PART TEST # 01 - Welcome to JEE Shikhar All India JEE ... : jeeshikhar.catalyser.in JEE TEST SERIES PART TEST # 01 PAPER – 1 (ADVANCE) SOLUTIONS SECTION – 1 : (Only One Option

India’s First ‘Rank Improving’ All India JEE Test Series

PART TEST # 01

ANSWERS & COMPLETE SOLUTIONS

PAPER – 1 (ADVANCED) & PAPER – 2 (MAIN)

TO KNOW YOUR EXPECTED ‘JEE RANK POTENTIAL’ & COMPLETE VIDEO SOLUTION OF THIS PAPER

Visit : jeeshikhar.catalyser.in

ANSWER KEY PAPER - 1 (ADVANCED PAPER) SET - A

Physics Chemistry Mathematics 1 A 21 D 41 C

2 D 22 C 42 B

3 C 23 B 43 A

4 B 24 B 44 B

5 A 25 D 45 A

6 CD 26 AB 46 ABCD

7 AC 27 BC 47 BCD

8 BD 28 ACD 48 ABC

9 ABC 29 AD 49 ABCD

10 ABD 30 AC 50 B

11 A 31 A 51 A

12 A 32 B 52 D

13 A 33 B 53 A

14 B 34 A 54 A

15 A 35 C 55 D

16 AC 36 BC 56 C

17 AC 37 C 57 B

18 AB 38 AD 58 B

19 D 39 AD 59 C

20 AB 40 CD 60 CD

ANSWER KEY PAPER - 2 (MAIN)

Physics Chemistry Mathematics 1 C 31 C 61 A 2 B 32 A 62 B 3 D 33 C 63 D 4 A 34 C 64 C 5 A 35 A 65 C 6 C 36 A 66 B 7 A 37 D 67 A 8 B 38 C 68 A 9 C 39 D 69 A

10 C 40 C 70 B 11 A 41 A 71 A 12 A 42 A 72 B 13 B 43 D 73 A 14 D 44 D 74 A 15 A 45 D 75 C 16 C 46 A 76 A 17 C 47 B 77 C 18 B 48 A 78 D 19 B 49 C 79 A 20 C 50 D 80 D 21 C 51 B 81 C 22 D 52 B 82 C 23 C 53 B 83 D 24 B 54 A 84 A 25 D 55 B 85 D 26 B 56 A 86 B 27 A 57 B 87 B 28 C 58 B 88 A 29 B 59 B 89 B 30 D 60 D 90 A

TO KNOW COMPLETE VIDEO SOLUTION OF THIS PAPER

Visit : jeeshikhar.catalyser.in

JEE TEST SERIES PAPER – 1 & 2 / SOLUTION / PAGE - 1

India’s First ‘Rank Improving’ All India JEE Test Series Website : jeeshikhar.catalyser.in

JEE TEST SERIES PART TEST # 01 PAPER – 1 (ADVANCE)

SOLUTIONS

SECTION – 1 : (Only One Option Correct Type)

Sol.1. (A) WD by friction force on a system is ref. frame independent. WDf = – µmgL

Sol.2. (D) vAx = vBx 2 2 2 2Ay Bx By Bxv v v v Ay Byv v R =

2 x yv v

g RA > RB

H = 22 yv

g HA > HB T =

2 yv

g TA > TB

Sol.3. (C) W = 2

0 00 02

00 0 ( )P t

Pdt dt P tt t

Sol.4. (B) WD = k = 60°

mgR (1–cos) = 12

mv2 – 0 – (1) v = gR

mg cos + kx = 2mv

R

x = 2mg

k = 1

2m Natural length = R – x = 1.5 m

Sol.5. (A) WD = k

mg = 21 2

1 .6 ( ) 02 6

m mv v

m m

v1 + v2 =73

g

SECTION – 2 : (One or More Than One Option Correct Type)

Sol.6. (CD) Work energy theorem : work done by all forces = change in kinetic energy for conservative force : |change in PE| = |change in KE|.

