RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / ANSWER KEY / CODE - 5
Date : 24-05-2015
IITJEE-ADVANCED - 2015 CODE - 5 : ANSWER KEY PAPER - I PHYSICS 1.
[2]
2.
[7]
3.
[2]
4.
[3]
5.
[3]
6.
[7]
7.
[6]
8.
[2]
9.
[ABD]
10.
[B]
11.
[AC]
12.
[BC]
13.
[ACD]
14.
[BD]
15.
[D]
16.
[C]
17.
[B]
18.
[ABC]
19.
[A - r ; B - p s ; C - q t ; D - p q r t]
20.
[A - p q r t ; B - q s ; C - p q r s ; D - p r t]
CHEMISTRY 21.
[8]
22.
[4]
23.
[4]
24.
[3]
25.
[4]
26.
[1]
27.
[2]
28.
[9]
29.
[AB]
30.
[B]
31.
[A]
32.
[BD]
33.
[A]
34.
[D]
35.
[A]
36.
[C]
37.
[ABC]
38.
[BCD]
39.
[A - p q s ; B - t ; C - q r ; D - r ]
40.
[A - r t ; B - p q s ; C - p q s ; D - p q s t ]
MATHEMATICS 41.
[8]
42.
[5]
43.
[2]
44.
[0]
45.
[4]
46.
[3]
47.
[8]
48.
[4]
49.
[AD]
50.
[ABC]
51.
[ACD]
52.
[CD]
53.
[BC]
54.
[BD]
55.
[AB]
56.
[AD]
57.
[AC]
58.
[BC]
59.
[A - p q ; B - p q ; C - p q s t ; D - q t]
60.
[A - p r s ; B - p ; C - p q ; D - s t] SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
66 6
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
PHYSICS =========================================== Q.1
1 mgH mV 2 H 2 1 R
H
H 2 1 V 2gH R V 2 2gH
H
g'
H 2 V R
V2 V2 2g R
GM GM 2 2 r R H
1 g' g 4
GM
R H
2
1 GM 4 R2
R H 2R HR
V gR Escape velocity V 2gR 2V N2
Q.2 Same final energy is same for both Difference in initial energy is equal to (30-27) mg
Initial energy of disc 2
1 Ip 22 2
p
1 3 3 mR 2 22 mV22 2 2 4 1 Initial energy of Disc 1 I p 12 2
11
2
V
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
1 3 3 2 2 2 mR 1 mV1 2 2 4
3 3 mV22 mV12 30 27 mg 4 4
V22 V12 4g V22 9 40 V2 49
V2 7
Q.3
R A 400R B
__ i
PA 104 PB
__ ii
Using Stefan’s Law P AT 4
PA 4R A2 TA4 PB 4R B2 TB4
Using (ii) 4R 2A TA4 104 4R 2BTB4
TA RB 1 10 1 10 10 TB RA 400 20 2 Using Weins’s displacement law
ATA BTB A TB 2 B TA
Q.4 Half life T years Amt of feul in the beginning is N 0 Total requirement N'
12.5 N0 0.125N0 100
After n half lifes 1 N N0 2
n
22
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5 n
1 0.125N0 N0 2
n=3
Q.5
S2P S1P optional
x 2 d 2 x 2 d 2 m
4 2 2 1 x d m 3
S1 d d S2
x 2 d2 m 2 2 9 x 2 9m 2 2 d 2
On comparing with x 2 P 2 m 2 2 d 2 We have P = 3
R.O.C=20 cm
F=100cm
2h
h
Q.6
I2
2h
15cm 50cm
1st reflectron at mirror 1 1 1 V 15 10
1 1 1 1 V 15 10 30
For lens :
1 1 1 1 1 V 20 V 20 10 V 20
m 1
M1 2 2 1 3 1 35 For case 2 1 Flens cm Flens 7 5 2 6 33
AlR x
P
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
R.O.C=20 cm
h
f 7 6
35 cm 2
14h
2h 20cm
15cm
140cm
50cm For lens : M lens
1 1 2 V 140 cm V 20 35
V 140 7 u 20
M 2 14 M 2 14 7 M1 2
Q.7 Consider the strip of length ‘dy’ and width L as shown in figure A L 3 a sec 2
Infinity wire = charge P.4 length
B
y
dy
3 a 2
D
L
C
d Ewire Ldy cos 3 a sec L dy cos 2 0 2
y 3 3 2 2 Since tan dy a sec d L sec cos d 2 3a 2 3 20 a sec 2 2 B
d 20
dy
Rectangle dstrip
wire
6
L 2o d
6
44
3 a sec 2 d 3 a 2
E wire
y
C
side view
