IB CHALLENGE - Differentiation Solutions

8/2/2019 IB CHALLENGE - Differentiation Solutions

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1

Differentiation – Hard solutions

1. (a) (i) f ( x) =

= 2 + by division or otherwise (M1)

Therefore as x f ( x) 2 (A1) y = 2 is an asymptote (AG)

OR = 2 (M1)(A1)

y = 2 is an asymptote (AG)

OR make x the subject

yx – 3 y = 2 x + 1

x( y – 2) = 1 + 3 y (M1)

x = (A1)

y = 2 is an asymptote (AG)

Note: Accept inexact methods based on the ratio of the

coefficients of x.

(ii) Asymptote at x = 3 (A1)

(iii) P(3, 2) (A1) 4

(b) f ( x) = 0 x = – (M1)(A1)

x = 0 f ( x) = – (M1)(A1) 4

Note: These do not have to be in coordinate form.

(c)

(A4) 4

Note: Asymptotes (A1)

Intercepts (A1)

“Shape” (A2).

3

12

x

x

3

7

x

3

12lim

x

x

x

2

31

y

y

0,2

1

2

1

3

1

,03

1

x

y

2

3



2

(d) f ( x) = (M1)

= (A1)

= Slope at any point

Therefore slope when x = 4 is – 7 (A1)And f (4) = 9 ie S(4, 9) (A1)

Equation of tangent: y – 9 = – 7( x – 4) (M1)

7 x + y – 37 = 0 (A1) 6

(e) at T , = – 7 (M1)

( x – 3)2

= 1 (A1)

x – 3 = ±l (A1)

(A1)(A1) 5

(f) Midpoint [ST ] =

= (3, 2)

= point P (A1) 1[24]

2. (a)

y2

= 9 x

62 = 9(4) (M1)36 = 36 (A1) 2

(4, 6) on parabola

(b) (i) y = 3

(M1)

= Slope at any point

Therefore at (4, 6), slope of tangent = (A1)

Slope of normal = – (A1)

Therefore equation of normal is y – 6 = – ( x – 4) (M1)

2)3(

)12()2)(3(

x

x x

2)3(

7

x

2)3(

7

x

)5,2(

)9,4(

5 – or9

2or4

T

S

y

x

2

59,

2

24

y

x

y x= 92

P

Q

x

x x

y

2

3

d

d

4

3

3

4

3

4



3

3 y – 18 = – 4 x + 16

4 x + 3 y – 34 = 0 (A1) 5

Notes: Candidates may differentiate implicitly to obtain

.

Answer must be given in the form ax + by + c = 0.

(ii) Coordinates of Q:

y = 0, 4 x = 34

x = (A1)

Q (A1) 2

(c) SP = (M1)

=

= (A1)

SQ = (M1)

=

= (A1) 4

(d) SP = SQ (M1)

But (alternate angles) (A1)

(A1) 3

[16] 3. (a) f (1) = 3 f (5) = 3 (A1)(A1) 2

(b) EITHER distance between successive maxima = period (M1)

= 5 – 1 (A1)

= 4 (AG)

OR Period of sin kx = ; (M1)

so period = (A1)

= 4 (AG) 2

(c) EITHER A sin + B = 3 and A sin + B = – 1 (M1) (M1)

A + B = 3, – A + B = – 1 (A1)(A1)

A = 2, B = 1 (AG)(A1)

y x

y

2

9

d

d

2

17

0,

2

17

2

2

)60(449

3616

49

4

25

4

9

2

17

4

9

4

34

425

PQSQPS ˆˆ

QP M PQS ˆˆ

QPSQP M ˆˆ

k

π2

2

π

π2

2

π

2

π3



4

OR Amplitude = A (M1)

A = (M1)

A = 2 (AG)

Midpoint value = B (M1)

B = (M1)

B = 1 (A1) 5

Note: As the values of A = 2 and B = 1 are likely to be quite

obvious to a bright student, do not insist on too detailed a proof.

(d) f ( x) = 2 sin + 1

f ( x) = + 0 (M1)(A2)

Note: Award (M1) for the chain rule, (A1) for , (A1) for

2 cos .

= cos (A1) 4

Notes: Since the result is given, make sure that reasoning is

valid. In particular, the final (A1) is for simplifying the result of

the chain rule calculation. If the preceding steps are not valid,

this final mark should not be given. Beware of “fudged” results.

