Solution Iit Jee
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2
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 2
(SOLUTION – 2010)
S OLUTIONS TO IIT–JEE 2 0 1 0 ( P APER 1 )
SECTION −I Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.
(A)
Br
OCH3 and
Br and
(C) 1.
⎯⎯ → the product are
OCH3
In the reaction
HBr
H2
CH3OH
(B)
Br and
CH 3Br
(D)
OH and
CH 3Br
(D) O
CH3
HBr ⎯−⎯ ⎯ → Br −
O
CH3
−
Br ⎯:⎯ ⎯ →
OH
+ CH 3Br
H
2.
Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is
(A)
k
(B)
k
T
(C)
k
(D) T
2.
T
(A) From Arrhenius equation: K Hence (A).
k
T
= Ae− Ea/RT , K will increase exponentially with temperature.
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
3
3.
3.
4.
4.
5.
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 3
(SOLUTION – 2010)
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is (A) Br 2(g) (B) Cl2(g) (C) H2O(g) (D) CH4(g) (B) The standard molar enthalpy of formation of an element is zero at 298K. Br 2 although an element, exists in the liquid state at 298K. Hence (B) The ionization isomer of [Cr(H2O)4Cl(NO2)]Cl is (A) [Cr(H2O)4(O2 N)]Cl2 (B) [Cr(H2O)4Cl2](NO2) (C) [Cr(H2O)4Cl(ONO)]Cl (D) [Cr(H2O)4Cl2(NO2)]H2O (B) [Cr(H2O)4Cl(NO2)]Cl ⎯→ [Cr(H2O)4Cl(NO2)]+ + Cl – [Cr(H2O)4Cl2]NO2 ⎯→ [Cr(H2O)Cl2]+ + NO2 – The correct structure of ethylenediaminetetraacetic acid (EDTA) is HOOC
CH2 N
(A) HOOC
CH2
HOOC
CH2 N
(C) HOOC
CH2 COOH
HOOC
CH CH N
HOOC
CH2 COOH CH2 CH2 N
CH2
N
(B) CH2 COOH
HOOC
(D) CH2 COOH
COOH CH2CH2 N COOH COOH CH CH2 2 H N CH CH N H HOOC
5.
CH2
CH2 COOH
(C) EDTA is HOOC
CH2 N
HOOC
CH2 COOH CH2 CH2 N
CH2
CH2 COOH
Hence (C) 6.
6.
7.
7.
The bond energy (in kcal mol –1) of C – C single bond is approximately (A) 1 (B) 10 (C) 100 (D) 1000 (C) The bond energy of C – C single bond is ∼ 100 kcals mol –1. Hence (C). Other data are absurd. The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are (A) BrCH2CH2CH2CH2CH3 and CH3CH2C≡CH (B) BrCH2CH2CH3 and CH3CH2CH2C≡CH (C) BrCH2CH2CH2CH2CH3 and CH3C≡CH (D) BrCH2CH2CH2CH3 and CH3CH2C≡CH (D) CH3 –CH2 –C ≡C – H
–
⊕
δ+
δ−
CH3 − CH 2 − CH 2 − CH 2 − Br
⎯ ⎯→CH 3 C − H2 C − ≡C : Na ⎯⎯ ⎯− ⎯ ⎯⎯ ⎯ ⎯ → H NaBr − N⎯ NaNH2 3
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4
PIE IIT JEE PAPER – I & II
CH 3 − CH 2
IIT-JEE 2010 SOLUTION - 4
(SOLUTION – 2010)
− CH 2 − CH 2 − C ≡ C − CH 2 − CH 3 3 − Octyne
Hence (D). 8.
The correct statement about the following disaccharide is CH2OH H
O
OH OH H
8.
H
H (a)
O
HOH2C
H
(b)
H
OCH2 CH2O
H
HO CH2OH
OH
OH
H
(A) Ring (a) is pyranose with α-glycosidic link (B) Ring (a) is furanose with α-glycosidic link (C) Ring (b) is furanose with α-glycosidic link (D) Ring (b) is pyranose with wit h β-glycosidic lilnk (A) Ring (a) is pyranose with α-glycosidic link but ring (b) is furanose with linkage. Hence (A)
β-glycosidic
SECTION −I I Multiple Correct Answer Type This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
9.
In the Newman projection for f or 2,2-dimethylbuane X H3C
CH3
H
H Y
9.
X and Y can respectively be (A) H and H (C) C2H5 and H (B, D)
(B) H and C2H5 (D) CH3 and CH3
CH3 H3C
CH2 C
CH3
CH3 2,2 Dimethyl buane
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5
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 5
(SOLUTION – 2010)
H H3C
CH3 CH3
H
H
H3C
CH3
H
H
C 2 H5
CH3
(I)
(II)
I. Considering rotation about C1 – C 2 bond one of the conformers will be as represented by the Newman projection formula (I). Hence (B). Also II. Considering rotation about about C2 –C3 bond, one of the conformers can be represented by the above Newman projection formula (II). Hence (D) 10.
10.
11.
11.
12.
12.
The reagent(s) used for softening the temporary hardness of water is (are) (A) Ca3(PO4)2 (B) Ca(OH)2 (C) Na2CO3 (D) NaOCl (B, C) Temporary hardness is due to dissolved Ca(HCO 3)2 Ca(HCO3)2 + Ca(OH)2 ⎯→ 2CaCO3↓ + 2H2O Ca(HCO3)2 + Na2CO3 ⎯→ CaCO3↓ + 2NaHCO3 Hence (B) and (C) Among the following, the intensive property is (properties are) (A) molar conductivity (B) electromotive force (C) resistance (D) heat capacity (A, B) By decreasing or increasing the amount of an electrolytic solution, neither molar conductance nor EMF is going to change. Aqueous solutions of HNO 3, KOH, CH3COOH, and CH 3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are) (A) HNO3 and CH3COOH (B) KOH and CH3COONa (C) HNO3 and CH3COONa (D) CH3COOH and CH 3COONa (C, D) OH
13.
In the reaction
⎯ NaOH(aq)/Br ⎯⎯ ⎯ ⎯⎯ ⎯→ 2
the intermediate(s) is(are)
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6
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 6
(SOLUTION – 2010)
O
O Br
(A)
(B) Br
Br O
Br
O
(C)
(D) Br Br
13.
(A, C) OH
O
O
O H
−
Br − Br
⎯⎯ ⎯→
⎯⎯ ⎯ → OH − H2 O
Br
Br
⎯⎯ →
Br − Br
O Br
O Br
Br
Br
Br
+
Br − Br ←⎯ ←⎯⎯ ⎯⎯
H Br
Br
Br
+
−H ←⎯ ⎯ ⎯
O
O
−H ←⎯ ⎯ ⎯
Br
H
Br
SECTION −I I I Linked Comprehension Type This section contains 2 paragraphs . Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct. Paragraph for Questions 14 to 16
Copper is the most noble of the first row transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite (CuSO4 5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu 2S) and malachite (Cu2(OH)2CO3). However, 80% of the world copper production comes from the ore chalcopyrite (CuFeS 2). The extraction of copper from chalcopyrite involves partial roasting, removal of iron and self-reduction.
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
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14.
14.
15.
15.
16.
16.
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 7
(SOLUTION – 2010)
Partial roasting of chalcopyrite produces (A) Cu2S and FeO (C) CuS and Fe 2O3 (B)
(B) Cu2O and FeO (D) Cu2O and Fe2O3
Iron is removed from chalcopyrite as (A) FeO (C) Fe2O3 (D)
(B) FeS (D) FeSiO3
In self-reduction, the reducing species is (A) S (C) S2– (C)
(B) O2– (D) SO2
Paragraph for Questions 17 to 18 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is: M(s) | M+ (aq; 0.05 molar) ||M + (aq; 1 molar) | M(s) For the above electrolytic cell the magnitude of the cell potential |Ecell| = 70 mV. 17.
17.
18.
18.
