A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

1. a) L = 1 + 0.0002
+ 0.000004
2

dL  dL 
= 0.0002 + 0.000008
mC-1 M1 
A1
d
 d 
dL
When
= 500C, = 0.0002 + 0.000008(500)
d

= 0.0002 + 0.004 = 0.0042 m/C = 0.42 cmC-1 B1
[3]

dL dL d

b) = M1
dt d
dt
d
-0.002t
= -e B1
dt
0.02
t = 10,
= 500e- M1
dL dL
 = 0.00412 M1 (sub into from a)
d
d

dL –0.02
So =-e  0.00412 A1
dt
= -0.00404 ms-1 A1
[7]

dy
2. a) = (2x2)(3 cos 3x) + (sin 3x)(4x) (product rule) M1 A1 A1
dx
= 6x2 cos 3x + 4x sin 3x or 2x(3x cos 3x + 2 sin 3x) A1 (simplified form)
[4]

2
   3 
b) x =  y = 2  sin 
M1
2
2  2 
 2
= A1
2

dy 4 3
= sin = - 2 M1 A1 f.t
dx 2 2

 2  
y  = - 2  x 
M1 (line equation)
2 

 2
2
y+ = - 2x + 2
2
2
y + 2x = A1 A1 (simplified form)
2
[7]

Page 1
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

3. f (
) = 6 cos 2
+ [(4
)(-3 sin 3
) + (cos 3
)(4)] (product rule) M1 A1 A1

= 6 cos 2
 12
sin 3
+ 4 cos 3

2

When
= , f (
) = 6 cos    12 
 sin  3
 + 4 cos  3
 M1
6  6  6  6   6 
= 3  2 A1
[5]

4. a) V = 100 sin 300t

dV
= (100)(300) cos 300t = 30000 cos 300t Vs-1 M1 A1
dt
dV
When t = 0.015 s, = 30000 cos(300  0.015) = 30000 cos 4.5 = -6324 Vs-1 M1 A1
dt
[4]

5

b) 300t = , M1 A1
2 2



t= , A1 (both)
600 120
[3]

sin 2x
5. a) y = tan 2x = M1
cos 2 x
dy (cos 2 x )(2 cos 2 x )  (sin 2 x )(2 sin 2 x ) 2(cos 2 2x
sin 2 2 x )
= = A1 A1
dx cos 2 2 x cos 2 2 x
2(1)
= = 2 sec22x A1
2
cos 2 x
[4]

 dy  2 
2
b) When x = , = 2 sec2 
= 2 =8 M1 A1
3 dx  3   2 
cos 

3 
[2]
dy
c) For turning point, =0
dx
0 = 2sec2 2x M1
2
0=
cos 2 2 x
2
no solution, since as 0 < cos22x  1, 2  A1
cos 2 2x
[2]

Page 2
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

3x
6. a) y =
2x 2  1
dy (2 x 2  1)(3)  (3x )(4 x )
Gradient of curve = = (quotient rule) M1 A1
dx

2
2x  1 
2

2
6 x  3  12 x
2
 6x  3
2
= = A1
(2x  1)
2 2


2x 2  1
2

dy  6( 2 ) 2  3
At the point ( 2 , 2 ), = =  53 B1 f.t
dx [2( 2 ) 2  1]2
[4]

1 3
b) M = = M1 A1
 53 5
3
(y  2)= (x  2 ) M1
5
5y  5 2 = 3x  3 2
5y – 3x = 2 2 A1 (simplified form)
[4]

Page 3
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

du
7. a) y = (3x2 – 1)5 Let u = 3x2 – 1, from which = 6x M1 (chain rule)
dx
dy du dy
Then y = u5 and
= 5u4 A1 ( and )
du dt dt
dy dy du
Hence =  = (5u4) (6x) = 30xu4 M1
dx du dx
= 30x(3x2–1)4 A1
[4]

b) y = (3x 2 5x  2)
du
Let u = 3x2 + 5x  2, then = 6x + 5 M1 A1
dx
u2
1 dy 1
1
So y = u = u 2 and = 2
du

Hence
dy dy du
= 
dx du dx
=
u (6x + 5)
1
2
 12
M1
6x 5
= A1
2 (3x 2 5x  2)
[4]

