c4 questions from past papers differentiation · city of london academy 1 c4 questions from past...

City of London Academy 1

C4 QUESTIONS FROM PAST PAPERS – DIFFERENTIATION

1. A curve C has equation

2x + y2 = 2xy

Find the exact value of at the point on C with coordinates (3, 2).

(Total 7 marks)

2.

The diagram above shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48π m3 min–1. At time t minutes, the depth of the water in the tank is h metres. There is a tap at a point T at the bottom of the tank. When the tap is open, water leaves the tank at a rate of 0.6 h m3 min–1.

(a) Show that t minutes after the tap has been opened

(5)

When t = 0, h = 0.2

(b) Find the value of t when h = 0.5 (6)

(Total 11 marks)

3. The curve C has the equation

cos2x + cos3y = 1,

(a) Find in terms of x and y.

(3)

The point P lies on C where x = .

x

y

d

d

ht

h54

d

d75

60,

44

yx

x

y

d

d

6


(b) Find the value of y at P. (3)

(c) Find the equation of the tangent to C at P, giving your answer in the form ax + by + cπ = 0, where a, b and c are integers.

(3) (Total 9 marks)

4. The area A of a circle is increasing at a constant rate of 1.5 cm2 s–1. Find, to 3 significant figures, the rate at which the radius r of the circle is increasing when the area of the circle is 2 cm2.

(Total 5 marks)

5. The curve C has the equation ye–2x = 2x + y2.


(5)

The point P on C has coordinates (0, 1).

(b) Find the equation of the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4) (Total 9 marks)

6. A curve C has the equation y2 – 3y = x3 + 8.


(4)

(b) Hence find the gradient of C at the point where y = 3. (3)

(Total 7 marks)

x

y

d

d

x

y

d

d


7.

A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm, as shown in the diagram above. Water is flowing into the container. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm3.

(a) Show that

(2)

[The volume V of a right circular cone with vertical height h and base radius r is given by the

formula ]

Water flows into the container at a rate of 8 cm3 s–1.

(b) Find, in terms of π, the rate of change of h when h = 12. (5)

(Total 7 marks)

8.

The diagram above shows a right circular cylindrical metal rod which is expanding as it is heated. After t seconds the radius of the rod is x cm and the length of the rod is 5x cm. The cross-sectional area of the rod is increasing at the constant rate of 0.032 cm2 s–1.

.27

4 2hV

.3

1 2hrV

5x

x

33


(a) Find when the radius of the rod is 2 cm, giving your answer to 3 significant figures.

(4)

(b) Find the rate of increase of the volume of the rod when x = 2. (4)

(Total 8 marks)

9. A curve has equation 3x2 – y2 + xy = 4. The points P and Q lie on the curve. The gradient of the

tangent to the curve is at P and at Q.

(a) Use implicit differentiation to show that y – 2x = 0 at P and at Q. (6)

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

(Total 9 marks)

10. A curve is described by the equation

x3 – 4y2 = 12xy.

(a) Find the coordinates of the two points on the curve where x = –8. (3)

(b) Find the gradient of the curve at each of these points. (6)

(Total 9 marks)

11. A curve has parametric equations

x = 7cos t – cos7t, y = 7 sin t – sin 7t, .

(a) Find an expression for in terms of t. You need not simplify your answer.

(3)

(b) Find an equation of the normal to the curve at the point where

Give your answer in its simplest exact form. (6)

(Total 9 marks)

12. A set of curves is given by the equation sin x + cos y = 0.5.

t

x

d

d

3

8

38

t

x

y

d

d

.6

t


(a) Use implicit differentiation to find an expression for .

(2)

For –π < x < π and –π < y < π,

(b) find the coordinates of the points where = 0.

(5) (Total 7 marks)

13. (a) Given that y = 2x, and using the result 2x = exln 2, or otherwise, show that = 2x ln 2.

(2)

(b) Find the gradient of the curve with equation at the point with coordinates (2,16). (4)

(Total 6 marks)

14. A curve C is described by the equation

3x2 – 2y2 + 2x – 3y + 5 = 0.

Find an equation of the normal to C at the point (0, 1), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 7 marks)

15. f(x) = (x2 + 1) ln x, x > 0.

(a) Use differentiation to find the value of f'(x) at x = e, leaving your answer in terms of e. (4)

(b) Find the exact value of

(5) (Total 9 marks)

16. A curve C is described by the equation

3x2 + 4y2 – 2x + 6xy – 5 = 0.

Find an equation of the tangent to C at the point (1, –2), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 7 marks)

17. The volume of a spherical balloon of radius r cm is V cm3, where V = r3.

(a) Find

x

y

d

d

x

y

d

d

x

y

d

d

)( 2

2 xy

e

1d)(f xx

34

r

V

d

d


(1)

The volume of the balloon increases with time t seconds according to the formula

(b) Using the chain rule, or otherwise, find an expression in terms of r and t for

(2)

(c) Given that V = 0 when t = 0, solve the differential equation , to obtain V in

terms of t. (4)

(d) Hence, at time t = 5,

(i) find the radius of the balloon, giving your answer to 3 significant figures, (3)

(ii) show that the rate of increase of the radius of the balloon is approximately 2.90 × 10–2 cm s–1.

(2) (Total 12 marks)

18. The value £V of a car t years after the 1st January 2001 is given by the formula

V = 10 000 (1.5)–t.

(a) Find the value of the car on 1st January 2005. (2)

(b) Find the value of when t = 4.

(3)

(c) Explain what the answer to part (b) represents. (1)

(Total 6 marks)

19. A curve has equation

x2 + 2xy – 3y2 + 16 = 0.

Find the coordinates of the points on the curve where = 0.

(Total 7 marks)

.0,)12(

1000

d

d2

ttt

V

.d

d

t

r

2)12(

1000

d

d

tt

V

t

V

d

d

x

y

d

d


20. The curve C with equation y = k + ln 2x, where k is a constant, crosses the x-axis at the point A

.

(a) Show that k = 1. (2)

(b) Show that an equation of the tangent to C at A is y = 2ex – 1. (4)

(c) Complete the table below, giving your answers to 3 significant figures.

x 1 1.5 2 2.5 3

1 + ln 2x 2.10 2.61 2.79

(2)

(d) Use the trapezium rule, with four equal intervals, to estimate the value of

. (4)

(Total 12 marks)

21. f(x) = x + , x .

(a) Find f (x). (2)

The curve C, with equation y = f(x), crosses the y-axis at the point A.

(b) Find an equation for the tangent to C at A. (3)

(c) Complete the table, giving the values of to 2 decimal places.

x 0 0.5 1 1.5 2

0.45 0.91

(2)

(d) Use the trapezium rule, with all the values from your table, to find an approximation for the value of

0,

e2

1

3

1

d)2ln1( xx

5

e x

5

e x

x

5

e x

x


. (4)

(Total 11 marks)

22. A drop of oil is modelled as a circle of radius r. At time t

r = 4(1 – e–t), t > 0,

where is a positive constant.

(a) Show that the area A of the circle satisfies

= 32 (e–t – e–2t).

(5)

In an alternative model of the drop of oil its area A at time t satisfies

, t > 0.

