IITJEE 2012 Paper1 solution

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C Question Paper FormatThe questions paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consist of three sections.

12. Section I contains 10 multiple choice questions. Each question has four choices. (A), (B), (C) and (D) out of

which ONLY ONE is correct.

13. Section II contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which

ONE or MORE are correct.

14. Section III contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both

inclusive).

D.Marking Scheme15. For each question in Section I, you will be awarded 3 marks if you darken the bubble corresponding to the correct

answer ONLY and zero marks if no bubbles are darkened. In all ohter cases, minus one (–1) mark will be awarded

in this section.

16. For each question in Section II, you will be awareded 4 marks if you darken ALL the bubble(s) corresponding to the

correct answer(s) ONLY. In all other cases zero (0) marks will be awarded. No negative marks will be awarded for

incorrect answers in this section.

17. For each question in Section III, you will be awarded 4 marks if you darken the bubble corresponding to the correct

answer ONLY. In all other cases zero (0) will be awarded. No negative marks will be awarded for incorrect answers

in this section.

Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS

Sri Chaitanya IIT AcademyAdmin Off : Plot No : 304, Kasetty Heights.Ayyappa Society - Madhapur , Pin : 500081

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Sri Chaitanya IIT Academy

PHYSICS SECTION – I

( SINGLE CORRECT CHOICE TYPE )This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct

1. A small block is connected to one end of massless spring of un–stretched length 4.9 m. The

other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless

surface. The block is stretched by 0.2 m and released from rest at t= 0. It then executes

simple harmonic motion with angular frequency /3

rad s . Simultaneously at t = 0, a

small pebble is projected with speed v from point P at an angle of 450 as shown in the figure.

Point P is at a horizontal distance to 10 m from O. If the pebble hits the block at t = 1s, the

value of v is (take g = 10 m/s2)

a) 50 /m s b) 51 /m s c) 52 /m s d) 53 /m s

Ans : A

Equation of SHM cosy A t

At 1t 0.2cos (1) 0.2cos 0.13

y m

Block is at a distance 5m

Range of pebble is R =5m

22sin90 50

10VR V

50 /V m s



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2. In the determination of Young’s modulus 24MLgY

ld

by using Searle’s method, a wire of

length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension

l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a

screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The num-

ber of divisions on their circular scale is 100. The contributions to the maximum probable

error of the Y measurement.

a) due to the errors in the measurements of d and l are the same

b) due to the error in the measurement of d is twice that due to the error in the measurement of l.

c) due to the error in the measurement of l is twice that due to the error in the measurement of d.

d) due to the error in the measurement of d is four times that due to the error in the measure-

ment of l.

Ans :A

2

2 2 4d d dd v

40.25

l l ll

3. A small mass m is attached to a massless string whose other end is fixed at P as shown in the

figure. The mass is undergoing circular motion in the x–y plane with centre at O and constant

angular speed . If the angular momentum of the system, calculated about O and P are de-

noted by 0L

and PL

respectively, then



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a) 0L

and PL

do not vary with time

b) 0L

varies with time while PL

remains constant

c) 0L

remains constant while PL

varies with time

d) 0L

and PL

both vary with time

Ans :C

Torque about O is zero angular momentum will be constant about ‘O’. Torque about P is nonzero. Angular momentum about P varies with time.

0L m r v

PL m OP v

direction of PL

changes with OP and PL OP

4. Young’s double slit experiment is carried out by using green, red and blue light, one color at

a time. The fringe widths recorded are ,G R and B , respectively. Then

a) G B R b) B G R c) R B G d) R G B

Ans :D

Dd

R G B

R G B



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5. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform

positive surface charge density. The variation of the magnitude of the electric field E r

and the electric potential V(r) with the distance r from the centre, is best represented bywhich graph ?

a) b)

c) d)

Ans : D

From centre to surface E = 0, then 1Er

From centre to surface V = k, then 1Vr

6. A bi–convex lens is formed with two thin plano–convex lenses as shown in the figure. Re-fractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curvedsurfaces are of the same radius of curvature R = 14 cm. For this bi–convex lens, for an objectdistance of 40 cm, the image distance will be

a) – 280. 0 cm b) 40.0 cm c) 21. 5 cm d) 13.3 cm



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Ans :B

1 2

1 1 1F f f =

1 21 1R

=

0.5 0.2 0.7 114 14 20

20F cm

40R cm

1 1 1 1 1 1 1 1 1 2 1 120 40 20 40 40 40F v u v v

40v cm

7. Two large vertical and parallel metal plates having as separation of 1 cm are connected to aDC voltage source of potential difference X. A proton is released at rest midway between thetwo plates. It is found to move at 450 to the vertical JUST after release. Then X is nearly

a) 51 10 V b) 71 10 V c) 91 10 V d) 101 10 V

Ans :C

mg Eq

45

mg

EqpVqd

910mgdV V approxq

8. Three very large plates of same area are kept parallel and close to each other. They are con-sidered as ideal black surfaces and have very high thermal conductivity. The first and thirdplates are maintained at temperatures 2T and 3T respectively. The temperature of the middle(i.e. second) plate under steady state condition is

a) 1465

2T

b) 1497

4T

c) 1497

2T

d) 1497 T

Ans : C

4 4 4 41 2 2 3A T T A T T

144 44 4

2 2 2972 32

T T T T T T



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9. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angularspeed , as shown in the figure. At time t = 0, a small insect starts from O and moves withconstant speed v with respect to the rod towards the other end. It reaches the end of the rod att = T and stops. The angular speed of the system remains throughout. The magnitude of the

torque of the torque on the system about O, as a function of time is best represented bywhich plot ?

a) b)

c) d)

Ans : B

dL d dIIdt dt dt

constantrodI

2red

d I mrdt

= 0 2 drm rdt

v

r = vt

22 2mr v m v t

t

when the insect reaches the end it stops v = 0

0



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10. A mixture of 2 moles of helium gas (atomic mass = 4amu) and 1 mole of argon gas (atomic

mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds heliumargon

rms

rms

vv

is

a) 0.32 b) 0.45 c) 2.24 d) 3.16Ans : D

rmsTVM

1 2

2 1

40 104

He

Ar

VV MV V M

SECTION – II( MULTIPLE CORRECT ANSWER TYPE)

This section contains 5 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer,out of which ONE or More are correct.

11. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle with the horizontal. A horizontal force of 1 N acts on the block through its center of mass asshown in the figure. The block remains stationary if (take 210 /g m s )

a) 045

b) 045 and a frictional force acts on the block towards P.

c) 045 and a frictional force acts on the block towards Q.

d) 045 and a frictional force acts on the block towards Q.

Ans : A (or) AC



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1cos sinmg

45

when 45

friction force is upwards

If friction force sufficient block will be in state of rest

12. Consider the motion of a positive point charge in a region where there are simulataneousuniform electric and magnetic fields 0

ˆE E j

and 0ˆB B j

. At time t = 0 , this charge has

velcoity v in the x – y plane, making an angle with the x – axis. Which of the followingoption(s) is(are) correct for time t > 0 ?

a) If 00 , the charge moves in a circular path in the x – z plane

b) If 00 , the charge undergoes helical motion with constant pitch along the y – axis.

c) If 010 , the charge undergoes helical motion with its pitch increasing with time, along the y – axis

d) If 090 , the charge undergoes linear but accelerated motion along the y–axis

Ans :C,D

when 90

0mF

when 10

path is helical and with increasing pitch.



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13. For the resistance network shown in the figure, choose the correct option(s).

a) The current through PQ is zero b) 1 3I A

c) The potential at S is less than that at Q. d) 2 2I A

Ans : A,B,C,D

112 34

I

212 318

I 2A

P QV V

S QalsoV V

14. A person blows into open – end of a long pipe. A s a result, a high – pressure pulse of airtravels down the pipe.When this pulse reaches the other end of the pipe,

a) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is open

b) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is open

c) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is closedd) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

Ans : B, D



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Reflected pressure pulse

Incident pressure pulse

Open end

At open end pressure should not change pressure wave will get phase change as



Closed end

At enclosed end pressure wave will get reflected without phase change15. A cubical region of side a has its centre at the origin. It encloses three fixed point charges,

q at 0, / 4,0 , 3a q at (0, 0, 0) and q at 0, / 4,0a . Choose the correctoption(s).

a) The net electric flux crossing the plane / 2x a is equal to the net electric flux crossingthe plane / 2x a

b) The net electric flux crossing the plane / 2y a is more than the net electric flux crossingthe plane / 2y a .

c) The net electric flux crossing the entire region is 0

q .

d) The net electric flux crossing the plane / 2z a is equal to the net electric flux crossingthe plane / 2x a



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Ans : ACD

Net flux passing the plane 2ax and 2

ax is same

Net flux through cube is 0

q

Flux through plane 2az and 2

ax is same

SECTION – III( INTEGER ANSWER TYPE)

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).16. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform

mass density and radius 2R, as shown in the figure. The moment of inertia of this laminaabout axes passing through O and P is 0I and PI , respectively. Both these axes are perpen-

dicular to the plane of the lamina. The ratio 0

PII to the nearest integer is

Ans : 3

2 2 2 2 2 20

1 12 22 2

I R R R R R R

4 4382

R R

4 213 132 2

R MR

2 2 2 21 14 2 4 2 52 2PI M R M R MR M R

2 2 2118 162

MR MR MR

2 211 37242 2

MR MR



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17. A circular wire loop of radius R is placed in the x - y plane centered at the origin O. A squareloop of side a a R having two turns is placed with its center at 3z R along the axis ofthe circular wire loop, as shown in figure. The plane of the square loop makes an angle of

45with respect to the z - axis. If the mutual inductance between the loops is given by 2

0/ 22p

aR

,

then the value of p is

Ans : 7

2

203/222

2 14 23

iRB ai R R

(n = 2)

22 20 0

9/21

2 22aM R a

18. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the

figure. Both the cylinder and the cavity are infinetely long. A uniform current density J flows

along the length. If the magnitude of the magnetic field at the point P is given by 012N aJ ,

then the value of N is



# 15


Ans : 5

01 2

jaB

2

200

22

3 1222

ajjaB

a a

1 2 01 12 12

B B ja

0 512ja

19. A proton is fired from very far away towards a nucleus with charge 120Q e , where e is the

electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wave-

length (in units of fm) of the proton at its start is:

(take the proton mass, 275 / 3 10 ;pm kg 15/ 4.2 10 . /H e j S c ;

9

0

1 9 10 / ;4

m F

151 10fm m )

Ans : 7

2 29

151209 10

10 10 2 p

e Pm

9 2 27

215

59 10 120 2 103

10 10

eP

17

153 2 10 3 2

10 10e eP

154.2 106

10

hp



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20. An infinitely long solid cylinder of radius R has a uniform volume charge density . It has a

spherical cavity of radius R/2 with its centre on the axis on the cylinder, as shown in the

figure. The magnitude of the electric field at the point P, which is at the distance 2R from the

axis of the cylinder, is given by the expression 0

2316

Rk . The value of k is

Ans: 6

3

2

2

42 3 8

2 2 p

RKk R ER R

4 113 32pE k R

0

1 114 24

R

0

234 24p

RE

24 4 616

K



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CHEMISRY SECTION – I

( SINGLE CORRECT CHOICE TYPE )This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct21. The number of aldol reaction(s) that occurs in the given transformation is

