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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 Marks 38 XII Marks 41
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
XI syllabus XII syllabus
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 Marks 25 XII Marks 54
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
XI syllabus XII syllabus
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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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.
Paragraph for questions 15 to 17
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
The number of elements in A : 1 Choices for B: 38
The number of elements in A : 2 Choices for B: 3
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.
Paragraph for questions 31 to 33
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
+ =