aakash-goiit-iit-jee-2010 paper-2 answer and solution with analysis

8/9/2019 Aakash-Goiit-IIT-JEE-2010 Paper-2 Answer and Solution With Analysis

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Answers by

IIT-JEE 2010(Division of Aakash Educational Services Ltd.)

PAPER - 2 : CODES

Q.No. 0 1 2 3 4 5 6 7 8 9

01. C D A C D C D C C D

02. A A C B C A C B A C

03. D C D C B D A D C B

04. B C D A D B D D D A

05. C B C D A C C A B D

06. D D B D C D B C D C

07. 7 2 7 2 7 2 2 3 2 3

08. 2 6 2 7 2 6 2 7 2 6

09. 2 3 2 6 3 2 7 6 7 7

10. 6 2 6 3 2 3 6 2 3 2

11. 3 7 3 2 6 7 3 2 6 2

12. B B B B B B B B B B

13. A C A C A C A C A C

14. D B D B D B D B D B

15. B B B B B B B B B B

16. C A C A C A C A C A

17. B D B D B D B D B D

18. A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s)

B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t)

C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q)

D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p)

19. A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s) A(p,s) A(r,s)

B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t) B(p,q,r,t) B(t)

C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q) C(p,q)

D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r) D(p) D(r)

20. A B C B D B D A B B

21. C C A D B C A D C B

22. B A B A D D C D A D

23. D B D C B D B B D C

24 B D B D C A B C D D


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Organic Chemistry Inorganic Chemistry Physical Chemistry Total

Easy 0 0 1 1

Medium 2 4 6 12

Tough 4 0 2 6

Total 6 4 9 19

Markwise 27 33 19 79

XI Marks 33 XII Marks 46

XI syllabus 8 XII syllabus 11

Topic-wise Analysis of CHEMISTRY - IIT-JEE 2010

(Paper-2)

5%

63%

32%

Distribution of Level of Questions in Chemistry

Easy Medium Tough

32%

21%

47%

Topic wise distribution in Chemistry

Organic Chemistry

Inorganic Chemistry

Physical Chemistry

42%

58%

Percentage Portion asked from Syllabus of Class XI& XII - Chemistry

XI syllabus XII syllabus

34%

42%

24%

Topic wise distribution of marks in Chemistry

Organic Chemistry

Inorganic Chemistry

Physical Chemistry

42%

58%

Percentage Portion asked (Mark-wise) from Syllabus of Class XI & XII -Chemistry

XI Marks XII Marks


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XII XI XI XI XII XII XII

Calculus Trigonometry Algebra2D-

GeometryProbability 3-D + Vector Algebra Total

Easy 2 0 0 3 0 2 0 7

Medium 2 1.25 2 1 0 0 1 7.25

Tough 1.25 0 2 0 1 0.5 0 4.75

Total 5.25 1.25 4 4 1 2.5 1 19

Markwise 19 5 21 12 5 14 3 79


XI syllabus 9.25 XII syllabus 9.75

Topic-wise Analysis of MATHEMATICS - IIT-JEE 2010

(Paper-2)

37%

38%

25%

Distribution of Level of Questions in Mathematics

Easy Medium Tough

28%

7%

21%21%

5%13%

5%

Topic wise distribution in MathsCalculus

Trigonometry

Algebra

2D- Geometry

Probability

3-D + Vector

Algebra

49%51%

Percentage Portion asked from Syllabus of ClassXI & XII


24%

6%

27%15%

6%

18%

4%

Topic wise distribution of marks in Maths

Calculus

Trigonometry

Algebra

2D- Geometry

Probability

3-D + Vector

Algebra

48%52%

Percentage Portion asked (Mark-wise) from Syllabus ofClass XI & XII

XI Marks XII Marks


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XII XI XII XI XII XI XII XII

ElectricityHeat &

Thermodynamics

Modern

PhysicsMechanics Optics

Oscillation

& WavesMagnetism

Experime

ntal Total

Easy 0 1 0 1 2 0 0 1 5

Medium 1 0 3 3 2 1 0 0 10

Tough 1 0 1 1 0 0 1 0 4

Total 2 1 4 5 4 1 1 1 19

Markwise 8 3 14 17 19 5 8 5 79


XI syllabus 7 XII syllabus 12

Topic-wise Analysis of PHYSICS - IIT-JEE 2010

(Paper-2)

