unit 2 multiple choice tests answers

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Unit 2 Answers: Multiple Choice Tests © Macmillan Publishers Limited 2013

Multiple Choice Test 1 1 D 2 B 3 A 4 A 5 B 6 D 7 A 8 A 9 B 10 B 11 A 12 C 13 C 14 B 15 A 16 D 17 A 18 C 19 B 20 D 21 B 22 B 23 A 24 C 25 A 26 B 27 B 28 B 29 A 30 B 31 B 32 C 33 B 34 C 35 A 36 A 37 B 38 B 39 C 40 B 41 B 42 C 43 B 44 C 45 B Multiple Choice Test 1 worked answers

1 1 3arg ( 3 3 ) tan3

i − − π −5π− − = − π = − π = − 6 6

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D 2 (x + iy)2 = –5 + 12i

x2 – y2 + i (2xy) = –5 + 12i x2 – y2 = –5

62 = 12xy yx

⇒ =

22

36 5∴ − = −xx

x4 + 5x2 – 36 = 0 (x2 + 9) (x2 – 4) = 0 x2 = 4, x = ± 2

When x = 2, = =6 32

y

62, 32

x y= − = = −−

∴ 5 12i− + = 2 + 3i or –2 – 3i B

3 (1 – i)5 − =1 2i

1arg (1 ) tan ( 1)4

i − −π− = − =

55 5 5(1 ) 2 cos sin 4 2 cos sin

4 4 4i i i

−π −π − π − π − = + = + 4

1 14 22 2

i = − + −

= –4 + 4i A

4 1 2 2 (1 2 ) 2z i z i− + = ⇒ − − = Circle centre (1, –2) radius 2 A

5 y = (x + 1) ln (x + 2) d 1ln ( 2)d 2y xxx x

+= + +

+

when = = +d 10, ln2d 2yxx

B

6 + += +sin (2 3) sin (2 3)d [ ] 2 cos (2 3)d

x xe x ex

D 7 x = t2 + t

y = 2t – 1 y = 2x – 3 2t – 1 = 2 (t2 + t) – 3 2t – 1 = 2t2 + 2t – 3 2t2 = 2

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t2 = 1, t = ± 1 A

8 x3y + 2xy = 4y Differentiate wrt x:

3 2d d d3 2 2 4d d dy y yx x y x yx x x+ + + =

When x = 0, y = 0 d 0dyx

⇒ =

A 9 y = x lnx

d 1 ln 1 lndy x x xx x

= + = +

B 10 f(x) = cos 2x

( )f x′ = –2 sin 2x ( )f x′′ = –4 cos 2x

B

11 2

2 24 3 ( 1) ( 3)x x x x

≡+ + + +

2( 1) ( 3) 1 3

A Bx x x x

≡ ++ + + +

2 ( 3) ( 1)A x B x⇒ = + + + When x = –1, 2 = 2A ⟹ A = 1 When x = –3, 2 = –2B ⟹ B = –1

2 1 1( 1) ( 3) 1 3x x x x

∴ ≡ −+ + + +

A

12 =+ − + −∫ ∫2 2 2

1 1d d4 ( 1) 2 ( 1)

x xx x

11 1tan2 2

x c− − = +

C

13 1 1

2 2 20 0

1 1d d1 1 1

x xx xx x x+

= ++ + +∫ ∫

11 2

0

1tan ( ) ln ( 1)2

x x− = + +

1 ln 24 2π

= +

C

14 1 21 1(ln ) d (ln )2

x x x cx

= +∫

B

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15 2 21d

2x xxe x e c= +∫

A

16 =

−∑50

1

(3 2)r

r

=

= −∑50

1

3 2 (50)r

r

= −3 (50) (51) 2 (50)

2

= 3725 D

17 (1 – 2x2)9 4th term = 9C3 (–2x2)3 = –672x6 A

18 ++

+ +! ( 1)!

( 1)! ( 2)!n n

n n

+ + +=

+( 2) ( !) ( 1)!

( 2)!n n n

n

+ + +=

+( 2) ! ( 1) !

