ANSWERS TO MOCK TEST PAPER - fullmarks.org€¦ · 1 MATHEMATICS–12 ANSWERS TO MOCK TEST PAPER 1....

1

MATHEMATICS–12

ANSWERS TO MOCK TEST PAPER 1. (a) 2. (c) 3. (c) 4. (b) 5. (b) 6. (d) 7. (a) 8. (b) 9. (c) 10. (d) 11. x + y – z = 2

12. ( ) ( )ˆ ˆˆ ˆ ˆ ˆ5 4 6 3 7 2r i j k i j k= − + + λ + + 13. (–1, 1) 14. –f (x) 15. 1 16. If A and B are skew-symmetric matrices, then A + B also skew-symmetric. Let A and B be two skew-symmetric matrices of the same order. Then, A′=–A and B′=–B Now (A+B)′=A′+B′=–A–B=–(A+B) ⇒ A + B is a skew-symmetric matrix. 17. sin2a + sin2b + sin2g = 1 – cos2a + 1 – cos2b + 1 – cos2g = 3 – (cos2a + cos2b + cos2g) = 3 – 1 = 2 [ cos2a + cos2b + cos2g = 1]

18. I = 3 2 2− −⋅∫ x xe dxI II

On Integrating by parts, we get

I = 3 3 3 22 2 2 2− − − −∫ ∫∫− −( )

x x x xe dx e dx dxlog

= 32

3 3 22

22

22

−−

−−

−

− −( )

−( )∫xx

xxe e dxlog

= 32

3 322

2 2−−

− −

−

− ⋅∫xx

x xe e dxlog = 32

322

−−

−

− ⋅xxe log I

I + log3 · I = 3

2

2 2− −⋅−

x xe

(1 + log3) I = 3

2

2 2− −⋅−

x xe

I = 32 1 3

2 2− −⋅− +( )

x xelog =

−( )+( )

−32 1 3

2e x

log

19. f (x) = x2 4− , a = 2, b = 4 Q f (x)hasuniqueanddefinitevalueforeachxintheclosedinterval[2,4]. \ f (x) is continuous in [2, 4].

f ′(x) = 12

4 221

2x x−( ) ×−

= x

x2 4−

which exists for all xintheopeninterval]2,4[ i.e., for 2 < x < 4], f (x)isderivablein]2,4[

1st Proof \ Anjum Azad \ Mock_ans_(Mathe-12) \ 30-01-20\17:36 Reader _______________________ Date __________

2 n MatheMatics – Xii


\ByLagrange’sMeanValueTheorem,wehave

f ′(c) = f b f ab a

f f( ) − ( )−

= ( ) − ( )−

4 24 2

c

c2 4− =

12 02

2 32

3− = = i.e., ⇒ c2 = 3(c2 – 4) ⇒ 2c2 = 12, \ c2 = 6 ⇒ c = ± 6 Rejecting c = – 6 which does not lie in ]2, 4[ \ c = 6 , which lies in ]2, 4[ Hence,theLMVTheoremisverified.

OR

f (x) = x −( )823

f (x) is continuous in [0, 16]

f ′ (x) = 23

813x −( )− =

2

3 813x −( )

Which does not exist at x = 8 Œ[0, 16] So, f (x)isnotderivablein(0,16).Hence,Rolle’stheoremisnotapplicable.

20. f (x) = x

xsin

\ f ′ (x) = sin cos

sinx x x

x1

2

( ) − ⋅( )

⇒ f ′ (x) = cos tan

sinx x x

x−( )

2 ...(i)

Now, cos x > 0 for 0 < x < π2

and sin x < x < tan x for 0 < x < π2

Q lim sinx

xx→

=

0

1

\ (tan x – x) > 0

\ From (i),wehavef ′(x) > 0 for 0 < x < π2

Thus, f (x)isanincreasingfunctionintheinterval0<x < π2

. 21. y = x2 + 2x + 3

dydx

= 2x + 2

dydx

( )at 0 3,

= 2(0) + 2 = 2

y – y1 = dydx x y

( )at 1 1,

(x – x1)