26%

53%

21%

Distribution of Level of Questions in Physics

Easy Medium Tough

11%5%

21%

27%

21%

5%5% 5%

Topic wise distribution in PhysicsElectricity

Heat &Thermodynamics

Modern Physics

Mechanics

Optics

Oscillation & Waves

Magnetism

Experimental Physics

37%

63%

Percentage Portion asked from Syllabus ofClass XI & XII - Physics


10%4%

18%

22%24%

6% 10% 6%

Topic wise distribution of marks in PhysicsElectricity

Heat &ThermodynamicsModern Physics

Mechanics

Optics

Oscillation & Waves

Magnetism

Experimental Physics

32%

68%

Percentage Portion asked (Mark-wise) from Syllabus ofClass XI & XII - Physics

XI Marks XII Marks


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Solutions

toIIT-JEE 2010

Time : 3 hrs. Max. Marks: 252

Regd. Office: Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472

Instructions :

1. The question paper consists of 3 parts (Chemistry, Mathematics and Physics) and each Part

consists of four Sections.

2. For each question in Section I,you will be awarded 5 marks if you have darkened only the bubble

corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,

minus two (2) mark will be awarded.

3. For each question in Section II,you will be awarded 3 marks if you have darkened the bubble

corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks

will be awarded for incorrect answers in this Section.

4. For each question in Section III,you will be awarded 3 marks if you have darkened only the bubble

corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,

minus one (1) mark will be awarded.

5. For each question in Section IV,you will be awarded 2 marks for each row in which you have

darkened the bubble(s) corresponding to the correct answer. Thus, each question in this section

carries a maximum of 8 marks. There is no negative marks awarded for incorrect answer(s) in this

Section.

DATE : 11/04/2010

PAPER - 2 (Code - 5)


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AakashIIT-JEE -Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, NewDelhi-75 Ph.: 011-47623417/23 Fax : 47623472

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IIT-JEE 2010 (Paper-2)

PARTI : CHEMISTRY

SECTION - I

Single Correct Choice TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

1. In the reaction H C3 C

O

NH2

(1) NaOH/Br2

(2) C

O

Cl

T, the structure of the product T is

(A)

H C3

C

O

O C

O

(B)

NH

C

O

CH3

(C)NH

C

O

H C3

(D)

H C3

C

O

NH C

O

Answer (C)

Hints :

H C3

C

O

NH2

NaOH + Br2H C

3NH

2

C

O

Cl

H C3

NH

O

C

2. Assuming that Hunds rule is violated, the bond order and magnetic nature of the diatomic molecule B2

is

(A) 1 and diamagnetic (B) 0 and diamagnetic

(C) 1 and paramagnetic (D) 0 and paramagnetic

Answer (A)

Hints :

Molecular orbital diagram for B2

where Hunds rule is violated.

2X2 2 2 2

0Y

2p1 *1 2 * 2

2ps s s s

Bond order = 1

and diamagnetic.


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3. The species having pyramidal shape is

(A) SO3

(B) BrF3

(C) 23SiO (D) OSF2

Answer (D)

Hints :

S

O FF

Pyramidal

4. The complex showing a spin-only magnetic moment of 2.82 B.M. is

(A) Ni(CO)4

(B) [NiCl4]2 (C) Ni(PPh

3)4

(D) [Ni(CN)4]2

Answer (B)

Hints : Ni2+ is sp3 hybridized and metal ion is connected with weak ligands.

Ni2+d8

sp3two unpaired electrons

s

= 2 (2 2) B.M.+

= 8 B.M. = 2.83 B.M.