. .3 4 conc aq NaOHCH CHO CHO

OH

OHHO

OH

a) 1 b) 2 c) 3 d) 4Ans : C

3 2OHCH CHO CH CHO

2 2 2H C H CH CHO O CH CH CHO

O

H

2 2OHCH CHO HO CH CH CHO H - C - H

OH

'HCHO Cannizaro sNaOH reactionCH CH

CHO

OH

OH CHO

HO

OH

OH OH OH

OHHO

Number of aldol reaction = 3



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22. A compound M PXP has cubic close packing (ccp) arrangement of X. Its unit cell structure isshown below. The empirical formula of the compound is

a) MX b) MX2 c) M2X d) M5X14

Ans : B

Corner

Face center 16 32

18 18

4

X :

Edge centre

Body center 1 1 1

14 14

2

M :

4 2X M

4 2X M 2MX

23. The carbonyl functional group (–COOH) is present ina) picric acid b) barbituric acid c) ascorbic acid d) aspirin

Ans : DAcetyle salicylic acid = Asprin

COOH

OCOCH3



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24. The colour of light absorbed by an aqueous solution of CuSO4 is :

a) organe-red b) blue-green c) yellow d) violetAns : A

Orange - red is absorbedBased up on Munshell wagon wheel

V

B

G

O

R

Y

25. The number of optically active products obtained from the complete ozonolysis of thegiven compound is :

CH3 - CH = CH - C - CH = CH - C - CH = CH - CH3

HCH3

H CH3

a) 0 b) 1 c) 2 d) 4Ans : A

As the ozonolysis products do not have chiral centres, number of optically active products = 0

26. As per IUPAC nomenclautre, the name of the complex 2 3 34 2Co H O NH Cl is

a) Tetraaqudiaminecobalt (III) chloride b) Tetraaquadiamminecobalt (III) chloridec) Diammineteraaquacobalt (III) chloride d) Diamminetetraaquacobalt (III) chloride

Ans : D

2 3 34 2Co H O NH Cl

Diammineteraaquacobalt (III) chloride

27. Which ordering of compound is according to the decreasing order of the oxidation state ofnitrogen ?

a) 3 4 2, , ,HNO NO NH Cl N b) 3 2 4, , ,HNO NO N NH Cl

c) 3 4 2, , ,HNO NH Cl NO N d) 3 4 2, , ,NO HNO NH Cl N



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Ans : B

3 5HNO

2NO

2 0N

4 3NH Cl

Decreasing order of

O -S = 3 2 4, , ,HNO NO N NH Cl

28. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. I/V plot is shownbelow. The value of the van der Waals constant a (atm. liter2 mol–2) is

a) 1.0 b) 4.5 c) 1.5 d) 3.0Ans : C

2

2

n aP v nb nRTv

’

n = 1, b = 0

2

qp v b RTv

apv RTv

apv RTv

2 221.6 20.1 1.5 .3 2

y c mx m a slop atm t mole



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29. In allene 3 4C H , the type(s) of hybridisation of the carbon atoms is (are)

a) sp and sp3 b) sp and sp2 c) only sp2 d) sp2 and sp3

Ans : B

2 2H C C CH

2 2sp sp sp

30. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohrradius]

a) 2

2 204

hma b)

2

2 2016

hma c)

2

2 2032

hma d)

2

2 2064

hma

Ans : C

mvr= 2nh

v= 2nh

mr

for second Bohr’s orbit

2r =4 0a

V=04

hma

22

2 2 2016

hVm a

K.E2

22 2

0

12 32

hmvma



# 22




31. Which of the following molecules, in prue form, is (are) unstable at room temperature ?

a) b) c)

O

d)

O

Ans : BC

cyclobutadiene and cyclopentadien are anti aromatic.

32. Which of the following hydrogen halides react(s) with 3AgNO aq to give a precipitate that

dissolves in 2 2 3Na S O aq ?

a) HCl b) HF c) HBr d) HI

Ans : ACD

3 3AgNO HX AgX HNO aq aq

2 2 3 3 2 3 22AgX Na S O Na Ag S O NaX aq aqs

AgF is soluble

33. Choose the correct reason(s) for the stability of the lyophobic colloidal particles >

a) Preferential adsorption of ions on their surface from the solution

b) Preferential adsorption of solvent of their surface the solution

c) Attraction between different particles having opposite charges on their surface

d) Potential difference between the fixed layer and the diffused layer of opposite chargesaround the coolidal particles

Ans : AD



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34. For an ideal gas, consider only P-V work in going from an

ini tial state X to the f inal state Z . The can be reached by

either of the two paths shown in the figure. Which of the

following choice(s) is (are) correct ? [Take S as change

in entropy and w as work done]

a) x z x y y zS S S b) x z x y y zW w w

c) x y z x yw w d) x y z x yS S

Ans : AC

Based up on the given graph

x y x y y zS S S

x y z x yW W

35. Identify the binary mixtrue(s) that can be separated into individual compounds, by differen-tial extraction,a s shown in the given scheme.

a) 6 5 6 5C H OH and C H COOH b) 6 5 6 5 2C H COOH and C H CH OH

c) 6 5 2 6 5C H CH OH and C H OH d) 6 5 2 6 5 2C H CH OH and C H CH COOH

Ans : BD

Benzyl alcohol is insoluble in both NaOH and 3NaHCO ,



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This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).

36. When the following aldohexose exists in its B-configuration, the total number C

CH2

CHOH

CHOH

CHO

CH2OH

OHH

of stereoisomrs in its pyranose form is :

Ans : 8

As the configration at C-5 is fixed, (D-configration), then only 4 stero isomers are possiblesolution

37. 29.2% (w/w) HCl stock solution has a densityh of 1.25 g mL–1. The molecular weight of HCl

is 36.5 g mol–1. The volume (mL) of stock solution required to prepare a 200 mL solution of

0.4 M HCl is :

Ans : 8

Molarity of stock solution =10 d x

GMW

=10 1.25 29.2

36.5

=10M

1 1 2 2

1

1

10 200 0.48 ( )

V M V MVV ML approx

38. The periodic table consists of 18 groups. An isotope of copper, on bombardment with pro-

tons, undergoes a nuclear reaction yielding element X as shown below. To which

group,element X belongs int he periodic table ?