Given that the area of the drop is 1 at t = 1,

(b) find an expression for A in terms of t for this alternative model. (7)

(c) Show that, in the alternative model, the value of A cannot exceed 4. (1)

(Total 13 marks)

23. The curve C has equation 5x2 + 2xy – 3y2 + 3 = 0. The point P on the curve C has coordinates (1, 2).

(a) Find the gradient of the curve at P. (5)

(b) Find the equation of the normal to the curve C at P, in the form y = ax + b, where a and b are constants.

(3) (Total 8 marks)

xxx

d5

e2

0

t

A

d

d

2

23

d

d

t

A

t

A


24.

The curve C with equation y = 2ex + 5 meets the y-axis at the point M, as shown in the diagram above.

(a) Find the equation of the normal to C at M in the form ax + by = c, where a, b and c are integers.

(4)

This normal to C at M crosses the x-axis at the point N(n, 0).

(b) Show that n = 14. (1)

The point P(ln 4, 13) lies on C. The finite region R is bounded by C, the axes and the line PN, as shown in the diagram above.

(c) Find the area of R, giving your answer in the form p + q ln 2, where p and q are integers to be found.

(7) (Total 12 marks)

25.

y

x

M

O

P

N

R

A

O x

y


The diagram above shows a graph of y = x sin x, 0 < x < . The maximum point on the curve is A.

(a) Show that the x-coordinate of the point A satisfies the equation 2 tan x + x = 0. (4)

The finite region enclosed by the curve and the x-axis is shaded as shown in the diagram above.

A solid body S is generated by rotating this region through 2 radians about the x-axis.

(b) Find the exact value of the volume of S. (7)

(Total 11 marks)

26. The function f is given by

f(x) = , x , x 2, x 1.

(a) Express f(x) in partial fractions. (3)

(b) Hence, or otherwise, prove that f(x) < 0 for all values of x in the domain. (3)

(Total 6 marks)

27. A curve has equation

x3 2xy 4x + y3 51 = 0.

Find an equation of the normal to the curve at the point (4, 3), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(Total 8 marks)

MARK SCHEME

1. B1

M1 A1 = A1

Substituting

M1

Accept exact equivalents M1 A17

[7]

)1)(2(

)1(3

xx

x

d

2 ln 2.2d

x x

x

d dln 2.2 2 2 2

d dx y y

y y xx x

3, 2

d d8ln 2 4 4 6

d d

y y

x x

d4ln 2 2

d

y

x


2. (a) M1 A1

B1

M1

Leading to cso A15

(b) separating variables M1

M1 A1

When ,

M1

When

awrt 10.4 M1 A1

Alternative for last 3 marks

= –15ln 1.5 + 15ln 3 M1 M1

= 15ln = 151n 2 awrt 10.4 A16

[11]

3. (a) –2 sin 2x – 3 sin 3y =0 M1 A1

Accept ,

d0.48 0.6

d

Vh

t

d d9 9

d d

V hV h

t t

d9 0.48 0.6

d

hh

t

d75 4 5

d

hh

t

75d 1d

4 5h t

h

15ln 4 5h t C

15ln 4 5h t C

0t 0.2h

15ln3 C

15ln3 15ln 4 5t h

0.5h

315ln 3 15ln1.5 15ln 15ln 2

1.5t

0.50.25h)– (4ln 15–t

5.1

3

x

y

d

d

y

x

x

y

3sin3

2sin2–

d

d

y

x

3sin3–

2sin2


A13

(b) At M1

A1

awrt 0.349 A13

(c) At , M1

M1

Leading to 6x + 9y–2π = 0 A13 [9]

4. B1

B1

When A = 2

M1

M1

awrt 0.299 A1

[5]

5. (a) e–2x A1 correct RHS *M1 A1

B1

y

x

3sin3

2sin2–

,6

x 13cos

6

2cos

y

2

13cos y

933

yy

9,

6

3

2

sin3

sin2

3sin3

2sin2–

d

d

3

3

9

6

x

y

6–

3

2–

9–

xy

5.1d

d

t

A

rr

ArA 2

d

d2

...797884.02

2 2

rr

t

r

r

A

t

A

d

d

d

d

d

d

t

rr

d

d25.1

299.02

5.1

d

d2

t

r

x

yyy

x

y x

d

d22e2–

d

d 2–

xxx yx

yx

2–2–2– e2–d

de)e(

d

d


(e–2x – 2y) *M1

A15

(b) At P, M1

Using mm′ = –1

M1

M1

x – 4y + 4 = 0 or any integer multiple A14

Alternative for (a) differentiating implicitly with respect to y.

e–2x – 2ye–2x A1 correct RHS *M1 A1

B1

(2 + 2ye–2x) e–2x – 2y *M1

A15

[9]

6. (a) C: y2 – 3y = x3 + 8 Differentiates implicitly to include either

±ky M1

Correct equation. A1

A correct (condoning sign error) attempt to

(2y – 3) combine or factorise their „ M1

xyx

y 2–e22d

d

y

y

x

yx

x

2–e

e22

d

d2–

–2

4–2–ºe

ºe22

d

d

x

y

4

1'm

)0–(4

11– xy

yy

x

y

x2

d

d2

d

d

y

xyy

yxxx

d

de2–e)e(

d

d 2–2–2–

y

x

d

d

x

x

y

y

y

x2–

2–

e22

2–e

d

d

y

y

x

yx

x

2–e

e22

d

d2–

2–

23d

d3–

d

d2 x

x

y

x

yy .)

d

dIgnore(.

d

d3or

d

d

x

y

x

y

x

y

23d

dx

x

y ,.

d

d3–

d

d2'

x

y

x

yy


Can be implied.

A1 oe4

(b) y = 3 9 – 3(3) = x3 + 8 Substitutes y = 3 into C. M1

x3 = – 8 x = – 2 A1

(–2, 3) Also can be ft using

their „x‟ value and y = 3 in the A1ft3

correct part (a) of

(b) final A1 . Note if the candidate inserts their x value and y = 3 into

, then an answer of their x2, may indicate

a correct follow through. [7]

7. (a) Similar triangles Uses similar triangles,

ratios or trigonometry to find either one of these M1 two expressions oe.

AG Substitutes

into the formula for the A12

volume of water V.