5. The compounds P, Q and S

COOH

HOP

OCH3

H C3

Q

C

S

O

O

were separately subjected to nitration using HNO3/H

2SO

4mixture. The major product formed in each case

respectively, is

(A)

COOH

HO

NO2

OCH3

H C3

NO2

C

O

O

O N2

(B)

COOH

HO NO2

OCH3

H C3NO

2

C

O

O

NO2


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(C)

COOH

HO

NO2

OCH3

H C3

NO2

C

O

O NO2

(D)

COOH

HO

NO2

OCH3

H C3

NO2

C

O

O

NO2

Answer (C)

Hints :

COOH

HO

HNO /H SO3 2 4

COOH

HO

NO2

OCH3

H C3

HNO /H SO3 2 4OCH3

H C3 NO2

C

O

O

HNO /H SO3 2 4C

O

O NO2

6. The packing efficiency of the two-dimensional square unit cell shown below is

L

(A) 39.27% (B) 68.02% (C) 74.05% (D) 78.54%

Answer (D)

Hints :

Packing efficiency =Area covered by particle

Total area

=

2

2

2 r

a

= ( )

2

2

2 r

42 2 r

=


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SECTION - II

Integer Type

This section contains a group of 5 questions. The answer to each of the questions is a Single-digit Integer, ranging from0 to 9. The correct digit below the equation number in the ORS is to be bubbled.

7. Among the following, the number of elements showing only one non-zero oxidation state is

O, Cl, F, N, P, Sn, Tl, Na, Ti

Answer (2)

Hints : F will only exhibit 1 oxidation state except zero.

and Na will exhibit +1 oxidation state.

8. The total number of diprotic acids among the following is

H3PO

4H

2SO

4H

3PO

3H

2CO

3H

2S

2O

7

H3BO

3H

3PO

2H

2CrO

4H

2SO

3

Answer (6)

Hints : H2SO

4, H

3PO

3, H

2CO

3, H

2CrO

4and H

2SO

3and H

2S

2O

7will behave as dibasic acid.

9. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines asshown in the graph below. If the work done along the solid line path is W

sand that along the dotted line path is

Wd, then the integer closest to the ratio W

d/W

sis

0.50.0 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

P(atm.)

V(lit.)

a

b

Answer (2)

Hints :

Solid line path work done (Ws) is isothermal beacuse PV is constant at each point & dash line path work done

(Wd) is isobaric.

Total work done on solid line path (Ws) = 2.303 nRT 2

1

Vlog

V

= 2.303 PV 2

1

Vlog

V

= 2.303 5.5

2log0.5

4.6 l atm

Total work done on dash line path (Wd) = 4 1.5 + 1 1 + 0.5 2.5

= 8.255 l atm.

d

s

W 8.252(closest integer)

W 4.6=


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10. Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH

3)] is

Answer (3)

Hints :

Rh Cl(CO)(PPh3)(NH

3)] is a square planar complex with four different ligands and hence it will have three

geometrical isomers

a b

cd

Rh

a b

dc

Rh

a c

bd

Rh

11. Silver (atomic weight = 108 g mol1) has a density of 10.5 g cm3. The number of silver atoms on a surface ofarea 1012 m2 can be expressed in scientific notation as y 10x. The value of x is

Answer (7)

Hints :

Volume of 1 Ag atom =34

r3

3 3

23

4 108r cm

3 6.023 10 10.5 =

r = 1.6 108 cm

r = 1.6 1010 m

Number of Ag atoms in 1012 m2

r2 n = 1012 m2

12

10 2

10

n 3.14 (1.6 10 )=

710 10n

8

=

n = 1.25 107

The value of x = 7

SECTION - III

Paragraph Type

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be

answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for questions 12 to 14

The hydrogen-like species Li2+ is in a spherically symmetric state S1

with one radial node. Upon absorbing light theion undergoes transition to a state S

2. The state S

2has one radial node and its energy is equal to the ground state

energy of the hydrogen atom.

12. The state S1

is

(A) 1s (B) 2s (C) 2p (D) 3s

Answer (B)

Hints :

S1 state is 2s

In 2s, one radial node is present


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13. Energy of the state S1

in units of the hydrogen atom ground state energy is

(A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50

Answer (C)

Hints :

Energy in S1 state

2

2

13.6 3 9E 13.6eV

42

= =

11( H )

E = 13.6 eV [Ground state]

So energy = 2.25 energy of e in ground state in 1H1.

14. The orbital angular momentum quantum number of the state S2

is

(A) 0 (B) 1 (C) 2 (D) 3

Answer (B)

Hints :

S2

state is 3porbital

Orbital angular momentum of 3pis 1.


Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO

3to give compound R, which upon treatment

with HCN provides compound S. On acidification and heating, S gives the product shown below :

H C3

H C3

OH

OO

15. The compounds P and Q respectively are

(A) H C3

CH3

C

CH H and H C3

O

C

O

H

(B) H C3

CH3

C

CH H and H

O

C

O

H

(C)

H C3

CH3

CH

H C3

C

O

HCH2

C

O

H

and

(D)

H C3

CH3

CH

H

C

O

HCH2

C

O

H

and


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SECTION - IV

Matrix Type

This section contains 2 questions. Each question has four choices (A), (B), (C) and (D) given in Column I and fivestatements (p), (q), (r), (s) and (t) in Column II. Any given statement in Column I can have correct matching withone or more statement(s) given in Column II. For example, if for a given question, statement B matches with the

statements given in (q) and (r), then for that particular question, against statement B, darken the bubbles correspondingto (q) and (r) in ORS.

18. All the compounds listed in Column I react with water. Match the result of the respective reactions with theappropriate options listed in Column II.

Column I Column II

(A) (CH3)2SiCl

2(p) Hydrogen halide formation

(B) XeF4

(q) Redox reaction

(C) Cl2

(r) Reacts with glass

(D) VCl5

(s) Polymerization

(t) O2 formation

Answer : A(p, s), B(p, q, r, t), C(p, q), D(p)

Hints :

(A) (CH ) SiCl + H O3 2 2 2 SiOH

CH3

CH3

OH

Polymerise SiO

CH3

CH3

O

n

(B)4 2 3 26XeF 12H O 4Xe 2XeO 24HF 3O+ + + +

(C) 2 2Cl H O HCl HOCl+ +

(D)5 2 3VCl H O VOCl HCl+ +

Note :Vanadium in (+V) oxidation state from only fluoride. Existence of VCl5

is doubtful.

19. Match the reactions in Column I with appropriate options in Column II.

Column I Column II

(A) N Cl +2 OH N=NNaOH/H O

0C

2

OH (p) Racemic mixture

(B) H C3

C C CH3

OH

CH3

OH

CH3

H SO2 4 H C3

CO

C

CH3

CH3

CH3

(q) Addition reaction

(C) C

O

CH3

1. LiAlH

2. H O

4

3

+ CH

OH

CH3

(r) Substitution reaction

(D)

Base

HS Cl

S

(s) Coupling reaction

(t) Carbocation intermediate


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Answer : A(r, s), B(t), C(p, q), D(r)

Hints :

(A) N N + O H

OH

N = N

H

O

N = N O

Cl

Coupling reaction and electrophilic substitution reaction.

(B) CH C C CH3 3

OH OH

CH3

CH3

CH C C CH3 3

OH OH2

CH3

CH3

+ H

CH C C CH3 3

O

CH3

CH3

CH C C3

OH

CH3

CH3

CH3

Carbocation intermediate

(C) C

O

CH3

LiAlH H4

Nucleophilic additionC

CH3

H

O

Racemic mixture

H O3

+

C

CH3

H

OH

(D)S Cl

H

B

Base S

Cl

Nucleophilicsubstitution S


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PARTII : MATHEMATICS

SECTION - I

(Single Correct Choice Type)

This section contains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLYONE is correct.

20. Let f be a real-valued function defined on the interval (1, 1) such that ( ) 4

0

2 1x

xe f x t dt = + + , for all

x (1, 1), and let f1 be the inverse function of f. Then (f1) (2) is equal to

(A) 1 (B)1

3(C)

1

2(D)

1

e

Answer (B)

Hints :

We have,

( ) 4

0

2 1x

xe f x t dt = + + x (1, 1)

Differentiating w.r.t. x, we get

( ) ( )( ) 4' 1xe f x f x x = +

( ) ( ) 4' 1 xf x f x x e = + +

f1 is the inverse of f

f1(f(x)) = x

f1(f(x)) f(x) = 1

( )( )( )

1 1''

f f xf x

=

( )( )( )

1

4

1'

1 xf f x

f x x e

=+ +

at x= 0, f(x) = 2

( )11 1

' 22 1 3

f = =

+

21. A signal which can be green or red with probability4

5and

1

5respectively, is received by station A and then

transmitted to station B. The probability of each station receiving the signal correctly is3

4. If the signal received

at station Bis green, then the probability that the original signal was green is

(A)3

5

(B)6

7

(C)20

23

(D)9

20Answer (C)


