Q is perpendicular to the tangent to C at P,

(c) show that the x-coordinate of Q is ).62(31

(Total 11 marks)


6. The curve C has equation

,)8)(3(x

xxy x > 0

(a) Find xy

dd

in its simplest form.

(4)

(b) Find an equation of the tangent to C at the point where x = 2(4)

(Total 8 marks)

7. Given that

0,23482

3

xx

xxxy

find xy

dd

.

(Total 6 marks)

8. Given that y = 3x2 + 4x, x > 0, find

(a)xy

dd

(2)

(b)2

2

dd

xy ,

(2)

(c) xyd .(3)

(Total 7 marks)

9. Find xxx d)743( 52 .(Total 4 marks)

10. The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C. Given that

f ′(x) = 3x2 – 6 – 2

8x ,

(a) find f(x).(5)

(b) Find an equation for the tangent to C at the point P, giving your answer in the formy = mx + c, where m and c are integers.

(4)(Total 9 marks)

11. A curve has equation y = f(x) and passes through the point (4, 22).

Given that

,733)(f 21

2 xxx'


use integration to find f(x), giving each term in its simplest form.(Total 5 marks)

12. The curve C has equation y = f(x), x > 0, and 2

864)(fx

xxx .

Given that the point P (4, 1) lies on C,

(a) find f(x) ands simplify your answer.(6)

(b) Find an equation of the normal to C at the point P (4, 1).(4)

(Total 10 marks)

13. The curve C with equation y = f(x) passes through the point (5, 65).

Given that f’(x) = 6x2 – 10x –12,

(a) use integration to find f (x).(4)

(b) Hence show that f(x) = x(2x + 3)(x – 4).(2)

(c) Sketch C, showing the coordinates of the points where C crosses the x-axis.(3)

(Total 9 marks)

Question Topic Marks marks you socred

1. January 2007 Basic differentiation 4 marks


3. June 2008 Basic differentiation 5 marks

4. June 2009 Indices & surds 8 marks



7. June 2010 Basic differentiation 6 marks

8. June 2007 Differentiation 7 marks

9. January 2008 Basic integration 4 marks




13. June 2007 Basic integration 9 marks


C1 Sequences

1. A girl saves money over a period of 200 weeks. She saves 5p in Week 1, 7p in Week 2, 9p in Week 3, and so on until Week 200. Her weekly savings form an arithmetic sequence.

(a) Find the amount she saves in Week 200.(3)

(b) Calculate her total savings over the complete 200 week period.(3)

(Total 6 marks)

2. A sequence x1, x2, x3, … is defined by

x1 = 1,

xn+1 = axn – 3, n > 1,

where a is a constant.

(a) Find an expression for x2 in terms of a.(1)

(b) Show that x3 = a2 – 3a – 3.(2)

Given that x3 = 7,

(c) find the possible values of a.(3)

(Total 6 marks)

3. A sequence of positive numbers is defined by

,3( 21 nn aa ,1n

21 a

(a) Find a2 and a3, leaving your answers in surd form.(2)

(b) Show that a5 = 4(2)

(Total 4 marks)

4. A sequence a1, a2, a3, … is defined by

a1 = k,

an+1 = 2an – 7, n ≥ 1,

where k is a constant.

(a) Write down an expression for a2 in terms of k.(1)

(b) Show that a3 = 4k – 21.Glyn Technology School 10

(2)

Given that ,434

1

r

ra

(c) find the value of k.(4)

(Total 7 marks)

5. A sequence a1, a2, a3 ... , is defined by

a1 = k,

an+1 = 3an + 5, n ≥ 1,

where k is a positive integer.

(a) Write down an expression for a2 in terms of k.(1)

(b) Show that a3 = 9k + 20.(2)

(c) (i) Find

4

1rra in terms of k.

(ii) Show that

4

1rra is divisible by 10.

(4)(Total 7 marks)

6. A sequence is given by:

x1 = 1,

xn+1 = xn(p + xn),

where p is a constant (p ≠ 0).