(b) From the question, B1

B1

Candidate‟s M1;

or or oe A1

3–2

3

d

d 2

y

x

x

y

3–2

3 2

y

x

rking.correct wo from4d

d

x

y

4d

d

3–6

)4(3

d

d

x

y

x

y

3–2

3

d

d 2

y

x

x

y

3–2

3

d

d 2

y

x

x

y

x

y

d

d

3

2

24

16 hr

h

r

27

4

3

2

3

1

3

1 322 h

hh

hrV

32hr

8d

d

t

V8

d

d

t

V

9

4

27

12

d

d 22 hh

h

V

9

4

27

12

d

d 22 hh

h

V

22

18

4

98

d

d

d

d

d

d

hhh

V

t

V

t

h

;

d

d

d

d

h

V

t

V

27

128

2h24

98

h

2

18

h


When h = 12, A1 oe isw5

Note the answer must be a one term exact value. Note, also you can ignore subsequent working

after

[7]

8. (a) From question, = 0.032

When x = 2 cm,

Hence, = 0.002546479... (cm s–1)

= 0.032 seen or implied from working. B1

2x by itself seen or implied from working B1

0.032 † Candidate‟s ; M1;

awrt 0.00255 A1 cso 4

(b) V = x2(5x) = 5x3

= 15x2

When x = 2 cm, = 0.24(2) = 0.48 (cm3 s–1)

V = x2(5x) = 5x3 B1

= 15x2 or ft from candidate‟s V in one variable B1ft

Candidate‟s ; M1ft

0.48 or awrt 0.48 A1 cso 4

8

1

144

18

d

d

t

h

8

1or

144

18

144

18

t

A

d

d

xxx

A

t

A

t

x

xx

AxA

016.0;

2

1)032.0(

d

d

d

d

d

d

2d

d2

2

016.0

d

d

t

x

t

x

d

d

t

A

d

d

x

A

d

d

x

V

d

d

}24.0{;016.0

.15d

d

d

d

d

d 2 xx

xt

x

x

V

x

V

t

V

d

d

x

V

d

d

t

x

x

V

d

d

d

d


[8]

9. (a) 3x2 – y2 + xy = 4 (eqn *)

not necessarily required.

giving –18x – 3y = 8x – 16y giving 13y = 26x Hence, y = 2x y – 2x = 0

Differentiates implicitly to include either

M1

Correct application of product rule B1

(3x2 – y2) → and (4 0) A1

Substituting into their equation. M1*

Attempt to combine either terms in x or terms in y together to give either ax or by. dM1*

simplifying to give y – 2x = 0 AG A1 cso 6

(b) At P & Q, y = 2x. Substituting into eqn * gives 3x2 – (2x)2 + x(2x) = 4 Simplifying gives, x2 = 4 x = ±2 y = 2x y = 4 Hence coordinates are (2, 4) and (–2, –4)

Attempt replacing y by 2x in at least one of the y terms in eqn * M1

Either x = 2 or x = –2 A1

Both (2, 4) and (–2, –4) A1 3 [9]

10. (a) x3 – 4y2 = 12xy (eqn *) x = –8 –512 – 4y2 = 12(–8)y –512 – 4y2= –96y

3

8

2

6

3

8

d

d

2

6

d

dor

2

6

d

d

0d

d

d

d26

yx

yx

x

y

xy

yx

x

y

yx

yx

x

y

x

yxy

x

yyx

)d

d (Ignore .

d

dor

d

d

x

y

x

yx

x

yky

)(

x

yyx

d

d26

3

8

d

d

x

y


Substitutes x = –8 (at least once) into * to obtain a three term quadratic in y. Condone the loss of = 0. M1

4y2 – 96y + 512 = 0 y2 – 24y + 128 = 0

(y – 16)(y – 8) = 0

y = 16 or y = 8.

An attempt to solve the quadratic in y by either factorising or by the formula or by completing the square. dM1

Both y = 16 and y = 8. or (–8, 8) and (–8, 16). A1 3

(b) 3x2 – 8y

@ (–8, 8), ,

@ (–8, 16), .

Differentiates implicitly to include either .

Ignore M1

Correct LHS equation; A1;

Correct application of product rule (B1)


Substitutes x = –8 and at least one of their y-values to attempt

to find any one of . dM1

One gradient found. A1

Both gradients of –3 and 0 correctly found. A1 cso 6

Aliter Way 2

2

)128(457624 y

x

yxy

x

y

d

d1212;

d

d

yx

yx

x

y

812

123

d

d 2

332

96

)8(8)8(12

)8(12)64(3

d

d

x

y

032

0

)16(8)8(12

)16(12)64(3

d

d

x

y

x

yx

x

yky

d

d12or

d

d

...d

d

x

y

x

y

d

d


@ (–8, 8), ,

@ (–8, 16), .


Ignore M1

Correct LHS equation; A1;

Correct application of product rule (B1)


Substitutes x = –8 and at least one of their y-values to attempt

to find any one of . dM1

One gradient found. A1

Both gradients of –3 and 0 correctly found. A1 cso 6

Aliter Way 3

x3 – 4y2 = 12xy (eqn *) 4y2 + 12xy – x3 = 0

x

y

xyy

x

yx 12

d

d12;8

d

d3 2

yx

yx

x

y

812

123

d

d 2

332

96

)8(8)8(12

)8(12)64(3

d

d

x

y

032

0

)16(8)8(12

)16(12)64(3

d

d

x

y

y

xy

y

xkx

d

d12or

d

d2

...d

d

y

x

y

x

x

y

d

dor

d

d

2

132

32

32

32

)9(2

1

2

3

8

9412

8

1614412

8

))(4(414412

xxxy

xxxy

xxxy

xxxy


@ x = –8

A credible attempt to make y the subject and an attempt to

differentiate either . M1

A1

A1

Substitutes x = –8 find any one of . dM1

One gradient correctly found. A1 Both gradients of –3 and 0 correctly found. A1 6

[9]

11. (a) x = 7cos t – cos 7t, y = 7 sin t – sin 7t,

Attempt to differentiate x and y with respect to t to give

in the form ± A sin t ± B sin 7t

in the form ± C cos t ± D cos 7t M1

Correct A1

Candidate‟s B1ft3

2

132

2

22

132

)9(4

318

2

3

d

d

)318(;)9(2

1

2

1

2

3

d

d

xx

xx

x

y

xxxxx

y

2

1

))512()64(9(4

)64(3)8(18

2

3

d

d

x

y

.0,32

3

2

3

d

d

32

48

2

3

)64(4

48

2

3

x

y

2

132 )9(

2

1or

2

3xxx

))g(()9(2

3

d

d2

132 xxxk

x

y

)318(;)9(2

1

2

1

2

3

d

d 22

132 xxxx

x

y

x

y

d

d

tt

tt

x

y

ttt

ytt

t

x

7sin7sin7

7cos7cos7

d

d

7cos7cos7d

d,7sin7sin7

d

d

t

x

d

d

t

y

d

d

t

y

t

x

d

d and

d

d

txty

dddd


Aliter Way 2

x = 7cos t – cos 7t, y = 7 sin t – sin 7t,

Attempt to differentiate x and y with respect to t to give

in the form ± A sin t ± B sin 7t

in the form ± C cos t ± D cos 7t M1

Correct A1

Candidate‟s B1ft3

(b) When t = ;

Hence m(N) = = awrt 0.58

When t = ,

N: y – 4 =

N:

or 4 =

Hence N:

ttt

tt

tt

tt

x

y

ttt

ytt

t

x

4tan)3sin4cos2(7

)3sin4sin2(7

7sin7sin7

7cos7cos7

d

d

7cos7cos7d

d,7sin7sin7

d

d

t

x

d

d

t

x

d

d

t

y

t

x

d

d and

d

d

txty

dddd

67

6

67

6

sin7sin7

cos7cos7

d

d)(m,

6

x

yT

1.73awrt3

7

37

27

27

237

237

3

1or

3

1

6

42

8

2

1

3

7

6

7sin

6sin7

342

38

2

3

2

37

6

7cos

6cos7

y

x

)34(3

1x

xyxyx

y 33or 3

3or

3

1

044)34(3

1 cc

xyxyxy 33or 3

3or

3

1


Substitutes t = or 30° into their expression; M1

to give any of the four underlined expressions oe (must be correct solution only) A1 cso

Uses m(T) to „correctly‟ find m(N). Can be ft from “their tangent gradient”. A1ftoe.