y – 3 = 2 (x – 0) ⇒ 2x – y + 3 = 0

answers to Mock test paper n 3


22. The equation of the lines are

xyp

z−−

=−( ) = −1

37 2

23

2 and

7 13

51

65

xp

y z−( )−

= − = −−

⇒ x y

pz−

−= − = −1

32

27

32

and xp

y z−− = − = −

−1

37

51

65

Since the lines are at right angle, therefore, a1a2 + b1b2 + c1c2 = 0

⇒ (–3) −

+

37

27

p p (1) + (2)(–5) = 0

⇒ 97

27

p p+ – 10 = 0 ⇒ 117p = 10 ⇒ p = 70

11OR

− +−

= − = +−

x y z22

17

33

and x y z+−

= − = −21

2 84

54

⇒ x y z− = − =

− −( )−

22

17

33

and x y z− −( )

−= − = −2

14

25

4 \ a1 = 2, b1 = 7, c1 = –3 and a2 = –1, b2 = 2, c2 = 4

cos q = a a b b c c

a b c a b c1 2 1 2 1 2

12

12

12

22

22

22

+ +

+ + + +

= 2 1 7 2 3 4

2 7 3 1 2 42 2 2 2 2 2

−( ) + ( ) + −( )( ) + ( ) + −( ) −( ) + ( ) + ( )

= − + −2 14 12

62 21 = 0

cos q = 0 ⇒ q = π2

23. limx

f x→ −

( )5

= limx

kx→ −

+( )5

1 = limh

k h→

−( ) + 05 1 = k(5 – 0) + 1 = 5k + 1

limx

f x→ +

( )5

= limx

x→ +

−( )5

3 5 = limh

h→

+( )03 5 –5 = 3(5 + 0) – 5 = 15 – 5 = 10

As f (x) is continuous at x = 5, So, limx

f x→ −

( )5

= limx

f x→ +

( )5

⇒ 5k + 1 = 10 ⇒ k = 95

24. Let 12

tan–1x = q

tan–1x = 2 q ⇒ x = tan 2 q ...(i)

R.H.S. = cos tantan

− + ++

12

2

1 1 22 1 2

θθ

= cos secsec

− +

1 1 2

2 2θ

θ

= cos cos− +

1 1 2

2θ

= cos cos−

122

2θ

= cos–1 cos q = q = 12

tan–1x [using (i)]



OR

Wehave 9 − −π8

94

13

1sin = 94

2 23

1sin−

⇒ 94

2 23

13

1 1sin sin− −+

=

98π

Now, L.H.S. = 94

2 23

13

1 1sin sin− −+

= 94

2 23

1 13

13

1 2 23

12 2

sin− −

+ −

[Q sin–1 x + sin –1 y = sin–1 ( )x y y x1 12 2− + − ]

= 94

2 23

1 19

13

1 89

1sin− − + −

=

94

2 23

2 23

13

13

1sin− × + ×

= 94

89

19

1sin− +

=

94

11sin− = 94 2

× π =

98π

= R.H.S.

Hence, 98 4

13

1π − 9 −sin = 9 −

42 2

31sin

25. Given: y log ydx – xdy = 0 ⇒ y log ydx = xdy

or dyy ylog

= dxx

Integration both sides, we get

1yydy

log∫ = dxx∫ ⇒ log (log y) = log x + log c

⇒ log (log y) = log cx ⇒ loge y = cx or y = ecx is the required solution. [Q elog f (x) = f (x)]

26. I = log tan xdx02π

∫ = log tan ππ

202 −

∫ x dx Q f x dx f a x dx

a a( ) = −( )

∫ ∫0 0

= logcot log tanxdx x dx02 1

02

π π

∫ ∫= ( )−

= −∫ log tan xdx02π

= –I ⇒ 2I = 0 or I = 0

27. D = b c c a a bc a a b b ca b b c c a

+ + ++ + ++ + +

= 0



Applying C1 ÆC1 + C2 + C3

D = 222

a b c c a a ba b c a b b ca b c b c c a

+ +( ) + ++ +( ) + ++ +( ) + +

= 0

⇒ D = 2111

a b cc a a ba b b cb c c a

+ +( )+ ++ ++ +

= 0

Applying R2 ÆR2 – R1 and R3 ÆR3 – R1

⇒ D = 2100

a b cc a a bb c c ab a c b

+ +( )+ +− −− −

= 0

⇒ D = 2(a + b + c)[1·(b – c)(c – b) – (b – a)(c – a)] = 0 D = 2(a + b + c)[ab + bc +ca – (a2 + b2 + c2)] = 0 D = –(a + b + c)[–2(ab + bc +ca) +2(a2 + b2 + c2)] = 0 D = –(a + b + c)[(a – b)2 + (b – c)2 + (c – a)2 = 0 ⇒ (a + b + c) = 0 or (a – b)2 + (b – c)2 + (c – a)2 = 0 ⇒ a + b + c = 0 or a = b = c