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
L L 06 20 6 0 6
n 6
Q.8 Energy incident radiation
1242 eV 13.8eV 90
Photon
Ionization energy for the electron = 13.8eV – 10.4 eV = 3.4 eV
E=13.8eV
e–
12 3.4 13.6 2 n 2 n 2nd orbital n2
Q.9 1 mole H 2 and 1 mole He. U
f1 f 5 3 n1R 1T1 2 n 2 RT2 1 R T 1 R T 4RT 2 2 2 2
A
T Average energy per mole = 2RT
M mixture
mixture
1 2 1 4 3gm / mole 2
C
p mix
C V mix
Speedmix Speed He
e–
7R 5R 1 2 2 3 5R 3R 2 1 1 2 2 1
mix RT M1 He RT M2
3 4 mix M2 2 5 He M2 3 3
55
K.E = 10.4eV
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
6 B 5
VRMS He VRMS H
2
Q.10 R fe fe
M H2 M He
2 1 4 2
D
fe 110 7 50 10 3 area 4 106
5 109 5 R fe 6 103 1.25 103 4 10 4 Al 2.7 10 8 50 10 3 R Al Al area 72 22 106
2.7 5 1010 0.3 104 6 45 10
0.03 103 Net resistance
1 1 1 R net R Al R Al
R net
R Al R fe R Al R fe
R net
1.25 0.03 103 1.25 0.03
R net 0.029 10 3
29 106 29
So close ans is B.
Q.11 eV0
hc 0
V0
hc 1 W0 e
y = m x –c
Option (C)
66
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
V0
hc W0 e
increases V0 decreases
V0
V0
dV0 1 hc 2 d 1
dV0 hc 1 2 d e
Options – A and C are correct. Q.12
1 One main scale division of vernir callipers is cm 8 5 V.S.D 4 M.S.D
1.V.S.D
4 M.S.D 5
least constnat of vernier callipers is L.C = 1 M.S.D – 1 V. S.D 4 1 M.S.D 5 1 1 M.S.D cm 5 40
1 m.m 0.25 mm 4
pitch of the screw guage is
2 Linear division (A) if pitch = 2 (L.C)
(A) 2 L.D 0.25 mm L.D 0.25mm So pitch of screw guage is
P
2 0.25mm 0.5mm 0.005mm 100 100
So A is wrong B is correct Option (C) : (L.C) = 2 (L.C)V.C. 1 mm 4
.d 2
So least count of screw guage is
77
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
1 2 mm 2 0.01mm L.C 100 So, D is wrong Q.13
M1L2T2 E hh E/ M1L2T1 1 T C LT 1 G M 1L3T 2 L h a Cb G c a
b
L ML2T 1 LT 1 M 1L3T 2
c
L M a L2a T a .Lb T b M c L3cT 2c L M a cL2a b3c .T a b2c a c 0
a=c a b 2c 0 3a b 0 b 3a b 3/ 2
2a b 3c 1 5a 3a 1 2a 1 1 a c 2
1
1
L A 2 .C3/ 2 .G 2 L
hG L h and L G C3
In the same way M h p cq G r a
b
M ML2 T 1 LT 1 M 1L3T 2 M1L0T 0 M a c
L2a b 3c
c
T a b 2c
a c 1, 2a b 3c 0, a b 2c 0
1 1 1 a , b ,c 2 2 2
M
hc G
M c
88
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
P 1, E1
Pmax
Q.14 A
x
amplitude A = a Pmax = b m. Vmax = b
ma1 b 1 E1
1 ma 212 2 2
P
2 , E1 x
R
Pmax R amplitude A R
Pmax R m. max R
m.A 2 R m R2 R m 2 1 3 E2
1 mR 222 4 2
from (2) & (4) E1 a 212 E 2 R 2 22 E1 n 212 2 1 E2 2 from (1) & (3) 99
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
a
ma1 b m2
1 b 2
1 b 1 2 a n 2
2 n 2 5 option (B) is correct 1
1 E1 ma 212 2 E2
1 mR 2 22 2
E1 a 2 12 2 2 2 n 12 E 2 R 2 2 2
E1 2 1 1 E 2 1 2 2
E1 E 2 1 2
So, Option (D) is correct. Q.15
Conservation of angular momentum of system
Li Lf 2 M 3 M 2 2 MR 0 MR R x f 8 5 8 2
M 9R 2 M 2 8 MR 20 MR 2 X 0 8 25 8 9 9 2 9R 2 X 2 R R2 8 200 8 9R 2 209R 2 X 2 8 200 8 225R 2 209R 2 X 2 200 200 8 110 0
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