The point B1

Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient)x + “c”. M1

Correct simplified EXACT equation of normal.

This is dependent on candidate using correct A1 oe6

Aliter Way 2

When t = ;

Hence m(N) = = awrt 0.58

When t = ,

N: y – 4 =

N:

or 4 =

Hence N:

Substitutes t = or 30° into their expression; M1

6

x

y

d

d

4) 6.9,awrt (or )4,34(

)4,34(

6

4tan

d

d)(m,

6

x

yT

1.73awrt3)1(2

)1(2

21

23

3

1or

3

1

6

42

8

2

1

2

7

6

7sin

6sin7

342

38

2

3

2

37

6

7cos

6cos7

y

x

)34(3

1x

xyxyxy 33or 3

3or

3

1

044)34(3

1 cc

xyxyxy 33or 3

3or

3

1

6

x

y

d

d


to give any of the four underlined expressions oe (must be correct solution only) A1 cso

Uses m(T) to „correctly‟ find m(N). Can be ft from “their tangent gradient”. A1ftoe.

The point B1

Finding an equation of a normal with their point and their normal gradient or finds c by using y = (their gradient)x + “c”. M1

Correct simplified EXACT equation of normal.

This is dependent on candidate using correct A1 oe6

Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) ,

but obtains M0 if they write y – 4 = (x – ).

If they write, however, N: x = , then they can score M1.

Beware: A candidate finding an m(T) = can obtain A1ft for m(N) = 0,

and also obtains M1 if they write y – 4 = 0(x – ) or y = 4. [9]

12. (a) sin x + cos y = 0.5 (eqn *)

cos x – sin y = 0 (eqn #)

Differentiates implicitly to include ± sin y . (Ignore ( =).) M1

A1 cso2

(b)

giving or

When x =

When x =

cos y = 1.5 y has no solutions

4) 6.9,awrt (or )4,34(

)4,34(

34

34

34

x

y

d

d

y

x

x

y

sin

cos

d

d

x

y

d

d

x

y

d

d

y

x

sin

cos

0cos0sin

cos0

d

d x

y

x

x

y

2

x

2

x

5.0cos2

sin,2

y

5.0cos2

sin,2

y


cos y = – 0.5 y =

In specified range (x, y) =

Candidate realises that they need to solve „their numerator‟ = 0

…or candidate sets = 0 in their (eqn #) and attempts to solve

the resulting equation. M1ft

both or x = 90° or awrt x = ± 1.57 required here A1

Substitutes either their into eqn * M1

Only one of or 120°

or –120° or awrt –2.09 or awrt 2.09 A1

Only exact coordinates of A1

Do not award this mark if candidate states other coordinates inside the required range. A15

[7]

13. (a) Way 1

y = 2x = ex ln 2

M1

Hence AG A1 cso2

Aliter Way 2

ln y = ln(2x) leads to ln y = x ln 2

Hence AG

Takes logs of both sides, then uses the power law of logarithms…

3

2or

3

2

3

2,

2 and

3

2,

2

x

y

d

d

2,

2

x

2or

2

xx

3

2-or

3

2 y

3

2,

2 and

3

2,

2

2lne.2lnd

d x

x

y

2ln2)2.(2lnd

d xx

x

y

2lnd

d1

x

y

y

2ln22lnd

d xyx

y


... and differentiates implicitly to give M1

2x ln 2 AG A1 cso2

(b)

When x = 2,

= 64 ln 2 = 44.3614...

M1

or 2x.y.ln 2 if y is defined A1

Substitutes x = 2 into their which is of the form M1

64 ln 2 or awrt 44.4 A14

Aliter Way 2

ln y = leads to ln y = x2 ln 2

When x = 2,

= 64 ln 2 = 44.3614...

M1

A1

Substitutes x = 2 into their which is of the form M1

64 ln 2 or awrt 44.4 A14 [6]

14.

2lnd

d1

x

y

y

2ln.2.2d

d2 )()( 22 xx x

x

yy

2ln2)2(2d

d 4x

y

x

y

d

d

)( 2

2 xAx

2ln.2.2 )( 2xx

x

y

d

d )()( 22

2or 2 xx Axk

)2ln(2x

2ln.2d

d1x

x

y

y

2ln2)2(2d

d 4x

y

x

y

d

d

2ln.d

d1Ax

x

y

y

2ln.2d

d1x

x

y

y

x

y

d

d )()( 22

2or 2 xx Axk

dydx = 0

d

d32

d

d46

x

y

x

yyx



(ignore .) M1



At (0, 1),

Substituting x = 0 & y = 1 into an equation involving dM1

to give or A1 cso

Hence m(N) or A1ftoe.

Uses m(T) to „correctly‟ find m(N).

Can be ft from “their tangent gradient”.

Either N: y – 1 = – (x – 0)

or: N:

+ with „their tangent or normal gradient‟; or uses y = mx + 1 with „their tangent or normal gradient‟ ; M1;

N: 7x + 2y 2 = 0 Correct equation in the form „ax + by + c = 0‟, A1 oe where a, b and c are integers. cso

[7]

Beware: does not necessarily imply the award of all the first four marks in this question.

So please ensure that you check candidates‟ initial differentiation before awarding the first A1

mark.

Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0.

Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) = , but obtains M0 if they write y 1 = ( x 0). If they write, however, N: x = 0, then can score M1.

Beware: A candidate finding an m(T) = can obtain A1ft for m(N) = 0, and also obtains M1 if they write y 1 = 0(x 0)or y = 1.

Beware: The final cso refers to the whole question.

x

yor

x

yky

d

d3

d

d

x

y

d

d

34

26

d

d

y

x

x

y

7

2

34

20

d

d

x

y

;d

d

x

y

7

2

7

2

2

7

72

1

2

7

127 xy

)0(1 xmy

7

2

d

d

x

y


Aliter Way 2

Differentiates implicitly to include either

(ignore .) M1



At (0, 1),

Substituting x = 0 & y = 1 into an equation involving ; dM1

to give A1 cso

Hence m(N) = or A1ftoe.

Uses m(T) or to „correctly‟ find m(N).

Can be ft using “1 . ”.

Either N: y 1 = M1

or N: y =

y 1 = m(x 0) with „their tangent, or normal gradient‟;

or uses y = mx + 1 with „their tangent, or normal gradient‟;

N: 7x + 2y 2 = 0 Correct equation in the form „ax + by + c = 0‟, A1oe where a, b and c are integers. cso

[7]

Aliter Way 3

2y2 + 3y 3x2 2x – 5 = 0

dxdy = 03

d

d24

d

d6

y

xy

y

xx

y

x

y

xkx

d

d2or

d

d

y

x

d

d

26

34

d

d

x

y

y

x

2

7

20

34

d

d

y

x

y

x

d

d

27

2

7

72

1

y

x

d

d

y

x

d

d

)0(27 x

127 x

y

x

d

d

y

x

d

d

25

23

169

43 22

xy x

43

1649

23 )

2

xy x


Differentiates using the chain rule; M1

Correct expression for ; A1 oe

At (0, 1),

Substituting x = 0 into an equation involving ; dM1

to give or A1 cso

Hence m(N) = A1ft

Uses m(T) to „correctly‟ find m(N). Can be ft from “their tangent gradient”.