( 2)!n n n n

n

+ + +=

+![ 2 1]

( 2)!n n n

n

=!n +

+ +(2 3)

( 2) ( 1) ( !)n

n n n

2 3( 2) ( 1)

nn n

+=

+ +

C 19 (2 + 3x)n

coefficient of x3: nC3 2n – 3 (3)3 coefficient of x4: nC4 2n – 4 (3)4

3 33

4 44

C (2 ) (3 ) 8C (2 ) (3 ) 15

n n

n n

−

− =

!2 8( 3)!3!

!3 15( 4)!4!

nn

nn

− = −

( 4)!n − 4!( 3) ( 4)!n n− −

453!

=

= −20 34

n

n = 8 B

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20 1 2( 1) ( 2)(1 ) 1 ( 1) ( ) ( )2!

x x x− − −+ = + − − + −

= 1 + x + x2 D

21 −

+

432 12

x

−31 < < 12

x

2 2< <3 3

x

B

22 = + + + +2 3

2 (2 ) (2 )1 22! 3!

x x xe x

∞

=

= ∑0

2!

n n

n

xn

B 23 f(x) = e4x

f1(x) = 4e4x when x = 1, f(1) = e4 f1(1)= 4e4 f(x) = f(1) + (x – 1) f1(1) e4x = e4 + e4 (x – 1) A

24 ex = 25x – 10 ex – 25x + 10 = 0 f(x) = ex – 25x + 10 f(0) = 10 f(1) = e – 25 + 10 = –ve Root in [0, 1] C

25 f(x) = 2 sinx – x f(1.5) = 0.49499 f(2) = –0.18141

+=

+1(1.5) (0.18141) 2 (0.49499)

0.49499 0.18141x

= 1.8659 A

26 B 27 2 – 6 + 18 – 54 + …

2 (–3)r-1 B

28 f(x) = x3 + 3x2 + 5x + 9 f'(x) = 3x2 + 6x + 5

− + − + − += − −

− + − +

3 2

2 2

( 2.5) 3 ( 2.5) 5 ( 2.5) 92.53 ( 2.5) 6 ( 2.5) 5

x

= –2.457

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B 29 a1 = 1, an + 1 = 2an + 4

a2 = 2a1 + 4 = 2(1) + 4 = 6 a3 = 2a2 + 4 = 2(6) + 4 = 16 1, 6, 16 A

30 an = 5 (2n – 1) – 4 B

31 P(exactly two heads) 38

=

B

32 126

52

125

2

×51

29

17

×

8

502

172

=

C 33 7! – 6! × 2!

= 5 × 6! = 3600 B

34 6! × 7P3 = 151 200 C

35 × = =6 93 2 15

5

720 240C C 720,1001C

A

36 532

A

37 =3 0.3

10

B 38 P(A) = 0.2, P(B) = 0.5

∩ = × = × =P (A B) P(A) P(B) 0.5 0.2 0.1 P (A B) 0.2 0.5 0.1 0.6∪ = + − =

( ) P(A) P(B)P (A B)P A BP(B)∩

= =P(B)

0.2=

( ) P(A B )P A B 0.2P(B )

′∩′ = =′

B 39 No. in committee = 5

6 seniors, 4 juniors 4C1 × 6C4 + 4C2 × 6C3 + 4C3 × 6C2 + 4C4 ×6C1 = 60 + 120 + 60 + 6 = 246 C

