T Enthusiast Course ALL Test Paper Solution

7/25/2019 T Enthusiast Course ALL Test Paper Solution

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PART-1 : PHYSICS SOLUTION

PAPER-1

SECTIONI

1. Ans. (A)

2. Ans. (B or C)

Sol. F E4 A, E

1, E

2, E

3

A .

3. Ans. (B)

4. Ans. (A)

Sol. B 2

l 60

\E

=2

2 32 30

l = p

5. Ans. (A)

Sol. .

E

= 0 .

6. Ans. (D)Sol. N , .

++

++++

++

++

++

7. Ans. (D)

Sol. f

=

0 0

6l=

e e

I

=0 0

6 3

8 4

l l=

e e

8. Ans. (A)

Sol.3 2

402 3 2 / 3

l= l = = =

ll

l l

20H

2 2

l = = = l

l

Path to Success

ALLENCAREER INSTITUTEKOTA (RAJASTHAN)

TM

CLASSROOM CONTACT PROGRAMME(ACADEMIC SESSION 2014-2015)

JEE (Advanced) # 01

Date : 04 - 05 - 2014TARGET : JEE (Main + Advanced) 2015

ENTHUSIAST COURSE (Phase-I)

HS-1/11

PAPER CODE 0 1 C T 2 1 4 0 0 1


2/132

Kota/01CT214001HS-2/11

Target : JEE (Main + Advanced) 2015/04-05-2014/Paper-1

9. Ans. (A,B,C)

Sol. .

10. Ans. (A,C,D)

11. Ans. (B, D)

Sol. A . A E

.

12. Ans. (A, D)

Sol. B

F F F+ =r r r

\3

F

3

pr=

13. Ans. (C)

14. Ans. (B)15. Ans. (B)

16. Ans. (D)

Sol.0

E

3

a=

\ =2

0

3

a= w

\ = 03

2

p

a

SECTIONIV

1. Ans. 3

Sol.3 7

3 7

3 7

A

B

0

1

2

E s

e

= = N/C \ . B A

V V E e Er q- = - =r r

2. Ans. 5

Sol. A 0 033

34

so un d m ed ium

sound medium source

v vf f f

v v v

- = = - +

B 0 0

35

34

sound mediu m

sound medium source

v vf f f

v v v

+ = = + -

\B = 1 2 035 33

534

f f f- - = =


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Enthusiast Course/Phase-I/04-05-2014/Paper-1

HS-3/11Kota/01CT214001

3. Ans. 6

Sol. /

2L

= , L . A

, L

,

. (..,)

=

1 1 1

2L 2 2

=

H D D < < , 1

2

D D =

I , = 100 H D = 3 H. H, 2 3

2 6%

100

D D = =

4. Ans. 5

Sol. A () =F 2 1

2 2 2 2= =

() =2 21 1 1 2 1

2 2 2= =

+x1

x2

1+ 2= 8 & =1 2

2

-

5. Ans. 2

Sol. A

=E

A

0.8 102 =q2

2

0.8cm

)45dE

45

F

=

q2 2

2

- q=

q

3

2

8 10 2 2

q= 45-

=38 10

4

2 103 =


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PART-3 : MATHEMATICS SOLUTION

SECTION-I

1. Ans. (C)

() = ( )21

2

2 1 +

+ -

2

2 1 01

02

1 1

2

+ - >

+ >

+

1

,2

2. Ans. (B)

Q

, 1 1 12 . 1

2 2 2

= +

2

1 1 0

2

- =

1 1

2

=

3. Ans. (D)

()

A, () = ( )1 2 2

- p-

\() = ( )1 2 2- - p-

PART2 : CHEMISTRY SOLUTION

SECTION-I

1. Ans (D)

= 2 (10) = 20 . = 2000

D = 1 5/2 (300) = 6250 = D = 8250

2. Ans. (B)

3. Ans. (A)

Sol. = 109, = 103

4. Ans. (D)

5. Ans. (D)

6. Ans. (A)

7. Ans. (C)

8. Ans. (B)9. Ans. (B)

10. Ans. (C, D)

11. Ans. (A,B,D)

12. Ans. (A,B, D)

13. Ans. (B)

14 Ans. (D)

15. Ans. (C)

16. Ans. (D)

SECTION-IV

1. Ans. 2

2. Ans. (1.4) [OMR Ans. 5]

3. Ans. 2

4. Ans. 05. Ans. 2

1 2 2

- p= - y=5 y=5

x=3x= 3 x

(0,4)

graph of y=(x)

(even function)

1 2 2 2

-p p = - -

1 2 ()2

-p= - = -

() .H, (), (), ()

, B () .