(a) Find x2 in terms of p.(1)

(b) Show that x3=1 + 3p + 2p2.(2)

Given that x3 = 1,

(c) find the value of p,(3)

(d) write down the value of x2008.(2)

(Total 8 marks)

7. Sue is training for a marathon. Her training includes a run every Saturday starting with a run of 5 km on the first Saturday. Each Saturday she increases the length of her run from the previous Saturday by 2 km.


(a) Show that on the 4th Saturday of training she runs 11 km.(1)

(b) Find an expression, in terms of n, for the length of her training run on the nth Saturday.(2)

(c) Show that the total distance she runs on Saturdays in n weeks of training is n(n + 4) km.(3)

On the n th Saturday Sue runs 43 km.

(d) Find the value of n.(2)

(e) Find the total distance, in km, Sue runs on Saturdays in n weeks of training.(2)

(Total 10 marks) 8. The first term of an arithmetic series is a and the common difference is d.

The 18th term of the series is 25 and the 21st term of the series is 2132 .

(a) Use this information to write down two equations for a and d.(2)

(b) Show that a = –7.5 and find the value of d.(2)

The sum of the first n terms of the series is 2750.

(c) Show that n is given by

.4055152 nn(4)

(d) Hence find the value of n.(3)

(Total 11 marks)

Question Topic Marks Marks you socred

1. June 2007 Arithmetic series 6 marks

2. June 2008 General sequences & series 6 marks




6. January 2008 General sequences & series 8 marks

7. June 2008 Arithmetic series 10 marks

8. January 2009 Simultaneous equations 11 marks


C1 Transformations of curves

1. (a) On the same axes sketch the graphs of the curves with equations

(i) y = x2(x – 2),(3)

(ii) y = x(6 – x),(3)

and indicate on your sketches the coordinates of all the points where the curves cross the x-axis.

(b) Use algebra to find the coordinates of the points where the graphs intersect.(7)

(Total 13 marks)

2.

y

(0 , 7 ) y = f( x )

O (7 , 0 ) x

The diagram above shows a sketch of the curve with equation y = f(x). The curve passes through the point (0, 7) and has a minimum point at (7, 0). On separate diagrams, sketch the curve with equation

(a) y = f(x) + 3,(3)

(b) y = f(2x).(2)

On each diagram, show clearly the coordinates of the minimum point and the coordinates of the point at which the curve crosses the y-axis.

(Total 5 marks)

3.


Above is a sketch of the curve C with equation y = f(x). There is a maximum at (0, 0), a minimum at (2, –1) and C passes through (3, 0).

On separate diagrams sketch the curve with equation

(a) y = f(x + 3)(3)

(b) y = f(–x).(3)

On each diagram show clearly the coordinates of the maximum point, the minimum point and any points of intersection with the x-axis.

(Total 6 marks)

4.

y

x

(2 , 5 )

1 4

The diagram above shows a sketch of the curve with equation y = f(x). The curve crosses the x-axis at the points (1, 0) and (4, 0). The maximum point on the curve is (2, 5). In separate diagrams sketch the curves with the following equations. On each diagram show clearly the coordinates of the maximum point and of each point at which the curve crosses the x-axis.

(a) y = 2f(x),(3)

(b) y = f(–x).(3)

The maximum point on the curve with equation y = f(x + a) is on the y-axis.

(c) Write down the value of the constant a.(1)

(Total 7 marks)

5. Given that f(x) = x1

, x ≠ 0,

(a) sketch the graph of y = f(x) + 3 and state the equations of the asymptotes.(4)

(b) Find the coordinates of the point where y = f(x) + 3 crosses a coordinate axis.(2)

(Total 6 marks)

6.


y

xO

The diagram above shows a sketch of the curve with equation .0,3 x

xy

(a) On a separate diagram, sketch the curve with equation ,2,2

3

x

xy showing the coordinates of

any point at which the curve crosses a coordinate axis.(3)

(b) Write down the equations of the asymptotes of the curve in part (a).(2)

(Total 5 marks) 7. (a) Factorise completely x3 – 6x2 + 9x

(3)