Ans : 863 1 1 4 1 5229 1 0 2 1 266 2CU H n He H X

26 Fe 8 th group



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39. An organic compound undergoes first-order decomposition. The time taken for its decom-

position to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the

Ans : 9

18

110

2.303 log

82.303 log

10

at ak

at ak

18

110

log8log10

t

t

18

110

3log 2t

t

18

110

10 9t

t

40. The substituents R1 and R2 for nine peptides are listed in the table given below. How many of

these peptides are positively charged at pH = 7.0 ?

Ans : 4

At pH below pI(Isoelectric point) Aminoacid exist as Cation. As BasicAminoacids of pHaround 9-8, these are expected to exists as cation of pH=7IV,VI,VIII,IX.



# 26


MATHEMATICSSECTION – I

( SINGLE CORRECT CHOICE TYPE )This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONLY ONE is correct

41. Let z be a complex number such that the imaginary part of z is nonzero and and 2 1a z z is real. Then a connot take the value

a) –1 b) 13 c)

12 d)

34

Ans. D

Let Z x iy 0y

2 1a z z becomes

2 1a x iy x iy

2 2 1 2 0x y x i xy y a i

Comparing real & imaginary parts

2 2 1x y x a ––––––(1)

2 0xy y –––––––(2)

From (2) 1

2x as 0y

Put value of x in (1)

221 1 1

2 2y a

21 14 2

y a

2 3 04

y a

34

a



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42. If 2 1lim 4

1x

x x ax bx

, then

a) a = 1, b = 4 b) a = 1, b = –4 c) a = 2, b = –3 d) a = 2, b = 3

Ans. B

2 1lim 41x

x x ax bx

2 1 1 14

1x

x a x a b blim

x

For existense of limit coefficient of 2 0x

1 – a = 0

a = 1

1 1lim 4

1x

x a b bx

1 4a b

b = – 4

43. Let ijP a be a 3 3 matrix and let ijQ b , where 2i jij ijb a for 1 , 3i j . If the deter-

minant of P is 2, then the determinant of the matrix Q is

a) 210 b) 211 c) 212 d) 213

Ans. D

as 2i jij ijb a

11 11 12 12 13 134 , 8 , 16b a b a b a

21 2 22 22 23 238 , 16 , 32b a b a b a

31 31 32 32 33 3316 , 32 , 64b a b a b a

11 12 13

21 22 23

31 32 33

2a a a

A a a aa a a



# 28


11 12 13

21 22 23

31 32 33

4 8 168 16 32

16 32 64

a a aB a a a

a a a

11 12 13

21 22 23

31 32 33

4 8 16 2 2 24 4 4

a a aa a aa a a

11 12 13

21 22 23

31 32 33

4 8 16 2 4a a aa a aa a a

4 8 16 2 4 2

132B

44. The ellipse 2 2

1 : 19 4x yE is inscribed in a rectangle R whose sides are parallel to the coor-

dinates axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangleR. The eccentricity of the ellipse E2 is

a) 22

b) 32

c) 12 d)

34

Ans. C

Given Ellipse 2 2

19 4x y

As ellipse is bounded by &x a y b

Sides of rectangle are

3 & 2x y

3,2 , 3,2C D

D(0,2)

(3,0)

(0,-2)

(-3,0)

A B

C

3, 2 , 3, 2A B

As new ellipse

Circumscribe the rectangle & passes through (0 ,4).



# 29


Let new ellipse is 2 2

2 2 1x ya b

Passes through 22

160,4 1 16bb

Passes through 3,2 2

9 4 116a

2 12a

So, Vertical ellipse 2 2

112 16x y

2 2 21a b e

212 16 1 e

2 12 1116 4

e

12

e

45. The function : 0,3 1, 29f , defined by 3 22 15 36 1f x x x x is

a) one-one and onto b) onto but not one-one

c) one-one but not onto d) neither one-one nor onto

Ans. B

: 0,3 1, 29f

3 22 15 36 1f x x x x

' 26 30 36f x x x

6 2 3x x

'' 6 2 5f x x

2x is pt of maxima.



# 30


3x is pt of minima.

& 2 29 0f

3 0f

so curve is

nf is not one one

2 3Min value 0 1f

Max value 2 29f

so nf is Onto.

46. The locus of the mid-point of the chord of contact of tangents drawn from points lying onthe straight line 4 5 20x y to the circle 2 2 9x y is

a) 2 220 36 45 0x y x y b) 2 220 36 45 0x y x y

c) 2 236 20 45 0x y x y d) 2 236 20 45 0x y x y

Ans. A

Let 1 1, & ,P h k Q x y

As P lies of 4x-5y=20

4 20,5

hP h

P

A

B

Q

Equation of chord of contact

1 0AB S

4 20 95

hxh y

5 4 20 45xh y h ------------------(1)

Equation of chord AB whose mid point is 1 1,Q x y



# 31


2 21 1 1 1xx yy x y ---------------(2)

Compare (1) & (2)

2 21 1 1 1

5 4 20 45h hx y x y

1 12 2 2 21 1 1 1

9 45& 4 20x yh hx y x y

On eliminating h

1 12 2 2 21 1 1 1

49 4520x yx y x y

2 21 1 1 120 36 45x y x y

Locus is

2 220 36 45 0x y x y

47. The point P is the intersection of the straight line joining the points Q (2, 3, 5) and

R (1, –1, 4) with the plane 5 4 1x y z . If S is the foot of the perpendicular drawn from thepoint T (2, 1, 4) to QR, then the length of the line segement PS is

a) 12 b) 2 c) 2 d) 2 2

Ans. A

Equation of 1 1 4

1 4 1x y zQR

Dr’s of 1, 4,1QR

Co-ordinates of 1 , 1 4 , 4P

As P is point of Intersection of QR & given plane, so

5 1 4 1 4 4 1

13

T(2,1,4)