Sol.7. (AC) WD = 0K

0spling frictionWD WD

21 0

22

K d mdg

mgd

K

PART I : PHYSICS

0

v

mgkx

v2

m



Sol.8. (BD) For equilibrium : F = 0 Stable : step of F – X curve is negative Unstable : step of F – X curve is positive Neutral : step of F – X curve is zero

Sol.9. (ABC)

P

(A)

P

(C)

A Bf f if cosP N

A C Bf f f if cosP N

Sol.10. (ABD)

SECTION – 3 : Matching List Type (Only One Option Correct)

Sol.11. (A) [(A) S (B) P (C) Q,R (D) T ] (P) There is tangential acceleration in ABCD s = vt, time is same, v is greater in ABCD s is more (Q) If s is same v is same a is same (since t is same) (R) No tangential acceleration in between (S) Curve is more flat at point D

Sol.12. (A) [(A) Q, S (B) P (C) R] . (A) aA sg But aBelt > sg arel 0 hence friction is kinetic (B) aA = kg > abelt = Clrel > 0 and Vrel < 0 Vrel will become zero after a long time, so friction will be static. (C) Car moving with banking velocities so no firction required. (D) No slippping between the block and wedge so friction is static, to prevent from sliding

down. SECTION – 4 : Comprehension Type (Only One Option Correct)

Paragraph for (Qs. 13 to 15) A block of mass M slides on ……………….. with force constant k. Answer following questions.

Sol.13. (A) Maximum Friction force (static) = s mg Maximum acceleration =

sg

Sol.14. (B) 12

(M + m) v02 = 1

2 kxm

2

kxm = µs(M + m)g xm = µs / k

12

(M + m) u02 = 1

2k

2( ) sM m g

k

k = 2

0

( )s gM m

v

]

P

(B)



PART II : CHEMISTRY

Sol.15. (A) aBox = µkg

aBlock = (µs – µk) mg

M M

kmgkxm

aBlock = m kkx mg

M

= (µs – µk)

mg

M

SECTION – 5 : (One Integer Value Correct Type)

Sol.16. 0005 1

2 2 21 2 2

1 1 12 2 2

kx m v m v ……. (1)

2 kg

4 m/s

3 m/s

1 1 2 2m v m v …….. (2)

5 /resV m s

Sol.17. 0005 Motion will start when F smg at t = 1s after that F – kmg = ma

3t2 – 2 = a or 3t2 – 2 = dV

dt V =

22

1

(3 2)t dt = 5 m/s]

Sol.18. 0004 No. ext. force, so m1a1 + m2a2 = 0 m2 = 1 1

2

m a

a

= 4 m/s2

Sol.19. 0003

Sol.20. 0004 21 sin37 ......... 12

s vt g t

20

16 cos17 sin372

s v t g t ……….. (2)

0 0sin sin cos37v v g t ………. (3)


Sol.21. (D) NO3– NO

N+5 N+2 change = 3 N+5 2 × N+2 total change = 6

n factor of per mole HNO3 = 68

= 34

Sol.22. (C)

Sol.23. (B) 22 2 3

1/2 5/2 1

5 / 23 25 / 2 4 / 2HN H NH X

3 2 411

NH H O NH OH

4 4 211NH OH HCl NH Cl H O

Sol.24. (B) BO of H2+ is 1

2 hence curve –2 shows bonding condition or decrease in potential energy

Sol.25. (D) Press. of the gas = 749 + 292 = 1041 mm Hg.




Sol.26. (AB) 2 1

2 4 2N H N O

1 4

2 2 4N O N O

Change = 3 (No. of Nitrogen) = 6 Change = 6

Sol.27. (BC) Out of Phase orbitals don’t overlap.

Sol.28. (ACD)

Sol.29. (AD) 6

322 7 2HK Cr O Cr

6 3 2 6n factor No. of eq= 0.6 6 3.6 eq

2 3

4 2 4 3Fe SO K SO

n = 1 No. of eq = 3.6 2 4Sn Sn n = 2 2 1.8 3.6qE

Sol.30. (AC)

SECTION – 3 : Matching List Type (Only One Option Correct)

Sol.31. (A) [(A) (i, iii, iv), (B) (ii) , (C) (iii) , (D) (i)] No. of moles doesn’t depend on temperature volume does.

Sol.32. (B) [A Q, B P, C S, D R] (A) (Q) ; CN– (6 + 7 + 1 = 14) 1s2 * 1s2

2s22s2

2px22py

22pZ

2

B.O. = 12

(10 – 4) = 62

= 3, Diamagnetic due to absence of unpaired electrons.

