10p14 adv 3 paper 1 solutions chem maths
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8/13/2019 10p14 Adv 3 Paper 1 Solutions Chem Maths
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Solutions to JEE Advanced - 3/JEE-2014/Paper - 1
[CHEMISTRY]
1.(B) 22 4 4Ag CrO 2Ag CrO
Addition of 24CrO results in decrease of Ag
.
2.(B) Sparingly soluble salt :
1 sp
spsp3
2 21 2 3
sp4
3
AB S K
Since K
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8.(D) (A) Na2O2decolourises acidified potassium permanganate.
2 2 4 2 4 2 4 2 4 4 2 25Na O 2KMnO 8H SO 5Na SO K SO 2MnSO 8H O 5O
(B) Sodium superoxide is prepared by reacting
Na2O2with O2at 450 C and 300 atm pressure
2 2 2 2Na O O NaO
(C)400 C
2 2 2Na O Na Na O
(D) 2 2 2 2 2Na O H O NaOH NaOH H O
2 2 2 4 2 4 2 2Na O H SO Na SO H O
9.(A) S1:2 2 2
2 2 2
BeCl . 4H O BeO HCl H O
MgCl . 6H O MgO HCl H O
BeCl2, MgCl2undergoes surface hydrolysis (and thus form
oxides on heating rather than simple halides) due to small size of Mg2+
, Be2+
as compared to Ca2+
.
S2: With the weakening of metallic bonding down the groups softness of alkali increases down the group.
S3: Plaster of Paris is 4 21CaSO . H O2
S4: True
10.(C)
11.(AC)2 2
2 4 2 2 3 2C O H O CO H O
n-factor = 2
22 4 2 3 2HC O H O CO H O
n-factor = 2
millimoles of H2O2used = n1+ n2
22 4 2 4 2HC O KOH C O H O
n-factor = 1
millimoles of KOH used = n2
12.(BD) Vapour Pressure of solution decreases on addition of a non-volatile solute
Only solvent molecules solidify at freezing point
13.(ABCD)
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14.(ABC)
(I) (II) (III) (IV)(A) (I) has centre of symmetry and (IV) has plane of symmetry so both are meso compounds.
(B) (II) and (III) are enantiomers.
(C) Since in (I), no hydrogen is anti with respect to the adjacent Br atom, so elimination is not feasible.
15.(ABCD)
SF4 2 SF bonds are equatorial
2 SF bonds are axial
ClF3 2 ClF bond axial
1 ClF bond equatorialBond angles not identical.
BF2Cl Due to un-symmetry, bond angles cannot be 120
16.(2) 12.2 g Benzoic Acid 0.1 mol.
100 gm water
Molality =0.1
1000 1 molal100
b bT iK m
0.27 1i
0.54 1 2
Benzoic acid must be dimerised.
17.(8)MgCl
MgCl2
f
f
H 125
H 642
(i) 21
Mg Cl MgCl2
H 125 KJ
(ii) 2 2Mg Cl MgCl H 642 KJ
22MgCl Mg MgCl H 392 KJ
ii 2 i 642 250
= 392
For one mole of MgCl , H 196 KJ
49x 196
x = 4
18.(7) Statement (1), (2) and (4) are correct.
19.(6) 2 4 3 3 2 3 2 2 7 2 4 2 3H SO , H PO , H CO , H S O , H CrO , H SO are diprotic acids.
20.(2) v Vq nC T
H Br
Br H
H Br
Br H
H
Br
Br
H
HBr
BrH
H Br
Br H
H
Br
Br
H
H Br
Br H
HBr
BrH
H
Br
Br
H
H
Br
Br
HH
Br
Br
H
H
BrBr
H
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6 f R12.5 10
22.4 2
6 f 212.5 10
22.4 2
28
56
Gas must be diatomic
Solutions to JEE Advanced - 3/JEE-2014/Paper - 1
[MATHEMATICS]
41.(A) If the robot is one unit away from its initial zero position, then it can be at either +1 or 1 . Since the robot has
taken 5 steps, it can be at +1 if it takes 3 steps right and 2 steps left. Similarly, it can be at 1 if it takes 3 steps left
and 2 steps right.