OR

D = p b ca q ca b r

= 0

Applying R1 ÆR1 – R2 and R2 ÆR2 – R3

D = p a b q

q b c ra b r

− −− −

00 = 0 ⇒ D =

p a q bq b r c

a b r

− − −( )− − −( )

00 = 0

Expanding along C1, we get D = (p – a) [(q – b)r +(r – c)b] + a[(q – b)(r – c) – 0] = 0 As a ≠ p, b ≠ q, c ≠ r,dividingthroughoutby(p – a) (q – b) (r – c), we get

rr c

bq b

ap a−

+−

+−

= 0

Adding 2 to both sides, we get

rr c

bq b

ap a−

+−

+

+−

+

1 1 = 1 + 1 ⇒ rr c

b q bq b

a p ap a−

+ + −−

+ + −−

= 2

⇒ rr c

qq b

pp a−

+−

+−

= 2

28. y = sin− − − −

1 21 1x x x x

Put x = sin q and x = sin f, we get

\ y = sin sin sin sin sin− − − −

1 2 21 1θ φ φ θ



\ y = sin sin cos sin cos− −

1 2 2θ φ φ θ

⇒ y = sin–1 [sin q cos f – sin f cos q] ⇒ y = sin–1 [sin (q – f )] = q – f \ y = sin–1 x – sin–1 x

⇒ dydx

= 1

11

11

22−−

−⋅

x x x

Hence, dydx

= 1

11

22 2−−

−x x x 29.Tofindthepointofintersectionofthecurves y2 = 4x ...(i) and 4x2 + 4y2 = 9 ...(ii) Weequatevaluesofy2 from these equations.

4x = 94

– x2 ⇒ x2 + 4x – 94

= 0

⇒ 4x2 + 16x – 9 = 0 ⇒ 4x2 + 18x – 2x – 9 = 0 ⇒ 2x(2x + 9) – 1(2x + 9) = 0 ⇒ (2x – 1) (2x + 9) = 0

⇒ x = 12

and x = – 92

which is rejected because if x = – 92

, then from the equation of the

parabola

y2 = 4 92

−

= –18 < 0 which is not possible.

Hence, x ≠ – 92

\ x = 12

When, x = 12

, y2 = 4 × 12

= 2 ⇒ y = ± 2

\ The points of intersection are B 12

2,

and C 12

2,−

Required area = Area of the region OCABO

= 2(Area of the region OMABO) [By symmetry]

= 2[(Area of the region OMBO) + (Area of the region BMAB)]

= 2 2 940

1 22

1 2

3 2

xdx x dx/

/

/

∫ ∫+ −

Y

Y′

X′

X

O

M

C

B

A

1

22,( (

1

22,−( (

3

20,( (

y

x2 =

4

4 + 4x2

y2

= 9



= 4 32

22

94

98 3

2

32

12

2 1

12

32

x x x x

+ − +

−sin

= 83

12

0 94

94

23

32 2

12

32

1

12

−

+ −

+

−x x xsin332

= 83

12 2

0 12

94

14

94

1 13

1 1× + − −

+ −

− −sin sin

= 43 2

94 2

13

12

21+ −

−−π sin = 98

94

13

43 2

12

1π − + −−sin

= 98

94

13

13 2

1π − +−sin = 98

94

13

26

1π − +−sin sq. units

30. Wehave, 1 1+

+ −

e dxdy

e xy

xy

xy = 0

dxdy

= e x

y

e

xy

xy

−

+

1

1 ...(i)

Which is a homogeneous equation of the form dxdy

= f xy

Put x = vy, so that dxdy

= v + y dvdy

\ The equation (i) becomes v + y dvdy

= e v

e

v

v

−( )+

11

⇒ y dvdy

= ve ee

v v

v

−+1

– v ⇒ y dvdy

= − −+e ve

v

v1

⇒ 1+

+e

v edv

v

v = – dyy

Integrating both sides, we get

1++

∫

ev e

dvv

v = – dyy∫

log (v + ev) = – log y + log C ⇒ logxyexy+

= log

Cy



⇒ xyexy+ =

Cy

⇒ x yexy+ = C is the required solution.