For the above cell (A) Ecell < 0; ΔG > 0 (B) Ecell > 0; ΔG < 0 (C) Ecell < 0; ΔG° > 0 (D) Ecell > 0; ΔG° < 0 (B) Ecell = –0.0591 log 0.05 = +ve as E cell is +ve so ΔG is less than zero. If the 0.05 molar solution of M + is replaced by a 0.0025 molar M + solution, then the magnitude of the cell potential would be (A) 35 mV (B) 70mV (C) 140mV (D) 700 mV (C) As –0.0591 log 0.05 = 70 mV ∴ –0.0591 log 0.0025 = 140 mV SECTION −I V
Integer Type This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 19. 19.
The value of n in the molecular formula Be nAl2Si6O is 3 The value of n in the Be nAl2Si6O18 is 3. PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
8
20.
PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 8
(SOLUTION – 2010)
The total number of basic groups in the following form of lysine is O CH2 CH2 CH2 CH2 CH C
H3 N
H2 N
O
20.
2 In lysine – NH 2 and –COO – are basic groups.
21. 21.
Based on VSEPR theory, the number of 90 degree F-Br-F angles in BrF 5 is 0 In BrF5 the shape is square pyramidal type but due to one lone pair no F-Br-F bond will be of 90° angle, it is around 85°.
22.
Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is KCN K 2SO4 (NH4)2C2O4 NaCl Zn(NO3)2 FeCl3 K 2CO3 NH4 NO3 LiCN
22.
3 The aqueous solutions of KCN, K 2CO3 and LiCN are alkaline in nature and hence will trun red litmus blue..
23.
Amongst the following, the total number of compounds soluble in aqueous NaOH is H3C
N
CH3
COOH
OCH2CH3
OH
CH2 OH
NO2
OH
CH2 CH3
COOH
CH2CH3
H3C
N
CH3
23.
4 All carboxylic acids and phenols are more acidic than water. Therefore 4.
24.
24.
A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 mL, and 25.0 mL. The number of significant figures in the average titre value is 3
25.
The number of neutrons emitted when and
90 38
235 92
U undergoes controlled controlled nuclear fission to
Sr is
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
142 54
Xe
9
25.
PIE IIT JEE PAPER – I & II
4 235 92
26.
IIT-JEE 2010 SOLUTION - 9
(SOLUTION – 2010)
U + 0n 1
⎯⎯ → 15442Xe + 3980Sr + 410 n
In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is 1. O 1. NaOH(aq.) NaOH(aq.) ⎯ ⎯⎯ ⎯→Y ⎯ ⎯ ⎯ ⎯⎯ → 2. Zn , H O 2. Heat 3
2
26.
1 O
O −
⎯OH ⎯⎯ →
O Δ ⎯−⎯ ⎯ → H O 2
O
OH
α, β unsa unsattura urated ted ketone one
As this is [2.2.0] type bicycle compound not [2.2.1]type so dehydration at Bridge head is possible, to form α, β-unsaturated ketone. 27.
27.
28.
28.
The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained: [R] (molar) 1.0 0.75 0.40 0.10 t(min.) 0.0 0.05 0.12 0.18 The order of the reaction is 0 x C −C . This is zero order reaction as data satisfy zero order equation k = or k = 0 t t 0.25 0.60 0.90 k= = = =5 0.05 0.12 0.18 The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is 5 The total cyclic isomers possible for C 4H6 is 5.
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
10 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 10
(SOLUTION – 2010)
SECTI ON – I Single Correct Choice Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 29.
⎢ ⎥ = ⎢0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣0⎥⎦
A y
29.
30.
× 3 matrices A whose entries are either 0 or 1 and for which the system
The number of 3 ⎡ x ⎤ ⎡1⎤
has exactly two distinct solutions, is
(A) 0 (B) 29 − 1 (C) 168 (D) 2 (A) There are 3 possible cases only either 1 solution or infinite solutions or no solution Hence 2 distinct solutions are not possible. The value of lim x →0
1 x3
x
t ln (1 + t )
∫ 0
t4
+4
dt is
(A) 0 (C) 30.
(B)
1
(D)
24
x →0
1 x3
x
t ln(1 + t )
∫ 0
t4
64
(
⎛ 0 form ⎞ ⎜ ⎟ ⎝ 0 ⎠
+4
Using L−Hospital’s rule ln (1 + x ) lim x →0 3x x 4 + 4 1 = lim x →0 (1 + x ) 3 x 4 + 4
⎛ 0 form ⎞ ⎜ ⎟ ⎝ 0 ⎠
)
{(
) + 3x 4x } 3
=
1 12
Let p and q be real numbers such that p ≠ 0, p3 ≠ q and p3 ≠ −q. If α and β are nonzero complex numbers satisfying α + β = − p and α3 + β3 = q, then a quadratic equation having
α β 31.
12 1
(B)
lim
31.
1
and
β α
as its roots is
(A) (p3 + q)x2 − (p3 + 2q)x + (p 3 + q) = 0 (C) (p3 − q)x2 − (5p3 − 2q)x + (p3 − q) = 0 (B) α + β = − p, α3 + β3 = q ⇒ (α + β)(α2 + β2 − αβ) = q
(B) (p3 + q)x2 − (p3 − 2q)x + (p3 + q) = 0 (D) (p3 − q)x2 − (5p3 + 2q)x + (p 3− q) = 0
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
11 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 11
(SOLUTION – 2010)
⇒ (α + β)2 − 3αβ = −
q
p
⇒ 3αβ = p2 + ⇒ αβ =
∴
p
p 3 + q 3 p
and α + β = 2
q
2
q p 3 + q
− + p
3 p
Sum of the roots =
=
p 3 − 2q 3 p
α 2 + β 2 p3 − 2q = 3 p + q αβ
Product of roots = 1 ∴ Equation is (p3 + q)x2 − (p3 − 2q)x + (p3 + q) = 0 32.
Equation of the plane containing the straight line plane containing the straight lines
32.
(A) x + 2y − 2z = 0 (C) x − 2y + z = 0 (C) Vector along the plane is ˆi jˆ k ˆ 3
4
2
4
2
3
x 3
=
y 4
=
x 2
=
y
=
z
3 4 z x y z and = = is 2 4 2 3 (B) 3x + 2y − 2z = 0 (D) 5x + 2y − 4z = 0
and perpendicular to the
ˆ = 8ˆi − jˆ − 10k
∴
Direction of normal vector to the plane is ˆi jˆ ˆ k ˆ = nr (say) 8 − 1 − 10 = 26ˆi − 52 jˆ + 26k 2
∴ ⇒ 33.
4 rr
Equation of plane is r .n = 0 x − 2y + z = 0.
If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the a c expression sin 2C + sin 2A is c a (A)
33.
3
1
(B)
2
(C) 1 (D) A + B + C = 180° ⇒ B = 60o and A + C = 120 o
3 2
(D) 3
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
12 PIE IIT JEE PAPER – I & II
a
c
sin 2C +
= 2sin(A + C) =
c a
IIT-JEE 2010 SOLUTION - 12
(SOLUTION – 2010)
sin 2A =
sin A sin C
× 2 sin C cos C +
sin C sin A
A cos × 2 sinNow,
3
34.
Let f g and h be real −valued functions defined on the interval [0, 1] by
34.
f (x ) = e x + e − x , g(x ) = xe x + e − x and h (x ) = x 2 e x + e − x . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then (A) a = b and c ≠ b (B) a = c and a ≠ b (C) a ≠ b and c ≠ b (D) a = b = c (D)
2
2
2
+ e − x , g(x ) = xe x + e − x ; h (x ) = x 2e x + e − x f (x ) ≥ g(x ) ≥ h( x ) ∀ x ∈ [0, 1]
f (x ) = e x Here
2
2
2
and maximum(f(x)) = e +
2
2
2
2
2
2
, x ∈ [0, 1]
1
at x = 1 e But at x = 1, f(x) = g(x) = h(x) ⇒ a= b=c
35.
35.
36.
36.
Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r 1, r 2 r r r and r 3 are the numbers obtained on the die, then the probability that ω 1 + ω 2 + ω 3 = 0 is 1 1 (A) (B) 18 9 2 1 (C) (D) 9 36 (C) Three such groups are possible as 3p, 3p + 1, 3p + 2 for 3p: possible no. on dice are (3, 6) for 3p + 1: possible no. on dice (1, 4) for 3p + 2: possible no. on dice (2, 5) 2 2 2 Probability = ∴ × × × 3! 6 6 6 2 = 9 Let P, Q, R and S be the points on the plane with position vectors − 3ˆi + 2 jˆ respectively. The quadrilateral PQRS must be a
− 2ˆi − jˆ, 4ˆi , 3ˆi + 3 jˆ
(A) parallelogram, which is neither nei ther a rhombus nor a rectangle (B) square (C) rectangle but not a square (D) rhombus, but not a square (A) PQ = 6ˆi + jˆ; QR = −ˆi + 3 jˆ RS = −6i − jˆ; SP = −i + 3 jˆ PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
and
13 PIE IIT JEE PAPER – I & II
Q
IIT-JEE 2010 SOLUTION - 13
(SOLUTION – 2010)
PQ || RS and QR || SP
⇒ PQRS is a parallelogram
But PQ.QR ≠ 0 and PR .SQ ≠ 0
⇒
Neither rectangle nor rhombus.
SECTION – II Multiple Correct Answer Type This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
37.
Let z1 and z2 be two distinct complex numbers and let z = (1 − t)z1 + tz2 for some real numbers t with 0 < t < 1. If Arg(w) denotes the principal argument of a nonzero complex number w, then (A) | z − z1 | + | z − z 2 |=| z1 − z 2 | (B) Arg(z − z1 ) = Arg(z − z 2 ) (C)
37.
z − z1
z − z1
z 2 − z1
z 2 − z1
(D) Arg(z − z1 ) = Arg(z 2
=0
− z1 )
(A, C, D) (1 − t )z1 + tz 2 As z = (1 − t ) + t
⇒ z lies on line segment joining z 1 & z2 z1
z2
z t : 1– t
⇒ z − z1 + z − z 2 = 38.
38.
z1 − z 2
⇒
z − z1 z2
− z1
z − z1 z2
− z1
= 0 and Arg(z – z 1) = Arg(z2 – z1)
Let A and B be two distinct points on the parabola y 2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be 1 1 (A) − (B) r r 2 2 (C) (D) − r r (C, D)
⎛ t 12 + t 22 ⎞ Centre C ≡ ⎜⎜ , t 1 + t 2 ⎟⎟ ⎝ 2 ⎠ ⇒ t1 + t2 = r Slope of AB =
2 t1
+ t2
=
2
(
2 B t 2 , 2t 2
t 12 , 2t 1
C A
r
Similarly if circle is below x-axis then −2 ⇒ slope of AB = r PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
)
14 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 14
(SOLUTION – 2010)
x
∞) by f(x) =
Let f be a real-valued function defined in the interval (0,
39.
lnx +
∫ 1 + sin t dt . 0
Then which of the following statement(s) is (are) true? (A) f ″(x) exists for all x ∈ (0, ∞) (B) f ′(x) exists for all x ∈ (0, ∞) and f ′ is continuous on (0, ∞), but not differentiable on (0, ∞) (C) there exists α > 1 such that f ′(x ) < f (x ) for all x ∈ (α, ∞) (D) there exists β > 0 such that f (x ) + f ′(x )
≤ β for all x ∈ (0, ∞)
(B, C)
39.
f(x) = lnx + f ′(x) =
1
f ″(x) =
−
x
+ 1 x2
x
∫
1 + sin t dt
0
1 + sin x
+
⇒ f ′(x) > 0 ∀ x > 0
cos x 2 1 + sin x
f ″(x) do not exist when sinx = –1 i.e., when x = 2n π –
π
.
2
So (A) is false and (B) is true. 1 + sin Now t ≥
⇒
x
∫
1 + sin t dt
0
0
≥ 0 ∀ x ∈ (0, ∞)
And ln x > 0 ∀ x ∈ (1, ∞) ⇒ f(x) > 0 ∀ x ∈ (1, ∞) For x ≥ e3 f(x) = ln x + 1
f ′(x) =
+
x Now for x
⇒
x
∫
1 + sin t dt ≥ 3
0
1 + sin x
≤
1 x
+
2
∀x>0
1
+
≥ e3 1
0 < f ′(x) ≤
x Hence (C) is correct. Note: lim f (x ) → ∞
+
2<
e
3
2 < 3 ∀ x ∈ (e3, ∞)
⇒ f ′(x ) < f (x )
x →∞
Then f (x ) + f ′(x ) is not bounded.
⇒ (D) is wrong. 1
40.
The value(s) of
∫ 0
(A)
22 7
(C) 0
−π
x 4 (1 − x )
4
1+ x2
dx is (are) (B) (D)
2 105 71 3π 15
−
2
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
15 PIE IIT JEE PAPER – I & II
40.
IIT-JEE 2010 SOLUTION - 15
(SOLUTION – 2010)
(A) 1
∫ 0
= = = = = = = = =
x 4 (1 − x )
4
1+ x2 4 4 1 x (x
− 4x 3 + 6x 2 − 4x + 1) dx ∫ 0 1+ x2 2 4 2 2 2 1 x (x + 1) + 4 x − 4 x (x + 1) dx ∫ 0 x2 +1 1 ⎛ 1 − 1 ⎞ ⎞⎟ dx 4 ⎛ 2 ⎜ + − + x x 1 4 x 4 ( ) ⎜ ⎟ ∫ 0 ⎝ ⎜ ⎝ x 2 − 1 ⎠ ⎠⎟ 1 ⎛ 4x 4 ⎞ 6 4 5 4 ∫ 0 ⎝ ⎜⎜ x + x − 4x + 4x − x 2 + 1 ⎠⎟⎟dx 1 ⎛ ⎛ (x 2 − 1) + 1 ⎞ ⎞⎟dx 6 4 5 ⎜ + − − x 5 x 4 x 4 ⎜ ⎟⎟ ∫ 0 ⎝ ⎜ x 2 + 1 ⎠ ⎠ ⎝ 1 ⎛ 4 6 5 4 2 − + − − + 4 ⎞⎟dx x 4 x 5 x 4 x ⎜ ∫ 0 ⎝ 2 x + 1 ⎠ π ⎞ 1 1 1 4 − 4 × + 5 × − − 4⎛ ⎜ ⎟+4 7 6 5 3 ⎝ 4 ⎠ 1
2
4
7 3 ⎛ 22
3
− +1− − π + 4
⎞ ⎜ − π⎟ ⎝ 7 ⎠
So, (A) is the correct option. 41.
Let ABC be a triangle such that
π
and let a, b and c denote the lengths of the 6 sides opposite to A, B and C respectively. The value(s) of x for which a = x 2 + x + 1, b = 2 x – 1 and c = 2x + 1 is (are)
(
(A) – 2 + 3 41.
∠ACB =
)
(C) 2 + 3 (B) π a 2 + b 2 − c 2 cos = 2ab 6
(2x (x
(B) 1 +
3
(D) 4 3
=
+ 2x 3 − 3x 2 − 2x + 1 (x 2 + x + 1)(x 2 − 1)
2x 4
+ 2x − 1)(x 2 − 1) ⇒ = 3 2 + x + 1)(x 2 − 1) ⇒ 2 − 3 )x 2 + 2 − 3 )x − 1 + 3 ) = 0 ⇒ x = – 2 + 3 ), 1 + 3 But as b + c > a ⇒ x > 1 ⇒ x = 1 + 3 2
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
16 PIE IIT JEE PAPER – I & II
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 16
SECTI SECTI ON – II I Linked Comprehension Type This section contains 2 paragraphs . Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct.