(B) (P) ; C2(12) 1s2 * 1s2

2s22s2

2px22py2

B.O. = 12

(8 – 4) = 42

= 2, Diamagneti

SECTION – 4 : Comprehension Type (Only One Option Correct)

Paragraph for (Qs. 33 to 35)

Sol.33. (B) Gases with larger value of ‘a’ are easily liquefied even at low pressure. V decreases. Z decreases.

Sol.34. (A) For positive deviation 0.V

For same P V is greater, 1PV

nRT

Sol.35. (C) In He Z > 1 so repulsive forces are dominating so molar volume is greater than 22.4 L



PART III : MATHEMATICS

SECTION – 5 : (One Integer Value Correct Type) Sol.36. (7) BrF4

– , XeF5–, IF5 , HCHO, BF3, CH3

+ , SO3 Sol.37. (2) = 4 × 10–8 m = 400 Å , so

12375400

= 13.6 Z2 11–4

Z2 = 3.04 Z = 1.74 ~ 2 (Helium)

Sol.38. (6) T

P 2

1

= 1

2

P

P 2 = 60 × 1

10 = 6

Sol.39. (6) Volume strength of H2O2 = 5.6 × Normality = 5.6 × 1.08 = 6.048

Sol.40. (9) Angular momentum for p-orbital e– s

= ( 1)2h

= 2

2 2h h

So we have to find out total no. of electrons in p-orbitals of phosphorous. 15P 1s2 2s2 2p6 3s2 3p3 Total p-orbital e– = 9


Sol.41. (C) 23 6 6 0f x x x x R

Hence 1 2 3f x is 1 2 3f f f

1 , 2 , 3let f f f 1 2 3 0g x

x x x

2 30

x x x x x x

x x x

2 3 0x x x x x x

0g 0g 0g g x has two roots

bet g g x

Sol.42. (B) Let 1 2, , ........... na a a be the sides of squarel 1 2, , ...........sns s respectively

1 10a 2 1 2102

2a a a

3

102

a and so on

1

10

2n n

a

Area

2

1 1

10 100 122

n n

12 100n 1 6 7 8n n n



Sol.43. (A)

Sol.44. (B) sec 1 tan .tan sec tan .tan2 2

sec 1 tan .tan 02

sec tan .tan 12

tan .tan 0,sec2

i.e. sec 1 cos 1 cos2 .n n z 0,2 ,4 ,6 No. of sol of is 201,

Sol.45. (A) 22 1/22 log 1 1 log 4x x for inequality to be defined

1 0x 24 0x 1x 2 2x

And 2 2

1log 14x

x

2

1 124

x

x

2 4 0x x 4,0x

Taking intersection 0,2x also new 1s


Sol.46. (ABCD) 21 tan 4tan 1 0k x roots are there. 0D 16 4 1 1 0k k

2 5K A

Also sum of roots 1 24 tan tan

1x x

k

Products of roots 1 21 tan ,tan

1k

x xk

Then

1 2

41tan 2

11

1

kx x Bk

k

If k = 2 , let tan x = y Equation is 3y2 – 4 y + 1 = 0

y = 13

or y = 1 tan x = 1 x1 = 4 (C)

If k = 1, let tanx y 22 4 0y y

0 2y or y 1 0x (D)

Sol.47. (BCD)



Sol.48. (A,B,C) Given 1 2 2 3 3 1cos tan , cos tan & cos tan

Taking 2 out of the above three, We get

1 2

3 3

sincos sinCos

…… (1)

32

1 1

sincos sinCos

…… (2)

3 1

1 2

sincos sinCos

…… (3)

Multiplying (1) & (2) we get 1 2sin sin similarly

1 2 3sin sin sin 1 2 3cos cos cos Now 1 2cos tan (Given) 1 2 2cos cos sin 2

1cos sin 2

1 1sin sin 1 0

11 5sin 2sin18

4

Hence A, B, C

Sol.49. (A,B,C,D) 2f x ax bx c

We have 1f x

Case – I : When a > 0, then

24 ,

4ac b

f xa

i.e.

241 1

4ac b

a

i.e.