Pr (3 steps right and 2 steps left) =3 25
33 2
5 5C
Pr (3 steps left and 2 steps right) =
3 25
3
2 3
5 5C
So, the required probability =
3 2 3 25 5
3 3
3 2 2 3
5 5 5 5C C
10 27 4 10 8 9 72
3125 3125 125
42.(B) Since the points of intersection lie on the hyperbolaxy= 1, let their co-ordinates be
11
1A t ,
t
and 22
1B t ,
t
Since the above points also lie on a variable line with slope2 1
2 1
1 1
t tm, m
t t
or1 2
1m
t t
or 1 2
1t t
m
.
LetP(x,y) be the point which dividesABin the ratio 1 : 2, then
2 12
3
t tx
and 2 1 1 2
1 2
1 2
2
3 3
t t t t y
t t
or 1 23 2x t t and 1 2 1 23 2t t y t t
Putting the value of t1t2in the above equation, we get :
1 23 2x t t and 1 23
2y
t tm
Solving the above 2 equations for t1and t2, we get :
1
33 6
yt x
m and 2
63 3
yt x
m
or 1 2 y
t xm
and 22y
t xm
Using 1 2
1
t t m
, we get :
2 1
2
y y
x xm m m
or2
2
2
4 2 12
xy y xyx
m m mm
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or 2 2 22 2 5 1 0m x y xy m
For the above second degree equation
2
2 2 2 2 52 2 2 0 0 02
mabc fgh af bg ch m m m
3 3 325 94 0
4 4
m m m as 0m
Also,2
2 225 44
mh ab m The curve is a hyperbola.
43.(C) u v w u . v w
2 1 1 3 7
1 0 3
i j k
v w i j k
2 2 23 7 3 7 1 u . v w u . i j k u cos
Where is the angle between u and 3 7
i j k
1 59u . v w cos
Maximum value of 59u . v w 44.(A) Direction ratios ofL1come from the following determinant
2 1 1
3 1 2
i j k
i j k or the direction ratios are 1 1 1, , .
(0, 0, 1) is a point which lies on both the planes (0, 0, 1) lies onL1.
So, the equation ofL1becomes1
1 1 1
x y zk
The equation ofL2is x y z
Take pointsPand QonL1andL2respectively. Then 1P k , k , k and Q , , .
Direction ratios ofPQare 1k, k, k .
Since we want the shortest distance betweenL1andL2, 1PQ L and 2PQ L
1 0k k k and 1 0k k k
3 1 0k and 3 1 0k
Solving the above two equations we get :1 1
4 4, k
1 1 3
4 4 4P , ,
and
1 1 1
4 4 4Q , ,
Shortest distance = 1
2PQ
45.(C)2 23 2 18 0x y xy xy
3 2 18 0xy x y
Sides of triangle are 0 0x , y and 3 2 18 0x y
Vertices of triangle are (0, 0), (0, 9) and (6, 0)
0 0 6 0 9 0
3 3G ,
or (2, 3).
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Since Glies on 1 0x y , the required circle touches the line 1 0x y at (2, 3) and passes through (1, 1).
The equation of family of circles touching 1 0x y at (2, 3) is
2 2
2 3 1 0x y k x y
Since the required circle passes through (1, 1)
2 2
1 2 1 3 1 1 1 0 5k k
So, the equation of circle is 2 2
2 3 5 1 0x y x y or 2 2 9 8 0x y x y
46.(A) As angle in a semi-circle is a right angle, 90ADB
2AD Rcos
180DCB asABCDis a cyclic quadrilateral.
CBA asABCDis a trapezium.
2BC AD Rcos
Now,AB= 2Rand from the figure we can see that
22 2 2DC AB ADcos BCcos R Rcos
Also 2h AD sin R cos sin
area 21 1 2 2 4 22 2
ABCD AB DC h R R R cos R sin cos
2 2 32 2 2 4R R cos R sin cos R sin cos f .
2 2 2 34 3f R sin cos sin sin
0 0f sin or 2 3tan
0 or 60
But a maxima occurs at 60
Maximum area 3
22 3 1 3 342 2 4
RABCD R
47.(D) Since the slope of tangent to the curvey=f(x) at (x,f(x)) is 2x+ 1
2 1f x x
2f x x x C , for some arbitrary constant C.