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40 1 1 21 1 3

0 5a

−

Cofactor of 3 = –a B

41 1 3 1 3 1 1

20 5 5 0a a

+ +

= 5 + 5 – 3a – 2a = 10 – 5a B

42 10 – 5a ≠ 0 a ≠ 2 C

43 PI is y = a cos 2x + b sin 2x B

44 − + =2

2

d d4 4 0dd

y y yxx

m2 – 4m + 4 = 0 (m – 2)2 = 0 m = 2 y = (Ax + B) e2x C

45 5d 2dyx y xx− =

− = 4d 2dy y xx x

22 d 2ln ln

2

1x x xxe e ex

−− −∫ = = =

= ∫ 22

1 dy x xx

32

1 13

y x cx

= +

5 213

y x cx= +

B Multiple Choice Test 2 1 C 2 A 3 D 4 D 5 A 6 C

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7 C 8 A 9 A 10 C 11 C 12 C 13 B 14 A 15 C 16 B 17 B 18 A 19 B 20 B 21 C 22 B 23 D 24 D 25 A 26 D 27 C 28 D 29 A 30 C 31 B 32 D 33 B 34 B 35 B 36 A 37 D 38 A 39 C 40 D 41 B 42 A 43 C 44 A 45 C Multiple Choice Test 2 worked answers 1 i49 = i48 i = (i2)24 (i) = (–1)24 i = i

C

2 + + + + + += × = = = +

− − +

23 3 2 6 5 5 5 12 2 2 5 5

i i i i i i ii i i

A 3 − − = =2 2 8 2 2i

arg 1 2( 2 2 ) tan2

i − − −3π − − = − π = − 4

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π−4− − =

3

2 2 2 2i

i e D

4 (2 + i) (2 + i) = 4 + 4i – 1 = 3 + 4i (3 + 4i) (3 + 4i) = 9 + 24i + 16i2 = –7 + 24i D

5 –x2 + y2 – 2xy – 2 = 0 Differentiate wrt x:

d d2 2 2 2 0d dy yx y x yx x

− + − − =

d (2 2 ) 2 2dy y x y xx

− = +

d 2 2d 2 2y y x x yx y x y x

+ += =

− −

A 6 y = ln (3x2 + 5)

2

d 6d 3 5y xx x=

+

C 7 y = sin–1 (3x)

2

d 3d 1 9yx x=

−

=− 2

91 9x

=− 2

119

x

C 8 y = tan–1 (x2 + 2)

=+ +2 2

d 2d 1 ( 2)y xx x

A 9

+ +

+ +

− −

2 2

2

13 2

3 2

3 2

x x xx x

x

∴ 2

2

3 213 2 ( 1) ( 2)x x

x x x x+

≡ −+ + + +

11 2

B Cx x

≡ − + + +

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3 2 ( 2) ( 1)x B x C x− − ≡ + + + When 1, 1x B= − =

2 1 41( 1)( 2) 1 2

xx x x x

∴ = + −+ + + +

A = 1, B = 1, C = 4− A

10 2

1 d9 4

xx−

∫

11 2sin2 3

x c− = +

C 11 sin 4 cos4 dθ θ θ∫

1 sin8 d2

θ θ= ∫

1 cos816

cθ= − +

C 12 y = 2t2 + 5

x = t + 3

= =d d4 , 1d dy xtt t

= ÷d d dd d dy y xx t t

= 4t C

13 2 cos , 2 2 sinx yθ θ= = + d d2sin , 2 cosd d

x yθ θθ θ= − =

d 2 cos cotd 2 sinyx

θ θθ

= = −−

2 23

2

d d d cosec 1[ cot ] cosecd d d 2 sin 2

y xx

θθ θθ θ θ

+= − ÷ = = −

−

B

14 ∫ cossin dxx e x = – e cos x + c A

15 Given that = + 1 d2 ,d

x yyx

is

ln y = (x + 1) ln 2 1 d dln 2 ln 2

d d= ⇒ =

y y yy x x

= 2x + 1 ln 2 C

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16 = + +2

12

x xe x

B 17 x = 0, y = 2

2 d d2 4 0, 1d dy yx x

+ = = −

22

2

d d d2 2 2 0d d dy y yy yx x x

+ + =

22

2

d2 (2) ( 1) + 4 2 ( 1) 0d

yx

− + − =

2

2

d 1d 2

yx

=

− +

2 1= 2 + ( ) ( 1)2! 2xy x +…

21= 2 + ...4

x x− +

B

18 −∑= 1

(5 4 )n

r

r

=

= −∑ ∑= 1 1

5 4n n

r r

r

−

( + 1)= 5 42

n nn

= 5n – 2n (n + 1) = n [5 – 2n – 2] = n (3 – 2n) A

19 =

−∑50

1

(5 4 )r

r

= =

= −∑ ∑50 50

1 1

5 4r r

r

= −

50 (51)5 (50) 42

= –4850 B

20 when x = 0, y = 2 2d 2 (2) 1 9

dyx= + =

2

2

d d4 4 2 9 72d d

y yyx x

= = × × =

y = y(0) + (x – 0) 2( 0)(0) (0) ...