2

1 14b

Ca

2 2

1 , 14 4b b

C Ca a

i.e. 1 1C 1C

When 1C , then 8a , 8b

Now

17a b c

Sol.50. (B)



SECTION – 3 : (Matching List Type (Only One Option Correct)

Sol.51. (A) (1) (S) , (2) – (P), (3) (R) , (4)(Q)

(1) 2

2

12 10012 100

x xy

x x

2 212 100 12 100x y xy y x x

2 1 12 1 100 1 0x y x y y x R

2 2144 1 400 1 0y y

12 1 20 1 12 1 20 1 0y y y y

4 1 4 0y y

1,44

y

Hence minimum 14

y

For which 10x (S)

(3) 5

1/4log1122

15

n

using 1

as

r

5

1log22 1

5n

5

1log52 1

2n

2 22 2n n (R)

Sol.52. (D) (1) (R), (2) (S), (3) (Q), (4) (P) (1) 90C & angles are in A.P. Other angles are 30 & 60 Hence answer is (R)

(2) cot 45 cot 17 28cot17 cot28 11cot17 cot28

cot17 cot28 cot17 cot28 1

2 cot17 1 cot28 1

2 cot17 cot28 cot17 cot28 1 =4 Hence Answer is (S)



(3)

6

6

66 6

6 6

6 6

cos12 cos48 sin12 sin48

2cos30 cos18 2 sin30 cos18

2 cos 18 cos30 sin30

2 cos 18 cos180 sin180

2 cos 18 Im 0

z i

z i

z i

i

z

Hence Answer is (Q)

(4) 1 16 6 , , 4 4X x x x are in A.P.

2 6 6 6 4 4x x x x

2 6 2 2 [Using 1 2xx

]

7 Hence Answer is (P)

SECTION – 4 : Comprehension Type (Only One Option Correct)

Paragraph for (Qs. 53 to 55)

Let rV denote the sum of first r terms of an …………

11 1,

4 1r r r r

r

Q T T WT r

and 3 rQ

rX for 1,2....r

Sol.53. (A) 2 1 2 12r

rV r r r

22 2 3 12r

rV r r r

23

2 2r

r rV r

21 2 2 2 2

2 6 2 2r

n n n n n n nV

1 2 2 2 12 2 2 2

n n n n n

22 3 3 4 2 32 6

n n n n n

21 1 3 1

12n n n n

Sol.54. (A) 2 2r r rT V V

2 1 2rT r 2 3 2 1 1rT r r Always odd

Sol.55. (D) 2 3rT r

1 2 31, 1, 3T T T

11 1 1 71

1 4 2 8 8W

21 1 7 1311 4 3 12 12

W

31 2 16 3 193 4 4 48 48

W

None of these (D)



SECTION – 5 : (One Integer Value Correct Type) Sol.56. (2)

Sol.57. (1) 2sin sin min 1, 4 5x y sinx siny 1

2sin cos 12 2

x y x y

& 0

2 2 2x y x y

Neglected

Sol.58. (1) let ,i R be the root Then, put in the equation 4 3 2

3 4a, a, a a 0i i

4 2 32 4 3a a a a, 0i

Which is possible, when 4 2

2 4 3a a 0& a a, 0 2 3

1

a0

aor (i)

Put in (i) 23 2 3

4211

a a aa 0

a a

2 23 1 2 3 1 4a a a a a 0a 3 1 4

1 2 2 3

a a a1

a a a a

Sol.59. (2) Let 2006

1

22007 cos2007K

Kx k

Replace 2007K K

2006

1

2cos2007K

Kx K

Add

2006

1

22 2007cos2007K

Kx

2 4 2 2002 2007 cos cos ........... cos

2007 2007 2007x

2 2005 20062007 cos sin2007 2007

2sin

2007

x

20072 2007 22 4

nx x n

Sol.60. (9)

2 2

2 2

2 2

2 2

4 14 1 4 3

4 3 4 118 4 1 4 3

1 1 18 4 1 4 3

n

xT

x x

x x

x x

x x

2 2 2 2

1 1 1 1 1 .......8 3 7 7 11nS

21 98 9nS K

END OF TEST PAPER



JEE TEST SERIES PART TEST # 01 PAPER –2 (MAINS)

SOLUTIONS

1. [C] The work done by man is negative of magnitude of decrease in potential energy of chain.

L/4

L/2

U = mg 2L –

2m g

4L = 3 mg

8L W = – 3

8mg

2. [B] W = (1,1)

(0,0)

.F dx

Here ds

= ˆdxi + ˆdyj + ˆdzk

W = (1,1)

2

(0,0)

( )x dy ydx = (1,1)

2

(0,0)

( . )x dy x dx (as x = y)

W = (1,1)3 2

(0,0)3 2y x

= 56

J

3. [D] Acceleration of system = 1 2

1 2

–m m

m m g Tension in string = 1 2

1 2

2m m

m m g

reading of spring balance is T.