3
c) x =

2x 2  45
du
Let u = 2x2 – 4, then = 4x M1 A1
dx
dy
y = 3u-5 so = -15u-6
du

dy dy du 15  60 x
Hence =  =  4x = M1 A1
6
dx du dx u (2 x 2  4) 6
[4]

Page 4
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dy
8. = - sin
 cos
B1
d

dy
=0 M1
d

- sin
 cos
= 0  sin
= - cos

tan
= -1 M1
3 7 

= , A1 A1 (-1 if in degrees)
4 4
3 2

= y=  =- 2 M1 A1
4 2

7 2

= y= = 2 A1 (-1 if not simplified)
4 2
[8]

dy
9. a) Gradient of curve y = x2 – x – 2 : m = = 2x – 1 M1 A1
dx
At the point (2, 0), x = 2 hence m =2(2) –1 = 3
For normal, m = - 13 M1 A1 f.t
y = - 13 (x –2) M1 (use of line equation)
3y = -x +2
3y + x = 2 A1 (simplified form)
[6]

b) 3y + x = 2
y = x2 – x – 2

3x2 – 3x – 6 = 2 – x M1 (attempt to solve

simultaneously)
3x2 – 2x – 8 = 0 A1 (correct quadratic)
(3x + 4) (x – 2) = 0
x = - 34 A1
11
y= 12
A1
[4]

Page 5
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

d d d d 3 d
10.a) (3y2)  (3x3) + (1) + (y ) = (0) M1
dx dx dx dx dx

dy dy
6y  9x2 + 0 + 3y2 = 0 A1 A1
dx dx

dy
Hence (6y + 3y2) = 9x2 M1
dx
dy 9x 2 3x 2
 = ( = ) A1
dx 6 y 3y 2 y ( 2 y)
[5]

b) y = -1  3 – 3x3 + 1 – 1 = 0 M1

3x = 3x3
1= x A1
dy 9
= M1
dx 6
3
= -3 A1
[4]

Page 6
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dx
11.a) x = 3t  =3 B1
dt
dy
y = 2t(t –1) = 2t2 – 2t  = 4t – 2 B1
dt
dy
dy dt 4t  2
= = M1 A1 f.t.
dx dx 3
dt
[4]

b) x = 8  3t – 1 = 8  t = 3 M1 A1
y = 12 B1
dy 10
= B1 f.t
dx 3
(y – 12) = 103 (x – 8) M1
3y – 36 = 10x – 80
3y – 10x + 44 = 0 A1 (simplified form)
[6]

c) If it meet again, have simultaneous solution of 3y – 10x + 44 = 0

x = 3t – 1
y = 2t(t –1)

 6t(t – 1) – 10(3t –1) + 44 = 0 M1 (substituting in)

6t2 – 6t – 30t + 10 + 44 = 0
6t2 – 36t + 54 = 0
t2 – 6t + 9 = 0 A1
(t – 3)2 = 0
 t = 3 only A1
So no further points
[3]

Page 7
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dx
12.a) =3 M1 A1
dt
dy 3
= A1
dt t2
dy
dy dt 1
= = M1 A1
dx dx t2
dt

dy 1
At (3p, 3
p
), = B1 f.t.
dx p2

Normal : m = p2 M1 A1

(y  p3 )= p2(x 3p) M1
y 3
p
= p x  3p
2 3
A1
[10]

b) p2 = 4 M1
p=2 A1 (both)
[2]

Page 8
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dy 2 cos 2 x 2 sin 2x
13.a) = M1 A1 A1
dx sin 2 x
cos 2x
2 cos 2 x
2 sin 2x 4 sin 2 x
= M1
sin 2 x
cos 2 x
4 sin 2x
=2 A1
sin 2 x
cos 2 x
[5]

dy
b) =0 B1
dt
4 sin 2x
2 =0
sin 2 x
cos 2 x

2sin2x + 2cos2x = 4sin2x M1 (attempt to solve)

cos2x = sin2x
1 = tan2x A1
[3]

c) tan2x = 1  2x = B1
4


x = B1 (-1 if in degrees)
8

y = ln 
1 1 
 M1

2 2
= ln
2 A1
1
= 2
ln2 A1

x 0.38 0.4 M1
8
dy
0.05 0 -0.03
dx

Maximum point A1
[7]