Paragraph for Questions 42 to 44 Let p be an odd prime number and T p be the following set of 2 × 2 matrices: ⎧ ⎫ ⎡a b ⎤ { } : a , b , c 0 , 1 , 2 ,..., p 1 T p = ⎨A = ⎢ ∈ − ⎬ ⎥ c a ⎣ ⎦ ⎩ ⎭ 42.
42.
43.
43.
44.
44.
The number of A in T p such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p is 2 (A) (p – 1) (B) 2(p – 1) 2 (C) (p – 1) + 1 (D) 2p – 1 (D) If A is symmetric then det(A) = a 2 – b2 = (a – b)(a + b) As it is divisible by p ⇒ either (a – b) = λ1 p or (a + b) = λ2 p ⇒ a = b or a + b = p hence a = b has p number of solution and a + b = p has (p – 1) number of solution Total solution = 2p – 1. The number of A in T p such that the trace of A is not divisible by p but det (A) is divisible by p is [Note: The trace of a matrix is the sum of its diagonal entries.] 2 3 2 (A) (p –1)(p – p + 1) (B) p – (p – 1) (C) (p – 1) 2 (D) (p – 1)(p2 – 2) (C) For the ordered pair (b, c) we have total (p – 1) 2 choices. 2 (p − 1) In choices there exists exactly two values of a for which (a 2 – bc) is a multiple of 2 (p − 1)2 p, and for remaining choices there exists no value of a for which (a 2 – bc) is a 2 multiple of p. 2 Hence total no. of choices for the triplet (a, b, c) such that a ≠ 0 and a – bc is multiple of (p − 1)2 × 2 = (p − 1) 2 . p, is 2 The number of A in T p such that det (A) is not divisible by p is (A) 2p2 (B) p3 – 5p 3 3 2 (C) p – 3p (D) p – p (D) 3 Total no. of matrices = p total no. of matrices, whose det (A) is divisible by p but trace not zero = (p – 1)
2
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
17 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 17
(SOLUTION – 2010)
Total no. of matrices having trace zero and whose det(A) is divisible by p = 2p – 1 As trace (A) = 0 ⇒ 2a = 0 ⇒ a = 0 now a2 – bc = λ p ⇒ –bc = λ p ⇒ bc = 0 ⇒ b = 0 or c = 0 Note for b = 0, c has p choices (namely 0, 1, 2, … p-1) Similarly for c = 0, b has the choices 1, 2, …., p − 1. Total choices for b & c = p + p – 1 = 2p – 1 Hence, total number of required matrices 3 2 3 2 p – (p – 1) – (2p – 1) = p – p Paragraph for Questions 45 to 46 2
2
The circle x + y – 8x = 0 and hyperbola 45.
45.
x2 9
−
y2 4
= 1 intersect at the points A and B.
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is (A) 2x – 5 y – 20 = 0 (C) 3x – 4y + 8 = 0 (B) Let the tangent be (x − 4 )cos θ + y sin θ = 4
⇒ ⇒
(B) 2x – 5 y + 4 = 0 (D) 4x – 3y + 4 = 0
y = − cot θ(x − 4 ) + 4 cos ecθ y = (− cot θ)x + 4(cot θ + cos ecθ)
Now, c = a m − b ⇒ 16(cotθ + cosecθ)2 = 9cot2θ − 4 ⇒ 16(cosθ + 1)2 = 9cos2θ − 4sin2θ= 13cos2θ − 4 2 ⇒ 3cos2θ + 32cosθ + 20 = 0 ⇒ cosθ = − 3 2 So, slope = − cot θ = 5 3 ⇒ cosecθ = 5 So, equation of tangent is 2 12 (x − 4 ) + y= 5 5 2
2
2
2
⇒ 2x − 5 y + 4 = 0 46.
46.
Equation of the circle with AB as its diameter is 2 2 2 2 (A) x + y – 12x + 24 = 0 (B) x + y + 12x + 24 = 0 (C) x2 + y2 + 24x – 12 = 0 (D) x2 + y2 – 24x – 12 = 0 (A) PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
18 PIE IIT JEE PAPER – I & II
2
2
Solving x + y
(SOLUTION – 2010)
x2
− 8x = 0 and
9
y2
−
4
IIT-JEE 2010 SOLUTION - 18
=1
4
⇒ x 2 + x 2 − 4 − 8x = 0 9
⇒
13 9
x 2 − 8x − 4 = 0 4 + 16 +
⇒
x=
52 9
=
4+
14
3 13
⎛ 13 ⎞ ⎜ ⎟ 9 ⎝ 9 ⎠ So, y = ± 8x − x 2 = ±2 3 So, A and B are (6, ± 2 3 )
=6
A
O
3
6
4
B
Centre of circle is the mid point of AB, which is (6, 0) and radius = 2 3 , so its equation is (x − 6)2 + y2 = 12 ⇒ x2 + y2 − 12x + 24 = 0 SECTION SECTION I V Integer Type
47.
This section contains TEN questions. The answer to each question is single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 2π 2π Let ω be the complex number cos + i sin . Then the number of distinct complex 3 3 z +1 numbers z satisfying
47.
ω ω2
ω z + ω2
ω2
1
z+ω
1
= 0 is equal to
1
ω = cos
2π 3
+ i sin
2π 3
, so ω3 = 1 and 1 +
ω + ω2 = 0.
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
19 PIE IIT JEE PAPER – I & II
z +1
ω z + ω2
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 19
ω2
1 =0 ω 1 z+ω ω2 ⇒ (z + 1) (z2 − z) − ω(zω) + ω2(−zω2) = 0 ⇒ z3 −z − zω2 − zω = 0 ⇒ z3 = 0 ⇒ z = 0 So number of distinct solution is 1. 48.
48.
The number of values of θin the interval
⎛ − π , π ⎞ such that θ ≠ nπ for n = 0, ±1, ±2 and ⎜ ⎟ 5 ⎝ 2 2 ⎠
tanθ = cot5θ as well as sin2 θ = cos4θ is 3 nπ tanθ = cot5θ, θ ≠ 5 cosθcos5θ − sin5θsinθ = 0 cos6θ = 0 5π 3π π π 3π 5π ⇒ 6θ = − , − , − , , , 2 2 2 2 2 2 5π π π π π 5π , ⇒ θ=− , − , − , 12 4 12 12 4 12 Again sin2θ = cos4θ = 1 − 2sin22θ ⇒ 2sin22θ + sin2θ − 1 = 0 1 ⇒ sin2θ = −1, 2 π π 5π ⇒ 2θ = − , , 2 6 6 π π 5π ⇒ θ=− , , 4 12 12 So, common solution are
θ= −
π π ,
4 12
and
5π 12
So number of solution is 3. 49.
For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [–10, 10] by ⎧ x − [ x ] if [ x ] is odd, f(x) = ⎨ ⎩1 + [ x ] − x if [ x ] is even Then the value of
49.
π2 10
10
∫
−10
f (x ) cos πx dx is
4
if ⎧{x} f (x ) = ⎨ ⎩1 − {x} if
[x ] = odd [x ] = even
Clearly f(x) is an even function and periodic with period 2. PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
20 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 20
(SOLUTION – 2010)
Also cosπx is periodic with period 2 So, f(x).cosπx is periodic with period 2 So,
π2
10
1
∫
f (x ) cos πx dx = 10 −10
π
2
∫ f ( x ) cos πxdx
−1
1
=
π2 × 2∫ f (x )cos πxdx
(As f(x) is even also)
0
1
= 2π
2
∫ (1 − x )cos πx dx 0
=
=
50.
50.