31. We know that 2 tan–1 x = cos− −+

12

2

11

xx

LHS = 22

1tan tan− −+

α βα β

A = cos

tan

tan

−

− −+

+ −+

1

2

2

12

12

α βα βα βα β

A

A

= costan tan

tan

−

−

+ +

+ 1

2 2

2

12

12

12

α β

α

A A

A

+ −

β 12

2tan A

= coscos sin cos sin

−

−

+

+

1

2 2 2 2

2 2 2 2α βA A A A

+

+

α βcos sin cos2 2 2

2 2 2A A A

−

sin2

2A

= cos coscos

− ++

1 α βα β

AAOR

Given, cos–1 x + cos–1 y + cos–1z = p ⇒ cos–1 x + cos–1 y = p – cos–1z

⇒ cos− − − −( )1 2 21 1xy x y = p – cos–1z

⇒ xy x y− − −1 12 2 = cos (p – cos–1z)

⇒ xy x y− − −1 12 2 = – cos (cos–1z) ⇒ xy x y− − −1 12 2 = – z

⇒ xy + z = 1 12 2− −x y ⇒ (xy + z)2 = (1 – x2) (1 – y2)

⇒ x2y2 + z2 + 2xyz = 1 – x2 – y2 + x2y2

⇒ x2 + y2 + z2 + 2xyz = 1

32. When a die is thrown, sample space = {1, 2, 3, 4, 5, 6}

p = probability of throwing six when a die is thrown = 16

q = probability of throwing six = 1 – p = 1 – 16

= 56

Captain of team A can win in 1st, 3rd, 5th .... throws



P(A win) = 16

56

16

56

16

2 4

+

+

+ .... =

16

1 56

2

−

Sum S∞ =−

ar1

= 16

3611

611

× =

P(B win) = 1 611

511

− =

As probability of winning the match by team A and team B are not the same

P A P B( ) = 6 > ( ) =

11

511

, so decision of referee was not fair.

33. A = 1 1 23 0 21 0 3

−−

|A| = 1 1 23 0 21 0 3

−− = 9 + 2 = 11

A11 = 0, A21 = 3, A31 = 2; A12 = –11, A22 = 1, A32 = 8; A13 = 0, A23 = –1, A33 = 3

Adj A = 0 3 211 1 80 1 3

−−

A(Adj A) = 1 1 23 0 21 0 3

−−

0 3 211 1 80 1 3

−−

=

0 11 0 3 1 2 2 8 60 0 0 9 0 2 6 0 60 0 0 3 0 3 2 0 9

+ + − − − +− − + + + −+ + + − + +

= 11 0 00 11 00 0 11

= 11I = |A|I

(Adj A)A = 0 3 211 1 80 1 3

−−

1 1 23 0 21 0 3

−−

=

0 9 2 0 0 0 0 6 611 3 8 11 0 0 22 2 240 3 3 0 0 0 0 2 9

+ + + + − +− + + + + − − +

− + + + + +

= 11 0 00 11 00 0 11

= 11I = |A|I

So, A(adj A) = A (adj A) = |A|IOR

Let cost of 1 pen = ` x cost of 1 pag = ` y cost of 1 instrument box = ` z Equations are 3x + 2y + 1z = 41 2x + 1y + 2z = 29 3x + 2y + 2z = 46



|A| = 3 2 12 1 23 2 2

= – 1

Let AX = B, here B = 412946

, X =

xyz

A11 = –2, A21 = –2, A31 = 3; A12 = 2, A22 = 3, A32 = –4; A13 = 1, A23 = 0, A33 = –1

A–1 = 11

2 2 32 3 41 0 1−

− −−−

X = A –1B ⇒ xyz

= 1

1

2 2 32 3 41 0 1−

− −−−

412946

⇒ xyz

=

2155

x = ` 2, y = ` 15, z = ` 5 34. Step I. Mathematical Formulation of L.P.P. Suppose the farmermixes x bags of brand P and y bags of brandQ.The given data is

condensed in the following table.Brand Number

of bagsCost

(`/bag)Element A(units/bag)