Page 9
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

14. a) Total area = rl + r2 M1

8
h

r
r = 8sin
B1

A = (8sin
) 8 + (64sin2
 M1
= 64(sin
+ sin2
)
= 64 sin
(1 + sin
) A1
[4]

dA
b) = 64(cos
+ 2sin
cos
) M1 A1
d

0 = 64(cos
+ 2sin
cos
) M1

0 = cos
(1+ 2sin
) M1
cos
= 0, or sin
=  12 (not valid, since it would require
> )

=
A1
2


But
= does not give a cone B1
2
[6]

Page 10
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dy
15.a) = cos t B1
dt
dx
= 2cos2t B1
dt
dy cos t
= B1
dx 2 cos 2 t
[3]


b) t = , x=1 B1
4
1
dy
= 2
dx 0

 vertical tangent M1
so x = 1
or x – 1 = 0 is equation A1
[3]

cos t
c) =0 M1
2 cos 2 t

cos t = 0
 3
t= , A1 A1
2 2

x = 0, 0 A1
y = 1,-1 A1 (both)
[5]

Page 11
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

16.a) 3

2

c
c x
= M1 A1
sin (  3
) sin

x sin(   3
)
c = A1
sin

x sin 3

= (since sin
= sin( 
) A1
sin

[4]

1
b) Area = 2
absinc M1
= 1
2
xc sin2

x sin 3 
= 1
x   sin2 
sin  
2

x2 sin 3
( 2 sin
cos
)

= M1
2 sin

x2
= 2sin3cos = x2sin3cos A1
2
[4]

dA
c) = x2 (3cos3cos  sin3sin) M1 A1 A1
d
0 = x2(3cos3cos  sin3sin)

sin3sin = 3cos3cos M1
sin 3 sin

=3 M1
cos 3 cos

tan3tan = 3 A1
[6]

d)  = 0.46 tan3tan = 2.57  = 0.47 tan3tan = 3.13 M1 A1 A1

 root in between
[3]

e) Use  = 0.46 –0.47 as initial value B1

If 1 = 0.46 If 1 = 0.465 If 1 = 0.47 M1
2 = 0.4690 2 = 0.4684 2 = 0.4677 A1 (any correct
3 = 0.4678 3 = 0.4679 3 = 0.4680 intermediate value)
4 = 0.4680  0.468 to 3dp  0.468 to 3dp A1 cao
 0.468 to 3dp
[4]

Page 12
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

cos x
17.a) y =
2 sin x

dy (2 sin x )( sin x ) cos x ( cos x )

= M1 A1 A1
dx ( 2 sin x ) 2
 2 sin x sin 2 x cos 2 x
=
(2  sin x ) 2
2 sin x
1
= M1 (sin2x+cos2x =1) A1
(2 sin x ) 2
[5]

dy
b) = 0 1 2sinx = 0 M1
dx
sinx = 12
x =
, 5
A1 A1
6 6
3
3
x=
y= 2
= M1 A1
6 3
3
2

 3
3
x = 5
y= 2
= A1
6 3
3
2
[6]

c)

3
( , )
6 3

(0, 12 )

G1 shape

3

G2 intersections with axes
2 2
(-1 e.e.o.o.)
G1 maximum shown

5
3
( , )
6 3

[4]

Page 13
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

18.a) V = x  3x  4x
= 12x3 B1

A = 2  x  3x + 2  x  4x + 2  3x  4x M1
= 38x2 A1

A
x= M1
38

3
 A
V = 12 
 M1

38 
3

 A 2
= 12  A1

38 
[6]

dA
b) = 0.2
dt

dV dV dA
= M1
dt dA dt
1
dV  A 2
= 12  3
2
 1
38  M1 A1
dA
38 
1
9  A 2
= 
19
38 

1
dV 9  152  2
=   0.2 M1
dt 19
38 
= 18
( = 0.189) cm3 s-1 A1
95
[5]

Page 14
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dy 2x dy
19. a) 2y + 2 + 2x + 2y = 0 M1 (use of implicit)
dx x 3 dx
M1 (diff. ln(x2 – 3))
A1 A1

2x
(2y + 2x) = - 2y  2
M1
x 3
2x  x 
 2y  y 2 
dy x 3
2

x  3 
= =- A1 f.t.
dx 2 y 2x x y
[6]

b) x = 2  y2 + ln1 + 4y = 5 M1
y2 + 4y – 5 = 0
(y + 5) (y – 1) = 0
y = 1, – 5 A1 A1