⎛ 1 ⎞ 2π × ⎜⎜ − ∫ x cos πxdx ⎟⎟ (Replace x by (1 − x)) ⎝ 0 ⎠ 1 1 ⎡ ⎤ 1 2 ⎞ ⎞ ⎛ ⎞ 2 ⎛ x − 2π ⎢⎜ sin πx ⎟ + ⎜ 2 cos πx ⎟ ⎥ = −2π2 ⎛ ⎜− 2 ⎟ = 4 ⎠ 0 ⎝ π ⎠ 0 ⎥⎦ ⎝ π ⎠ ⎢⎣⎝ π 2
If the distance between the plane Ax – 2y + z = d and the plane containing the lines x −1 y − 2 z − 3 x −2 y−3 z−4 and is 6 , then then d is is = = = = 2 3 4 3 4 5 6 ˆi jˆ k ˆ 2
3
4
3
4
5
ˆ (8 − 9) ) = ˆi (15 − 16) − jˆ(10 − 12) + k
ˆ (normal vector) − ˆi + 2 jˆ − k Equation of plane is −1(x − 1) + 2(y − 2) − 1(z − 3) = 0 −x + 1 + 2y − 4 − z + 3 = 0 −x + 2y − z = 0 x − 2y + z = 0 and Ax − 2y + z − d = 0 =
∴
distance =
|d| 1+ 4 +1
=
6
|d| = 6. 51.
51.
x2
y2
= 1. If this line passes through the − a 2 b 2 point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is 2 1 a 1 e y = 0, x = , = , a = 2 e 2 2 2 2 2 1 = 4a − (a (e − 1)) = 4a2 − a2e2 + a2 1 = 5a2 − a2e2 The line 2x + y = 1 is tangent to the hyperbola
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
21 PIE IIT JEE PAPER – I & II
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 21
4 = 5e2 − e4 ⇒ e4 − 5e2 + 4 = 0 e2 = 1, e2 = 4 but e2 ≠ 1 ∴ e = 2. 52.
Let Sk , k = 1, 2, …, 100, denote the sum of the infinite geometric series whose first term is k − 1 1 100 2 100 2 and the common ratio is . Then the value of k − 3k + 1 Sk is + 100! k =1 k ! k 4 100 k −1 1 1 ( k 2 − 3k + 1)Sk (where Sk = k ! / ⎛⎜1 − k ⎞⎟ = ( k − 1)! ⎝ ⎠ k =1
∑(
52.
)
∑
100
=
∑
(k
k =1
= 2+ = 4−
∴
53.
53.
100 − 3k + 1) ( k − 1) ( k − 2 ) − 1 =1+1+ ∑ ( k − 1)! ( k − 1)! k=3
2
(for k ≥ 3, k 2 − 3k + 1 > 0)
⎛ 1 1 ⎞ − ⎜ ⎟⎟ ∑ ⎜ k 3 ! k 1 ! − − ( ) ( ) k =3 ⎝ ⎠ 100
1
98! 100 2
1
−
99!
+4−
1
−
1
100! 98! 99! 100 1 1 = − +4− 99! 99! 98! 1 1 = 4+ = 4. − 98! 98!
Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to 9 dy The equation of tangent : Y − y = (X − x) dx dy y intercept = y − x = x3 dx dy y − = −x 2 dx x I.F. = e y x
=−
−
x2 2
1
∫ x dx =
1 x
+c
f(1) = 1, c =
3 2
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
22 PIE IIT JEE PAPER – I & II
y
x2
=−
+
3
⇒ y=−
x 2 2 f(−3) = 9. 54.
2
+
3x 2
r
If a and b are vectors in space given by a = r
r
r
(2ar + b)⋅ [(ar × b )× (ar − 2 b )] 54.
x3
r
r
5
r
IIT-JEE 2010 SOLUTION - 22
(SOLUTION – 2010)
ˆi − 2 jˆ 5
r
and b =
ˆ 2ˆi + jˆ + 3k 14
, then the value of
is
r
r
b b . =1
a b . =0
(2ar + b).((arr×r b )× (ar − 2r b )) rr
a.a = 1 r
r
r
r
r
r
r
r
r
r
r
(a × br )× (a −r 2 b ) = (a × br)× a − 2(a × b )× b r r r = 2 b × (a × b ) − a × (a × b ) rr r rr r rr r rr r = 2(( b b . )a − (a b . b ) ) − (a b. )a + (a.a ) b b r r
= 2(a ) + b r
r
r
r
r
r
(2a + b).r((a × b )×r(a − 2 b )) r r = (2a + b ).(2a + b ) r r rr rr
r
r
= 4a.a + 2a b . + 2 b.a + b.b =4+1=5
55.
55.
The number of all possible values of θ, where 0 < θ < π, for which the system of equations (y + z) cos3θ = (xyz) sin3 θ 2 cos 3θ 2 sin 3θ x sin3θ = + y z (xyz) sin3θ = (y + 2z) cos3 θ + y sin3θ have a solution (x 0, y0, z0) with y0 z0 ≠ 0, is 3 xyz sin3θ = (y + z)cos3θ (1) xyz sin3θ = 2zcos3θ + 2ysin3θ (2) xyz sin3θ = (y + 2z)cos3θ + ysin3θ (3) ⇒ (cos3θ − 2sin3θ)y − (cos3θ)z = 0 ((1) − (2)) and (sin3θ)y + (cos3θ)z = 0 ((3) − (1)) for y ≠ 0, z ≠ 0 cos 3θ − 2 sin 3θ cos 3θ
⇒
=−
sin 3θ cos 3θ cos3θ − 2sin3θ = −sin3θ tan3θ = 1 3θ = nπ +
θ=
nπ 3
+
π 4
π 12
,n∈I ,n∈I
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
23 PIE IIT JEE PAPER – I & II
5π 9π , 12 12 12 number of values of θ = 3.
∴ θ= ∴
π
,
56.
The maximum value of the expression
56.
2 f (θ) = sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ
3
= 1 + 4cos 2θ + = 3 + 2cos2 θ + f(θ)min = 3 −
∴
IIT-JEE 2010 SOLUTION - 23
(SOLUTION – 2010)
5 2
=
2 3 2
1 sin
2
θ + 3 sin θ cos θ + 5 cos 2 θ
is
sin2θ sin2θ
1 2
maximum value of
1 f (θ)
= 2.
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
24 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 24
(SOLUTION – 2010)
SECTION - I This Section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
57.
57.
58.
An AC voltage source of variable angular frequency ω and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased (A) the bulb glows dimmer (B) the bulb glows brighter (C) total impedance of the circuit is unchanged (D) total impedance of the t he circuit increases i ncreases (B) Vo I rms = 2 1 ⎞ ⎛ 2 2 R + ⎜ ⎟ ⎝ ωC ⎠ A thin flexible wire of length L is connected to two adjacent fixed points and carries a current I in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength B going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× ××××××××××××× IBL (B)
(A) IBL (C) 58.
π
IBL
(D)
2π
(C) 2Tsinθ = Fm For small θ 2Tθ = B × I × Δ l = B × I × 2 θR BIL T = BIR = 2π
IBL 4π
Δl T cosθ cosθ
θ
Fm θ
θ θ
T Tsinθ Tsinθ
T cosθ cosθ T
Tsinθ Tsinθ
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
25 PIE IIT JEE PAPER – I & II
59.
IIT-JEE 2010 SOLUTION - 25
(SOLUTION – 2010)
A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and tanθ > μ. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P 1 = mg(sinθ – μ cosθ) to P2 = mg(sinθ + μcosθ), the frictional force f versus P graph will look like f
P2 P1
(B)
P
f
P1
(D)
P
59.
(A) P±f
60.
A real gas behaves like an ideal gas if its (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high (D)
61.
P1
P2
P1
P2
P
f
P2
60.
θ
f
(A)
(C)
P
P
= mg sin θ = constant
Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is
t L
(A) directly proportional to L (C) independent of L
(B) directly proportional to t (D) independent of t
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26 PIE IIT JEE PAPER – I & II
61.
(C)
ρ× L ρ = L× t t
R =
62.