Either N: y 1

or N: y M1

y 1 = m(x 0) with „their tangent or normal gradient‟; or uses y = mx + 1 with „their tangent or normal gradient‟

N: 7x + 2y 2 = 0 A1 oe Correct equation in the form „ ax + by + c = 0‟, where a, b and c are integers.

[7]

15. (a) f(x) = (x2 + 1) × + ln x × 2x M1 A1

f(e) = (e + 1)× + 2e = 3e + M1 A1 4

(b) M1 A1

=

= A1

= M1 A1 5

[9]

16. Differentiates

)13(2

1

d

d 21

2

1649

23

xxx

y x

x

y

d

d

7

2

7

4

2

1

16

49

2

1

d

d 21

x

y

x

y

d

d

72

72

2

7

)0(27 x

172 x

x1

e1

e1

dxx

xxxxx 1)3

(1n)3

(33

dxxxxx )13

(1n)3

(33

e

xxxxx

1

33

)9

(1n)3

(

910

92 3 e


to obtain: 6x + 8y – 2,

.................................. + (6x + 6y) = 0 +(B1)

Substitutes x = 1, y = –2 into expression involving , to give = – M1, A1

Uses line equation with numerical „gradient‟ y – (– 2) = (their gradient) M1 (x – 1) or finds c and uses y = (their gradient) x + “c”

To give 5y + 4x + 6 = 0 (or equivalent = 0) A1ft [7]

17. (a) B1 1

(b) Uses in any form, = M1, A1 2

(c) V = (2t + l)–2 dt and integrate to p(2t + 1)–1,

= – 500(2t + 1)–1 (+c) M1, A1

Using V = 0 when t = 0 to find c , (c = 500, or equivalent) M1

V = 500(1 – ) (any form) A1 4

(d) (i) Substitute t = 5 to give V,

then use r = to give r , = 4.77 M1, A1 3

(ii) Substitutes t = 5 and r = „their value‟ into „their‟ part (b) M1

(≈ 2.90 × 10–2) (cm/s) AG A1 2

[12]

18. (a) Substitutes t = 4 to give V, = 1975.31 or 1975.30 or 1975 or 1980 (3 s.f) M1, A1 2

(b) = –ln1.5 × V; = –800.92 or –800.9 or –801 M1 A1; A1 3

M1 needs ln 1.5 term

(c) rate of decrease in value on 1st January 2005 B1 1 [6]

x

y

d

d

x

y

d

d

yxyx

xy

86662

dd

x

y

d

d

x

y

d

d

10

8

24 rdrdV

dVdr

dtdV

dtdr .

22 )12(41000

tr

1000

121t

343

V

0289.0dd

tr

t

V

d

d


19. 2x + = 0 M1 (A1) A1

= 0 x + y = 0 (or equivalent) M1

Eliminating either variable and solving for at least one value of x or y. M1 y2 – 2y2 – 3y2 + 16 = 0 or the same equation in x y = 2 or x = 2 A1 (2 – 2), (–2, 2) A1 7

Note:

Alternative:

3y2 – 2xy – (x2 + 16) = 0

y =

M1 A1 A1

= 0 = 1 M1

64x2 = 16x2 + 192 M1 A1 x = 2 A1 7 (2, –2), (–2, 2)

[7]

20. (a) 0 = k + ln 2 0 = k – 1 k = 1 (*) M1 A1 2

(Allow also substituting k = 1and x = into equation and

showing y = 0 and substituting k = 1 and y = 0 and showing x =

.)

(b) B1

At A gradient of tangent is = 2e M1

Equations of tangent: y = 2e M1

Simplifying to y = 2ex – 1 (*) cso A1 4

(c) y1 = 1.69, y2 = 2.39 B1, B1 2

x

yyy

x

yx

d

d62

d

d2

x

y

d

d

xy

yx

x

y

3d

d

6

)19216(2 2 xx

)19216(

8.

3

1

3

1

d

d2

x

x

x

y

x

y

d

d

)19216(

82 x

x

e2

1

e2

1

e2

1

xx

y 1

d

d

e211

e2

1x


(d) B1

… × (1.69 + 2.79 + 2(2.10 + 2.39 + 2.61)) ft their (c) M1 A1ft 4.7 A1 4

accept 4.67 [12]

21. (a) Differentiating; f(x) = 1 + M1;A1 2

(b) A: B1

Attempt at y – f(0) = f(0)x; M1

y – x or equivalent “one line” 3 termed equation A1 ft 3

(c) 1.24, 1.55, 1.86 B2(1,0) 2

(d) Estimate = ; (×) [(0.45 + 1.86) + 2(0.91 + 1.24 + 1.55)] B1 M1 A1 ft

= 2.4275 A1 4

[11]

22. (a) A = r2, = 4e–t B1, B1

= 2r , = 2 × 4(1 – e–t) × 4e–t M1, M1

= 32( e–t – e–2t) A1cso 5

(b) = M1

Separation

= (+ c) M1, A1

–2 = –1 + c Use of (1, 1) M1 c = –1 A1

So 2A = + 1 = Attempt = or A = M1

i.e. A = (or equivalent) A1 7

3

1(...)

2

1

2

1d)2ln1( xx

5

e x

5

1,0

5

6

5

1

2

5.0

2.43,

2.429

2.428

t

r

d

d

t

A

d

d

t

r

d

d

t

A

d

d

t

A

d

d

AA d23

tt d2

21

21

A

1

1

t

21

t

1A

t

t

1

2A

2

2

)1(

4

t

t


(c) Because < 1 or t2 < (1 + t)2 ( A < 4) B1 1

[13]

23. (a) 10x, +(2y + 2x ), –6y = 0 M1, (B1), A1

At (1 , 2) 10 + (4 + 2 ) – 12 = 0 M1

= 1.4 or or 1 A1 5

(b) The gradient of the normal is M1

Its equation is y – 2 = (x – 1) M1

(allow tangent)

y = x + 2 or y = x + A1cao 3

[8]

24. (a) M is (0, 7) B1

= 2ex M1

Attempt

gradient of normal is – M1

ft their y(0) or = –

(Must be a number)

equation of normal is y – 7 = – (x – 0) or x +2y – 14 = 0

x + 2y = 14 o.e. A1 4

(b) y = 0, x = 14 N is (14, 0) (*) B1 cso 1

(c)