16R 2 X 2 200 8 16R 2 4R X2 X 200 5 Q.16
0
0 0 q
r0
0 –q
r0
(1)
(2)
Fnet F 2 F1
2K 0q 2K 0q r0 x r0 x 2X 2K 0 r 2 2 r0 x
4K 0q X r02 x 2
Fnet
4 0 q X r02
In case – I motion is SHM for small displacement But in case – 2 charge particle moves in direction of displacement. In second case
F1
(1)
F2
(2)
F2 > F1 so particle move in direction of displacement
111 1
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
R=10cm
Q.17 S1 =1.5
R=10cm =1.5 S2
50cm
For Ist spherical refraction 1 1.5 1 1.5 0.5 V 50 10 10 1 1.5 0.5 V 50 10
1 3 1 V 100 20 1 3 5 V 100 100 1 5 3 2 1 V 100 100 100 50
V 50 for second speherical refraction 1.5 1 1.5 1 0.5 1 V 50 d 10 10 20 1 1 d 50 20
d 50 20 d 70
Q.18
F dF I dL B
F I Length B for constant B
from Diagram Length is along + X - axis If B is along Z axis then F = I Length B sin 90 Length = L + R + R + L 112 2
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
Length =2 (L + R) F = I 2( L + R)B
F L R So option (A) If B is along X axis the F I Length B
then F 0 Option B is correct. If B is along y-axis F I dL B
I length Bsin 90 I 2L 2R B F 2I L R B F LR So, option C is correct
Q.19
AR B P,S C Q,T D P,Q, R
Q.20
(A) U U 0 2
F
F
x2 1 2 a
2
dU u 2 x 2 2x 2 1 2 2 dx 2 a a
2u 0 x 2 1 X a 2 a 2
F0
X=0 X=a X=–a 113 3
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / PHYSICS / SOLUTIONS / CODE - 5
for X a
0 x a F positive but, a x 0 F negative (B) U
U0 2 X 2a 2
F
du 2U 20 X dx 2a
F
U0 X a2
F = 0 when X = 0 (C) x2
U0 X2 a2 U3 e 2 a2
2
2
x x 2 dU U 0 2 2x a 2 F 2 X 2 e e a 2X dx 2a a
x2
x2 U0 a2 F 2 2x e 2 1 2a a
When X = 0
F0
X=a X=–a 2
x U0 a2 x 2 F 2 e X 1 2 a a
(D)
U4
U 0 X X 3 2 a 3a 3
dU U 0 1 3X 2 F dx 2 a 3a 3
F0
when X = a X=–a
F is always act along negative x-axis. 114 4
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
CHEMISTRY =========================================== Q.21
O=N
O
N=O
or O=N–N
O
O Q.22
Fe SCN 6
3
Fe CN 6
3
SCN– is weak ligand
CN– is strong ligand
Fe3 3d 5
Fe3 3d 5
5 unpaired electrons
1 unpaired electron
n n 2 B.M
n n 2 B.M
35 5.9B.M
3 1.73B.M
So, Difference in magnetic movement = 5.92 – 1.73 4 Q.23
BeCl2 –
N3
Cl – Be – Cl –
+ – N=N=N
N2 O
NNO
+ NO2
+ ONO
O3
SCl2
ICl2
Cl | I | Cl
Hybridisation shape/ structure Linear sp Hybridisation shape/ structure Linear sp
O
Cl
Hybridisation shape/ structure 3 sp V- shape, bent or angular
S Cl
Hybridisation shape/ structure Linear sp
Hybridisation shape/ structure 2 bent or angular sp
O O
Hybridisation shape/ structure Linear sp
Hybridisation shape/ structure 3 sp d Linear
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
15 1 5 15
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
I3
I | I | I
XeF2
Hybridisation shape/ structure 3 sp d Linear
F | Xe | F
Hybridisation shape/ structure 3 sp d Linear
Ans = 4 Q.24
When the electron is excited to IInd excited state (H=3) all the orbitals 3s, 3p and 3d are degenerated. Total number of orbital is 9.