IIT-JEE 2010 SOLUTION - 26
(SOLUTION – 2010)
A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is
P
4R
3R 4R
2GM
(4 7 R
(A)
(D)
4R
V p
=
−2GM ⎡ 4 7R 2 ⎣
Wext
7R 2GM 5R
(
(4
2 −5
)
)
2 −1
16R 2 + r 2
2 − 5⎤ R
⎦
= (0 − VP ) =
2GM
⎡4 ⎣
2 − 5⎤
⎦ 7R Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistance R 100 100, R 60 60 and R 40 40, respectively, the relation between these resistance is 1 1 1 (A) (B) R 100 = + 100 = R 40 40 + R 60 60 R 100 R 40 R 60 (C) R 100 100 > R 60 60 > R 40 40
(D)
1 R 100
>
1 R 60
>
1 R 40
(D)
R = R 100 1 R100 64.
2GM
rdr
= −2πGσ ∫ 3R
63.
−
(A) 4R
63.
(B)
GM
(C) 62.
)
2 −5
V2 P < R 60
>
1 R 60
< R 40 >
1 R 40
To verify Ohm’s law, a student is provided with a test resistor R T, a high resistance R 1, a small resistance R 2, two identical galvanometers G1 and G2, and a variable voltage source V. The correct circuit to carry out the experiment is PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
27 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 27
(SOLUTION – 2010)
G1
G1
R 2
(A)
R 1 G2
R T
R 1
(B)
V G1
(C)
R T
G2
R T
V R 1
G1
G2
(D)
R 2
R T
R 2 V
64.
R 2
G2 R 1
V
(C) When high resistance R 1 is connected in series with G 1, it acts as a voltmeter. SECTION – II Multiple Correct Answer Type This section contains 5 multiple correct questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.
65.
65.
A point mass of 1 kg collides elastically with a stationary point mass of 5kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms –1. Which of the following statement(s) is (are) correct for the system of these two masses? (A) total momentum of the system is 3kg ms –1 (B) momentum of 5kg mass after collision is 4 kg ms –1 (C) kinetic energy of the t he centre of mass is 0.75 J (D) total kinetic energy ener gy of the system is 4 J (A, C) u= 5v–2 (1) v+2=u (2) Solving equations (1) and (2), we get v = 1m/s, u = 3 ms −1 1× 3 1 vcm = = 6 2 1 1 3 KE cm = × 6 × = J 2 4 4 1 9 TE = ×1 × 9 = = 3.5J 2 2
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
28 PIE IIT JEE PAPER – I & II
66.
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 28
A few electric field lines for a system of two charges Q 1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that (A) Q1 > Q 2 (B) Q1 < Q 2 (C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero.
66.
(A, D)
67.
A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60° (see figure). If the refractive index of the material of
67.
B o
60
C
o
135
the pris prism m is 3 , which which of of the follow following ing is (are) (are) corr correct ect?? (A) the ray gets totally t otally internally reflected refle cted at face fac e CD (B) the ray comes out through face AD (C) the angle between the incident ray and the emergent ray is 90° (D) the angle between bet ween the incident ray and the emergent ray is 120° (A, B, C) sin sin 60º = 3 sinr 1 sin r = ; r ⇒ 30º 2 ∴ Ray PQ is || to BC TIR will take place at the face CD since i > θc 1 where i = 45o and θc = sin−1 3 1 sinθ <
o
75o
90
D
A
B 60
60
o
o
30
o
C
o
135
45
o
45
o
o
30
o
o
75
90
D
A 60
o
μ
⎛ θ < sin −1 ⎜ ⎝ 68.
⎞ < 45o ⎟ 3⎠
1
One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P 0. Choose the correct option(s) from the following (A) Internal energies at A and B are the same (B) Work done by the gas in i n process AB is P 0V0 ln 4 P (C) Pressure Pressure at C is 0 4 T (D) Temperature at C is 0 4
V B
4V0
V0
C
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A T0
T
29 PIE IIT JEE PAPER – I & II
68.
(SOLUTION – 2010)
(A, B) AB→ Isothermal and AC→ Isobaric ∴ U A = UB
WAB = PV ln
IIT-JEE 2010 SOLUTION - 29
V B
4V0
⎛ V2 ⎞ ⎜ V ⎟ = PoVoln4 ⎝ 1⎠
V0
A
C
T0
69.
69.
T
A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec. for this and records 40 seconds for 20 oscillations. For this observation, which of t he following statement(s) is are true? (A) Error ΔT in measuring T, the time ti me period, is 0.05 seconds (B) Error ΔT in measuring T, the time period, is 1 second (C) Percentage error in the determination of g is 5% (D) Percentage error in the determination of g is 2.5% (A, C) least count 1 = = 0.05s Error in time period, ΔT = n 20 T= g=
Δg g
t n
= 2π
l
g
4π2 ln 2 t2
×100 = −
2Δt t
×100 =
−2 ×1 40
×100 = −5
SECTION −I I I Linked Comprehension Type This section contains 2 paragraphs . Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct. Paragraph for Questions 70 to 72
When a particle of mass m moves on the x-axis in a potential of the form V(x) = kx2, it performs simple harmonic motion. The corresponding time period is proportional to
V(x)
m
, as can be seen k easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx 2 and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
V0
x X0
30 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 30
(SOLUTION – 2010)
energy is V(x) = αx4 (α > 0) for |x| near the origin and becomes a constant equal to V 0 for |x| ≥ X0 (see figure). 70.
70. 71.
If the total energy of the particle is E, it will perform periodic motion only if (A) E < 0 (B) E > 0 (C) V0 > E > 0 (D) E > V0 (C) For periodic motion of small amplitude A, the time period T of this particle is proportional to (A) A (C) A
71.
(B)
α α
(D)
m
1
m
A
α
m
A 1
α α
A
m
The acceleration of this particle for |x| > X0 is (A) proportional to V0 (C) proportional to
72.
1
(B) For small amplitude (mω2A2) ∝ (α A4) ⎛ m ⎞ T2 ∝ ⎜ ⎟ ⎝ αA 2 ⎠
T∝ 72.
m
(B) proportional to V0
mX 0
V0 mX0
(D) zero
(D) For, |x| ≥ xo V(x) = constant ∴ F = 0, acceleration of particle = zero
Paragraph for Questions 73 to 74 Electrical resistance of certain materials, known as TC (B) superconductors, changes abruptly from a non zero value to zero as their temperature is lowered below a TC (0) critical temperature TC(0). An interesting property of superconductors is that their critical temperature becomes smaller than TC(0) if they are placed in a magnetic field, i.e., the critical temperature TC(B) is a O function of the magnetic field strength B. The dependence of TC(B) on B is shown in the figure.
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B
31 PIE IIT JEE PAPER – I & II
73.
IIT-JEE 2010 SOLUTION - 31
(SOLUTION – 2010)
In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic fields B1 (solid line) and B 2 (dashed line). If B 2 is larger than B1, which of the following graphs shows the correct variation of R with T in these fields? R
R B2
(A)
(B) B1 B2
B1
O
T
T
O
R
R B1
(C)
(D) B1 O
73.
(A) Q B2
B2
B2 T
T
O
> B1
∴ (Tc)2 < ( Tc)1 74.
74.
A superconductor has T C(0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its T C decreases to 75K. For this material one can definitely say that when (A) B = 5 Tesla, T C(B) = 80 K (B) B = 5 Tesla, 75 K < TC(B) < 100 K (C) B = 10 Tesla, 75 K < TC (B) < 100 K (D) B = 10 Tesla, T C(B) = 70K Tc (B) (B) As shown in graph for B = 5 75 K < Tc (B) < 100 K Tc (0) T 75
O
5
7.5
B
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
32 PIE IIT JEE PAPER – I & II
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 32
SECTION −I V
Integer Type This section contains TEN questions. The answer to each question is a single digit integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled. 75.
75.
A binary star consists of two stars A (mass 2.2M S) and B (mass 11M S), where MS is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is 6 CM Total angular momentum of binary MA MB star, Ltotal = Itotalω Itotal = MA (5d/6)2 + MB(d/6)2 d/6 5d/6 angular momentum of star B LB = IBω IB = MB (d/6)2 L total LB
2.2M ( 5d / 6 ) ( = s
2
+ 11M s (d / 6) 2 ) ω
11M s (d / 6) 2 ω
=6
76.
