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Vidyamandir Classes

VMC | JEE Mains-2019 1 Solutions | Morning Session

SOLUTIONS Joint Entrance Exam | IITJEE-2019

08th APRIL 2019 | Morning Session

Vidyamandir Classes


Joint Entrance Exam | JEE Mains 2019

PART-A PHYSICS

1.(3) Refer NCERT

2.(1) Let us assume x grams of water vaporizer. amount of water frozen Heat gained by vaporized water = Heat lost by frozen water

3.(3) Let us assume and

4.(1) When catapult is released, elastic potential energy stored in cord gets converted to KE of stone. EPE lost by cord = KE gained by stone

SI units

5.(3)

Given

6.(4)

Þ (150 )= - x

Þ (150 )´ = - ´V fx L x L5 521 10 (150 ) 3.36 10´ ´ = - ´ ´x x

Þ 21 504 3.36= -x x Þ504 20.7 2024.36

= = »x g g

0bV V= =aV xV

1 2 3 0+ + =i i i

Þ 31 2

1 2 10

2 2x Ex E x E

R R R-- =

+ + =

2 4 4 02 2 2- - -

+ + =x x x

Þ 3 10 0- =x Þ 3.3=x V0 3.3- = - =a bV V x V

2 21 12 2æ öD =ç ÷è ø

!!

YA mv

Þ( )23

2 23 10

(0.2) 0.02 (20)0.42

-p ´´ ´ = ´Y Þ 6

0.02 400 0.429 10 0.04

Y -´ ´

=p´ ´ ´

63 10» ´

2max 2 1

min 2 1

I a aI a a

æ ö+= ç ÷-è ø

1

2

13

=aa

Þ 2 1

2 1

3 1 23 1

+ += =

- -a aa a

Þ max

min4I

I=

VdRn

r=

3100 / min 0.1 / min= =Q l m30.1 /

60= m s

20.1 160 (0.05)

= = ´p´

QVA

Þ 32

3

0.110 0.160 (0.05)

10-

= ´ ´´p´

R4

410

60 25 10-=

´p´ ´42 10» ´

Vidyamandir Classes


7.(1)

8.(1)

9.(3)

Conserver momentum,

along x :

along y :

Squaring and adding,

10.(2) Rate of heat dissipation

Rate of energy storage in inductor

Assuming initially current was zero, after time t,

,

11.(3) KE of particle = work done by force = area under F–s curve = 6.5J

2loadStressarea

mgr

= =p

3 24 3.1(2 10 )-´ p

=p´ ´

6 23.1 10 /= ´ N m

m Bt = ´ˆ=B Bi

2 ˆ[ ( ) ]= pm N r i j

Þ 2 ˆ[ . ]N r i B Rt = - p

Þ 2| | r NiBt = p

10 cos+ = q

l lh h

20 sin+ = q

l lh h

2

22

2 21 2

1æ ö=ç ÷ç ÷ ll + lè ø

hh Þ 2 22 1 2

1 1 1= +

l l l

2= i RdiiLdt

=

2 dii R iLdt

= Þ =di R idt L

1-æ ö

ç ÷= -ç ÷è ø

RtLEi e

R-

=RtLdi E e

dt L

Þ 1-

-æ öç ÷= ´ -ç ÷è ø

Rt RtL LE R Ee e

L L R

Þ12

-=

RtLe Þ ln 2=

RtL

202 210

Lt ln lnR

= =

2 2ln=

12 2 (2 3) 12

= ´ + ´ + ´

Vidyamandir Classes


12.(4)

(1) For first refraction by lens

(2) For reflection by mirror

(3) For second refraction by lens

image is at same position as object image is of same size and inverted.