Q.25
X Y;
Gº 193KJ/ mol
M M 3 2e ;
Eº 0.25volt
Since, Gº nFE ºcell where n = number of moles of electrons.
193 103 Joule n 96500 0.25 on solving, n
193000 2 8 96500 0.25 0.25
Since, 1 mole of M+ loses 2 moles of electrons 4 moles of M+ loses 8 moles of electrons. Q.26
Molality = 0.01, Van’t Hoffs’ Factor = i
Tf i K f molality i
Tf 0.0558 3 K f molality 1.86 0.01
Now, For dissociated electrolyte i 1 n 1 (Assuming 100% dissociation) i = n = 3 Complex is Co NH 3 5 Cl Cl 2 , hence these are 2Cl– ions in Ionisation sphere while 1Cl– in coordination sphere.
116 6
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
Q.27
M=
can be written as
CH3
*
O
O
*
It contains two chiral carbon but it exists as two stereomers only. CH3
H
H
O
CH3
O –
O–H
O NaOH
Q.28
N
–
O
–
O
O
O –
–
–
O
O
–
–
–
O
O
–
O
Total are ‘9’
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
17 1 7 17
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
Q.29
OH H 2 O 2 O2 2H
H2O2 shows better reducing nature in basic medium. (A) H2O2 in presence of NaOH. (B) Na 2 O2 2H 2O NaOH H 2O2 Ans. (A) & (B) Q.30
yield Rate Temperature
Since the process is exothermic so the % yield will decrease with increase in temperature. But there is certain temperature required to activated complex state. Q.31
For C.C.P Arrangement of O–2 ions (oxide), number of atoms / ions per unit cell = 4. In spinal structure, AB2O4 general formula A+2 ions occupy 1 B ions occupy 2 +3
H Q.32
1 8
th
th
of octahedral voids. H
Br
Br
H2 Catalyst
(A)
Optically active H
H
Br
Br
H2 Catalyst
(B)
Optically Inactive
H
H
Br
Br
H2 Catalyst
(C) CH2
CH3 Optically active
Br (D)
Br
H
H
H2 Catalyst
Optically inactive
118 8
of tetrahedral voids.
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
O
O
H
–
O
–
OH
Q.33 O
H
OH
O
–
O
O
O
H
+ –
H Br
Q.34
+
+
Br – Br + 1, 2-Product minor
D and L are enantiomers.
H HO H H
CHO OH H OH OH CH 2 – OH
HO H HO HO
CHO H OH H H CH 2 – OH
:
Q.35
Br 1, 4-Product major
O
O:
C–O–H
Q.36
C–O–H
NaNO2 dil HCl
NH2
N2Cl
NGP
–
+
+
O
:
C – OH H2O
2
(SN attack) O C – OH
OH SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
19 1 9 19
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
Q.37
Cr 2
Cr 3 e
3d 4
3d 3
(A) Cr+2 is a reducing agent, since it undergo self oxidation to Cr3+.
–
Mn 3
+e
3d 4
Mn 2 3d5
(B) Mn3+ is an oxidising agent, since it undergo self reduction to Mn2+ and attains half filled configuration. (C) Both have same configuration i.e.g 3d4. Q.38
(A) Pure Cu used as cathode not an impure copper. (B) Acidified CuSO4 used as electrolyte CuSO 4 and H 2SO 4 (C) pure metal always obtained at cathode. (reduction) (D) Impurities settle as anode mud
Q.39
(A) Carbonate (P, Q, S)
(P) siderite (FeCO3) (Q) malachite (CuCO3. Cu(OH)2) (S) calamine (ZnCO3)
(B) Sulphide (T)
(T) argentite (Ag2S)
(C) Hydroxide (Q)
(Q) Malachite (CuCO3. Cu(OH)2)
(D) Oxide (R)
(R) Bauxite (Al2O3. 2H2O)
220 0
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / CHEMISTRY / SOLUTIONS / CODE-5
Q.40
(A) Freezing of water at 273 K and l atm
H 2 O l H 2 O s ; q 0; S 0; G 0;
Exchange of heat energy takes place ; q O liquid to solid; S 0 ; at equilibrium; G 0 As the process occurs; W 0 (B) 1 mole of an ideal gas expanded in to vaccum under isolated conditions. q 0; W 0; U 0;
Expansion is spontaneous; S 0; G 0 (C) Mixing of ideal gas at constant T and pressure in an isolated container. Isolated ; q = 0, As temperature constant, U 0 W = 0, As mixing is spontaneous, G 0 , S 0
(D) Isobaric heating from 300k 600k reversible and cooling back to 300k reversibly As it is cyclic process, U, S, G {state function} are zero.