(b) Sketch the curve with equation

y = x3 – 6x2 + 9x

showing the coordinates of the points at which the curve meets the x-axis.(4)

Using your answer to part (b), or otherwise,

(c) sketch, on a separate diagram, the curve with equation

y = (x – 2)3 – 6(x – 2)2 + 9(x – 2)

showing the coordinates of the points at which the curve meets the x-axis.(2)

(Total 9 marks)

Question Topic Marks marks you scored

1. January 2007 Transformations & graphs 13 marks

2. June 2008 Transformations & graphs 5 marks




6. June 2007 Transformations & graphs 5 marks

7. June 2009 Quadratics 9 marks


C1 Answers


C1 Algebra

1. (a) 6√3 (a = 6) B1 1

±6√3 also scores B1.

(b) Expanding (2 – √3)2 to get 3 or 4 separate terms M17, –4√3 (b = 7, c = –4) A1, A1 3.

[4]

2.)32()32(

)32()35(

M1

...33710

...)3(353210 2

M1

13713 Allow)3713( 13 (a = 13) A1

)7(37 b A1 4 [4]

3. (a) 5 (± 5 is B0) B1 1

(b) 2)5their (1

or 2

5their 1

M1

= 251

or 0.04 )A0is251( A1 2

[3]

4. 75 27 5 3 3 3 M1

= 2 3 A1

[2]

5. (a) Attempt 3 8 or 3 4 )8( M1= 16 A1 2

(b) 31

5x 5, 31

x B1, B1 2

[4]

6. (a) 2 B1 1

Negative answers:

Allow –2. Allow ±2. Allow ‘2 or –2’.

(b) x9 seen, or (answer to (a))3 seen, or (2x3)3 seen. M1

8x9 A1 2Glyn Technology School 17

[3]

7. 32 = 25 or 2048 = 211, 21

21

)2048(2048or22 B1, B1

211

a

5.5or

215or B1

[3]

8. x(x2 – 9) or (x ± 0)(x2 – 9) or (x – 3)(x2 + 3x) or (x + 3)(x2 – 3x) B1

x(x – 3)(x + 3) M1A13.

[3]

9. y = 3x – 2 (3x – 2)2 – x – 6x2 (= 0) M1

9x2 – 12x + 4 – x – 6x2 = 0

3x2 – 13x + 4 = 0 (or equiv., e.g. 3x2 = 13x – 4) M1 A1cso

(3x –1)(x – 4) = 0 x = …31

x (or exact

equivalent) x = 4 M1 A1

y = –1 y = 10 (Solutions need not be “paired”) M1 A1[7]

10. (x – 2)2 = x2 – 4x + 4 or (y + 2)2 = y2 + 4y + 4 M: 3 or 4 terms M1(x – 2)2 + x2 = 10 or y2 + (y + 2)2 = 10 M: Substitute M12x2 – 4x – 6 = 0 or 2y2 + 4y – 6 = 0 Correct 3 terms A1(x– 3)(x + 1) = 0, x = ... or (y + 3)(y – 1) = 0, y = ... M1(The above factorisations may also appear as (2x – 6)(x + 1) or equivalent).

x = 3 x= –1 or y= –3 y = 1 A1y = 1 y= –3 or x= –1 x = 3 M1M1 7

(Allow equivalent fractions such as: x = 26

for x = 3).

[7]

11. (a) 5x > 10, x > 2 [Condone x > 210 = 2 for M1A1] M1, A1 2

(b) (2x + 3)(x – 4) = 0, ‘Critical values’ are – 23

and 4 M1, A1

423– x M1 A1ft 4

(c) 2 < x < 4 B1ft 1 [7]

12. Use of b2 – 4ac, perhaps implicit (e.g. in quadratic formula) M1(–3)2 – 4 × 2 × –(k + 1) < 0 (9 + 8(k + 1) < 0) A1

8k < –17 (Manipulate to get pk < q, or pk > q, or pk = q) M1


k < 8

17

125.2or

812 :equivOr kk A1cso 4

[4]