(2ex + 5) dx = [2ex + 5x] M1

some correct

2

2

)1( t

t

x

y

d

d

x

y

d

d

x

y

d

d

x

y

d

d

10

14

d

d

x

y

5

752

7

5

7

5

7

5

7

5

7

5

7

19

x

y

d

d

x

y

d

d

21

2

1

21

R T1


R1 = (2ex + 5) dx = (2 × 4 + 5 ln 4) – (2 + 0) M1

limits used

= 6 + 5 ln 4 A1

T = × 13 × (14 – ln 4) B1

Area of T

T = 13(7 – ln2) ; R1 = 6 + 10 ln 2 B1

Use of ln 4 = 2ln 2

R = T + R1, R = 97 – 3 ln 2 M1, A1 7 [12]

25. (a) M1, A1

At A = 0 dM1

sin x + cos x = 0 (essential to see intermediate line before given answer)

2tan x + x = 0 (*) A1 4

(b) V = y2dx = x2 sin xdx M1

= M1 A1

= M1

= A1

= [2 – 2 – 2] M1 = [2 – 4] A1 7

[11]

26. (a) , and correct method for finding A or B M1

A = 1, B = 2 A1, A1 3

(b) f(x) = M1 A1

Argument for negative, including statement that square terms are positive for all values of x. (f.t. on wrong values of A and B) A1 ft 3

[6]

27. Differentiates w.r.t. x to give

4ln

0

21

xxx

xx

ycos)(sin

2sin

d

d 2

1

xxx

x cos)(sin2

sin 2

1

2

x

0

2 dcos2cos xxxxx

0

2 dsin 2sin2cos xxxxxx

02 cos 2sin2cos xxxxx

12)1)(2(

)1(3

x

B

x

A

xx

x

22 )1(

2

)2(

1

xx


3x2, 2x + 2y, 4 + 3y2 = 0 M1, B1, A1

At (4, 3)

48 (8y + 6) 4 + 27y = 0 M1

y = = 2 A1

Gradient of normal is M1

y 3 = (x 4) M1

i.e. 2y 6 = x 4

x – 2y + 2 = 0 A18 [8]

============================================================================= EXAMINERS‟ REPORTS

1. This question was also well answered and the general principles of implicit differentiation were well understood. By far

the commonest source of error was in differentiating 2x; examples such as 2

x, 2

x lnx and x2

x–1 were all regularly seen.

Those who knew how to differentiate nearly always completed the question correctly, although a few had

difficulty in finding correctly. A minority of candidates attempted the question by taking the logs of both

sides of the printed equation or a rearrangement of the equation in the form 2x = 2xy – y

2. Correctly done, this leads to

quite a neat solution, but, more frequently, errors, such as ln(2x + y

2) = ln 2

x + ln y

2, were seen.

It was noteworthy that a number of correct solutions were seen using partial differentiation, a topic which is not in the A level Mathematics or Further Mathematics specifications. These were, of course, awarded full marks.

2. Many found part (a) difficult and it was quite common to see candidates leave a blank space here and proceed to solve part (b), often correctly. A satisfactory proof requires summarising the information given in the question in an equation,

such as = 0.48π – 0.6πh, but many could not do this or began with the incorrect = 0.48π – 0.6πh. Some

also found difficulty in obtaining a correct expression for the volume of water in the tank and there was some confusion

as to which was the variable in expressions for the volume. Sometimes expressions of the form were differentiated with respect to r, which in this question is a constant. If they started appropriately, nearly all candidates could use the chain rule correctly to complete the proof.

Part (b) was often well done and many fully correct solutions were seen. As noted in the introduction above, some poor algebra was seen in rearranging the equation but, if that was done correctly, candidates were nearly always able to demonstrate a complete method of solution although, as expected, slips were made in the sign and the constants when integrating. Very few candidates completed the question using definite integration. Most used a constant of integration (arbitrary constant) and showed that they knew how to evaluate it and use it to complete the question.

3. As has been noted in earlier reports, the quality of work in the topic of implicit differentiation has improved in recent

years and many candidates successfully differentiated the equation and rearranged it to find . Some, however,

x

y

d

d

x

y

d

d

19

38

2

1

2

1

2x

)2(d

dxy

x

T

V

d

d

t

h

d

d

2V r h

x

y

d

d


forgot to differentiate the constant. A not infrequent, error was candidates writing = –2sin 2x – 3sin3y and

then incorporating the superfluous on the left hand side of the equation into their answer. Errors like

(cos3y) = – .

were also seen. Part (b) was very well done. A few candidates gave the answer 20° , not recognising that the question required radians. Nearly all knew how to tackle part (c) although a few, as in Q2, spoilt otherwise completely correct solutions by not giving the answer in the form specified by the question.

4. Connected rates of change is a topic which many find difficult. The examiners reported that the responses to this question were of a somewhat higher standard than had been seen in some recent examinations and the majority of candidates attempted to apply the chain rule to the data of the question. Among those who obtained a correct relation,

1.5 = 2πr or an equivalent, a common error was to use r = 2 , instead of using the given A = 2 to obtain r =

.

Unexpectedly the use of the incorrect formula for the area of the circle, A = 2πr2

, was a relatively common error.

5. As noted above work on this topic has shown a marked improvement and the median mark scored by candidates on this

question was 8 out of 9. The only errors frequently seen were in differentiating implicitly with respect to x. A

few candidates failed to read the question correctly and found the equation of the tangent instead of the normal or failed to give their answer to part (b) in the form requested.

6. A significant majority of candidates were able to score full marks on this question. In part (a), many candidates were able to differentiate implicitly and examiners noticed fewer candidates differentiating 8 incorrectly with respect to x to give 8. In part (b), many candidates were able to substitute y = 3 into C leading to the correct x-coordinate of –2. Several candidates either rearranged their C equation incorrectly to give x = 2 or had difficulty finding the cube root of –8. Some

weaker candidates did not substitute y = 3 into C, but substituted y = 3 into the expression to give a gradient of x2

.

7. A considerable number of candidates did not attempt part (a), but of those who did, the most common method was to

use similar triangles to obtain and substitute r into V = to give Some candidates

used trigonometry to find the semi-vertical angle of the cone and obtained from this. A few candidates

correctly used similar shapes to compare volumes by writing down the equation

Part (b) discriminated well between many candidates who were able to gain full marks with ease and some candidates

who were able to gain just the first one or two marks. Some incorrectly differentiated V = to give

Most of the successful candidates used the chain rule to find by applying The

final answer was sometimes carelessly written as . Occasionally, some candidates solved the differential

equation and equated their solution to and then found or differentiated implicitly to find

.

8. At the outset, a significant minority of candidates struggled to extract some or all of the information from the question.

These candidates were unable to write down the rate at which this cross-sectional area was increasing, = 0.032;

or the cross-sectional area of the cylinder A = x2

and its derivative = 2πx; or the volume of the cylinder V = 5πx3

x

y

d

d

x

y

d

d

x

y

d

d

x

y

d

d

y3sin3

1

t

r

d

d

2

xey 2–

xy

dd

32hr hr 2

31 3

274 hV

32hr

.

242416

3

231

hV

hr 231

.3

1

d

d 2rh

V

t

h

d

d.

d

d

d

d

h

V

t

V

8

1

8

1

8d

d

t

V

27

4 3h

h

t

d

d

t

h

d

d

t

A

d

d

t

A

d

d


and its derivative = 15πx2

.