13.(3) Using Snell’s law,

Length of fibre transverse in one refraction Length of wire = 2m

Number of refractions

14.(3)

Direction of magnetic field will be perpendicular to both electric field and direction of propagation of wave

4 40 2= - = -cm fÞ 2 40= + = +v f cm

1vmu

= = -

(60 20) 40 2= - - = - = -u cm fÞ 2 40v f cm= - = -

1vmu

= - = -

(60 20) 40 2u cm f= - - = - = -

2 40v f cmÞ = =

1vmu

Þ = = -

Þ ( 1) ( 1) ( 1) 1netm = - × - × - = -

Þ

21 sin 40 1.31sin× ° = q

Þ 2sin 0.5q » Þ 2 30q = °6 5

2cot 20 10 3 3.46 10d m- -= × q = ´ ´ » ´

Þ 52 57000

3.4 10-= »

´

=E cB

Þ 86 2 103 108

-= = ´´

B T

Vidyamandir Classes


15.(1)

16.(2)

slope of A will greater in magnitude at t = 0

17.(4) We know

18. BONUS Force per molecule = 2mv

Total force

19.(3) Energy from radiation

For helium

2 to 4

20.(2)

,

21.(4) Assuming no diode potential drop across resistor = 7.2 V

Zener breakdown has occurred Current through resistor

Current through resistor

Current through Zener diode

1 1 2 2 3 3 4 4

1 2 3 4

+ + +=

+ + +cMm a m a m a m aa

m m m m

! !( ) 2 ( ) 3 ( ) 4 ( )2 3 4

- + + + -=

+ + +

! !m ai m a j m ai m a jm m m m

!!2 2 ( )

10 5-

= = -!

!ai a j a i j

28 10= ´ ´AMass V 310= ´Bmass V

\ >B Amass mass

( )0 0msd d kkdt dt msq q= q-q Þ = q-q

28 10 2000= ´ ´ ´A Am s V 310 4000= ´ ´B Bm s V

\ >B B A Am s m s \

00

00 0

1c c Î= Þ Î =

Î µµ

21 2

0

14

q qFr

=Îp

2 21 3 4 21 2

0 2 2 2[ ] [ ][ ] [ ][ ][ ] [ [ ]q q A T M L T Af r MLT L

- --Î = = =

\ 1 1 3 4 2 1 2 3 20[ ] [ ][ ] [ ]- - - - -Î = =c LT M L T A M L T A

26 4 222 10 10 2 10- -= ´ ´ = ´22 22 210 2 10 2 /-= ´ ´ = N m

2 21 113.6 (1) 10.21 2æ ù= ´ - =ç úè û

eV

(2 to )n

2 21 110.2 13.6 (4) 42

æ ö= ´ - Þ =ç ÷è ø

nn

\

cosT mgq =6

3200 5 10 1tan

22 10 10Eqmg

-

-´ ´

q = = =´ ´

sinT Eqq = Þ 1tan 0.5-q =

800W

Þ Þ 800W 5.6800

A=

Þ 200W 9 5.6 3.4200 200-

= =

Þ3.4 5.6 8 10200 800 800

mA= - = Þ

Vidyamandir Classes


22.(1) means means Red, Black, Brown. If Red replaced by green

23.(4) Frequency will be same in both

Tension will be same in both but mass per unit length will be 4 times in 2 (as radius is twice)

24.(3) radius of the circle

FBD of any one particle

25.(4) If velocity of strip is v emf in it is

Current in it is

Resistive force exerted by

Amplitude reduces as If amplitude reduces by factor of e

The new frequency

New time period Number of oscillations

26.(3)

Using parallel axis theorem

200W 120 10´ 150 10 500´ = W

1 21 22 2

pc qc pc qcl l= Þ =

µ

\ 2 14µ = µ \1 2

T Tp q=µ µ

\ 1

2

12

pq

µ= =

µ

2a r= =

2 2 2

2 22 1

2 2 / 2GM GM Mva a a

+ = Þ1 1 1.162 2

GM GMa a

+ =

BvlBvlR

\2 2B vlB BilR

= =

20( )

b tmA t A e

-=

34

2 2 2 22 2 2 50 10 101 10

2 (0.1) (0.1)bt m mRt sm b B l

-´ ´ ´= = = = =

2 2 2

3 4 40.5 1 1' 10 10

2 50 10 10 10k bm m -

æ ö æ ö æ öw = - = - = -ç ÷ ç ÷ ç ÷´è ø è ø è ø!