13. (a) 23 2x or 63 or 2

p B1

2q B1 2

(b)

U shape with min in 2nd quad(Must be above x-axis and not on y = axis) B1

U shape crossing y-axis at (0, 11) only(Condone (11,0) marked on y-axis) B1 2

(c) 2 24 6 4 11b ac M1

= 8 A1 2 [6]

14. (a) Attempt to use discriminant b2 – 4ac M1k2 – 4(k + 3) > 0 k2 – 4k – 12 > 0 (*) A1cso 2

(b) k2 – 4k – 12 = 0 (k a)(k b), with ab = 12

or (k =)2

12444 2 or (k – 2)2 22 – 12 M1

k = –2 and 6 (both) A1k < –2, k > 6 or (– , –2);(6, ) M: choosing “outside” M1 A1ft 4

[6]

15. (a) b2 – 4ac > 0 16 – 4k(5 – k) > 0 or equiv., e.g. 16 > 4k(5 – k) M1A1

So k2 – 5k + 4 > 0 (Allow any order of terms,e.g. 4 – 5k + k2 > 0) (*) A1cso 3

(b) Critical Values (k – 4)(k – 1) = 0 k = . . . . M1

k = 1 or 4 A1

Choosing “outside” region M1

k < 1 or k > 4 A1 4 [7]

16. (a) (x + 2k)2 or 2

24

kx M1

(x ± F)2 ±G ± 3 ± 11k (where F and G are any functions of k,not involving x) M1


(x + 2k)2 – 4k2 + (3+11k) Accept unsimplifiedequivalents such as A1 3

,1132

4–2

4 22

kkkx

and i.s.w. if necessary.

(b) Accept part (b) solutions seen in part (a).

“4k 2 –11k – 3”= 0 (4k +1)(k – 3) = 0 k = …, M1

[Or, ‘starting again’, b2 – 4ac = (4k)2 – 4(3 +11k)

and proceed to k = …] 341– k (Ignore any inequalities

for the first 2 marks in (b)). A1

Using b2 – 4ac < 0 for no real roots, i.e. “4k2 –11k – 3”< 0,to establish inequalities involving their twocritical values m and n M1

(even if the inequalities are wrong, e.g. k < m, k < n).

341– k (See conditions below) Follow through

their critical values. A1ft 4

The final A1ft is still scored if the answer m< k < nfollows k <m, k < n.

(c)

Shape (seen in (c)) B1

Minimum in correct quadrant, not touching the x-axis, not on they-axis, and there must be no other minimum or maximum. B1

(0, 14) or 14 on y-axis.Allow (14, 0) marked on y-axis.

n.b. Minimum is at (–2,10), (but there is no mark for this). B1 3 [10]


C1 Coordinate geometry

1. (a) m = 147or

147,

)6(843or

86)3(4

21

M1, A1

Equation: )8(21)3(or ))6((

214 xyxy M1

x + 2y – 2 = 0 (or equiv. with integer coefficients… must have ‘= 0’) A1 4(e.g. 14y + 7x – 14 = 0 and 14 – 7x – 14y = 0 are acceptable)

(b) (–6 – 8)2 + (4 –(–3))2 M1142 + 72 or (–14)2 + 72 or 142 + (–7)2 (M1 A1 may be implied by 245) A1AB = 245or )12(7or 714 22222 A1cso 3

57[7]

2. (a) Putting the equation in the form y = mx (+c) and attempting to extract the m or mx (not the c) ,M1

or finding 2 points on the line and using the correct gradient formula.

Gradient = 53– (or equivalent) A1 2

(b) Gradient of perp. line = "53–"1–

(Using m1– with the m from part (a)) M1

3–"35"1– xy

M1

4–35– xy (Must be in this form... allow y=

312–

35 x but not y= )

312–5x

A1 3

This A mark is dependent upon both M marks. [5]

3. (a) y = 23

x(+ 4) Gradient = 23

M1A1 2

(b) 3x + 2 = 23

x + 4 x = ..., 94

M1, A1

y = 3

94

+ 2 =

313

310

A1 3

(c) Where y = 1, l1: xA = 31

l2: xB = 2 M: Attempt one of these M1A1

Area = 21

(xB – xA)(yP – 1) M1

= 18132

1849

37

37

21

o.e. A1 4

[9]


4. (a)

4 0 4 7 2 5ABm M1

Equation of AB is: 40 ( 2)5

y x or 44 75

y x (o.e.)