Element B(units/bag)

Element C(units/bag)

P x 250 3 2.5 2Q y 200 1.5 11.25 3

Total cost = 250x + 200y Let Z = 250x + 200y Wehavethefollowingmathematicalmodelforthegivenproblem: Minimise Z = 250x + 200y ...(i) subjecttotheconstraints: 3x + 1.5y ≥ 18 [Given: Minimum requirement of nutritional element A is 18 units i.e., ≥ 18 units]

or 3x + 1510

y ≥ 18

Multiplyingby10anddividingby15, or 2x + y ≥ 12 (Nutritional element A constraint) ...(ii) 2.5x + 11.25y ≥ 45 [Given: Minimum requirement of nutritional element B is, 45 units i.e., ≥ 45 units]

or 2510 x +

1125100

y ≥ 45

or 2x + 9y ≥ 36 (Nutritional element B constraint) ...(iii) 2x + 3y ≥ 24 (Nutritional element C constraint) ...(iv) [Given: Minimum requirement of nutritional element C is 24 units i.e., ≥ 24 units] x, y ≥ 0 (Qnumberofbagscan’tbenegative) ...(v)



Step II. Constraint (v) x, y ≥ 0 ⇒Feasibleregionisinfirstquadrant. Table of values for the line 2x + y = 12 of constraint (ii)

x 0 6y 12 0

Draw the straight line joining the points (0, 12) and (6, 0). Let us test for origin (x = 0, y = 0) in constraint 2x + y ≥ 12,wehave0≥ 12 which is not true. \ Region for constraint (ii) 2x + y ≥ 12 is the half-plane not containing the origin i.e., region

on the non-origin side of the line 2x + y = 12. Table of values for the line 2x + 9y = 36 for constraint (iii)

x 0 18y 4 0

Let us draw the line joining the points (0, 4) and (18, 0). Let us test for origin (x = 0, y = 0) in constraint (iii) 2x + 9y ≥ 36,wehave0≥ 36 which is not

true. \ Region for constraint (iii) is the region on the non-origin side of the line 2x + 9y = 36. Table of values for the line 2x + 3y = 24 for constraint (iv)

x 0 12y 8 0

Draw the line joining the points (0, 8) and (12, 0). Let us test for origin (x = 0, y = 0) in constraint (iii) 2x + 3y ≥ 24,wehave0≥ 24 which is not

true. \ Region for constraint (iii) 2x + 3y ≥ 24 is again the region on the non-origin side of the

line 2x + 3y = 24. Theshadedregioninthefigureisthefeasibleregiondeterminedbythesystemofconstraints

from (ii) to (v). The feasible region is unbounded. Step III.ThecoordinatesofthecornerpointsAandDare(18,0)and(0,12)respectively. Corner point B: It is the point of intersection of the lines 2x + 3y = 24 and 2x + 9y = 36

Subtracting – 6y = – 12 ⇒ y = −−126

= 2

Putting y = 2 in 2x + 3y=24,wehave 2x + 6 = 24 ⇒ 2x = 18 ⇒ x = 9 \ Corner point B is (9, 2). Corner point C: It is the point of intersection of the lines 2x + y = 12 and 2x + 3y = 24

Subtracting – 2y = – 12 ⇒ y = −−122

= 6 Putting y = 6 in 2x + y=12,wehave 2x + 6 = 12 ⇒ 2x = 6 ⇒ x = 3 \ Corner point C is (3, 6).



2O 4

2

4

6

8

10

12

14

16

6 8 10 12 14 16 18 20

XX¢

Y¢

Y

D(0, 12)

C(3, 6)

B(9, 2)

A(18, 0)

2x+ 9y = 36

5x+ 4y = 39

2x+ y = 12

2x+ 3y = 24

(12, 0)(6, 0)

(0, 4)

(0, 8)

Feasible Region

Step IV.Now,weevaluateZateachcornerpoint.