Q S R



# 32


4 1 13, ,3 3 3

P

Let 1 , 1 4 ,4S

Dr’s of 1, 4 2,TS

TS QR So,

1 1 4 4 2 0

12

3 9,1,2 2

S

2 2 24 3 1 13 913 2 3 3 2

PS

12

48. Let 2 cos , 0

0, 0

x xf x x

x

, x IR , then f is

a) differentiable both at x = 0 and at x = 2

b) differentiable at x = 0 but not differentiable at x = 2

c) not differentiable at x = 0 but differentiable at x = 2

d) differentiable neither at x = 0 nor at x = 2

Ans. B

2 cos ; 0

0 ; 0

x xf x x

x

At x=0



# 33


2

0 0

cos 00 0lim limh h

hf h f hRHDh h

0lim cos 0h

hh

2

0 0

cos 00 0

lim limh h

hf h f hLHD

h h

0lim cos 0h

hh

Differentiable at x=0

At x=2

2

0 0

2 cos 4 cos2 2 2 2lim limh h

hf h f hRHDh h

2

0 0


hf h f hLHDh h

Not Differentiable at x=2.

49. The total number of ways in which 5 balls of different colours can be distributed among 3persons so that each person gets at least one ball is

a) 75 b) 150 c) 210 d) 243

Ans. B

5 Balls to 3 persons, so that all persons get atleast one ball.

Case-I

Total ways 2 2 1

2 1 2

1 2 2

5 3 12 2 1 3C C C



# 34


10 3 1 3 90

Case-II

Total ways

5 2 13 1 1 3C C C

11

1

13

3

1

3110 2 1 3 60

Total ways = 90 + 60 = 150.

50. The integral

2

9/ 2sec

sec tanx dx

x x equals (for some arbitrary constant K)

a) 2

11/ 2

1 1 1 sec tan11 7sec tan

x x Kx x

b) 2

11/ 2


x x Kx x

c) 2

11/ 2


x x Kx x

d) 2

11/ 2


x x Kx x

Ans. C

9

2

sec

sec tan

x dxx x

Let sec tanx x t -------------(1)

So, 2sec tan secx x x dx dt

sec dtx dxt

------------(2)



# 35


1sec tanx xt

----------(3)

From (1) & (3)

1 1sec2

x tt

So, Integral becomes

9

2

sec .sec

sec tan

x x dxx x

92

1 12

dtt ttt

2

132

1 12

t dtt

9 132 21

2t dt t dt

7 112 21

7 112 22

t t K

7 112 2

7 11t t K

7 11

22

1 1

11 sec tan7 sec tan x xx x

2

112


x x Kx x



# 36


SECTION – II( MULTIPLE CORRECT CHOICE TYPE )

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for itsanswer, out of which ONE OR MORE is/ are correct

51. A ship is fitted with three engines E1 , E2 and E3. The engines function independently of each

other with respective probabilities 1 1,2 4 and

14 . For the ship to be operational at least two of

its engines must function. Let X denote the event that the ship is operational and let X1, X2and X3

denote respectively the events that the engines E1, E2 and E3 are functioning. Which ofthe following is (are) true ?

a) 13|

16cP x X b) P [Exactly two engines of the ship are functioning | X]

78

c) 25|

16P X X d) 1

7|16

P X X

Ans. B,D

A)

1

1 1 112 4 4

1 84

cP X XP X

1 1 3 1 1 1 12 22 4 4 2 4 4 4

P X

B) 1 1 3 1 1 12

72 4 4 2 4 4/ 1 84

P Exactly twoengines of the ship are functioning X

C) 2

22

1 1 3 1 1 1 252 4 4 2 4 4/ 1 8

4

P X XP X X

P X

D)

11

1

1 1 3 1 1 1272 4 4 2 4 4/ 1 16

2

P X XP X X

P X



# 37


52. Tangents are drawn to the hyperbola 2 2

19 4x y

, parallel to the straight line 2 1x y . The

points of contact of the tangents on the hyperbola are

a) 9 1,

2 2 2

b) 9 1,

2 2 2

c) 3 3, 2 2 d) 3 3,2 2

Ans. A, B

2 2

19 4x y

Tangent to the hyperbola is of the form2y x

2 9 4 4 36 4 32

4 2

Tangent is : 2 4 2 0x y

Point of contact is 2 2 9 1 9 1, , , ,

2 2 2 2 2 2a l b mn n

53. Let S be the area of the region enclosed by 2xy e , y = 0, x = 0, and x = 1. Then

a) 1Se

b) 11Se

c) 1 114

Se

d)

1 1 112 2

Se

Ans. A, B, D

21 1

0 0

11x xI e dx e dxe

11Ie

21 1xe e

21 1 1

1

0 0 0

xe dx e dx dx

21

0

1xe dxe



# 38


54. If y (x) satisfies the differential equation ' tan 2 secy y x x x and y (0) = 0, then

a) 2

4 8 2y

b) 2

'4 18

y

c) 2

3 9y

d) 24 2'

3 3 3 3y

Ans. A,D

tan 2 secdy y x x xdx

tan. cos

dxI F e x

General solution is cos 2 sec cosy x x x x dx 2x c

2 sec secy x x c x

0,0

0 0 0c c .

2 secy x x

2

24 16

y

2

8 2

2

23 9

y

2 sec tan 2 secdy x x x x xdx

.

2

1 2 3 2 23 9 3

y

22 433 3



# 39


55. Let , 0, 2 be such that

22cos 1 sin sin tan cot cos 12 2

,

tan 2 0 and 31 sin2

.

