10p14 Adv 3 Paper 1 Solutions Chem Maths

8/13/2019 10p14 Adv 3 Paper 1 Solutions Chem Maths

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Vidyamandir Classes

VMC/JEE-2014/Paper-1 1 JEE ADV-3/Solutions/Chemistry-Maths

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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Vidyamandir Classes


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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Vidyamandir Classes


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 2 and 3 = 600



Number of integers divisible by 2, 3 and 5 = 120

Number of integers divisible by 2, 3 or 5


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Vidyamandir Classes


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


3 41 1 5 3 4 120 1440C C


2 51 1 5 2 5 120 1200C C


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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Vidyamandir Classes


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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Vidyamandir Classes


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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Vidyamandir Classes


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. . . . . . .


12/13


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Vidyamandir Classes


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

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10p14 Adv 3 Paper 1 Solutions Chem Maths