City of London Academy 1 C4 QUESTIONS FROM PAST PAPERS – DIFFERENTIATION 1. A curve C has equation 2 x + y 2 = 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 m 3 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 m 3 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 h t h 5 4 d d 75 6 0 , 4 4 y x x y d d 6
39
Embed
C4 QUESTIONS FROM PAST PAPERS DIFFERENTIATION · City of London Academy 1 C4 QUESTIONS FROM PAST PAPERS – DIFFERENTIATION 1. A curve C has equation 2x + y2 = 2xy Find the exact
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
City of London Academy 2
(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.
(a) Find in terms of x and y.
(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.
(a) Find in terms of x and y.
(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
City of London Academy 3
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
City of London Academy 4
(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
City of London Academy 5
(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
City of London Academy 6
(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
City of London Academy 7
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
City of London Academy 8
. (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
City of London Academy 9
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
City of London Academy 10
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
City of London Academy 11
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
City of London Academy 12
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
City of London Academy 13
(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
City of London Academy 14
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
City of London Academy 15
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
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
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)
not necessarily required.
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
City of London Academy 18
@ (–8, 8), ,
@ (–8, 16), .
Differentiates implicitly to include either .
Ignore M1
Correct LHS equation; A1;
Correct application of product rule (B1)
not necessarily required.
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
City of London Academy 19
@ 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
City of London Academy 20
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
City of London Academy 21
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
City of London Academy 22
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
City of London Academy 23
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
City of London Academy 24
... 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
City of London Academy 25
Differentiates implicitly to include either .
(ignore .) M1
Correct equation. A1
not necessarily required.
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
City of London Academy 26
Aliter Way 2
Differentiates implicitly to include either
(ignore .) M1
Correct equation. A1
not necessarily required.
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
City of London Academy 27
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.
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
City of London Academy 29
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
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
City of London Academy 34
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
City of London Academy 35
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
City of London Academy 36
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
City of London Academy 37
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
City of London Academy 38
(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
City of London Academy 39
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