The focal length of a thin t hin biconvex lens is 20cm. When an object is moved from a distance of 25 cm in front of it to50 cm, the magnification of its image changes from m 25 to m50. m The ratio 25 is m50
76.
6
m=
⎛ f ⎞ ⎜f +u⎟ ⎝ ⎠
m 25
=
m 25
=6
m50 77.
77.
20 20 − 25
and m50
=
20 20 − 50
A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10 –7m2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s –1. If the Young’s modulus of the material of the wire is n × 109 Nm –2, the value of n is 4 By Hooke's law F x = −Y A L ⎛ YA ⎞ x F = −⎜ ⎟ ⎝ L ⎠ PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
33 PIE IIT JEE PAPER – I & II
IIT-JEE 2010 SOLUTION - 33
(SOLUTION – 2010)
YA
ω=
Lm
Y=
mLω2
A Y = 4 ×10 1 09 n=4
( 0.1) × (1) × (140) 2 = 7 4.9 ×10 − = n × 109
78.
When two progressive waves y 1 = 4 sin (2x – 6t) and y 2 = 3 sin
78.
superimposed, the amplitude of the resultant wave is 5 A 2 = A12 + A 22 + 2A1A 2 cos φ
79.
79.
are
given that A1 = 4, A2 = 3 and φ = - π/2 A=5 Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T 1and T2 respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B? 9 λmT = const T1 λ 2 = =3 T2 λ1 E1 E2
80.
⎛ 2x − 6 t − π ⎞ ⎜ ⎟ 2⎠ ⎝
2
4
σA1T 4 ⎛ r1 ⎞ ⎛ T1 ⎞ ⎛ 6 ⎞2 = = ⎜ ⎟ × ⎜ ⎟ = ⎜ ⎟ × (3)4 =9 4 σA 2 T2 ⎝ r2 ⎠ ⎝ T2 ⎠ ⎝ 18 ⎠ 1
Gravitational acceleration on the surface of a planet is
6 11
g , where g is the gravitational
acceleration on the surface of the earth. The average mass density of the planet is
2
times 3 that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms –1, the escape speed on the surface of the planet in kms –1 will be 80.
3
g= gp ge
GM R2
ve
4 3
πR 3
R2 2
4
= πρGR 3
ρp × R p 3 × R p = = = ρe × R e Re
⇒ R p vp
=
G ×ρ×
=
=
3 6 22
11
R e
2g p R p 2g e R e
6
=
6 11
×
3 6 22
=
3 11
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34 PIE IIT JEE PAPER – I & II
v p
= ve
(SOLUTION – 2010)
IIT-JEE 2010 SOLUTION - 34
3
11 v p = 3 km/sec 81.
81.
A stationary source is emitting sound at a fixed frequency f 0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f 0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms –1. 7 Frequency received by car is ⎛ v + vc ⎞ f1 = f o ⎜ ⎟ ⎝ v ⎠ Reflected frequency received by observer 2
⎛ v + vc ⎞ = f 2 = f o ⎜ ⎟ ⎝ v ⎠ v ⎞ ⎛ f 2 = f o ⎜1 + 2 c ⎟ v ⎠ ⎝ 2f ( Δv ) Δf 2 = o c
⎛ v ⎞ f o ⎜1 + c ⎟ ⎝ v⎠
2
(since vc 0, is 5,
1 5 ⎞ , ⎟ 3 2 ⎠
(A) x + 2y − 2z − α = 0 Perpendicular distance from P(1, −2, 1) is |1− 4 − 2 − α | =5 1+ 4 + 4 α = 10 x −1 y + 2 z −1 = = Equation of PQ is = k 1 2 −2
P(1, −2, 1)
Q
Coordinates of Q is (k + 1, 2k − 2, −2k + 1) satisfying this, in equation of plane 5 We get, k = 3 8 4 7 ⎞ ⇒ Q is ⎛ ⎜ , ,− ⎟ ⎝ 3 3 3 ⎠ 21.
21.
A signal which can be green or red with probability
4
and
1
respectively, is received by 5 5 station A and then transmitted to station B. The probability of each station receiving the 3 signal correctly is . If the signal received at station B is green, then the probability that 4 the original signal was green is 3 6 (A) (B) 5 7 20 9 (C) (D) 23 20 (C) Original A B R ⎯⎯ R ⎯ → →G → →G R ⎯⎯ G ⎯ G ⎯⎯ G ⎯ → →G → →G G ⎯⎯ R ⎯ Required probability ∴ PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
45 PIE IIT JEE PAPER – I & II
4 = 40
=
46
=
1
5 20
IIT-JEE 2010 SOLUTION - 45
(SOLUTION – 2010)
3
1
5 1
4
4
5
3
3
4
1
1
4 1
4 3
5 4
4 3
4 3
4
1
1
4
4
5
4
4
5
4
4
× × + × ×
× × + × × + × × + × ×
23 uuur
22.
)
Two adja adjace cent nt sid sides of a par paral alle lelo logr gram am ABCD BCD are are giv given by AB = 2i
)
)
+ 10j + 11k
and
ˆ AD = −ˆi + 2 jˆ + 2k The sides AB is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD′. If AD′ makes a right angle with the side AB, then the cosine of the angle α is given by 8
(A)
9 1
(C) 22.
(D)
9
(B) α + θ = 90o α = 90o − θ cosα = sinθ
sinθ = r
23.
9 4 5 9 D
D′
r
r
r
r
C
α
a × b
θ
| a || b |
r ˆ a × b = −2ˆi − 15 jˆ + 14k
∴
17
(B)
A
B
17
sinθ=
9
For r = 0, 1, …., 10, let A r , Br and Cr denote, respectively, the coefficient of xr in the expansions of (1 + x) 10, (1 + x)20 and (1 + x) 30. Then 10
∑ A (B r
10
B r − C10 A r )
r =1
is equal to (A) B10 − C10 23.
2 (B) A10 B10
(D) C10 − B10
(C) 0 (D) (1 + x)10 (1 + x) 20 (1 + x) 30 A r = 10C r Br = 20 C r C r = 30C r 10
∑
10
(
C r
20
C10
20
C r − 30 C10
10
− C10 A10
C r
)
r =1
=
20
10
C10
∑
10
C10−r C r − C10 20
30
r =1
=
20
C10
(
30
C10 − 1
10
∑
10
C r 2
r =1
)−
30
C10
(
20
)=
C10 − 1
30
C10 − 20C10
= C10 − B10
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
46 PIE IIT JEE PAPER – I & II
24.
IIT-JEE 2010 SOLUTION - 46
(SOLUTION – 2010)
Let f be a real valued function defined on the interval ( −1, 1) such that x
−x
e f(x) = 2 +
∫ t
4
+ 1 dt , for all x ∈ (−1, 1) and let f −1 be the inverse function of f. Then
0
−1
(f )′(2) is equal to (A) 1 (C) 24.
(B)
1
(D)
2
1 3 1 e
(B) x
−x
e f (x ) = 2 +
∫ 1 + t dt 4
∀ x ∈ (−1, 1)
0
At x = 0, f(0) = 2 Now −e−xf(x) + e−x f ′(x) = 0 +
⇒ − e0f(0) + e0f ′(0) = ⇒ −2 + f ′(0) = 1 ⇒ f ′(0) = 3
1+ x4
1+ 0
Now f -1(f(x)) = x ⇒ (f −1f(x))′ f ′(x) = 1 ⇒ (f −1f(0))′f ′(0) = 1 1 1 ⇒ (f −1(2))′ = = f ′(0 ) 3 25.
25.
Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to (A) 25 (B) 34 (C) 42 (D) 41 (D) S = {1, 2, 3, 4} Let P and Q be disjoint subsets of S ai ∈ P and ai ∉ Q ai ∉ P and ai ∈ Q ai ∉ P ad ai ∉ Q for every element there are three option hence total option are 3 4 = 81 which includes the case in which P and Q are φ 81 − 1 Now after removing the order in P and Q we get the total Number = + 1 = 41. 2 SECTION II (Integer Type) This section contains 5 questions. The answer to each question is single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.