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Hints :

From the tree-diagram it follows that

BG

BR

BG

BR

BG

AG

AR

AR

AG

RG

S

45

34

34

34

34

34

34

14

14 1

4

14

14

15

14

46( )

80=GP B

( )10 5

|16 8

GP B G = =

( )5 4 1

8 5 2 = =GP B G

( )

11 80 202|

( ) 2 46 23G

G

P G BP B

= = =

22. Let S= {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of Sis equal to

(A) 25 (B) 34 (C) 42 (D) 41

Answer (D)

Hints : Let ,A B = ,A B S

The number of elements in A : 0 Choices for B: 24



Total number of possible subsets is 41

23. For r = 0, 1,..., 10, let Ar, B

rand C

rdenote, respectively, the coefficient of xr in the expansions of

(1 + x)10, (1 + x)20 and (1 + x)30. Then ( )10

10 101

r r rr

A B B C A=

is equal to

(A) B10

C10

(B) ( )210 10 10 10A B C A (C) 0 (D) C10 B10Answer (D)

Hints : Ar

= Coefficient of xr in (1 + x)10 = 10Cr

Br

= Coefficient of xr in (1 + x)20 = 20Cr

Cr= Coefficient of xr in (1 + x)30 = 30C

r

( )10 10 10

10 10 10 101 1 1

r r r r r r r r r r

A B B C A A B B A C A= = =

= 10 10

10 20 20 10 30 10

10 101 1

r r r r r r

C C C C C C = =

=

10 1010 20 20 10 30 10

10 10 10 101 1

r r r r r r

C C C C C C = =

=

10 1020 10 20 30 10 10

10 10 10 101 1

.r r r r

r r

C C C C C C = =

=

( ) ( )

20 30 30 20

10 10 10 101 1C C C C =

30 20

10 10 10 10C C C B = =


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24. If the distance of the point P(1, 2, 1) from the plane x+ 2y 2z= , where > 0, is 5, then the foot of theperpendicular from Pto the plane is

(A)8 4 7

, ,3 3 3

(B)

4 4 1, ,

3 3 3

(C)

1 2 10, ,

3 3 3

(D)2 1 5

, ,3 3 2

Answer (A)

Hints : Distance of point Pfrom plane = 5

1 4 25

3

=

= 10

P(1, 2, 1)

Foot of perpendicular

1 2 1 5

1 2 2 3

x y z + = = =

8 4 7, ,

3 3 3x y z= = =

Thus the foot of the perpendicular is

8 4 7, ,

3 3 3f

25. Two adjacent sides of a parallelogram ABCDare given by

2 10 11AB i j k = + + and 2 2AD i j k = + + .The side ADis rotated by an acute angle in the plane of the parallelogram so that ADbecomes AD. If ADmakes a right angle with the side AB, then the cosine of the angle is given by

(A)8

9(B)

17

9(C)

1

9(D)

4 5

9

Answer (B)

Hints :

2 10 11AB i j k = + +2 2AD i j k = + +

Angle '' between AB

and AD

A B

CD

= ( )( )( )

. 2 20 22 8cos

15 3 9

AB AD

AB AD

+ + = = =

( )17

sin9

=

90 + =

( ) ( ) ( )17

cos cos 90 sin9

= = =


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SECTION - II

(Integer Type)

This Section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correctdigit below the question no. in the ORS is to be bubbled.

26. Let fbe a function defined on R(the set of all real numbers) such thatf'(x) = 2010 (x 2009) (x 2010)2 (x 2011)3 (x 2012)4, for all xR.

If gis a function defined on Rwith values in the interval (0, ) such that f(x) = ln (g(x)), for all xR, then thenumber of points in Rat which ghas a local maximum is

Answer (1)

Hints : ( ) ( ) ,f x

g x e x R =

( ) ( ) ( )' . 'f x

g x e f x =

f(x) changes its sign from positive to negative in the neighbourhood of x= 2009

f(x) has local maxima at x= 2009

So, the number of local maximum is one.