2 2' 2' 10

T p p= = » »

w\

410 50002

= =

52 2 0

00

2(2 )( )5

R

CMRI dm r r dr r r pr

= = p r =ò ò3

00

0

223

R RM dm rdr r pr= = p r =ò ò

2CMI I MR= +

5 5 50 0 0

2 2 165 3 15

R R R= pr + pr = pr

5 20

3 16 2 82 15 3 5

R MR= ´ ´ pr =

Vidyamandir Classes


27.(4) Initial

Final

28.(1) Time period

Time to go from

29.(4)

Position of B as function of time

Position of A as function of time

Relative position of B as fraction of time

Taking magnitude square Time derivative

30.(3)

PART-B CHEMISTRY 1.(1) Ellingham diagram tells us about values (feasibility) of thermal reduction of an ore using suitable

reducing agents.

2.(4) From given data

Rate

Now, … (i)

… (ii)

… (iii) From equation (i) and (ii) we get,

1 2

KQ KQpd VR R

= - =

1 2 2 2

4 4KQ KQ KQ KQpdR R R R

æ ö æ ö= - - -ç ÷ ç ÷è ø è ø

1 2

KQ KQ VR R

= - =

2 2 1100 50

Tp p= = = =w p

1 3.32 4 12 6 300A T T Tto A s ms= - = = =

ˆ ˆ ˆ80 150 10i j ti= + -

ˆ ˆ30 50t i t j+

\ ˆ ˆ ˆ ˆ80 150 40 50i j t i t j= + - -2 2(80 40 ) (150 50 )t t= - + -

2(80 40 )( 40) 2(150 50 )( 50) 0t t= - - + - - =

Þ (8 4 )( 4) (15 5 )( 5) 0t t- - + - - = Þ 16 25 32 75t t+ = +

Þ 41 107t = Þ107 2.641

t hrs= =

max maxE d V=

4500 5 10106

d m-= = ´

12 40

12 40

15 10 5 10 8.58.86 10 10

k A CdC kd A

- -

- -

Î ´ ´ ´= Þ = = »

Î ´ ´

GD

a bk[A] [B]=a b0.045 k[0.05] +=

a b a0.090 k[0.05] 2+=a b 2a b0.72 k[0.05] 2+ += ×

a1 1 a 12 2

Þ = Þ =

Vidyamandir Classes


And from equation (i) and (iii) we get

3.(3)

No. of 3 0 1 2 Unpaired electrons (all complex are inner orbital complex because ligands are strong) as, Magnetic moment

n = no. of unpaired electrons More the number of unpaired electron more will the value of spin only magnetic moment

4.(1) For iso-electronic species size is governed by proton to electron ratio i.e. ratio

More the value of ratio, smaller will be its size as more number of proton will have an ability to

hold electrons more strongly resulting in decrement in ionic size. Hence, size is affected by nuclear charge i.e. no. of proton.

5.(4) In acidic medium reduces to

oxidizes to

oxides to For

gm equivalents of = gm equivalents of gm equivalents of gm

equivalent of gm of

6.(1) ; moles

0.01 moles in 100 ml water. equivalent in 100 ml water 0.02 equivalent of in 100 ml water

0.01 moles of in 100 ml water of

Hardness

2 b0.045 10.72 2 +Þ = 2 b

45 1720 2 +Þ =

(2 b)720 245

+Þ = 4 2 b2 2 +Þ = B 2Þ =

4 4 3 26 6 3 6 3 6[V(CN) ] , [Fe(CN) ] , [Ru(NH ) ] , [Cr(NH ) ]- - + +

Þ 2V + 2Fe + 3Ru + 2Cr +

Þ 3s d° 6s d° 5s d° 4s d°

n(n 2),µ = +

Pe-æ öç ÷è ø

Pe-æ öç ÷è ø

4KMnO 2Mn +

22 4C O -

2CO2Fe + 3Fe +

4KMnO 2 4FeC O + 2 2 4 3Fe (C O ) +

4FeSO + 2 4 3Fe (SO ) .