Then cannot satisfy

a) 0 2 b)

42 3 c)

4 33 2 d)

3 22

Ans. A, C, D

3tan 2 0, 1 sin2

3 5 1, 0 cos2 3 2

______ (1)

22cos 1 sin sin tan / 2 cot / 2 cos 1

1sin cos2

1 sin 12

(from (1))

13 176 6

2 73 6 (from (1))

cannot satisfied by A, C, D



# 40


SECTION –III( INTEGER ANSWER TYPE)

This section contains 5 questions . The answer to each of the questions is a single digit integer, ranging from 0 to9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened.

56. Let p (x) be a real polynomial of least degree which has a local maximum at x = 1 and a localminimum at x =3. If p (1) = 6 and p (3) = 2, then ' 0p is

Ans. 9

3 2P x ax bx cx d

1 6 6P a b c d

3 2 27 9 3 2P a b c d

1 1 0 3 2 0P a b c

1 3 0 27 6 0P a b c

1, 6, 9a b c

1 23 2P x ax bx c

1 0 9P c

57. If ,a b and c are unit vectors satisfying

2 2 2 9a b b c c a ,then 2 5 5a b c

is\

Ans. 3

sol. 1a b c

2 2 2 9a b b c c a

3.

2a b b c c a

3 2 . . 0a b b c c a

22 2 2 . . . 0a b c a b b c c a

2

0a b c



# 41


0a b c ––––––(1)

Taking dot product with (1) by , &a b c

then 1 1 1. , . , .

2 2 2a b b c c a

2 5 5 3a b c

58. The value of 32

1 1 1 16 log 4 4 4 ...3 2 3 2 3 2 3 2

is

Ans. 4

sol.1 4

3 2x x

3 2 4x x

squaring on both sides218 4x x 218 4 0x x 218 9 8 4 0x x x

9 2 1 4 2 1 0x x x .

2 1 9 4 0x x

1 4,2 9

x

49

x 0x

Required answer is 2

3/ 236 log2

= 6 – 2 = 4

59. Let S be the focus of the parabola 2 8y x and let PQ be the common chord of the circle2 2 2 4 0x y x y and the given parabola. The area of the triangle PQS is

Ans. 4



# 42


sol.P

Q

S(0, 0)

(2,4)

(2, 0)2

(0,4)

2 8 2 4 2 0x x x x

0,2 0,0 , 2,4x P Q

area of SPQ

= 1 2 42

= 4

60. Let :f IR IR be defined as 2 1f x x x . The total number of points at which f attainseither a local maximum or a local minimum is

Ans. 5sol. 2 1, 1f x x x x

21 , 1 0x x x

21 , 0 1x x x

2 1, 1x x x

' 2 1, 1f x x x

2 1, 1 0x x

0-1 1x

y

1 2 , 0 1x x

1 2 , 1x x

differentiable at 1 1,2 2

x ,

not differentiable at –1, 0, 1

The number of points = 5

Sri Chaitanya IIT Academy., A.P.

IIT-JEE 2012Paper - 1 along with Key

& Solutions

Admin Off : Plot No : 304, Kasetty Heights.Ayyappa Society - Madhapur , Pin : 500081



invigilators.


of this booklet.
























inclusive).



in this section.






in this section.



# 3












a) 50 /m s b) 51 /m s c) 52 /m s d) 53 /m s

Ans : A


At 1t 0.2cos (1) 0.2cos 0.13

y m



22sin90 50

10VR V

50 /V m s



# 4



ld











ment of l.

Ans :A

2

2 2 4d d dd v

40.25

l l ll




noted by 0L

and PL

respectively, then



# 5


a) 0L

and PL


b) 0L


remains constant

c) 0L


varies with time

d) 0L

and PL

both vary with time

Ans :C


0L m r v

PL m OP v

direction of PL





Ans :D

Dd

R G B

R G B



# 6





a) b)

c) d)

Ans : D




a) – 280. 0 cm b) 40.0 cm c) 21. 5 cm d) 13.3 cm



# 7


Ans :B

1 2

1 1 1F f f =

1 21 1R

=

0.5 0.2 0.7 114 14 20

20F cm

40R cm

1 1 1 1 1 1 1 1 1 2 1 120 40 20 40 40 40F v u v v

40v cm


a) 51 10 V b) 71 10 V c) 91 10 V d) 101 10 V

Ans :C

mg Eq

45

mg

EqpVqd

910mgdV V approxq


a) 1465

2T

b) 1497

4T

c) 1497

2T

d) 1497 T

Ans : C

4 4 4 41 2 2 3A T T A T T

144 44 4

2 2 2972 32

T T T T T T



# 8




a) b)

c) d)

Ans : B

dL d dIIdt dt dt

constantrodI

2red

d I mrdt

= 0 2 drm rdt

v

r = vt

22 2mr v m v t

t


0



# 9




rms

rms

vv

is

a) 0.32 b) 0.45 c) 2.24 d) 3.16Ans : D

rmsTVM

1 2

2 1

40 104

He

Ar

VV MV V M




a) 045




Ans : A (or) AC



# 10


1cos sinmg

45

when 45




ˆE E j

and 0ˆB B j







Ans :C,D

when 90

0mF

when 10




# 11





Ans : A,B,C,D

112 34

I

212 318

I 2A

P QV V

S QalsoV V





Ans : B, D



# 12




Open end




Closed end






q .