(2, 1):
dy
=–

1
2 = – 1 B1 f.t
dx 1
2
 m= 1 B1
y – 1= x – 2 M1
y= x  1 A1

(2, -5):
dy
=-

5
2 = – 1 B1 f.t
dx 5
2
m=1 B1
y + 5 = x –2
y=x –7 A1
[10]

Page 15
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

dy
20.a) = 2e2x(x + 1)2 + e2x2(x + 1) M1 A1 A1
dx
= 2e2x(x + 1) (x + 1 + 1)
= 2e2x(x + 1) (x + 2) A1
[4]

dy
b) =0 M1
dx
x = -1 , x = -2 A1 A1
y = 0 , e-4 A1 (both)
[4]

c)

(0,1)

G1 intercept
G1 both turning points
G1 shape

(-2,e-4)
(-1,0)
[3]

dy
d) y = ax  = axlna B1
dx
at x = 1 : a lna = 2e2(2) (3) = 12e2 M1 A1
[3]

e) a = 26: a lna – 12e2 = -3.96 a = 27: a lna – 12e2 = 0.319 M1 A1

change of sign  root
[2]

f) Use of 26 – 27 for a1 B1
a1 = 26 a1 = 26.5 a1 = 27 M1
a2 = 27.21 a2 = 27.06 a2 = 26.90 A1 (at least 1
a3 = 26.84 a3 = 26.89 a3 = 26.93 intermediate
a4 = 26.95 a4 = 26.94 answer correct)
a5 = 26.92  26.9 to 1dp  26.9 to 1dp A1 cao
a6 = 26.93
 26.9 to 1dp
[4]

Page 16
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

21.a) y = 12 sin-13x  2y = sin-1 3x

 sin2y = 3x M1
 x = 13 sin2y A1
[2]

1 dy
b) 1 =  2cos2y M1 (implicit) M1(chain)
3 dx
dy 3
= A1
dx 2 cos 2 y
[3]

c) cos2y = (1  sin 2 2 y) M1

= (1  (3x ) 2 ) M1
dy 3
= A1
dx 2 1  9 x 2
[3]

dy
22.Increasing function if >0 M1
dx
dy x
= lnx + = lnx +1 M1 A1 (product)
dx x

So lnx + 1 > 0
lnx > -1 A1
x > e-1 A1
[5]

Page 17
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

23. a) x  -k B1 () B1 (-k)

[2]

b) t = a  x = a2 – k, y = 3a – a3 B1
t = -a  x = a2 – k, y = -3a + a3 (= -(3a – a3) B1

Putting in values of t of opposite signs gives the same x-value, but y-values of
opposite signs, so curve is symmetrical about x-axis. B1
[3]

dx dy
c) =2t; = 3 – 3t2 M1 A1 (both)
dt dt
dy dy dt 3 3t 2
=  = M1 A1 ft
dx dt dx 2t
[4]

3 3t 2
d) = 0  3 – 3t2 = 0 M1
2t
Giving t = 1 A1 (both)
t = 1  (1k, 2) A1
t = -1 (1k, -2) A1
[4]

dy 3
e) t = 0  = M1
dx 0
So tangent is vertical A1
t = 0  x = -k
So equation is x = -k A1
[3]

f) x = 0  t = k M1
 y =  (3k – (k)3) M1
=  (3k)k A1

G1 (shape)
(1-k, 2) (0, (3-k)k) G1 (symmetry)
G1 ((-k,0))
(-k, 0)
(1-k, -2) (0, -(3-k)k)

[6]

Page 18
 A LEVEL MATHEMATICS QUESTIONBANKS

DIFFERENTIATION 2

x 2x 2 A Bx C
24.a)  M1
(2  x )(1 x 2 ) 2  x 1 x2

x + 2x2  A (1+x2) + (Bx + C)(2 – x)

x = 2: 10 = 5A A=2 A1
Coeff. x2: 2=A–B B=0 A1
Const: 0 = A + 2C  C = -1 A1
[4]

b) f(x) = 2(2 – x)-1 – (1+x2)-1

f ( x ) = 2(-1)(-1)(2x)-2 – (-1)(2x)(1+x2)-2 M1 (chain rule) A2

f (0) = 2(-1)(-1)(20)-2 – (-1)(20)(1+02)-2 M1
f (0) = 12 A1
[5]

Page 19