4 2 4 2 2 4 3 4 2 4 3f KMnO f FeC O f Fe (C O ) f FeSO f Fe (SO )(moles n ) (moles n ) (moles n ) (moles n ) (moles n )´ = ´ + ´ + ´ + ´

4KMnO(moles) 5 (1 3) (1 6) (1 1) (1 0)´ = ´ + ´ + ´ + ´

4KMnO(moles) 2=

3 2 3 2Ca (HCO ) Ca(HCO )0.81 1g 0.81g, n162 200

= = = 3 2 3 2Mg(HCO ) Mg(HCO )1g 0.73g, n200

= =

T1 1n 0.01200 200

= + =

(0.01 2)´ \ 3CaCO\ 3CaCO

0.01 100g´ 3CaCO

Þ 31 10 mg 1000 10,000ppm100L

´ ´ =

2 3 3 42 4

3 42 2 4 3

2 24

4 3

Fe Fe , C CFeC O 1 (1 2)Fe (C O ) 2 3 6 C CFeSO 1 Fe FeFe(SO ) 0 No oxidation

+ + + +

+ +

+ +

® ®+ ´´ = ®

®

Vidyamandir Classes


7.(3) Given :

As, [hence solution formed is ideal solution]

And Using Raoult’s law:

, , And in vapor phase:

[from Raoult’s law] …(i)

[from Dalton’s law of partial pressure] Equating (i) and (ii), we get

8.(1)

9.(1)

10.(3) value E : Energy of electron in a particular subshell n : principal quantum number l : azimuthal quantum number (I) (II) (III) (IV) According to Aufbau’s rule, lower the value of (n + l), lower will be its energy. In case if (n + l) value

are same for two different subshell then subshell having lower value of n will have lower energy. So, correct increasing order of energy is: (IV) < (II) < (III) < (I)

oAP 400 mmHg=oBP 600mmHg=

A B solutionV V V+ =

B Ax 0.5, x 0.5= =

o oT A A B BP x P x P= + TP 200 300= + TP 500mmHg=

oA A AP x P=

A A v TP (x ) P=

A v0.5 400 4(x ) 0.5500 5´

= = ´2 0.410

= =

B v(x ) 1 0.4 0.6= - =

4 33 4 4 4

3s 4sa sZr (PO ) 3Zr 4PO+ -

-+

4 3 3 4sp 4K [Zr ] [PO ]+ -=

3 4[3s] [4s]=3 427s 256s= ´

76912 s=1/7

spKS6912æ ö

= ç ÷è ø

E (n l)µ +

n l 6+ = n l 5 (n 3)+ = =

n 1 5(n 4)+ = = n l 4+ =

Vidyamandir Classes


11.(3) As alkylation of nitrogen increases, Basicity of amines increase due to (+I) effect of Alkyl groups which results in more electron cloud density over nitrogen atom (available toward donation). Hence correct order is [Gaseous phase]

12.(3) Gabriel pthalimide reaction is used to prepare only 1° aliphatic amines.

In above reaction, R – Br undergoes reaction, Hence electrophilicity of α-carbon should be high as

well as its hindrance should be low for greater ease of reaction.

Hence, ease of formation of n-butyl amine is higher. Therefore, it is most probable answer

13.(1) x should be a weak acid as it is soluble in 10% NaOH only.