W = 0 as initial and final volume remain same and q = 0.
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
21 2 1 21
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
MATHEMATICS =========================================== n
Q.41
n
1 1 1 C0 nC1 n nC2 n ....... 1 2 2 2
1
1 n 0.96 2n 2 n
1 0.96
0.04
n1 2n
n 1 2n
4 n 1 n 100 2
1 n 1 n 25 2
n8 Q.42
n 6! 5! m 5C4 . 6C2 .2!.4!.5! n 6!.5! =5
Q.43
y x 2 1 y x 3 y x 3 0
centre 3, 2
3 2 3 r 2
2 r r2 2
2222
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
x x 2
2
Q.44
I
2 x 1
1
0
1
2
I 0 0 1
I
0
x2 4
2
1
2
dx
x
20 1
2 1 1 4
4
4I 1 0 Q.45
V r 2h 2
2
Vm r 2 h r 2 h r 2 2 2
4 r 1 h r 2 2
4 r 1
dVm dr
V 2 r 2 2 2 r
1 2 4 2 3 V 4 r 2 0 r r r 10 r 10
2
h 2 1 4 12 4V 100 1000
2
V 12 12 V 4 V 1000 250 1000 x2
Q.46
f x
6
2 cos 2 t.dt............ i
x
a
as per given information f x dx f ' a 2........... ii 0
diff. (ii) w.r.t. a f a f '' a ...... 3
2 2 2 diff.(i) f ' x 2 cos x 2 x 2 cos x 6 2 again diff. f '' x 4 2 cos x sin x cos x sin x cos x 6 6 6 3 Now f 0 f '' 0 4. 3 4 SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
23 2 3 23
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
Q.47
Given equation 2 5 cos 2 2 x sin 2 x cos 2 x 2sin 2 x cos 2 x sin 2 x cos 2 x sin 4 x cos 4 x sin 2 x cos 2 x 2 4
5 cos 2 2 x 1 2 sin 2 x cos 2 x 1 3sin 2 x cos 2 x 2 4
5 cos 2 2 x 5sin 2 x cos 2 x 0 4
5 5 cos 2 2 x sin 2 2 x 0 4 4
5 cos 4 x 0 4
cos 4 x 0 number of solutions in 0, 2 4 2
8 Q.48
Let h, k be image of t 2 , 2t in x y 4 0 mid point of PQ will be on L
y
t 2 2t 8 h k 0....... i
2
y =4x
Q mPQ 1
2
(t ,2t)
t 2 2t h k 0........ ii x
0
h4 2
i ii t
2
Putting t in ii x 4 y 4
(h,k)
(-4,0)
2
Now y 5 x 4 4
L=x+y+4=0 A
x 4 2 6, 2
y=-5
AB 4 units
Q.49
f ' 0 lim h 0
lim h 0
B
f h f 0 h
g h h
2244
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
f ' 0 lim
g h g h g 0 lim g ' 0 0 h 0 h h
f ' 0 lim
g h g h g 0 lim g ' 0 0 h0 h h
h0
h 0
f is differentialble at x = 0, A is correct
ex h x x e
x0
h' 0 1 h ' 0 1
x0
h is not differentiable at x 0 B is not true Now f h x g h x x
g e
x
f ' h 0 g' 1 f ' h 0 g' 1 g' 1 as g ' 1 0 C is false h f x e f x
e
f x
f x 0
f x
e
f x 0
h ' f x f ' x e f ' x e
f x
f x
f x 0 f x 0
h' f 0 f ' 0 e
f 0
0 h' f 0
f f x is differentiable at x =0 D is true. Q.50
1 sin x 1
sin x g x , 2 2 2 2 2
1 sin sin x 1 2 SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
25 2 5 25
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
sin sin x 6 6 2 6
1 1 sin sin sin x 2 2 2 6 1 1 f x , Ais true 2 2 1 1 g x , fog x , 2 2 2 2
B is correct.