4. [A] K = 4

2

Pdt = [t3 – t2 + t]24 = 46J

5. [A] F = – dU

dx F = 3x2 – 12x Now F = min. d F

d x = 0 x = 2 m

Ui + Ki = Uf + Kf 15 + 12

× 2 × 80 = [– (2)3 + 6 × (2)2 +15] + 12

× 2 × v2

v2 = 64 v = 8m/sec

6. [C] 7. [A] For mass and block system -

Pm + Pf + Pmg = 0 k = 0 Pm = – (Pf + Pmg)

= 4 3. . ( ) . .5 5

mg v m M g v

[m = mass of block, M = mass of man] = 0.65 kW

8. [B]

PART 1 : PHYSICS



9. [C]

O R

reference d R

idU = – m

Rd

× g × R[1 – cos ] idU =

2mgR

[1 – cos]d

Ui = 2mgR

sinR R

and Uf = 0 Wext = U

10. [C] 11. [A] Limiting friction on block B = 0.5 × (2 + 8) × 10 = 50 N block B will not move.

12. [A] Since the spring balances are massless, readings of both springs will be same.

13. [B] ac = k2 rt2 or 2v

r= k2 rt2 or v = krt

Therefore, tangential acceleration, at =dv

dt= kr

or Tangential force, Ft = mat = mkr

Only tangential force does work. Power = Ft v = (mkr) (krt)

or Power = mk2 r2 t

14. [D] 15. [A] In this case spring force is zero initially F.B.D of A and B

A

m

mg

B

2m

2mg

aB=g aA=g 16. [C]

17. [C] Power = .F V

= ˆ ˆ ˆ(10 10 20 )i j k . ˆ ˆ ˆ(5 3 6 )i j k = (50 – 30 + 120) watt = 140 watt 18. [B]

u

v

m fr

fr = mg sin Work done by fr = – mg sin cos vt

= – sin22

mg (v ×t)

19. [B]

20. [C] Retardation = g (sin + cos ) = 5 ( 3 ) Now v = u – at

a = u

t as v = 0 5( 3 ) = 10 = 0.27

21. [C] 22. [D] 23. [C] 24. [B] 25. [D]



26. [B] Man should hold the umbrella in direction in which rain appears to come i.e., rmv

rmv

= rv

– mv

; vrm Velocity of rain with respect to man.

rv

= rmv

+ mv

vrm vr

vm tan = r

m

v

v = 60

20 = 3 = tan–1 3

27. [A] Let direction of river flow be along x-axis.

Angle made by MGV

with x-axis tan = RG

MR

V

V

28. [C] Component of velocity in vertical should be same.

29. [B] 2 sin2u

g

= 23

2v

g sin2 = 3

2 = 30º

2 2sin2

u

g

= 2

8u

g sin = 1

2 or = 30º

30. [D] H = 2 21 1sin

2u

g

=

2 22 2sin

2u

g

Sol.31. (C) Paramagnetic nature of oxygen is explained by molecular orbital theory.

Sol.32. (A) moles

Molarity =volume(l)

Consider, 1l water

Mass of 1l water = 1 kg.

Moles of 1kg water 1000

18moles .

So, Molarity

1000

1855.555 5.6

1M

Sol.33. (C) Debloglie wavelength h

P

Now, KE is 9 times, so velocity must have increased 3 time as

1

2m 2

1

1

2

v

m 2

2

1

9v

2 13v v

So, new debroglie wavelength 2 1' 2

h h h

P mv mv .

PART 2 : CHEMISTRY



Sol.34. (C) 2 2Ba H PO

So, 12

2 2and ;Ba H PO let ox no. of Phosphorus

he x So, 2 1 2 2 1x

2 4 1x 1x Sol.35. (A) Stable halt filled p-orbital gives extra stability.

Sol.36. (A) 8

27c

aT

Rb

B

aT

Rb &

2i

aT

Rb .

So, 3i cT T T

Sol.37. (D) 4.2 g of 3N

So, moles of 3 4.2

14N

No. of 3N particles 4

4.2

14N

N has 5 valence e So, 3N has 8 valence electrons.