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47 PIE IIT JEE PAPER – I & II
26.
26.
Let a1, a2, a3, ….., a11 be real numbers satisfying a1 = 15, 27 − 2a2 > 0 and ak =2ak −1 − ak −2 for k = 3, 4, …., 11. 2 a + a 2 + ... + a 11 a 12 + a 22 + ..... + a 11 If = 90, then the value of 1 is equal to 11 11 0 a1 = 15 27 1 27 − 2a2 > 0 ⇒ a2 < = 13 2 2 ak = 2ak −1 − ak −2 ⇒ ak − ak −1 = ak −1 − ak −2 ⇒ a1, a2, a3, …. a11 are in A.P. 3 1 Let d be the common difference as a 2 < 13 and a1 = 15 , so, d < − 2 2 a 12
⇒ ⇒ ⇒
+ (a 1 + d )2 + (a1 + 2d ) + ...... + (a1 + 10d )2 11a 12
+
11 2 2 1 + 2 + ...... + 10 2 d 2
7d2 + 30d + 27 = 0 (7d + 9) (d + 3) = 0 9 ⇒ d = −3, − 7 3 But d < − ⇒ d = −3 2 a + a 2 + a 3 + ..... + a11 So, 1 11 = a1 + 5d = 15 + 5( −3) = 0 27.
27.
IIT-JEE 2010 SOLUTION - 47
(SOLUTION – 2010)
=
= 90
+ 2a1d(1 + 2 + 3 + ..... + 10)
× ⎛⎜ 2 ⎝
11
= 11 × 90
2a1 + 10d ⎞ 11
⎟ ⎠
Let f be a function defined on R (the set of all real numbers) such that f ′(x) = 2010(x − 2009) (x − 2010)2(x − 2011)3(x − 2012)4, for all x ∈ R. If g is a function defined on R with values in the interval (0, ∞) such that f(x) = ln(g(x)), for all x ∈ R, then the number of points in R at which g has a local maximum is 1 f ′(x ) = 2010(x − 2009 )(x − 2010 ) (x − 2011) (x − 2012 ) f(x) = ln g(x) ⇒ g(x) = ef(x) ⇒ g′(x) = ef(x) . f ′(x) + + + − − 2009
2010
2
3
2011
2012
4
Local maximum at x = 2009 Local minimum at x = 2011
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
48 PIE IIT JEE PAPER – I & II
28.
Let k be a positive real number and let
⎡ 2k − 1 ⎢ A = ⎢ 2 k ⎢− 2 k ⎣
28.
IIT-JEE 2010 SOLUTION - 48
(SOLUTION – 2010)
2 k 2 k ⎤ 1 2k
⎥ − 2k ⎥ − 1 ⎥⎦
⎡ 0 2k − 1 ⎢ B = ⎢1 − 2k 0 ⎢ − k − 2 k ⎣
and
k ⎤
⎥
2 k ⎥ . 0
⎥ ⎦
If det(adj A) + det(adj B) = 10 6, then [k] is equal to [Note : adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k]. 4 2k − 1 | A |= 2 k
−2
2k − 1
− 2k −1
= 2 k
1
k
2k − 1
2 k 2 k 2k 0
= 2 k
1 + 2k 2k
−4
0
k
2 k
−2
0
2 k
1 + 2k
− 2k −1
k 2k + 1
−2k 2k − 1
= (1 + 2k) ((2k − 1)2 + 8k) 4k 2 − 4k + 1 + 8k = (1 + 2k) 2 ⇒ |A| = (1 + 2k) (2k + 1) 2 = (2k + 1)3 |B| = 0 (as B is skew symmetric of odd order) det (adj A) + det(adjB) = |A|2 + |B|2 = 106 ⇒ (2k + 1)6 = 106 ⇒ 2k + 1 = 10 ⇒ k = 4.5 [k] = 4.
(
29.
29.
)
Two parallel chords of a circle of radius 2 are at a distance 3 + 1 apart. If the chords π 2π subtend at the centre, angles of and , where k > 0, then the value of [k] is k k [Note [k] denotes the largest integer less than or equal to k] 3
π ⎞ ⎟ = ( 3 + 1) k 2k ⎠ ⎛ 3 + 1 ⎞ π π ⎟ + cos − 1 = ⎜⎜ 2 cos 2 ⎟ 2k 2k ⎝ 2 ⎠ 3 − π − 1 ± 13 + 4 3 cos = = , ⎛ ⎝
2⎜ cos
π
+ cos
2k
cos
⇒
π
4
= cos
2k k = 3.
2
2π/k 3 +1 2
π/k
π 6
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
49 PIE IIT JEE PAPER – I & II
30.
IIT-JEE 2010 SOLUTION - 49
(SOLUTION – 2010)
Consider a triangle ABC and let a, b and c denote 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 ∠ ACB is obtuse and if r denotes the radius of the incircle of triangle, then r 2 is equal to
30.
(3)
Δ=
A 1 2
ab sin C 1
⇒
15 3
⇒
sin C =
c
= × 6 ×10 × sin C
10
2
3
⇒ C = 120o
B
2 2 2 So, c = a + b2 − 2ab cosC
= 36 + 100 − 120 ×
⇒ c = 14 Δ
So, r =
⇒
31.
⎛ − 1 ⎞ = 196 ⎜ ⎟ ⎝ 2 ⎠
15 3 ⎛ 6 + 10 + 14 ⎞
⎜ ⎝
2
C
=
3
⎟ ⎠
r 2 = 3.
SECTION – III (Paragraph Type) This section contains 2 paragraphs . Based upon the first paragraph 3 multiple choice questions have to be answered. Each of these questions has four choices A), B), C), D) out of which ONLY ONE is correct. Paragraph for Questions 31 to 33 Consider the polynomial f(x) = 1 + 2x + 3x 2 + 4x3 Let s be the sum of all distinct real roots of f(x) and let t = |s|. The real numbers s lies in the interval 3 ⎞ ⎛ 1 ⎞ ⎛ (A) ⎜ − , 0 ⎟ (B) ⎜ − 11, − ⎟ 4 ⎠ ⎝ 4 ⎠ ⎝
(C) 31.
s
=
6
Δ = 15 3
⎛ − 3 , − 1 ⎞ ⎜ ⎟ ⎝ 4 2 ⎠
⎛ ⎝
(D) ⎜ 0,
1 ⎞
⎟
4 ⎠
(C) f(x) = 4x3 + 3x2 + 2x + 1 f ′(x) = 12x2 + 6x + 2 > 0 ∀ x ∈ R ⇒ Only one root is real 1 ⎛ 1⎞ 1 ⎛ 3 ⎞ Also f ⎜ − ⎟ = > 0 and f ⎜ − ⎟ = − < 0 2 ⎝ 4 ⎠ ⎝ 2⎠ 4
⇒
s∈
⎛ − 3 , − 1 ⎞ ⎜ ⎟ ⎝ 4 2 ⎠
PIE EDUCATION, Corporate Office: 44A/2, Sarvapriya Vihar, New Delhi - 16. Ph: 011–41828585. Fax: 011− 26520144 26520144 Head Office− AXIS 117/N/581, Kakadeo Kanpur − − Visit us : www.pieeducation.com
50 PIE IIT JEE PAPER – I & II
32.
The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval ⎛ 3 ⎞ ⎛ 21 11 ⎞ (A) ⎜ , 3 ⎟ (B) ⎜ , ⎟ ⎝ 4 ⎠ ⎝ 64 16 ⎠
⎛ ⎝
(D) ⎜ 0,
(C) (9, 10) 32.
IIT-JEE 2010 SOLUTION - 50
(SOLUTION – 2010)
21 ⎞
⎟
64 ⎠
(A) Required area, A t
=
∫ (4x
2
+ 3x 2 + 2 x + 1)dx
0
⇒ A = t4 + t2 + t2 + t As
33.
1 2
3
15
4
16
< t <
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