sin sin sin x f x 6 2 lim lim x 0 g x x 0 sin x 2 sin sin 6 lim 0 6 C is correct 1 1 1 1 f x . g f x sin , sin 2 2 2 2 2 2
Now sin
Q.51
1 1 1 sin 1 gof x 1 2 2 2 2 4
D is false. abc 0 a b c
Q
...(i)
a
(i), (i)
130°
R
a 2 b 2 c2 2b c c 2 48
c 4 3 b ; isosceles
So
c 120°
a a b c b c
30°
c2 48 a 12 24 12 12 : A is correct 2 2 2266
b
P
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
c2 48 a 12 36 : Bis incorrect 2 2 3 a b ab cos 12 4 3 6 2
72 :D is correct a b c a ab sin xˆ ca sin xˆ 6 6 1 b c 12 2
Q.52
8 3 6 48
3 C is correct
(A) P y3 z 4 z 4 y3 P ' z 4 y3 y3 z 4
P' P symmetric. (B) P x 44 y 44 P ' x144 y144 x 44 y 44
P' P is symmetirc. (C)
P x 4 z 3 z3 x 4 P ' z13 x14 x 41z13 z 3 x 4 x 4 z3
P' P skew symmetric matrices (D) P x 23 y 23 P ' x 23 y123 P ' x 23 y 23
P' P
Hence P is skew symmetric
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
27 2 7 27
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
Q.53
1 2 2
1 4 4 2
1 6 9 2
4 4 2
4 8 4 2
4 12 9 2
9 6 2
9 12 4 2
9 18 9
1 2 1 1 4 4 1 6 9
2
4 2 2
9 2 3
1
1
1
1 1
1
1 4 9
2 1 2 1 3
4 9
1 2 3 1 1 1
3
2
2 3
2
1 2 2 3 3 1
2
83 648 0 or 9 Q.54
P3 x z 1 y 0 distance from 0,1, 0 1
1
1
1 2 1
1 2 2
squaring both side and solving
1 2
P3 2x y 2z 2 0
Now, distance from , , 2 2 2 2 4 1 4 2 2 2 6 2 2 2 6 (or) 2 2 2 6 2 2 8 0 (or) 2 2 4 0
Q.55
L will be parallel to the line of intersection of P1 and P2 L is parallel to n1 n 2
where n1 normal to P1
n 2 normal to P2 2288
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
ˆi ˆj kˆ n1 n 2 1 2 1 ˆi 3jˆ 5kˆ 2 1 1
L is
x 0 y0 z0 1 3 5
Any point on L is P , 3, 5 foot of perpendicular P to the Plane P1 : x 2y z 1 0 is M x, y, z
x y 3 z 5 6 5 1 1 2 1 6 1 1 3 x ; 6 3 y
1 6 5
z
Locus M is
1 1 z 1 3 6 x 6 3 5 y
Option (A) & (B) Satisfy the equation of locus 1 P t2 , t 2
Q.56
(0,0) Q
y 2 2x
1 2 2 ,
Circle as PQ diameter is 1 2 1 2 x t x y t y 0 2 2
...(1)
(i) passes through (0,0,)
1 2 2 t t 0 4 SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
29 2 9 29
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
1 t t 1 0 4 t 4
Area OPQ 3 2
0 1 1 2 t 2 2 1 2 2
0 1 t 1 3 2 1
1 2 1 t 2 t 6 2 2 2 t t 12 2
t 4
t 3 2
t
2
2
t 4t
18 16 t
2
t 2
t 3 2
On solving , t 2 2, 2 I Q
Q.57
P 4, 2 2 ; 1, 2
y e 1 e dy dx x
x
1
dy x dy dx e dx y 1 Inteprating both sides we has
y y ex x c
dy d y ex 1 dx dx
...(1)
As y 0 2 hence c 4 Now equation of (1) become y So
y 4 0
x4 ex 1
...(ii)
....(A)
Differentiate (ii) w.r.t. x we have
3300
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5 x x dy e 1 x 4 e 2 dx ex 1
dy 0 dx
ex
y ex
1 x 3
y
From graph there is a intersection point
–3
x in between y e & y
Q.58
–1
1 x 3
O
1 where x 1, 0 x 3
Let centre of the circle is (a,a,) and radius ‘r’ Now equation of circle is