M1

4 x – 5 y – 8 = 0 ( o.e.) A1 3

(b) 22 0427 AB M1

= 41 A1 2

(c) Using isos triangle with AB = AC then t = 2 × yA = 2 × 4 = 8 B1 1

(d) Area of triangle = 12 (7 2)t M1

= 20 A1 2 [8]

5. (a) y – 5 = )2–(21– x or equivalent,

e.g. 21–

2–5–

xy

y = 621– x M1A1,A1cao 3

(b) x = –2 y = –21

(–2) + 6 = 7 (therefore B lies on the line) B1 1

(c) (AB2 =) (2 – –2)2 + (7 – 5)2, = 16 + 4 = 20, AB = 5220 M1, A1, A1 3

C is (p, 21– p + 6), so AC2 = (p – 2)2 +

2

5–621–

p M1

(d) Therefore 25 = p2 –4p + 4 + 2

41 p – p + 1 M1

25 = 1.25p2 – 5p + 5 or 100 = 5p2 – 20p + 20 (or better, RHS simplified A1to 3 terms) Leading to: 0 = p2 – 4p – 16 (*) A1cso 4

[11]

6. (a)

5

-2 .5

y

x

B1M1A1 3

B1 for curve of correct shape i.e 2 branches of curve, in correct quadrants, of roughly the correct shape and no touching or intersections with axes.


M1 for a straight line cutting the positive y-axis and the negative x-axis. Ignore any values.

A1 for (0, 5) and (–2.5, 0) or points correctly marked on axes.Do not give for values in tables.

(b) 2x + 5 = x3

M1

2x2 + 5x = 3[= 0] or 2x2 + 5x = 3 A1(2x – 1)(x + 3)[= 0] M1

x = –3 or 21

A1

y = 3

3

or 2 × (–3) + 5 or y = 213

or 2 × (21

) + 5 M1

Points are )6,21( and )1,3( (correct pairings) A1ft 6

[9]


C1 Differentiation and Integration

1. 4x3 kx2 or 21

21

2

kxx (k a non–zero constant) M1

12x2, 21

x ...., (–1 0) A1, A1, B1 4

[4]

2. x4 → kx3 or 03or32–

31

kxx (k a non-zero constant) M1

34dd x

xy

.........., with ‘3’ differentiated to zero (or ‘vanishing’) A1

xy

dd

............. + 32–

31 x or equivalent, e.g. 233 2 3

1or3

1

xxA1

[3]

3. (a) [f(x)= ] 3 + 3x2 M1A1 2

(b) 3 + 3x2 = 15 and start to try and simplify M1x2 =k → x = √k (ignore ) M1x = 2 (ignore × = –2) A1 3

[5]

4. (a) xxxx 22 )4(–12–12–9)4–3( M1

24–169 21

21

–xx A1, A1 3

(b) f’(x) = – 21

–23

–

216,

29 xx M1 A1, A1ft 3

(c) f ′(9) = 25

616

61–

31

216

271

29– M1 A1 2

[8]

5. (a) x = 2: y = 8 – 8 – 2 + 9 = 7 (*) B1 1

(b) 1–4–3dd 2 xx

xy

M1 A1

x = 2: )3(1–8–12dd

xy

A1ft

y – 7 = 3(x – 2), y = 3 x +1 M1, A1 5

(c)31–m (for

m1

with their m) B1ft

)o.e.(092–

34–or02–12–9,

31–1–4–3 222 xxxxxx M1, A1

6663621618

7214412

x or (3x – 2)2 6 → 3x = 2 6 M1

A1cso 5Glyn Technology School 24

6231

x

[11]