# 13


Ans : ACD


ax is same


q


ax is same






Ans : 3

2 2 2 2 2 20

1 12 22 2

I R R R R R R

4 4382

R R

4 213 132 2

R MR

2 2 2 21 14 2 4 2 52 2PI M R M R MR M R

2 2 2118 162

MR MR MR

2 211 37242 2

MR MR



# 14




0/ 22p

aR

,


Ans : 7

2

203/222

2 14 23

iRB ai R R

(n = 2)

22 20 0

9/21

2 22aM R a







# 15


Ans : 5

01 2

jaB

2

200

22

3 1222

ajjaB

a a

1 2 01 12 12

B B ja

0 512ja





9

0

1 9 10 / ;4

m F

151 10fm m )

Ans : 7

2 29

151209 10

10 10 2 p

e Pm

9 2 27

215

59 10 120 2 103

10 10

eP

17

153 2 10 3 2

10 10e eP

154.2 106

10

hp



# 16






2316


Ans: 6

3

2

2

42 3 8

2 2 p

RKk R ER R

4 113 32pE k R

0

1 114 24

R

0

234 24p

RE

24 4 616

K



# 17





OH

OHHO

OH

a) 1 b) 2 c) 3 d) 4Ans : C

3 2OHCH CHO CH CHO


O

H


OH


CHO

OH

OH CHO

HO

OH

OH OH OH

OHHO




# 18




Ans : B

Corner

Face center 16 32

18 18

4

X :

Edge centre

Body center 1 1 1

14 14

2

M :

4 2X M

4 2X M 2MX



COOH

OCOCH3



# 19





V

B

G

O

R

Y



HCH3

H CH3

a) 0 b) 1 c) 2 d) 4Ans : A




Ans : D








# 20


Ans : B

3 5HNO

2NO

2 0N

4 3NH Cl

Decreasing order of

O -S = 3 2 4, , ,HNO NO N NH Cl


a) 1.0 b) 4.5 c) 1.5 d) 3.0Ans : C

2

2

n aP v nb nRTv

’

n = 1, b = 0

2

qp v b RTv

apv RTv

apv RTv

2 221.6 20.1 1.5 .3 2




# 21




Ans : B

2 2H C C CH

2 2sp sp sp


a) 2

2 204

hma b)

2

2 2016

hma c)

2

2 2032

hma d)

2

2 2064

hma

Ans : C

mvr= 2nh

v= 2nh

mr


2r =4 0a

V=04

hma

22

2 2 2016

hVm a

K.E2

22 2

0

12 32

hmvma



# 22





a) b) c)

O

d)

O

Ans : BC





Ans : ACD



AgF is soluble






Ans : AD



# 23









Ans : AC


x y x y y zS S S

x y z x yW W




Ans : BD




# 24





CH2

CHOH

CHOH

CHO

CH2OH

OHH


Ans : 8




0.4 M HCl is :

Ans : 8


GMW

=10 1.25 29.2

36.5

=10M

1 1 2 2

1

1

10 200 0.48 ( )

V M V MVV ML approx




Ans : 863 1 1 4 1 5229 1 0 2 1 266 2CU H n He H X

26 Fe 8 th group



# 25




Ans : 9

18

110

2.303 log

82.303 log

10

at ak

at ak

18

110

log8log10

t

t

18

110

3log 2t

t

18

110

10 9t

t



Ans : 4




# 26





a) –1 b) 13 c)

12 d)

34

Ans. D

Let Z x iy 0y

2 1a z z becomes

2 1a x iy x iy

2 2 1 2 0x y x i xy y a i


2 2 1x y x a ––––––(1)

2 0xy y –––––––(2)

From (2) 1

2x as 0y


221 1 1

2 2y a

21 14 2

y a

2 3 04

y a

34

a



# 27


42. If 2 1lim 4

1x

x x ax bx

, then

a) a = 1, b = 4 b) a = 1, b = –4 c) a = 2, b = –3 d) a = 2, b = 3

Ans. B

2 1lim 41x

x x ax bx

2 1 1 14

1x

x a x a b blim

x


1 – a = 0

a = 1

1 1lim 4

1x

x a b bx

1 4a b

b = – 4



a) 210 b) 211 c) 212 d) 213

Ans. D

as 2i jij ijb a

11 11 12 12 13 134 , 8 , 16b a b a b a

21 2 22 22 23 238 , 16 , 32b a b a b a

31 31 32 32 33 3316 , 32 , 64b a b a b a

11 12 13

21 22 23

31 32 33

2a a a

A a a aa a a



# 28


11 12 13

21 22 23

31 32 33

4 8 168 16 32

16 32 64

a a aB a a a

a a a

11 12 13

21 22 23

31 32 33

4 8 16 2 2 24 4 4

a a aa a aa a a

11 12 13

21 22 23

31 32 33

4 8 16 2 4a a aa a aa a a

4 8 16 2 4 2

132B

44. The ellipse 2 2



a) 22

b) 32

c) 12 d)

34

Ans. C

Given Ellipse 2 2

19 4x y



3 & 2x y

3,2 , 3,2C D

D(0,2)

(3,0)

(0,-2)

(-3,0)

A B

C

3, 2 , 3, 2A B

As new ellipse




# 29



2 2 1x ya b

Passes through 22

160,4 1 16bb


9 4 116a

2 12a


112 16x y

2 2 21a b e

212 16 1 e

2 12 1116 4

e

12

e




Ans. B

: 0,3 1, 29f

3 22 15 36 1f x x x x

' 26 30 36f x x x

6 2 3x x

'' 6 2 5f x x

2x is pt of maxima.



# 30


3x is pt of minima.