(e) (i) y = k – x is a tangent – = cos (M1)

– 1 = cos (A1)

x = or 3 or ...

x = 2 or 6 ... (A1)

Since 0 x 5, we take x = 2, so the point is (2, 1) (A1)

(ii) Tangent line is: y = – ( x – 2) + 1 (M1)

y = (2 + 1) – x

k = 2 + 1 (A1) 6

(f) f ( x) = 2 2 sin + 1 = 2 (A1)

sin (A1)

x = (A1)(A1)(A1) 5

[24] 4. (a) y = e

2 xcos x

2

4

2

)1(3

2

2

2

)1(3

x

2

π

x

2

π

cos22

π

2

π

x

2

π

x

2

π

x2

π

x

2

π

2

π

x

2

π

2

1

2

π

x

6

13πor

6

5πor

6

π

2

π

x

3

13or

3

5or

3

1



5

= e2 x

( – sin x) + cos x (2e2 x

) (A1)(M1)

= e2 x

(2 cos x – sin x) (AG) 2

(b) = 2e2 x

(2 cos x – sin x) + e2 x

( – 2 sin x – cos x) (A1)(A1)

= e2 x

(4 cos x – 2 sin x – 2 sin x – cos x) (A1)= e

2 x(3 cos x – 4 sin x) (A1) 4

(c) (i) At P, = 0 (R1)

3 cos x = 4 sin x (M1)

tan x =

At P, x = a, ie tan a = (A1)

(ii) The gradient at any point e

2 x

(2 cos x – sin x) (M1)Therefore, the gradient at P = e

2a(2 cos a – sin a)

When tan a = , cos a = , sin a = (A1)(A1)

(by drawing a right triangle, or by calculator)

Therefore, the gradient at P = e2a

(A1)

= e2a

(A1) 8[14]

5. (a) = 30 – at => s = 30t – a + C (A1)(A1)(A1)

Note: Award (A1) for 30t, (A1) for a , (A1) for C.

t = 0 => s = 30(0) – a + C = 0 + C => C = 0 (M1)

=> s = 30t – at 2

(A1) 5

(b) (i) vel = 30 – 5(0) = 30 m s – 1

(A1)

(ii) Train will stop when 0 = 30 – 5t => t = 6 (M1)

Distance travelled = 30t – at 2

= 30(6) – (5) (62) (M1)

= 90m (A1)

90 < 200 => train stops before station. (R1)(AG) 5

(c) (i) 0 = 30 – at => t = (A1)

(ii) 30 – (a) = 200 (M1)(M1)

x

y

d

d

2

2

d

d

x

y

2

2

d

d

x

y

4

3

4

3

4

3

5

4

5

3

5

3

5

8

t

s

d

d

2

2t

2

2t

2

02

2

1

2

1

2

1

a

30

a30

21

2

30

a



6

Note: Award (M1) for substituting , (Ml) for setting equal

to 200.

=> = 200 (A1)

=> a = = 2.25 m s – 2 (A1) 5

Note: Do not penalize lack of units in answers.[15]

6. (a) x = 1 (A1) 1

(b) (i) f ( – 1000) = 2.01 (A1)

(ii) y = 2 (A1) 2

(c) f ( x) = (A1)(A1)

= (A1)

= (AG) 3

Notes: Award (M1) for the correct use of the quotient rule, the

first (A1) for the placement of the correct expressions into the

quotient rule.

Award the second (A1) for doing sufficient simplification to make

the given answer reasonably obvious.

(d) f ′(3) = 0 stationary (or turning) point (R1)

f (3) = > 0 minimum (R1) 2

(e) Point of inflexion f ( x) = 0 x = 4 (A1)

x = 4 y = 0 Point of inflexion = (4, 0) (A1)

OR

Point of inflexion = (4, 0) (G2) 2 [10]

7. (a) (A1)

(A1) 2

(b)

(must be a correct expression) (A1)

(AG) 1

(c) (A2) 2

(d) (i) tangent gradient (A1)

gradient of L

(A1) (N2) 2

a

30

aaa

450450 –

900

49

200450

4

22

)1(

)20132)(1(2)134()1(

x

x x x x x

3

22

)1(

)40264()13174(

x

x x x x

3)1(

279

x

x

16

18

3h

2k

( ) f x2

( 3) 2 x

26 9 2 x x

26 7 x x

( ) 2 6 f x x

2

1

2



7

(ii) EITHER

equation of L is (M1)

. (A1)

OR

(A2) (N2) 2

(iii) EITHER

(M1)

(may be implied) (A1) (may be implied) (A1)

(A1) (N3) 4

OR

(or a sketch) (M1)

(A3) (N3) 8

[13]

8. (a) (i) p = (10 x + 2) – (1 + e2 x

) A2 2

Note: Award (A1) for (l + e2x

) – (10x + 2).