6. (a) 1–2

24–5–24–5– xxxxxy (or equiv., e.g. )24–8–3

xx M1 A1

2–241dd x

xy

or 2241

dd

xxy

M1 A1 4

(b) x = 2: y = –15 Allow if seen in part (a). B1

74241

dd

xy

B1ft

This must be simplified to a “single value”. y +15 = 7(x – 2) (or equiv., e.g. y = 7x – 29)

Allow 72–

15

xy

M1 A1 4

[8]

7.2

13 2 3 2x x xx

M1 A1

1

2 2224 , 2 , 3 2y x x x M1 A1

12 2224 2 3 2

x x x

A1 A1[6]

8. (a) 21

1

246

dd

xx

xy

or

21

26 xx M1A1 2

(b) 23

6

x or 23

16

x M1A1ft 2

(c) Cxx 23

3

38

A1: 33

x3 or

23

4 23

x A1: both, simplified and +C M1A1A1 3

[7]

9. 3x2 kx3 or 4x5 → kx6 or –7 → kx (k a non-zero constant) M1

64or

33 63 xx (Either of these, simplified or unsimplified) A1

x3 + 3

2 6x – 7x or equivalent unsimplified, such as 1

63

76

43

3 xxx A1

+ C (or any other constant, e.g. + K) B1 4[4]

10. (a) 3x2 cx3 or –6 cx or –8x–2 cx–1 M1

f(x) =

xxxCxxx 86)(

186

33 3

12

A1A1

Substitute x = 2 and y = 1 into a ‘changed function’ to form an


equation in C. M11 = 8 – 12 + 4 + C C = 1 A1cso

(b) 3 × 22 – 6 – 228

M1

= 4 A1Eqn. of tangent: y – 1 = 4(x – 2) M1y = 4x – 7 (Must be in this form) A1 4

[9]

11. (f(x) =) )(7–

23

3–3

3 23

3

cxxx M1

= )(7–2– 23

3 cxxx A1A1

f(4) = 22 22 = 64 – 16 – 28 + c M1

c = 2 A1cso 5 [5]

12. (a) 4x kx2 or 1

22

3 8or 6 kxx

kxx (k a non-zero constant) M1

f(x) = 2x2, 23

4x , –8x–1 (+ C)(+ C not required) A1, A1, A1

At x = 4, y = 1: 1 = (2 × 16) – C

12

34844 Must be in part (a) M1

C = 3 A1 6

(b) f(4) = 16 – (6 × 2) + 29

168

(= m)

(b)part in beMust

f given with the(4)fAttempt :M M1

Gradient of normal is

)(f their of use on theDependent rule. grad. perp.Attempt :M1

92

xm M1

Eqn. of normal: y – 1 = )4(92

x

or any equiv. form, e.g. 92

41

xy

) M1A1 4

[10]

13. (a) f(x) = 2

103

6 23 xx – 12x (+C) M1A1

x = 5: 250 – 125 – 60 + C = 65 C = 0 M1A1 4

(b) x(2x2 – 5x – 12) or (2x2 + 3x)(x – 4) or (2x + 3)(x2 – 4x) M1= x(2x + 3)(x – 4) (*) A1cso 2

(c)

– 4 – 3 – 2 – 1 1 2 3 4 5 6

4

2

– 2

– 4

– 6

x

y

Shape B1 Through origin B1


0,

23

and (4, 0) B1 3 [9]


C1 Sequences

1. (a) Identify a = 5 and d = 2 May be implied B1(u200 =)a + (200 – 1)d (= 5 + (200 – 1) × 2) M1= 430(p) or (£)4.30 A1 3

(b) (S200 =)2

200 [2a + (200 – 1)d] or 2

200 (a + “their 403”) M1

= 2

200[2 × 5 + (200 – 1) × 2] or

2200

(5 + “their 403”) A1

= 40 800 or £408 A1 3 [6]

2. (a) [x2 =] a – 3 B1 1

(b) [x3 = ] ax2 – 3 or a(a – 3) – 3 M1= a(a – 3) – 3 (*)= a2 – 3a – 3 (*) A1cso 2