In part (a), some candidates wrote down the volume V of the cylinder as their cross-sectional area A. Another popular error at this stage was for candidates to find the curved surface area or the total surface area of a cylinder and write

down either A = 10πx2

or A = 12πx2

respectively. At this stage many of these candidates were able to set up a correct

equation to find and usually divided 0.032 by their and substituted x = 2 into their expression to gain 2 out

of the 4 mark available. Another error frequently seen in part (a) was for candidates to incorrectly calculate as

0.0251. Finally, rounding the answer to 3 significant figures proved to be a problem for a surprising number of candidates, with a value of 0.003 being seen quite often; resulting in loss of the final accuracy mark in part (a) and this sometimes as a consequence led to an inaccurate final answer in part (b).

Part (b) was tackled more successfully by candidates than part (a) – maybe because the chain rule equation

is rather more straight-forward to use than the one in part (a). Some candidates struggled by

introducing an extra variable r in addition to x and obtained a volume expression such as V = πr2

(5x). Many of these candidates did not realise that r ≡ x and were then unable to correctly differentiate their expression for V. Other

candidates incorrectly wrote down the volume as V = 2πx2

(5x). Another common error was for candidates to state a

correct V, correctly find , then substitute x = 2 to arrive at a final answer of approximately 188.5.

About 10% of candidates were able to produce a fully correct solution to this question.

9. This question was generally well done with a majority of candidates scoring at least 6 of the 9 marks available.

In part (a), implicit differentiation was well handled with most candidates appreciating the need to apply the product

rule to the xy term. A few candidates failed to differentiate the constant term and some wrote “ = ...” before starting

to differentiate the equation. After differentiating implicitly, the majority of candidates rearranged the resulting

equation to make the subject before substituting as rather than substituting for in their

differentiated equation. Many candidates were able to prove the result of y – 2x = 0. A surprising number of candidates

when faced with manipulating the equation , separated the fraction to incorrectly form two equations 6x

+ y = 8 & 2y – x = 3 and then proceeded to solve these equations simultaneously.

Some candidates, who were unsuccessful in completing part (a), gave up on the whole question even though it was still possible for them to obtain full marks in part (b). Other candidates, however, did not realise that they were expected to substitute y = 2x into the equation of the curve and made no creditable progress with this part. Those candidates who

used the substitution y = 2x made fewer errors than those who used the substitution x = . The most common errors in

this part were for candidates to rewrite – y2

as either 4x2 or –2x

2; or to solve the equation x

2 = 4 to give only x = 2 or

even x = ±4. On finding x = ±2, some candidates went onto substitute these values back into the equation of the curve, forming a quadratic equation and usually finding “extra” unwanted points rather than simply doubling their two values

of x to find the corresponding two values for y. Most candidates who progressed this far were able to link their values of x and y together, usually as coordinates.

10. This question was generally well done with many candidates scoring at least seven or eight of the nine marks available.

In part (a), the majority of candidates were able to use algebra to gain all three marks available with ease. It was disappointing, however, to see a significant minority of candidates at A2 level who were unable to correctly substitute y = –8 into the given equation or solve the resulting quadratic to find the correct values for y.

In part (b), implicit differentiation was well handled, with most candidates appreciating the need to apply the product rule to the 12xy term although errors in sign occurred particularly with those candidates who had initially rearranged the given equation so that all terms were on the LHS. A few candidates made errors in rearranging their correctly

x

V

d

d

t

x

d

d

x

A

d

d

4

032.0

t

x

x

V

t

V

d

d

d

d

d

d

x

V

d

d

x

y

d

d

x

y

d

d

x

y

d

d

3

8

3

8

x

y

d

d

3

8

2

6

xy

yx

2

y


differentiated equation to make the subject. Also some candidates lost either one or two marks when

manipulating their correctly substituted expressions to find the gradients.

11. In part (a), many candidates were able to apply the correct formula for finding in terms of t. Some candidates

erroneously believed that differentiation of a sine function produced a negative cosine function and the differentiation of a cosine function produced a positive sine function.

Other candidates incorrectly differentiated cos 7t to give either sin 7 t or –sin 7t and also incorrectly differentiated

sin 7t to give either cos 7t or cos 7t.

In part (b), many candidates were able to substitute into their gradient expression to give , but it was

not uncommon to see some candidates who made errors when simplifying their substituted expression. The majority of

candidates were able to find the point ( , 4). Some candidates, however, incorrectly evaluated cos ( ) and sin

( ) as and respectively and found the incorrect point ( , 3). Some candidates failed to use the

gradient of the tangent to find the gradient of the normal and instead found the equation of the tangent, and so lost valuable marks as a result. It was pleasing to see that a significant number of candidates were able to express the equation of the normal in its simplest exact form.

12. In part (a), the majority of candidates were able to successfully differentiate the given equation to obtain a correct

expression for , although there were a small proportion of candidates who appeared to “forget” to differentiate the

constant term of 0.5. Some candidates, as was similar with Q3, produced a sign error when differentiating sin xand cos

y with respect to x. These candidates then went on to produce the correct answer for , but lost the final accuracy

mark. A few candidates incorrectly believed that the expression could be simplified to give cot x.

In part (b), the majority of candidates realised that they needed to set their numerator equal to zero in order to solve

= 0. Most candidates were then able to obtain at least one value for x, usually x = , although x = was not

always found. A surprising number of candidates did not realise that they then needed to substitute their x value(s) back into the original equation in order for them to find y. Of those who did, little consideration was given to find all the

solutions in the specified range, with a majority of these candidates finding y = , but only a minority of candidates

also finding y = . Therefore it was uncommon for candidates to score full marks in this part. Some candidates

also incorrectly set their denominator equal to zero to find extra coordinates inside the range. Also another small

minority of candidates stated other incorrect coordinates such as or in addition to the

two sets of coordinates required. These candidates were penalised by losing the final accuracy mark.

13. In part (a), candidates either replaced 2x with e

xln 2 and applied the chain rule; or took logs of both sides of the given

equation and then differentiated implicitly. A majority of the candidates were equally likely to correctly apply either one of these two methods. Weaker candidates, however, seemed oblivious to the fact that x ln 2and ln 2xare, in fact, different, and wrote them almost interchangeably.

x

y

d

d

x

y

d

d

x

y

d

d

7

1

7

1

6

t 3

346

7

6

7

2

3

2

133

x

y

d

d

x

y

d

d

y

x

sin

cos

x

y

d

d

2

2

3

2

3

2

3

2,

2

3

2,

2


Part (b) proved challenging for a significant number of candidates. Those candidates who used implicit differentiation in parts (a) and (b) were more likely to achieve the correct gradient. Such an approach avoided the errors seen when

candidates were trying to handle indices. Such errors included either = 2x.x

= 2x.2

x or =2

x.2

2 = 4.2

x or

. Another common error was for some candidates to argue that since the derivative of 2xis 2.ln 2, then

the derivative of must be ln 2.

14. This question was successfully completed by the majority of candidates. Whilst many demonstrated a good grasp of the idea of implicit differentiation there were a few who did not appear to know how to differentiate implicitly. Candidates

who found an expression for in terms of x and y, before substituting in values of x= 1 and y = 1, were prone to

errors in manipulation. Some candidates found the equation of the tangent and a number of candidates did not give the equation of the normal in the requested form.