(ii) = 10 – 2e2 x

A1A1

= 0 (10 – 2e2 x

= 0) M1

x = (= 0.805) A1 4

(b) (i) METHOD 1

x = 1 + e2 x

M1

1n( x – 1) = 2 y A1

f – 1

( x) = A1 3

METHOD 2

y – 1 = e2 x

A1

= x M1

f – 1( x) = A1 3

1

2 y x c

1c

1

12 y x

11 ( 4)

2 y x

2 16 7 1

2 x x x

2

2 11 12 0 x x (2 3)( 4) 0 x x

1.5 x

2 16 7 1

2 x x x

1.5 x

x

p

d

d

x

p

d

d

2

5n1

2

)1(n1 Allow

2

)1(n1 x y

x

2

)1ln( y

2

)1(n1 Allow

2

)1(n1 x y

x



8

(ii) a = M1

= × 21n2 A1

= 1n 2 AG 2[11]

9. (a) (i) (or ) (A1)(A1) (N1)(N1)

(ii)

(M1)

(A1) (N1)

(iii)

(A1) (N1) 5

(b) (i) (A1) (N1)

(ii) (M1)

(A1)(A1) (N2) 4

(c) (i) when x = 4, gradient of tangent is 4 – 1 = 3 (may be implied) (A1)

gradient of normal is (A1)

(A1) (N3)

(ii) (or sketch/graph) (M1)

(may be implied) (A1)

(A1) (N2) 6

[15]

10. METHOD 1

l + 2w = 60 (M1)

l = 60 2w (A1)

A = w(60 2w) (= 60w 2w2) (A1)

22n1

2

1

2

)15(n1

21

2 p 4q 4, 2 p q

( 2)( 4) y a x x

8 (6 2)(6 4)a

8 16a

1

2a

1( 2)( 4)

2 y x x

21 ( 2 8)2

y x x

214

2 y x x

d1

d

y x

x

1 7 x

8, 20 P is (8, 20) x y

1

3

1 1 40 ( 4)

3 3 3 y x y x

21 1 44

2 3 3 x x x

21 2 160

2 3 3 x x

23 4 32 0 x x (3 8)( 4) 0 x x

8or 4

3 x x

8( 2.67)

3 x



9

= 60 4w (A1)

Using = 0 (60 4w = 0) (M1)

w = 15 (A1) (C6)

METHOD 2

w + 2l = 60 (A1)

w = 60 2l (A1)

A = l(60 2l) (= 60l 2l2) (A1)

= 60 4l (A1)

Using = 0 (60 4l = 0) (M1)

l = 15

w = 30 (A1) (C6)[6]

11. (a) METHOD 1

f ( x) = 6 sin 2 x + 2 sin x cos x A1A1A1

= 6 sin 2 x + sin 2 x A1

= 5 sin 2 x AG N0

METHOD 2

(A1)

f ( x) = 3 cos 2 x + A1

f ( x) = A1

f ( x) = A1

f ( x) = 5 sin 2 x AG N0

(b) k = A2 N2

[6]

12. (a) f ( x) = x2

+ 4 x 5 A1A1A1 N3

(b) evidence of attempting to solve f ( x) = 0 (M1)

evidence of correct working A1

eg ( x + 5) ( x 1), sketch

x = 5, x = 1 (A1)

so x = 5 A1 N2

w

A

d

d

w

A

d

d

l

A

d

d

l

A

d

d

2

2cos1

sin

2 x

x

x2cos2

1

2

1

2

12cos

2

5 x

x2sin2

52

57.12

,2

20164



10

(c) METHOD 1

f ( x) = 2 x + 4 (may be seen later) A1

evidence of setting second derivative = 0 (M1)

eg 2 x + 4 = 0

x = 2 A1 N2

METHOD 2

evidence of use of symmetry (M1)

eg midpoint of max/min, reference to shape of cubic

correct calculation A1

eg

x = 2 A1 N2

(d) attempting to find the value of the derivative when x = 3 (M1)

f (3) = 16 A1

valid approach to finding the equation of a line M1

eg y 12 = 16( x 3), 12 = 16 3 + b y = 16 x 36 A1 N2

[14]

,2

15

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IB CHALLENGE - Differentiation Solutions