Oleic acid > Benzamide > o-toluidine > m-cresol Order of decreasing acidic strength x is m-cresol (weakest of all)

2 5 2 2 5 2 3(C H ) NH C H NH NH> >

2SN

\

Vidyamandir Classes


14.(4) Electronic configuration of following lanthanide ions are given below. (colorless, no unpaired electron)

(colorless, no unpaired electron) colourless (yellow, 5 unpaired electrons) Ions having f electrons show colour. of these ions may be attributed to presence of f electrons.

15.(3)

Carboxylic group has high priority than hydroxyl group so numbering starts from carbon of carboxylic group. 3-hydroxy-4-methylpentanoic acid.

16.(1)

17.(1) Benzene diazonium chloride react with that ring of 1-naphthol which contain group as it is activating and also it will undergo coupling at p-position w.r.t. group of 1-napthol.

18.(2) … (i)

Slope of v/s graph gives value of 1/n

From graph

Putting in (i) ,

19.(3) Four donor atoms are present in it hence it act as a tetradentate.

3 141u : 4f3 01a : 4f+

3 7Gd : 4f+

3 5Sm : 4f+

3

3 2

CH OH5 4 | 3| 2 1

H C CH CH CH COOH- - - -

OH-OH-

1nx k (P)

m=

x 1log logk logPm n= +

xlogm

logP

\1 2n 3=

1n

2/3x (P)mµ

Vidyamandir Classes


20.(2) According to first law of thermodynamics:

For adiabatic process, q = 0 Hence,

21.(2) Consider the following reactions:

22.(4)

23.(4) Maltose is a disaccharide made up two D-glucose units. On treatment with dil. HCl is undergoes hydrolysis to give two D-glucose units. (Monosaccharide)

24.(2)

Here 1 and 2 are conc. HBr sensitive regions Alkene, will undergo electrophilic addition reaction [Markonikov’s addtion] Ether, will undergo forced [Acid catalysed] nucleophilic substitution reaction mechanism]

U q wD = +

U wD =

Oxidation2 6 2 2 3 2

3B H O B O 3H2

+ ¾¾¾¾® +

Hydrolysis2 6 2 3 3 2B H 6H O 2H BO 6H+ ¾¾¾¾¾® +

PH nC dTD = ò1000

300

3 (23 0.01T)dT= +ò10002

300

0.01T3 23T2

é ù= +ê ú

ë û

220.01 0.01(300)3 23000 (1000) 23(300)

2 2é ù

= + - -ê úë û

[ ]3 23000 5000 6900 450 61950J= + - - =

62kJ!

1º2 º 2[SN

Vidyamandir Classes


25.(2)

26.(3) Let number of B atoms = N

Number of atoms of

Number of oxygen atoms = 2N A : B : O

is the formula

27.(1) Greater is the reduction potential, stronger is the oxidizing agent.

28.(2) Fact (refer of NCERT) Chemistry ; Class XI, Page No – 405 & 406.

29.(3) Hydration enthalpy charge on an ion

Hence, correct order of hydration enthalpy is:

30.(3) Using plastic bags is wrong as plastic bags cause environmental pollution.

NA2

=

N :N : 2N21 :1: 221: 2 : 4

2 4AB O

µ1

size of an ionµ

Li Na K Rb Cs+ + + + +> > > >

Vidyamandir Classes


PART-C MATHEMATICS

1.(2)

Liner differential equation

Integrating factor

Let

,

2.(1)

For non-trivial solutions

D = 0

, Greatest value of ‘c’ is 1/2.

3.(4)

4.(2) Case I

2 2 2( 1) 2 ( 1) 1dyx x x ydx

+ + + =

2 2 22 11 ( 1)

dy x ydx x x

æ öÞ + =ç ÷+ +è ø

221x dxxe +=

ò2 1 2x t xdx dt+ = Þ =

ln 2 1dt

tte e t x= = = = +ò

2 12( 1) tan1

dxy x x Cx

-Þ + = = ++ò 2 1( 1) tany x x C-Þ + = +

0 0x y= Þ =!