x a
2
2
y a r2
x 2 y 2 2ax 2ay 2a 2 r 2 0 Diff. w.r.t. x we get
...(i)
x yy1 a ay1 0
....(ii)
x yy1 a 1 y 1
...(iii)
Diff. once again equation (ii) w.r.t x we get 1 yy 2 y12 ay 2 0 using (iii) is (iv) we have
1 yy
2
...(iv)
x yy1 y12 y2 0 1 y 1
2 1 1 y1 y1 y1 y x y 2 0
Hence P y x Q.59
&
Q 1 y1 y12
(A) Projection of ˆi ˆj on 3 ˆi ˆj is 3
3
2
3
or
2 3
2 3 4 2 3
0
or
3
If
0
then
2 2
If
3
then
1 1
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
31 3 1 31
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
(B) 2 3a x 2 x 1 f x , x 1 bx a
As f x is differentiable for all x R , then f x also contineous for all x R 3a 2 b a 2
...(i)
a 2 3a 2 b 0
As f x is diff. at x 1 , hence LHD RHD
at
6ax b
in (i) we have a 2 3a 2 0
using b 6a
...(ii)
x 1 b 6a
a 1 or 2
(C)
2 3w 3w 2 3 2w 3w 2 3 3w 2w w w2 2
Hence 3 3w 2w 2 3 2w 2w 2
2 3w 3w 2
4n 3
1 w 4n 3 w 2 4n 3
4n 3
1 w w
3 2w 2w 2
0
4n 3
2
if
4n 3
3 2w 3w 2
n 3k
Hence value of n are 1,2,4 and 5
(D) 2ab 4 ab
ab 2 ab
...(i)
a,5,q,b are in A-P. ab q5
a a 3d a 2d 5 ad 5
...(ii)
5 a d q a 2d q a 2d
...(iii) 3322
4n 3
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
a a 3d 2 a a 3d
By putting value b a 3d in the equation (i)
5 d 5 2d 2 2 5 d 3d
Replace a 5 d from the equation (ii)
5 5d 2d 2 2 10 d 25 5d 2d 2 20 2d 2d 2 3d 5 0 2d 2 5d 2d 5 0
d 1 2d 5 0 d 1, Q.60
5 2
(A) 2 a 2 b 2 c 2 2 2 2 2 sin x sin y sin z
2sin x y sin x y sin 2 z
2sin z sin x y sin 2 z
sin x y 1 sin z 2
as x y z
n as cos n 0 cos 0 2
n is an odd integer (B) 1 cos 2x 2 cos 2y 2 sin x sin y 2 2 2 cos x 2 1 2 sin y 2 sin x sin y
2sin 2 x 4sin 2 y 2sin x sin y sin 2 x sin x sin 2 y 2sin 2 y 0 a 2 ab 2b2 0 a 2b a b 0
a a 2 or 1 b b SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
33 3 3 33
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
which is not possible Hence
a 1 b
(C) Angle bisector of
x & y is r 0 k xˆ yˆ
, 1 3 1 3
k 1 3 ˆi ˆj
As line passes through origin hence acute angle bisector will be y = x
3,1
and Pt z is , 1 Distance from pt z on bisector is
1
2
3 , hence 2 or 1 2
2 or 1
(D) If 0 then y x 1 x 2 x
3 F(0) = Area of rectangle C
0 ABC – Area of under parabola
B
(3,0)
2
2
y=
2 3 2 x dx
x
D (2,2)
0
O b/ 2 2
62
6
x 3 / 2 0
8 2 3
Hence f 0
8 2 6 3
If 1 then
(T)
y x 1 x 2 x
3 x x 0,1 1 x x 1, 2
3344
1
A
RAO IIT ACADEMY / IIT-JEE / ADVANCED EXAM - 2015 / PAPER - I / MATHEMATICS / SOLUTIONS / CODE -5
f 6
8 2 1 2 2 3 2
C
B
(3,0)
2
8 2 f 1 5 3
y=
O
8 Hence f 1 2 5 3
1
x
D (2,2) A
SANTACRUZ | ANDHERI | GOREGAON | KANDIVALI (E) | KANDIWALI (W) | BORIVALI | BHAYANDER | VASAI | POWAI | DADAR | SION | THANE | LOKPURAM (THANE) | DOMBIVLI | KALYAN | PANVEL | KAMOTHE | NERUL | SANPADA | KHARGHAR |
35 3 5 35