So, no. of electrons.3 4.2

14

4 8N 42.4N

Sol.38. (C) O

H H

N

H HH

Xe

F

FH

NN

N171

Shape is bent

Shape is Pyramidol distorted Shape is

linear

Every molecule has a different shape than its geometry so every molecule is distorted. Sol.39. (D) Genetral knowledge (factual) Sol.40. (C) Here, while going from energy is absorbed due to repulsion however overall 2O O

energy is released. Sol.41. (A) H C N N is more electronegative than carbon so, ox no. of N is 3. So, let x be the ox no. of C.

1 3 0x 2x

Sol.42. (A) 3rms

RTV

M 2

3

64rms

RTV SO ...........(1)

3 300

4rms

RV He ...........(2)

So, 3 3003 1

64 2 4

RRT

Sol.43. (D) Radial node 1n l and angular node l So for 9" "S orbital. Radial node 5 4 1 0 angular 4l



Sol.44. (D) 3 8 2 2 25 3 4C H O CO H O

So, 1 ml of propane combines with 5 ml of 2.O

So, 20 ml of propane will require 5 20 100ml ml

Sol.45. (D) We’ll see electronic configuration.

2 1 1 2 2 * 2 2zs s s s P PzN * *

2Px

2Py

2Px

2Py *

*

5

.O 2.52

B .B Oof 1 3N

2 1 1 2 2 2 2zs s s s P PxO *

2Px

2Py

2Px

2Py *

*

* *

5

.O 2.52

B

.B Oof 2O is 2.

Sol.46. (A) NO2

O H

Due to intra molecular H-bonding in o-nitrophenol. Sol.47. (B) Here, 2H & 2Cl reacts at room temperature. Sol.48. (A) Due to high electron density.

Sol.49. (C) ;PV

nRT

pressure, vol. and temp. is same so, moles are also same.

So, no. of atoms : of 2 2 2H n n

of 1eH n n

of 2 2 2O n n

of 3 3 3O n n Sol.50. (D)

Sol.51. (B) (a) (b) (c)

0

(d)

0



Sol.52. (B) 2 1 1 2 1 1s s s sH H * *

.0 1B .0 0.5B

As e is removed from 2 ,eH density decreases.

Sol.53. (B) Trick : See electrons configuration of 2 2 3, , , , , , ,Mg Al Na Mg Mg Al Al Al and so on.

Sol.54. (A) (a) 2CS S C S Linear

(b)

S

O O

2SO

bent

(c) 3BF B

F

F

F Trigonal planar.

(d) 3NH N

H

H

H Trigonal pyramidal

Sol.55. (B) ( )PM

Density DRT

also, P RT

D M

Sol.56. (A) 2 2

4 2 4 2 22 5 16 2 10 8nMnO C O H M CO H O

Sol.57. (B) +

+-

-y axis

2P

Sol.58. (B) 3 2KClO KCl O

5 1Cl Cl 2

2O O Sol.59. (B) Let 1, , 1A B Cn n n n n n

2 2

1

1 1 1

B C

RZn n

22

1 1

1RZ

n n

2 2

2

1 1 1

1RZ

nn

2 2

3

1 1 1

1 1RZ

n n

1 2 3

1 1 1

So, 1 2

3

1 2

Sol.60. (D) .0 3B C O

2BO O C O

1.33B O

C O O

or O

C O O



61. (A) 0.3 0.3010310 2 10

30 100 30103 3110 2 10 10

62. (B) 2 1x px x a x b

Then c a c b a d b d

c a c b d a d b

22 1 1c pc d pd

2 21 1c pc d pd a b p

But 2 1c qc and 2 1 0d qd qc pc qd pd cd q p q p

2 2 2 2cd q p q p 1cd

63. (D) 2 2 22f x x b c b

2 2min ( ) 2cf x b Also 2 22g x x cx b

22 2b c x c

max 2 2g x b c

As min max gf x x 2 2 2 22c b b c

2 22 | c | 2 | b |c b

64. (C) Let 2f x ax bx c

0 0f c

1 0f a b c

1 0f a b c

Thus 2 2y ax bx c represent parabola that open downwards

Thus 1 and 0

1

65. (C) 10log log log 5a aA x a 10log log 5ax 10log 5 loga x

10log 15 5xAa 10log 1x

Let 10 10log log 1 110x

B x

13 3B 33

100 4log log 2C x

21 1

10 2log 2 log 2x

10

1 1 1log2 2 2x

1129 9 3 3.3c

PART 3 : MATHEMATICS

1 -1



According to question

6 35 3.35 3

1 16.5 3 33

1 16.5 3 10 2 25 3

Which is possible when 2

10log 2x = 210 100x

66. (B) 2 2 0x xe a e a

2 1 2 2 2 1 1x x x x xe a e e e e

2

1 4 1 3 1x x xe e a e

31 4

1x

xe a

e

2 3

31 41

x

xe a

e

4 2 2a (by AM GG.M)