10 tan 0 0C C-= + Þ =1

2 12

tan( 1) tan1xy x x y

x

--+ = Þ =

+

(1)8

y p= (1)

32a y p

=

1 18 32 4 16

a a ap pÞ ´ = Þ = Þ =

0x cy cz- - =

0cx y cz- + =

0cx cy z+ - =

0x y zD D D= = =! \

2 2 21

1 1(1 ) ( ) ( )1

c cD c c c c c c c c c

c c

- -= - = - + - - - +

-2 2 3 3 21 c c c c c= - - - - -3 22 3 1 0c c= - - + =

3 22 3 1 0c c+ - =2(2 1)( 1) 0c cÞ - + = 1, 1, 1/ 2c = - -

\

2

2

2

212 1ln 21 11

xx xf xx

x

æ ö-ç ÷æ ö += ç ÷ç ÷+è ø ç ÷+ç ÷+è ø2

2( 1) 1ln 2ln

1( 1)x x

xx

æ ö- -= =ç ÷ç ÷ ++è ø

12ln 2 ( )1x f xx

-æ ö= =ç ÷+è ø

Vidyamandir Classes


Case II

or (Not possible since )

5.(1)

Sum of coefficients of all even degree term is

6.(4)

… (i)

… (ii)

Adding (i) and (ii) we get,

7.(3) are roots

,

Least value of n = 4

2x ³4xÞ ³

2 4 2 0x x x- + - + =

3xÞ =9x =

2x <4xÞ <

2 4 2 0x x x- + - + =

4 5 0x x+ - =

( )( )4 1 0x x- - =

1x = 4x = 2x <1xÞ = Sum 9 1 10Þ = + =

( ) ( )6 63 31 1x x x x+ - + - -

6 4 3 2 3 2 3 30 2 4 62 ( 1) ( 1) ( 1)C x C x x C x x C xé ù= + - + - + -ë û

[ ]0 2 4 4 6 62 3C C C C C C- + + - -

[ ]2 1 15 15 15 4 24- + + - =

4tan( )3

a +b =5tan( )12

a -b =

1 4( ) tan3

- æ öÞ a +b = ç ÷è ø

1 5( ) tan12

- æ öÞ a -b = ç ÷è ø

1 14 52 tan tan3 12

- -æ ö æ öÞ a = + ç ÷ç ÷è øè ø

4 5633 12tan 2 20 161

36

+a = =

-

2 2 2 0x x- + = &a b

2 4 12

i+ -a = = +

2 4 12

i- -b = = -

11i ii

a += =

b -

( ) 1n

niaæ ö = =ç ÷bè ø

Vidyamandir Classes


8.(4)

9.(4)

Parallel vector to line is

10.(1) Vector perpendicular to plane containing vectors and is parallel to vector

Required magnitude of projection is

11.(4) Then contrapositive of this is If you are not a citizen of India, then you are not born in India.

12.(3) ;

;

Vertex = (2,0) ;

Tangent of slope 1 on parabola

;

Distance of this line from y = x

2020

0(3 2) r

rr C

=

+å20 20

20

0 03 2r rr rr C C

= =

= +å å20

19 201

03 20 2 2rr

C -=

= ´ + ´å19 203 20(2) 2 (2)= ´ + ´

21 252 (15 1) 2= + =

ˆˆ ˆ5 3 4AP i j k= - +!!!"

ˆˆ ˆ10 7i j k- +

ˆ ˆˆ ˆ ˆ ˆ(5 3 4 ) (10 7 )cos150

i j k i j k- + × - +q =

75 3 1cos sin2 250 3

Þ q = = Þ q =

5 2 sinPN = q

5 2 1 52 2´

= =

ˆˆ ˆi j k+ + ˆˆ ˆ2 3i j k+ +

ˆˆ ˆˆˆ ˆ1 1 1 2

1 2 3

i j ki j k= - +

\ˆ ˆˆ ˆ ˆ ˆ(2 3 ) ( 2 ) 3

26i j k i j k+ + × - +

=

p qÞ

q pÞ! !