27. Two parallel chords of a circle of radius 2 are at a distance 3 1+ apart. If the chords subtend at the center,

angles ofk

and

2

k

, where k> 0, then the value of [k] is

[Note : [k] denotes the largest integer less than or equal to k]

Answer (3)

Hints : Let 2k

=

cos2

x =

2

2

2

x

C

3 + 1 x

3 1cos2

2

x+ =

23 1

2cos 12

x+ =

2 3 12 1

4 2x x + =

23 3 0x x + =

1 1 12 4 3

2x

+ +=

1 13 4 3

2

+=

1 2 3 13

2

+ += =


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3cos

2 6

= =

Required angle 23k

= = =

k= 3

28. Let kbe a positive real number and let

2 1 2 2 0 2 1

2 1 2 and 1 2 0 2 .

2 2 1 2 0

k k k k k

A k k B k k

k k k k

= =

If det (adj A) + det(adj B) = 106, then [k] is equal to

[Note : adj Mdenotes the adjoint of a square matrix Mand [k] denotes the largest integer less than or equal tok]

Answer (4)

Hints :

|A| = (2k+ 1)3, |B| = 0

det (adj A) det (adj B) = (2k+ 1)6 = 1069

.2

k=

[k] = 4

29. Consider a triangle ABCand let a, band cdenote the lengths of the sides opposite to vertices A, B and C

respectively. Suppose a= 6, b= 10 and the area of the triangle is 15 3. If ACBis obtuse and if rdenotes

the radius of the incircle of the triangle, then r2 is equal to

Answer (3)

Hints :

3sin and

2C C= is given to be obtuse.

2

3C

=

30. Let a1, a

2, a

3,......, a

11be real numbers satisfying

a1

= 15, 27 2a2

> 0 and ak

= 2ak 1

ak 2

for k= 3, 4, ...., 11.

If2 2 21 2 11.... 90,

11

a a a+ + +=

then the value of1 2 11....

11

a a a+ + +is equal to

Answer (0)


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SECTION-III (Paragraph Type)

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be

answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Tangents are drawn from the point P(3, 4) to the ellipse2 2

19 4

x y+ = touching the ellipse at points A and B.

31. The coordinates of A and Bare

(A) (3, 0) and (0, 2) (B)8 2 161 9 8

, and ,5 15 5 5

(C)

8 2 161, and (0, 2)

5 15

(D)

9 8

(3, 0) and ,5 5

Answer (D)

Hints :

Figure is self explanatory

B

DP

(3, 4)

A(3, 0)

F

32. The orthocentre of the triangle PABis

(A)8

5,7

(B)7 25

,5 8

(C)11 8

,5 5

(D)8 7

,25 5

Answer (C)

Hints :

Equation of ABis

8850 ( 3) ( 3)

9 243

5

y x x = =

95

, 85

P(3, 4)

(3, 0)

AB

1

( 3)3

y x=

x+ 3y= 3 ...(i)

Equation of the straight line perpendicular to ABthrough Pis 3x y= 5

Equation of PA is x 3 = 0


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The equation of straight line perpendicular to PA through9 8

,5 5

B

is y =8

5.

Hence the orthocentre is 11 8,5 5

33. The equation of the locus of the point whose distances from the point Pand the line ABare equal, is

(A) 9x2 + y2 6xy 54x 62y+ 241 = 0

(B) x2 + 9y2 + 6xy 54x+ 62y 241 = 0

(C) 9x2 + 9y2 6xy 54x 62y 241 = 0

(D) x2 + y2 2xy+ 27x+ 31y 120 = 0

Answer (A)

Hints :

Equation of1

0 ( 3)3

AB y x = =

x+ 3y 3 = 0

|x+ 3y 3|2 = 10[(x 3)2 + (y 4)2]

(Look at coefficient of x2 & y2 in the answers)

Paragraph for question 34 to 36

Consider the polynomial f(x) = 1 + 2x + 3x2

+ 4x3

. Let s be the sum of all distinct real roots of f(x) = 0 and lett= |s|.

34. The real number slies in the interval

(A) 1, 04

(B) 311,

4

(C) 3 1,

4 2

(D)

10,

4

Answer (C)

Hints :

f(x) = 2(6x2 + 3x+ 1) = 9 24 < 0

Hence f(x) = 0 has only one real root.

1 3 41 1 0

2 4 8f

= + >

3 6 27 1081

4 4 16 64f

= +

64 96 108 108

064

+ =

aakash-goiit-iit-jee-2010 paper-2 answer and solution with analysis

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