2!xy y−′ ′′+ +

y = 2 + 9x + 36x2 +…= 5.24

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x = 0.3, y = 2 + 9(0.3) = 4.700 B

21 Third iteration C

22 u1 = 2, u n + 1 = 2 u n, n ≥ 1 u 2 = 2 u 1 = 4 u 3 = 2 u 2 = 2(4) = 8 u 4 = 16 2, 4, 8, 16 B

23 Divergent D 24 u n = 2n D

25 − − −+ = + − +3 2( 3) ( 4)(1 2 ) 1 ( 3) (2 ) (2 )

2!x x x

= 1 – 6x + 24x2 A

26 −− < < < <

1 11 2 1,2 2

x x

D 27 Sn = pn + qn2

S3 = 6 ⇒ 3p + 9q = 6 p + 3q = 2 [1] S5 = 11 ⇒ 5p + 25q = 11

1155

p q+ = [2]

[2] – [1] ⇒ = =1 12 ,5 10

q q

3 210

p + =

1710

p =

17 1,10 10

p q= =

C

28 = + 217 1S10 10n n n

− = − + − 21

17 1S ( 1) ( 1)10 10n n n

117S S10n n nu n−= − = 21

10n+

1710

−1710

n + − 2110

n + −1 15 10

n

= +16 110 5

n

= +1 (2 16)

10n

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D 29 2 ln x + x – 2 = 0

f(x) = 2 ln x + x – 2 f(1) = 1 – 2 = –1 f(2) = 2 ln 2 + 2 – 2 = 2 ln 2 [1, 2] A

30 = − + −2 4 6(2 ) (2 ) (2 )cos 2 1

2! 4! 6!x x xx

Coefficient of −= − =

66 2 4

6! 45x

C

31 − + = +2

2

d d3 2 1dd

y y y xxx

AQE: m2 – 3m + 2 = 0 (m – 1) (m – 2) = 0 m = 1 or 2 y = Aex + Be 2x B

32 2

2

d d, , 0d dy yy ax b ax x

= + = =

Substituting into the differential equation: –3a + 2 [ax + b] = x + 1 2a = 1

12

a =

–3a + 2b = 1 3 2 1

2b−

+ =

=522

b

54

b =

1 52 4

y x= +

D

33 y = Aex + Be2x 1 52 4

x+ +

When x = 0, y = 1

⇒ = + + ⇒ + = −5 11 A B A B4 4

[1]

2d 1A 2Bd 2

x xy e ex= + +

When x = 0, d 2dyx=

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12 A 2B2

⇒ = + +

+ =3A 2B2

[2]