As the curve passes through (1, 3) f(1) = 3
C = 1 2 1f x x x .
1 2
1
y
y x sin sin x
. . . .(i)
Let 1 1y
g x log g y log x
1d d
log g y log xdx dx
or 1
11
dg y dylog x
g dx x dx
1 11
ydg y dyx log x
dx x dx
So, differentiating (i) w.r.t.xwe get :
411 1 21 1ydy y dyx log x sin x cos x
dx x dx sin x
Atx= 0,y= 1.
DC
BA O
h
2R
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1dy
dx slope of normal = 1 .
Equation of normal is 1 1y x or 1y x
For the area bounded by y f x and 1y x , we get the points of intersection as 2 1 1x x x
or 2 2 0x x or 2 0x x
or 0 2x ,
Area bounded 0
2
2
1 1x x x dx
00 3
2 2
22
4 42 0
3 3 3
xx x dx x
48.(B) Let
2
2 1
1
x sin xI dx
cos x
2 2
0
2 1 2 1
1 1
x sin x x sin xI dxcos x cos x
2
0
4
1
x sin xdx
cos x
. . . . (i)
2 2
0 0
4 4
1 1
x sin x x sin xI dx dx
cos x cos x
. . . .(ii)
Adding (i) and (ii), we get :
2 20 0
44
2 1 1
sin x dxsin xdx
I cos x cos x
2
2
0
42 2
1
sin xdxI
cos x
or
2 2 2
2 22
0 0 0
24 4 4
1 11
2
sin x dxsin xdx cos x dx
Icos x sin x
cos x
Put sin x t
cos xdx dt
1
11 2
2 00
4 4 4 041
dtI tan t
t
49.(A)4 2 23600 2 3 5
Number of integers divisible by 2 = 1800
Number of integers divisible by 3 = 1200
Number of integers divisible by 5 = 720
Number of integers divisible by 2 and 3 = 600
Number of integers divisible by 3 and 5 = 240
Number of integers divisible by 2 and 5 = 360
Number of integers divisible by 2, 3 and 5 = 120
Number of integers divisible by 2, 3 or 5
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1800 1200 720 600 240 360 120 2640
Number of integers not divisible by 2, 3 or 5 3600 2640 960
But these 960 integers include 1.
Excluding 1, we get there are such 959 integers.
50.(B) DECISION
I I, D, E, C, S, O, N
Case 1 : There is one letter between Is
Similarly,Case 2 : There are 2 letters between 2 Is
5 21 1 5 5 2 120 1200C C
Case 3 : There are 3 letters between 2 Is
4 31 1 5 4 3 120 1440C C
Case 4 : There are 4 letters between 2 Is
3 41 1 5 3 4 120 1440C C
Case 5 : There are 5 letters between 2 Is
2 51 1 5 2 5 120 1200C C
Case 6 : There are 6 letters between 2 Is
1 61 1 5 1 6 120 720C C
Total number of words = 2 (720 + 1200 + 1440) = 6720 =8
3
51.(AD) 1
1
4 2 1
xe , if x
f xx , if x
1
4 2 1
1 1
4 2 1
x
x , x
e , x
x , x
1
1
4 2 6 4
4 2 2 4 1
1 0
0 1
4 2 2 1 4
4 2 6 4
x
x
x x x
x x x
e x
e x
x x x
x x x
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1
1
6 6
6 6 4
2 4 2
2 2 1
1 0
0 1
2 1 2
2 2 4
6 4 6
6 6
x
x
x x
x x
x x
x x
e x
e x
x x
x x
x x
x x
Plotting the graph off(x), we get :
From the graph of f(x), we can see that f(x) is continuous everywhere and f(x) is not differentiable at the points
6 4 2 0 2 4 6x , , , , , , .
Note that 1 1x xd e edx
which is equal to 1 at 1x . Similarly, 1 1x xd e edx
which is equal to 1 at
x= 1.
52.(BD) The point 2, lies on 2y x
Solve 2 0x y and 2y x to get :
22x x
or 2 2 0x x or 2 1 0x x
or 1 2x ,
Similarly, solve 2x y and 2y x to get :
22 x x or 2 2 0x x
or 2 1 0x x or 2 1x ,
So, from the graph we can see that for 2, to lie inside the required triangle 1 1, .