(c) a2 – 3a – 3 = 7a2 – 3a – 10 = 0 or a2 – 3a = 10 M1(a – 5)(a + 2) = 0 dM1a = 5 or – 2 A1 3

[6]

3. (a) 2 4 3 7a B1

3 "their 7" 3 10a B1ft 2

(b) 4 10 3 13a M1

5 13 3 4a * A1 cso 2 [4]

4. (a) (a2 =)2k – 7 B1 1

(b) (a3 =)2(2k – 7) – 7 or 4k – 14 – 7, = 4k – 21 (*) M1, A1cso 2

(c) (a4 =)2(4k – 21) –7 (= 8k – 49) M1

4

1rra = k + “(2k – 7)” + (4k – 21) + “(8k – 49)” M1

k + (2k – 7) + (4k – 21) + (8k – 49) = 15k – 77 = 43 k = 8 M1 A1 4 [7]

5. (a) (a2 =)3 k + 5 [must be seen in part (a) or labelled a2 =] B1 1

(b) (a3 =)3(3k + 5) + 5 M1= 9 k + 20 (*) A1cso 2

(c) (i) a4 = 3(9k + 20) + 5 (= 27k + 65) M1

4

1rra = k + (3k + 5) + (9k + 20) + (27k + 65) M1


(ii) = 40k + 90 A1= 10(4 k + 9) (or explain why divisible by 10) A1ft 4

[7]

6. (a) 1(p + 1) or p +1 B1 1

(b) ((a))(p + (a)) [(a) must be a function of p]. [( p +1)(p + p + 1)] M1=1+ 3p + 2p2 (*) A1cso 2

(c) 1+ 3p + 2p2 =1 M1p(2p + 3) = 0 p = ... M1

p = 23

(ignore p = 0, if seen, even if ‘chosen’ as the answer) A1 3

(d) Noting that even terms are the same. M1

This M mark can be implied by listing at least 4 terms, e.g. 1, ,...21,1,

21

x2008 = 21

A1 2

[8]

7. (a) 5, 7, 9, 11 or 5 + 2 + 2 + 2 = 11 or 5 + 6 = 11use a = 5, d = 2, n = 4 and t4 = 5 + 3 × 2 = 11 B1 1

(b) tn = a + (n – 1)d with one of a = 5 or d = 2 correct(can have a letter for the other) M1= 5 + 2(n – 1) or 2n + 3 or 1 + 2(n + 1) A1 2

(c) Sn =2n [2 × 5 + 2(n – 1)] or use of

2n (5 + “their 2n + 3”)

(may also be scored in (b)) M1A1= {n(5 + n – 1)} = n(n + 4) (*) A1cso 3

(d) 43 = 2n + 3 M1[n] = 20 A1 2

(e) S20 =20 × 24, = 480 (km) M1A1 2[10]

8. (a) a +17d = 25 or equiv. (for 1st B1),a + 20d = 32.5 or equiv. (for 2nd B1), B1, B1 2

(b) Solving (Subtract) 3d = 7.5 so d = 2.5 M1a = 32.5 – 20 × 2.5 so a = –17.5 (*) A1cso 2

(c) 2750 = )]1–(2535[–

2nn

M1A1ft

{ 4 × 2750 = n(5n – 75) }

4 × 550 = n(n – 15) M1

n 2 – 15 n = 55 × 40 (*) A1cso 4

(d) n2 – 15n – 55 × 40 = 0 or n2 – 15n – 2200 = 0 M1(n – 55)(n + 40) = 0 n = . . . M1

n = 55 (ignore – 40)A1 3 [11]


C1 Transformations of curve

1. (a)2 0

1 0

– 1 0

– 2 0

y

x

– 3 – 2 – 1 1 2 3 4 5 6 7

(i) Shape or or B1

Max. at (0, 0). B1(2, 0), (or 2 shown on x-axis). B1 3

(ii) Shape B1(It need not go below x-axis)Through origin. B1(6, 0), (or 6 shown on x-axis). B1 3

(b) x2(x – 2) = x(6 – x) M1x3 – x2 – 6x = 0Expand to form 3-term cubic (or 3-term quadratic if divided by x),with all terms on one side. The “= 0” may be implied. M1x(x – 3)(x + 2) = 0 x = ... Factor x (or divide by x), and solve quadratic. M1x = 3 and x = –2x = –2 : y = –16 Attempt y value for a non-zero x value by M1

substituting back into x2 (x – 2) or x(6 – x).x = 3 : y = 9 Both y values are needed for A1. A1(–2,–16) and (3,9)(0, 0) This can just be written down. Ignore any ‘method’ shown.