15. The product rule was well understood and many candidates correctly differentiated f(x) in part (a). However, a significant number lost marks by failing to use ln e = 1 and fully simplify their answer.

Although candidates knew that integration by parts was required for part (b), the method was not well understood with

common wrong answers involving candidates mistakenly suggesting that and attempting to use u = x2

+ 1 and in the formula .

Candidates who correctly gave the intermediate result often failed to use

a bracket for the second part of the expression when they integrated and went on to make a sign error by giving

rather than .

16. `This question was generally well answered with most candidates showing good skills in differentiating explicitly.

Candidates who found an expression for in terms of x and y, before substituting in values, were more prone to

errors in manipulation. Some candidates found the equation of the normal and a number of candidates did not give the equation of the tangent in the requested form. It was quite common to see such statements as

, but often subsequent correct working indicated that this was just poor

presentation.

17. The fact that this question had so many parts, with a good degree of independence, did enable the majority of candidates to do quite well. All but the weakest candidates scored the first mark and the first 3 were gained by most. The integration in part (c ) did cause problems: examples of the more usual mistakes were to write

, or to omit the constant of integration or assume it

equal to zero; two of the mistakes which came more into the “howler” category were

and

.

Many candidates were able to gain the method marks in parts (d) and (e).

18. (a) Most understood the context of this problem and realised that they needed to use t = 4, although t = 0, 1 or 5 were often seen.

)( 2

2 x )( 2

2 x

2)( )2(22 xx

)( 2

2 x )( 2

2 x

dx

dy

x

xx1

dln

xx

vln

d

d x

x

uvuvx

x

vu d

d

dd

d

d

xx

xx

xxx

d1

3ln

3

e

l

3e

l

3

xx

9

3

xx

9

3

y

x

d

d

06d

d62–

d

d86

d

d

y

x

yx

x

yyx

y

x

32 )12(or

12

2or

12

1or

12

1d

)12(

1

t

k

tttt

t

1000ln1000lnd1000

1VorVV

........d1

1

4

1

4

1d

144

1d

)12(

1222

t

ttt

ttt

t


(b) Very few had any idea at all about how to differentiate V (many gave their answer as

–t(1.5)-t-1

, or had a term (1.5)-2t

).

(c) The comments made in answer to the request to interpret their answer to part (b) were usually too generalised

and vague. The examiners required a statement that the value of which had been found represented the

rate of change of value on 1st

January 2005

19. Almost all candidates could start this question and the majority could differentiate implicitly correctly. This is an area

of work which has definitely improved in recent years. Many, having found , could not use it and it was disturbing

to find a substantial number of students in this relatively advanced A2 module proceeding from to x + y

= 3y – x. Those who did obtain y = –x often went no further but those who did could usually obtain both correct points, although extra incorrect points were often seen.

20. In part (a), the log working was often unclear and part (b) also gave many difficulty. The differentiation was often

incorrect. was not unexpected but expressions like were also seen. Many then failed to substitute

into their and produced a non-linear tangent. Parts (c) and (d) were well done. A few did, however,

give their answers to an inappropriate accuracy. As the table is given to 2 decimal places, the answer should not be given to a greater accuracy.

21. For many candidates this was a good source of marks. Even weaker candidates often scored well in parts (c) and (d). In

part (a) there were still some candidates who were confused by the notation, often interpreted as , and

common wrong answers to the differentiation were and 1 + . The most serious error, which occurred far too

frequently, in part (b) was to have a variable gradient, so that equations such as = x were common.

The normal, rather than the tangent, was also a common offering.

22. There were two common approaches used in part (a); substituting for r to obtain a formula for A in terms of t or using

the chain rule. The inevitable errors involving signs and were seen with both methods and the examiners were disappointed that some candidates did not seem to know the formula for the area of a circle:

were common mistakes. Part (b) proved more testing. Most could separate the variables but

the integration of negative powers caused problems for some who tried to use the ln function. Many did solve the differential equation successfully though sometimes they ran into difficulties by trying to make A the subject before finding the value of their arbitrary constant. The final two marks were only scored by the algebraically dexterous. There

was some poor work here and seeing followed by was not

uncommon. The final part eluded most. Those who had a correct answer to part (b) sometimes looked at the effect on A

of t but only a small minority argued that since t >0 then , and therefore A <4.

23. This was usually well done, but differentiation of a product caused problems for a number of candidates. Many still

insisted on making the subject of their formula before substituting values for x and y. This often led to

unnecessary algebraic errors.

24. Whilst the majority of answers to part (a) were fully correct, some candidates found difficulties here. A small number failed to find the coordinates of M correctly with (0, 5) being a common mistake. Others knew the rule for perpendicular gradients but did not appreciate that the gradient of a normal must be numerical. A few students did not show clearly

that the gradient of the curve at x = 0 was found from the derivative, they seemed to treat y = 2ex + 5 and assumed the

gradient was always 2. Some candidates failed to obtain the final mark in this section because they did not observe the

dV

dt

d

d

y

x

03

x y

y x

1

2x

1x

x

1

2ex

d

d

y

x

f -1f

5

e xxe

5

1y

5

e1

x

2 2122 , and 4r r r

2 11

tA 2

4 11 or 1

2

At

A t

2 2(1 )t t

dy

dx


instruction that a, b and c must be integers.

For most candidates part (b) followed directly from their normal equation. It was disappointing that those who had made errors in part (a) did not use the absence of n = 14 here as a pointer to check their working in the previous part. Most preferred to invent all sorts of spurious reasons to justify the statement.

Many candidates set out a correct strategy for finding the area in part (c). The integration of the curve was usually correct but some simply ignored the lower limit of 0. Those who used the simple “half base times height” formula for

the area of the triangle, and resisted the lure of their calculator, were usually able to complete the question. Some tried to find the equation of PN and integrate this but they usually made no further progress. The demand for exact answers

proved more of a challenge here than in 6(c) but many candidates saw clearly how to simplify 2eln4

and convert ln 4 into 2 ln 2 on their way to presenting a fully correct solution.

25. Most candidates made some attempt to differentiate x√sin x, with varying degrees of success. √sin x + x√cos x was the most common wrong answer. Having struggled with the differentiation, several went no further with this part. It was surprising to see many candidates with a correct equation who were not able to tidy up the √ terms to reach the required result.

Most candidates went on to make an attempt at ∫πy2

dx. The integration by parts was generally well done, but there were many of the predictable sign errors, and several candidates were clearly not expecting to have to apply the method twice in order to reach the answer. A lot of quite good candidates did not get to the correct final answer, as there were a number of errors when substituting the limits.

26. Most answered part (a) correctly, and errors were usually because candidates had miscopied a sign, or written 3x +1 instead of 3(x +1).

The differentiation was usually correct, though a minority integrated and some misunderstood the notation and found the inverse function instead of the derived function. Those who returned to the original expression and differentiated gave themselves more work. Many found it difficult to answer the final part of this question. The examiners were looking for a statement that square terms are always positive

27. No Report available for this question.

c4 questions from past papers differentiation · city of london academy 1 c4 questions from past...

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