2 2y x= - y x=

2 ( 2)y x= - 2 14 ( 2)4

y x= ´ -

14

a =

11 ( 2)4

y x= ´ - +74

y x= -

7 74 1 1 4 2

d = =´ +

Vidyamandir Classes


13.(2)

14.(1) For point to be equidistant from axes it must lie on the or line

2nd quadrant

15.(4) Odd digits are 1, 1, 3 Even places 2nd , 4th , 6th, 8th

No. of ways

16.(2)

Sign scheme of is Local minimum at –2 and 1 Local maximum at 0

17.(1) Let the point P be (h, k)

5sin2

sin2

x

dxxò

52 sin cos2 2

2sin cos2 2

x x

x x

×Þ ò

52 sin cos2 2sin

x x

x

×Þ ò

sin3 sin 2sinx xx

+Þ ò

33sin 4sin 2sin cossin

x x x xx

- +Þ ò

2(3 4sin 2cos )x x dxÞ - +ò(3 2(1 cos2 ) 2cos )x x dx= - - +ò(2cos2 2cos 1)x x dx= + +òsin 2 2sinx x x C= + + +

y x= y x= -

3 5 15x y+ =

8 15y x x= Þ =

1515 158 ,

15 8 88

xPt

y

ü= ïï æ öý ç ÷

è øï=ïþ

y x= - 3 5 15x y+ =

3 5( ) 15x x+ - =

2 15x- =15 15,2 2

x y-= =

15 15,2 2-æ ö

ç ÷è ø

33! 6!4 1802! 4!2!C= ´ ´ =

3 2'( ) 36 36 72f x x x x= + -3 236( 2 )x x x= + -236 ( 2)x x x= + -

36 ( 2)( 1)x x x= + -

'( )f x

1 2{ 2,1}, {0}S S= - =

Vidyamandir Classes


(required locus)

18.(1) Mean

Variance = 16

19.(3) Slope of the tangent at for ellipse is

Slope of tangent at (1,2) is –2 Perpendicular tangents slope will be 1/2 Slope of tangent at

For is

;

20.(2)

2 2 2 21 ( 1) 4h k h k\ + + + + - =

2 2 2 2( 1) 3h k h kÞ + - = - +

2 2 2 2 2 22 1 9 6h k k h k h kÞ + - + = + + - +

2 23 4h k kÞ + = +2 28 9 8 16 0k h kÞ + - - =2 29 8 8 16x y yÞ + - =

2 4 10 12 14 428 87 7

a b a b+ + + + + + + += = Þ =

42 56a b+ + =2

2xVNå

= -µ 14a b+ =

2 2 2 2 2 2 222 4 10 12 14 8

7a bV + + + + + +

= -

2 2 46016 647

+ += -a b

2 2 460807

a b+ +=

2 2560 460a b= + +2 2 100a b+ =2 2 100a b+ =

2( ) 2 100a b ab+ - =2(14) 2 100ab- =

196 2 100ab- =196 100 2ab- =96 482

ab ab= Þ =

1 1( , )x y2 2

12 8x y

+ = 1 1

1 1

8 42x xy y

- ´ = -

\

( cos , sin )a bq q2 2

2 2 1x ya b

+ = cot 2cotba-

q = - q

12cot2

Þ - q =1 1cot cos4 17-

q = Þ q = ±

( )22 2 22 cos 2cos17

aÞ = q = q =

3 4cos sin5 5

a = a =

Vidyamandir Classes


21.(2) n is between 100 & 200 HCF (91, n) > 1 (Sum of natural no divisible by 7) + (Sum of natural no divisible by 13) – (Sum of no divisible by 91)

22.(3)