67. (A) Since 0f x x R , a > 0 and 2 4 0b ac

' 2f x ax b and "f x a

2 2 2g x ax bx c ax b a

2 2 2ax a b x a b c

We have a > 0 and 22 4 2a b a a b c

2 2 24 4 8 4 4a ab b a ab ac 2 24 4 0b ac a

2 4 0b ac

68. (A) /b a = /c a

2 2

1 1

b

2 2

2

2

2

2b b ac

a c

2 2 22 . .bc ab ac AP 69. (A)

2 2 2ax b x c 2 2 20 4 4 0b ac

Now 0a parabola upward

32 2 2 0a x bx c

If 0a parabola down ward.

32 2 2 0a x b x c



70. (B) 1

1 1 11 1

n

n

k

an k n k

1

2 11

n

kn k

For 2n

1

11 1

1 1 1 1 12 1 2

n n

n n

k k

a an k n k

1

1 1 1 11 2 1 2

n

kn n k n n

2

1 1 01 2

n

kn n k

1n na a

71. (A) Let d be common difference

log 1y x d 1 dx y

log 1 2z y d 1 2dy z

1 3

15 log 1 315x

dz d z x

1 2 11 d ddx y z

1 1 2 1 3 /15d d dx

1 1 2 1 3 15d d d 3 26 11 6 16 0d d d

22 6 8 0d d d 2d

26 8d d complex Roots 1 1 1 4,dx y y y z

13 3x z z

72. (B) We know that

1 + 3 + 5 + …. + 22 1k k Given equation written as

2 2 21 1 12 2 2

p q r

There for

1, 1, 1p q r from a Pythagorean triplet 6, 1 7p p

We way take 1 8, 1 6, 1 10p q r

7 5 9 21p q r 73. (A) 74. (A)

2

111....1 22......2n times n times

21 22 1 10 19 9

n n

21 10 210 19

n n

2 21 10 1 33.....3

9n

n times



75. (C) Let odd one were number | 1na

, to the given sequence. 21 0a 2 22 1 12 1a a a 2 23 2 22 1a a a

2 21 12 1n n na a a

2 2

1 2 1n n na a a

Adding we get 1 2 2

1 1 12

n n n

l l l

a i a i ai n

22 2

11 1 1

2n n n

i i i n

l l l

a i a a a

21

12

n

i i

l

a n a n

1/ 2

n

i

l

a n

1 21......a

2na a

76. (A) 77. (C) 78. (D) 79. (A) 80. (D) 81. (C) 82. (C) 83. (D) 84.

(A) Since is fifth root of unity, we have 2 3 41 0

Given expression is :

3 3 3 34 3 22 2 2 2

12 9 6 38

2 4 9 38 [Since 5 1 ]

8 1 8

85. (D) 91 2 81 1 ..........z z z z z

9

1 2 81 ..........1

zz z z

z

9 8

1

1log log1e e k

k

zz

z

Put z = i

98

1

1log log1e k e

k

ii

i

1log log 1 01e e

i

i



86. (B) 13 12sin 5cos 13x x 2& 2 8 21 13y y

12sin 5cos 13x x

2& 2 8 21 13y y

13y

87. (B) Change to sin & cos, to get cos4 5sin2 3x x

21 2sin 2 5sin2 3x x

22sin 2 5sin2 2 0x x

sin2 2x reject

1&sin22

x

Since 3 3, , 2 ,4 2 2

x x

88. (A) Let 12 5 60

1cos5 cos602

2cos5 1 0 Hence the polynomial equation in cos , would be a degree 5 equation. 89. (B) Use property

2 2 2 22 21 2 1 2 1 2az bz az bz a b z z

90. (A) cot2

i ie e

log 2 3log 2 3

cos 22

1 2 32 3

22 3 2 3 2

2

ei ii

e ei

Documents

PART TEST # 01 - Welcome to JEE Shikhar All India JEE ... : jeeshikhar.catalyser.in JEE TEST SERIES PART TEST # 01 PAPER – 1 (ADVANCE) SOLUTIONS SECTION – 1 : (Only One Option