[2] – [1] ⇒ 3 1 7B2 4 4

= + =

A = –2 27 1 52

4 2 4x xy e e x= − + + +

B

34 4d 2 2d

xy y ex− =

IF = 2 2dx xe e− −∫ = 2 4 22 dx x xye e e x− −= ∫ 2 22 dx xye e x− = ∫

ye–2x = e2x + c y = 1 + ce2x B

35 when x = 0, y = 4, 4 = 1 + c ⇒ c = 3 y = 1 + 3e2x B

36 8! = 40 320 A 37 6! × 3! = 4320 D 38 5! × 6P3 = 14 400

A 39 10C2 × 10C3 × 2 = 10 800 C

40 × × =3 2 1 16 5 4 20

D

41 1 52 4 14 6 7

x pay qz r

− =

− −

− +4 1 2 1 2 4

56 7 4 7 4 6

a

= –34 – 5 (10) + a (28) = –84 + 28a B

42 a ≠ 3 A

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43 − − −

1 5 3 182 4 1 74 6 7 29

R2 → R2 – 2R1 R3 → R3 – 4R1

181 5 30 14 5 430 14 5 43

− − −

− − +

3 3 2R R R→ +

181 5 30 14 5 430 0 0 0

− −

Infinite set of solutions. C

44 = −

1 3 4M 2 5 1

3 8 4

− −= − +

5 1 2 1 2 5M 3 4

8 4 3 4 3 8

= 28 – 33 + 4 = –1 A

45 Matrix of cofactors −

= + − + − + −

28 11 120 8 123 9 1

1

28 20 23M 11 8 9

1 1 1

−

− = − − − −

− − = − − −

28 20 2311 8 9

1 1 1

C Multiple Choice Test 3 1 B 2 C 3 A 4 C 5 B 6 B 7 B

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8 A 9 D 10 D 11 C 12 C 13 B 14 D 15 B 16 C 17 A 18 B 19 C 20 A 21 A 22 B 23 A 24 C 25 D 26 B 27 B 28 D 29 A 30 C 31 D 32 A 33 C 34 D 35 D 36 C 37 D 38 A 39 C 40 C 41 D 42 A 43 A 44 B 45 B Multiple Choice Test 3 worked answers

1 2 2 33 3 3

i i ii i i

+ + += ×

− − +

+ +=

26 510i i

+=

5 510

i

1 12 2

i= +

B

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2 − −

− − = − π − 1 3arg ( 1 3 ) tan

1i

− π=

32

C

3 2 22 cos sin 22 2

i i π π + = + 4 4

= +2 2i A

4 π π + 6 6

6

cos sini

= π + πcos sini π= ie

C 5 − + =1 4 3z i

(1 4 ) 3z i⇒ − − = Circle centre (1, – 4) radius 3 B

6 y = ecos(2x) cos 2d 2 sin (2 )

d= − xy x e

x

B

7 = + =+

3 12,2

x t yt

−= = − +2 2d d3 , ( 2)d dx yt tt t

= ÷d d dd d dy y xx t t

−= ×

+ 2 2

1 1( 2) 3t t

2= −+2

13 ( 2)t t

B

8 −= +

− + − +3 4 A B

( 3) ( 2) 3 2x

x x x x

1 23 2x x

= +− +

A

9 2

1 d4 4 5

xx x+ +∫

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= + + −

∫ 2

1 d14 5 12

xx

2

1 d14 42

xx

= + +

∫

2

1 1 d4 1 1

2

xx

= + +

∫

11 1tan4 2

x c− = + +

11 2 1tan4 2

x c− + = +

D

10 +

= + 1ln2

xyx

= ln (x + 1) – ln (x + 2) d 1 1d 1 2yx x x= −

+ +

+ − +=

+ +2 ( 1)

( 1) ( 2)x xx x

=+ +

1( 1) ( 2)x x

D

11 ∫ 2sin (2 ) dx x

= −∫1 1 cos 4 ) d2

x x

1 1 sin 42 8

x x c= − +

C

12 ∫ 3 dxxe x

3 31 13 9

x xxe e c= − +

C 13 xy + 2y2 + 3x = 3

d d4 3 0d dy yx y yx x+ + + =

When x = 1, y = 0 ⇒ d 3dyx= −

B

14 + +=

+ + + +∫ ∫2 2

2 1 2 4d d24 5 4 5

x xx xx x x x

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21 ln 4 52

x x c= + + +

D 15 y = tan–1(x2)

4

d 2d 1y xx x=

+

B 16 Divergent C 17 un = (–1)n (n + 1)2

u6 = (–1)6 (6 + 1)2 = 49 A

18 = =

−∑ ∑ 2

1 1

1n n

r r

r

+ += −

( 1) (2 1)6

n n nn

[ ]= − + +6 ( 1) (2 1)6n n n

= − − −2[6 2 3 1]6n n n

= − + −2(2 3 5)6n n n

B

19 + − −+

=−

( 1) ( ) ( 1) ( 2)!( 1)!( 2)!