& 2 29 0f

3 0f

so curve is

nf is not one one

2 3Min value 0 1f

Max value 2 29f

so nf is Onto.


a) 2 220 36 45 0x y x y b) 2 220 36 45 0x y x y

c) 2 236 20 45 0x y x y d) 2 236 20 45 0x y x y

Ans. A

Let 1 1, & ,P h k Q x y


4 20,5

hP h

P

A

B

Q


1 0AB S

4 20 95

hxh y

5 4 20 45xh y h ------------------(1)




# 31


2 21 1 1 1xx yy x y ---------------(2)

Compare (1) & (2)

2 21 1 1 1

5 4 20 45h hx y x y

1 12 2 2 21 1 1 1

9 45& 4 20x yh hx y x y

On eliminating h

1 12 2 2 21 1 1 1

49 4520x yx y x y

2 21 1 1 120 36 45x y x y

Locus is

2 220 36 45 0x y x y



a) 12 b) 2 c) 2 d) 2 2

Ans. A

Equation of 1 1 4

1 4 1x y zQR

Dr’s of 1, 4,1QR



5 1 4 1 4 4 1

13

T(2,1,4)

Q S R



# 32


4 1 13, ,3 3 3

P

Let 1 , 1 4 ,4S

Dr’s of 1, 4 2,TS

TS QR So,

1 1 4 4 2 0

12

3 9,1,2 2

S

2 2 24 3 1 13 913 2 3 3 2

PS

12

48. Let 2 cos , 0

0, 0

x xf x x

x

, x IR , then f is





Ans. B

2 cos ; 0

0 ; 0

x xf x x

x

At x=0



# 33


2

0 0

cos 00 0lim limh h

hf h f hRHDh h

0lim cos 0h

hh

2

0 0

cos 00 0

lim limh h

hf h f hLHD

h h

0lim cos 0h

hh


At x=2

2

0 0


hf h f hRHDh h

2

0 0


hf h f hLHDh h



a) 75 b) 150 c) 210 d) 243

Ans. B


Case-I

Total ways 2 2 1

2 1 2

1 2 2

5 3 12 2 1 3C C C



# 34


10 3 1 3 90

Case-II

Total ways

5 2 13 1 1 3C C C

11

1

13

3

1

3110 2 1 3 60

Total ways = 90 + 60 = 150.

50. The integral

2

9/ 2sec

sec tanx dx


a) 2

11/ 2


x x Kx x

b) 2

11/ 2


x x Kx x

c) 2

11/ 2


x x Kx x

d) 2

11/ 2


x x Kx x

Ans. C

9

2

sec

sec tan

x dxx x

Let sec tanx x t -------------(1)


sec dtx dxt

------------(2)



# 35


1sec tanx xt

----------(3)

From (1) & (3)

1 1sec2

x tt


9

2

sec .sec

sec tan

x x dxx x

92

1 12

dtt ttt

2

132

1 12

t dtt

9 132 21

2t dt t dt

7 112 21

7 112 22

t t K

7 112 2

7 11t t K

7 11

22

1 1


2

112


x x Kx x



# 36









a) 13|


78

c) 25|

16P X X d) 1

7|16

P X X

Ans. B,D

A)

1

1 1 112 4 4

1 84

cP X XP X

1 1 3 1 1 1 12 22 4 4 2 4 4 4

P X

B) 1 1 3 1 1 12

72 4 4 2 4 4/ 1 84


C) 2

22

1 1 3 1 1 1 252 4 4 2 4 4/ 1 8

4

P X XP X X

P X

D)

11

1

1 1 3 1 1 1272 4 4 2 4 4/ 1 16

2

P X XP X X

P X



# 37



19 4x y



a) 9 1,

2 2 2

b) 9 1,

2 2 2

c) 3 3, 2 2 d) 3 3,2 2

Ans. A, B

2 2

19 4x y


2 9 4 4 36 4 32

4 2



2 2 2 2 2 2a l b mn n


a) 1Se

b) 11Se

c) 1 114

Se

d)

1 1 112 2

Se

Ans. A, B, D

21 1

0 0

11x xI e dx e dxe

11Ie

21 1xe e

21 1 1

1

0 0 0

xe dx e dx dx

21

0

1xe dxe



# 38



a) 2

4 8 2y

b) 2

'4 18

y

c) 2

3 9y

d) 24 2'

3 3 3 3y

Ans. A,D


tan. cos

dxI F e x


2 sec secy x x c x

0,0

0 0 0c c .

2 secy x x

2

24 16

y

2

8 2

2

23 9

y


.

2

1 2 3 2 23 9 3

y

22 433 3



# 39




,

tan 2 0 and 31 sin2

.

Then cannot satisfy

a) 0 2 b)

42 3 c)

4 33 2 d)

3 22

Ans. A, C, D

3tan 2 0, 1 sin2

3 5 1, 0 cos2 3 2

______ (1)


1sin cos2

1 sin 12

(from (1))

13 176 6

2 73 6 (from (1))




# 40





Ans. 9

3 2P x ax bx cx d

1 6 6P a b c d

3 2 27 9 3 2P a b c d

1 1 0 3 2 0P a b c

1 3 0 27 6 0P a b c

1, 6, 9a b c

1 23 2P x ax bx c

1 0 9P c


2 2 2 9a b b c c a ,then 2 5 5a b c

is\

Ans. 3

sol. 1a b c

2 2 2 9a b b c c a

3.

2a b b c c a

3 2 . . 0a b b c c a

22 2 2 . . . 0a b c a b b c c a

2

0a b c



# 41


0a b c ––––––(1)


then 1 1 1. , . , .

2 2 2a b b c c a

2 5 5 3a b c

58. The value of 32

1 1 1 16 log 4 4 4 ...3 2 3 2 3 2 3 2

is

Ans. 4

sol.1 4

3 2x x

3 2 4x x


9 2 1 4 2 1 0x x x .

2 1 9 4 0x x

1 4,2 9

x

49

x 0x


3/ 236 log2

= 6 – 2 = 4


Ans. 4



# 42


sol.P

Q

S(0, 0)

(2,4)

(2, 0)2

(0,4)

2 8 2 4 2 0x x x x

0,2 0,0 , 2,4x P Q

area of SPQ

= 1 2 42

= 4



21 , 1 0x x x

21 , 0 1x x x

2 1, 1x x x

' 2 1, 1f x x x

2 1, 1 0x x

0-1 1x

y

1 2 , 0 1x x

1 2 , 1x x


x ,





& Solutions