; ;

23.(4) will be intersecting

if

Length (Distance of from origin)

(radius)

24.(4)

3 1cos sin10 10

b = b =

sin( ) sin cos cos sina -b = a b- a b

12 35 10 5 10

-9

5 10=

1 9sin5 10

- æ öa -b = ç ÷è ø

14 8

1 1(98 7 ) (91 13 ) (182)

r rr r

= =

+ + + -å å1372 735 728 468 (182) 3121+ + + - =

2

1cos

62 cotcos

3

xy

x

-

æ öæ öpæ ö-ç ÷ç ÷ç ÷è øç ÷ç ÷=pæ öç ÷ç ÷+ç ÷ç ÷ç ÷è øè øè ø

2

1cos

62 cotsin

6

xy

x

-

æ öæ öpæ ö-ç ÷ç ÷ç ÷è øç ÷ç ÷=pæ öç ÷ç ÷-ç ÷ç ÷ç ÷è øè øè ø

212 cot cot

6y x-æ öæ öpæ ö= -ç ÷ç ÷ç ÷è øè øè ø

22

6y xpæ ö= -ç ÷

è ø2 ' 2

6y xpæ ö= - -ç ÷

è ø'

6y x p= -

x y n+ =2 2 16x y+ = 1,2,3,4,5n = ( )n NÎ

2nOL = x y n+ =

4OM =2

2 162nMN = -

22( ) 4 16

2nMN

æ ö= -ç ÷

è ø5 2

1

5 6 114 16 4 16 5 320 110 2102 2 6

=

æ ö ´ ´é ù- = ´ - = - =ç ÷ ê úè ø ´ë ûån

n

cos sinsin cos

Aa - aé ù

= ê úa aë û

2 cos sin cos sinsin cos sin cos

Aa - a a - aé ù é ù

= ê ú ê úa a a aë û ë û

Vidyamandir Classes


25.(2) is function If is decreasing If is increasing

26.(3)

27.(1)

Required area

28.(4)

29.(2) Plane passing through given two planes can be written as

2 cos2 sin 2sin 2 cos2

Aa - aé ù

= ê úa aë û

32 cos32 sin32 0 1sin32 cos32 1 0

Aa - a -é ù é ù

= =ê ú ê úa aë û ë û

322p

a =64p

Þ a =

''( ) 0 (0,2)f x x> " Î

'( )f xÞ '( ) '( ) '(2 )x f x f xf = - -

(0,1)xÎ'( ) 0 ( )x xf < Þ f

(1,2)xÎ'( ) 0 ( )x xf > Þ f

2

0

sinlim2 1 cos

2x

xx® æ ö-ç ÷

è ø

2 2

0 2

4sin cos2 2lim

2 2 sin4

x

x x

x®=

2

220 0

4 sin4lim cos lim 12

2 216

x x

xx x

xx® ®

é ù= =ê úë û4 2=

12

0

11 59( 3 ) 4 2 86 6

x x dx= + + ´ = + =ò

( ) ( )( / )( ) ( )

P A B P AP A BP B P BÇ

= =

( ) ( )( )P A P AP B

³

(2 4) ( 2 4) 0x y y z- - + l + - =

3 ( 3) 0- + l - =

1l = -

Vidyamandir Classes


30.(3)

2 4 1( 2 4) 0x y y z- - - + - =

0x y zÞ - - =

2 cos( ( )) ln2 cosx xg f xx x

-æ ö= ç ÷+è ø

4

4

2 cosln2 cosx xI dxx x

p

p-

-æ ö= ç ÷+è øò

4

4

2 cosln2 cosx xI dxx x

p

p-

+æ ö= ç ÷-è øò

( )4

4

2 ln 1 0I dx

p

p-

Þ = =ò 0 log 1eIÞ = Þ

solutions - vidyamandir...joint entrance exam | jee mains 2019 part-a physics 1.(3) refer ncert...

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