n n n nnn −( 2)!n

= (n2 – 1) n = n3 – n C

20 9

2 2xx

−

− − 9 2 9 2C ( )

rr

r xx

18 – 3r = 0 ⇒ r = 6 Term independent of x: 9C6 (–2)6 = 5376 A

21 finite A (22) (3r + 1)

B

(23) +∑= 1

(3 1)n

r

r

A

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24 −

+12(1 2 )x

2 3

1 3 1 3 51 2 2 2 2 21 (2 ) (2 ) (2 )2 2! 3!

x x x

− − − − − = + − + + +

Fourth term: − 352

x

C

25 = + + +2 3

2 (2 ) (2 )1 22! 3!

x x xe x

= + + +2 341 2 23

x x x

D 26 f(x) = ex, f '(x) = ex, f ''(x) = ex

21 1 1( 1)( 1)

2!x xe e x e e−= + − + +…

2( 1)1 ( 1)2!

xe x −

= + − +

B 27 ln (1 + x) is valid for

–1 < x ≤ 1 B

28 f(x) = x3 + 10x2 + 10x – 4 f(0) = – 4 f(1) = 1 + 10 + 10 – 4 = 17 Root lies in the interval [0, 1] D

29 = − +2 4

cos 12! 4!x xx

= − +2 4(0.1) (0.1)cos (0.1) 1

2 24

= 0.9950041667 0.995 A

30 2S 13

n

n = −

= − =12 1S 13 3

24 5S 19 9

= − =

25 1 29 3 9

u = − =

C 31 5!

D

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32 OE P W R 4! × 2! = 4! × 2 A

33 =5 4 3 2 1 4 80

3!

C 34 18C9 × 2C2 = 18C9

D 35 5C4 + 5C3

+ 5C2 = 5 + 10 + 10 = 25 D

36 0.75 × 0.40 = 0.30 C

37 1P (A) =3

1P (A B )6

′ ′∩ =

1P (A B)6

′∪ =

1 5P (A B) 16 6

∪ = − =

D

38 = − −

1 2 1 1 0 4AB 1 0 1 1 1 1

3 2 2 2 1 0

1 3 61 1 45 4 14

= −

A

39 −1 1 01 2 1

1 3 2

− −= − +

2 1 1 1 1 20

3 2 1 2 1 3

= 1 – (–3) = 4 C

40-42 2

2

d 4d

y y xx

− =

CF: 2

2

d 0d

y yx

− =

AQE: m2 – 1 = 0 m = ± 1 ∴ y = Aex + Be– x

of 23


PI: Let y = ax + b

=ddy ax

2d 0d

yx=

Substituting into the differential equation: –(ax + b) = 4x a = –4 b = 0 ∴ y = –4x General solution is y = Aex + Be–x – 4x

40 y = Aex + Be–x C 41 y = –4x D 42 When x = 0, y = 2 ⇒ A + B = 2 [1]

xd A Be 4d

xy ex

−= − −

When d0, 0dyxx

= =

A B 4⇒ − = [2] Adding [1] and [2]: 2A = 6, A = 3 A = 3, B = –1 ∴ y = 3ex – e–x – 4x A

43 1 2 1

A 1 1 13 3 a

= −

1 2 11 1 1

3 3 a−

1 1 1 1 1 12

3 3 3 3a a− −

= − +

= (a – 3) – 2 (–a – 3) + (–3 – 3) = a – 3 + 2a + 6 – 6 = 3a + 3 A 0 1a= ⇒ =

A

44 2d 1dy y xx x− =

11 d ln ln 1x x xxe e e

x−− −∫ = = =

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= ∫ 21 1 dy x xx x

1 dy x xx

= ∫

21 12

y x cx

= +

312

y x cx= +

B

45 when x =1, y = 2 ⇒ 122

c= +

122

c = −

=32

31 32 2

y x x= +

B

unit 2 multiple choice tests answers

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