(But must be seen in part (b)). B1 7[13]

2. (a)

1 0

(7 , 3 ) B1B1B1 3

1st B1 for moving the given curve up. Must be U shaped curve, minimum in first quadrant, not touching x-axis but cutting positive y-axis. Ignore any values on axes.

2nd B1 for curve cutting y-axis at (0, 10). Point 10(or even (10, 0) marked on positive y-axis is OK)

3rd B1 for minimum indicated at (7, 3). Must have both coordinates and in the right order.

(b)


(3 .5 , 0 )

7

y

x

B1B1 2

1st B1 for U shaped curve, touching positive x-axis and crossing y-axis at (0, 7)[condone (7, 0) if marked on positive y axis] or 7 marked on y-axis

2nd B1 for minimum at (3.5, 0) or 3.5 or 27

or marked on x-axis.

Allow “stopping at” (0, 10) or (0, 7) instead of “cutting”[5]

3. (a)

Shape , touching the x-axis at its maximum. M1

Through (0, 0) & – 3 marked on x-axis, or (–3,0) seen. A1

Allow (0, – 3) if marked on the x-axis. Marked in the correct place, but 3, is A0.

Min at (–1, –1) A1 3

(b)

Correct shape (top left – bottom right) B1

Through – 3 and max at (0, 0). Marked in the correct place, but 3, is B0. B1

Min at (–2, –1) B1 3 [6]


4. (a)(2 , 1 0 )

1 4

Shape: Max in 1st quadrant and 2 intersections on positive x-axis B1

1 and 4 labelled (in correct place) or clearly stated as coordinates B1

(2, 10) labelled or clearly stated B1 3

(b)

(– 2 , 5 )

– 4 – 1

Shape: Max in 2nd quadrant and 2 intersections on negative x-axis B1

–1 and –4 labelled (in correct place) or clearly stated as coordinates B1

(–2, 5) labelled or clearly stated B1 3

(c) (a = ) 2 May be implicit, i.e. f (x + 2) B1 1[7]

5. (a)

– 4 – 3 – 2 – 1 1 2 3 4x

y8

6

4

2

– 2

– 4Shape of f(x) B1Moved up ↑ M1Asymptotes: y = 3 B1x = 0 (Allow “ y –axis”) B1 4

(b) 31

x = 0 No variations accepted. M1

x = 31

(or – 0.33 …) Decimal answer requires at least 2 d.p. A1 2


(a) This scores B0 (clear overlap with horiz. asymp.) M1 (Upward translation… bod that both branches have been translated).

B 0 M 1 B 0 M 1 B 0 M 0

No marks unless the original curve is seen, to show upward translation.[6]

6. (a)6

4

2

– 2

– 4

– 6

– 6 – 4 – 2 2 4 6

Translation parallel to x-axis M1

overlap obvious Noonsintersecti no hasbranch Lower axis ve intersectsbranch Top y

A1

23,0 or

23

marked on y-axis B1 3

(b) x = –2, y = 0 B1, B1 2 [5]

7. (a) x(x2 – 6x + 9) B1= x(x – 3)(x – 3) M1 A1 3

(b)

Shape

Through origin (not touching) B1Touching x-axis only once B1Touching at (3, 0), or 3 on x-axis B1ft 4

(c)

Moved horizontally (either way) M1(2, 0) and (5, 0), or 2 and 5 on x-axis A1 2

Allow a fully correct graph (as shown above) to score M1A1 whatever they have in (b)[9]


c1 revision workbook

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