Corner Point Z = 250x + 200yA(18, 0) 4500B(9, 2) 2650C(3, 6) 1950 = m ← Smallest

D(0, 12) 2400 Fromthistable,wefindthat1950isthesmallestvalueofZatthecornerC(3,6).Sincethe

feasibleregionisunbounded,1950mayormaynotbetheminimumvalueofZ. Step V. To decide this, we graph the inequality Z < m i.e., 250x + 200y < 1950 or 5x + 4y < 39. Table of values for the line 5x + 4y = 39 corresponding to constraint Z < m i.e., 5x + 4y < 39.

x 0 395

= 7.8

y 394

= 9.75 0

Let us draw the dotted line joining the points (0, 9.75) and (7.8, 0).



The line is to be shown dotted because equality sign is absent in the constraint Z < m i.e., in 5x + 4y < 39.

Let us test for origin (x = 0, y=0)inthisconstraint,wehave0<39whichistrue. \ Region for constraint Z < m i.e., 5x + 4y < 39 is towards the origin side of the line. WeobservethatthehalfplanedeterminedbyZ<m has no point in common with the feasible

region.Hencem=1950istheminimumvalueofZattainedatthepointC(3,6). \ Minimum cost is ` 1950 when 3 bags of brand P and 6 bags of brand Q are mixed. 35.Letrbetheradiusofthegivenrightcircularconeofgivenheighth. Let the radius of the inscribed cylinder be x and its

height be y. InsimilartrianglesAPQandARC,wehave

PQRC

= APAR

i.e., xr

= h yh−

[Q AP = AR – PR = h – y] Cross-multiplying, hx = rh – ry \ ry = rh – hx = h(r – x)

\ y = hr

(r – x)...(i)(givencondition)

Letzdenotethevolumeofthecylinder \ z = px2y ...(ii) [Here z is to be maximised] Puttingthevalueyfrom(i) in (ii),

z = px2 hr

(r – x) = πhr

(rx2 – x3) ...(iii)

\ dzdx

= πhr

(2rx – 3x2), d zdx

2

2 = πhr

(2r – 6x)

For max. or min. put dzdx

= 0

\ πhr

(2rx – 3x2) = 0 or πhr

x (2r – 3x) = 0

But x ≠ 0, \ 2r – 3x = 0 or 3x = 2r or x = 23r

At x = 23r

, d zdx

2

2 = πhr

2 123

r r−

=

πhr

(– 2r) = – 2ph whichisnegative.

\ z is maximum at x = 23r

.

Putting x = 23r

in (i), y = hr

r r−

23

= hr

·r3

= h3

Putting x = 23r

in eqn. (iii),

Maximumvolumeofcylinder=πhr

r r r⋅ −

49

827

2 2

y

x

α

A

B C

R

P Q



= πhr

r3 49

827

−

= phr2 12 8

27−

= 427

phr2 = 427

ph (h tan a)2

Q In ARC RCAR

∆ , tan . ., tan tan= = ∴ =

α α αi e rh

r h

= 427

ph3tan2 a.

OR AC = 2R and BC = x Radius of the cylinder

= 12

4 2 2R − x

= V = π 12

4 2 22

R −

x x

⇒ V = π4

4 2 3R x x−( )

Find ddxVandsolve

ddxV

= 0 ⇒ x = 2

3R

Vmax = π4

4 2 3R x x−( ) = 43 3

3πR

\ Vmax = 13

43

3πR

= 13

(Volume of sphere)

36.Cartesianequationofgivenplanesare 2x + y – 3z + 4 = 0 ...(i) and 3x + 4y + 8z – 1 = 0 ...(ii) Let the equation of the required plane be (2x + y – 3z + 4) + l(3x + 4y + 8z – 1) = 0 ⇒ (2 + 3l)x + (1 + 4l)y + (–3 + 8l)z + (4 – l) = 0 ...(iii) For point A(a,0,0), equation (iii) becomes (2 + 3l)a + (4 – l) = 0 ...(iv) For point B(0,a,0), equation (iii) becomes (1 + 4l)a + (4 – l) = 0 ...(v) By equation (iv) and equation (v), we get (1 + 4l)a – (2 + 3l)a = 0 ⇒ l = 1 Put l = 1 in equation (iii), we get [2 + 3(1)]x + [1 + 4(1)]y + [–3 + 8(1)]z + 4 – 1 = 0 i.e., 5x + 5y + 5z + 3 = 0

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CD

x2R

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X

Z

Y

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