53.(ABCD) 1 2c
z za
or 1 21
11
cz z
a
or 21 1 z or 2 1z
1 2
11
1
b bbz z
a a a
As 2
1 2 1 2 1 2z z z z z z we get :
1 2 1 21 z z z z
or 1 21 2
1 11 z zz z
as 21 1 11z z z and 22 2 21z z z
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or
21 2
1 2
1z z
z z
2
1 2 1 2z z z z 2
2b c b aca a
1 2 1 1iz z z z e as 2 1z and POQ is given in the question.
or 2 2
1 2 1 1 1 1iz z z e cos sin
22 2 4 2
2 2cos cos cos
As 1 2 1 2 1 602 2
z z , cos
or2
1203
1 2 1 1 1 1i iPQ z z z z e z e
2 21 1 2 2 60 32
cos sin sin sin
54.(ABCD) 21 1k k k k k x x x x x
1
11 1 1 1
1 1 1
k k
k k k k k k k
x x
x x x x x x x
1
1 1 1
1k k kx x x
1 2 100 1 2 2 3 100 101 1 101
1 1 1 1 1 1 1 1 1 1 1
1 1 1. . . . . . . .
x x x x x x x x x x x
1 2 1 1 3 2 21 1 3 3
1 12 2 2 4
x , x x x , x x x
3 4 5 101
3 7 211 1 1 1
4 4 16x x x . . . . . x
101
10 1
x
1 101 101
1 1 12 1
x x x
Similarly, we can prove that
1 2 101 102
1 1 1 12 1
1 1 1. . . .
x x x x
1 2 102 103
1 1 1 12 1
1 1 1
. . . .
x x x x
and1 2 103 104
1 1 1 12 1
1 1 1. . . .
x x x x
55.(BD) 12 2x sin tan
Assume that 12
22 2
1
tantan x sin
tan
2
2 2 4
51 2
11 42 3
y sin tan
Assume that14 4
3 3tan tan
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4
5sin and
3
5cos
2 2 21 22 2 2
cos cos sin sin
2 23 1 1
1 25 2 2 5 2 5
sin sin sin
Since 14
0 03 2 2
tan , sin
or1
2 5sin
1
5y
Putting the values ofxandyin the options, we get 2 1y x and x y .
56.(8) Question is Cancelled
Question should had been that aand bare relatively prime and cand dare relatively prime.
3 3
2 22
3 3 3 3
2 2 2
nx nx nx nx nx nxf xnx nx nx nx
Put nx t to get :
2
2 2
3 3 3 3 3 3
2 22 2
t t t t tg t y
t tt t
23 3 2 3 2 0y t y t y . . . .(i)
For the domain off(x),x> 0 and 2 2 2 0n x nx .
Taking nx t , we get : 2 2 2 0t t
or 21 1 0t which is true for all real values of t.
Asx> 0, nx t takes all real values.
So, we have to find out all possible values ofyin (i) where tcan take all real values.
As t R , discriminant 0
2
3 2 4 3 3 2 0y y y or 3 2 3 2 4 3 0y y y
or 3 2 9 2 0y y or 2 3 2 9 0y y
3 9
2 2y ,
.
As 3 9 3 2 9 22 2
a c, , , a , b , c , db d
3 2 9 2
82 2
a b c d
57.(1) Let0 1 2
0
0 2
3 3 3 3n
n
cos nx cos x cos x cos xS . . . .
Now, let0 1 2
0
0 2
3 3 3 3n
n
sin nx sin x sin x sin xR . . . .
Consider0 0 1 1 2 2
0 0 2 2
3 3 3 3 3 3
cos x i sin x cos x i sin x cos x i sin xS iR . . . .
0 1 20 2
0 1 2 3 3 33 3 3
i x ix i x ix ix ixe e e e e e. . . . . . .
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0x
xlim
cos x
0As 1
x
sin xlim
x
00
1
Hence,
0
0 0
0 1x
x x
lim x n tan x lim tan x e
Now, consider 0
1
xn lim n sec x
x
Apply L Hospital Rule to get :
0 0
1
01x x
sec xtan xsec xn lim lim tan x
Hence, 1 0
0 0
10 1
x
x xlim n sec x lim sec x e
x
1 1 2