4. Ans. (C)

0

()

2

0

1 2 .(1 )(1 2)...(1 2 )2 = + + +

( 1)

1 1. .(2 2 2 ....2)

2 2+

= 1442443

1 1

2

22

2

+-= =

5. Ans. (C)

G 3() () = l

. ..

3(). + () =1

.

4p= ,


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HS-5/11Kota/01CT214001

3 1 1 1 4

2 2 2 2

+ = p

1

2

=2

p

6. Ans. (D)

5 3 0

2 .

12 2

12

+ - = -

+ +

5 3 0 1

22( ) 12

-= -

+ +

3

3

5 3 0

.

1

12

- = -

+ +

3

5 3 0

1 1

6 12- = -

+ +

0 2

3

1 1

C 2

=

+ +

\ N C = 0, = 2, .

7. Ans. (C)

l(+ ) =

2+

2....(1) = 0

l(1 + (0)) =

2

(0) ....(2)Q l(1 + ) =

2

( )

= 0

\ (0) = 0 (2)N . (1) .. .

1 . 2 2

+ = + + = = 0

1(1 + (0)) = 0 (0) = 18. Ans. (B)

2 3

2 0

......

2 3

- + - -

2 3

2 0

(1 ) ....

2 3

- - + -=

C 1 = 0 .. = 1

1

2= -l

\ 12

+ =l

9. Ans. (A,D)

L

3

= ,

1 2() 3

3 3

= + + + + =

\ [ ]

() 3. 3

= = .

A, () + () = 2

2 = 2 = 1 1 < 2, .

A, () + () = +

.

10. Ans. (A,B,D)

C () = 1 " I, () I.\ () .

11. Ans. (A,B,D)

1 0 2()

( 1) 2 3

-


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Kota/01CT214001HS-6/11


13. Ans. (D)

() = (2)

+

2

2

22

2

1 0 1

1

11 1


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PAPER-2

SECTIONI

1. Ans. (A)

2. Ans. (D)

Sol.

3. Ans. (A)

Sol.

( )3/ 2

2 2

KE

=

+

rr

4. Ans. (B)

Sol.

5. Ans. (B,C)

Sol. F ()

=

2

0

2 2

0 0

,

3

, 0

2 6

r

r r -


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Kota/01CT214002HS-8/11


SECTION-I

1. Ans. (C)

2. Ans. (D)

3. Ans. (C)

4. Ans. (C)

5. Ans. (B, C)6. Ans. (A,B,C)

7. Ans. (B, D)

8. Ans. (A,B,C,D or A,B,C or D)

9. Ans. (A,B,D)

10. Ans. (A,C,D)

11. Ans. (D)

13. Ans. (C)

Sol. 01

v Em

SECTIONII

1. Ans.A-P,R,S,T; B-P,S,T; C-P,R,S,T;

D-P,S,T

SECTIONIII

1. Ans. 003

Sol. E0=

( )2 4

/ 2

l l=

E >> l.

E =( )

2 2

l=

l

02

E E

4

=

l

2. Ans. 100

Sol. 6 6

7 10 10 7T N N-= =

( )

15 7 8B

F N\ = - =

\ = 2 \ A = 100 2

\2 = 100 2 = 100

3. Ans. 008

Sol. =1

22

(82 02) 109 5000 =1

2

1032

64 106= 2

4. Ans. 002

Sol.( )

1

2 1

4

+ l=l ...()

( )2

2 1

4

- l=l ... ()

L l1 l

2=

2= ...() p2 =

2

2

p=

12. Ans. (B)

13. Ans. (C)

SECTION-II

1. Ans. (A) (S); (B) (R,T); (C) (Q,R,T);

(D) (Q,T)

SECTION-III

1. Ans (0.2) [OMR Ans. 200]

2. Ans. (1020) [OMR Ans. 102]

3. Ans. 003

4. Ans. 004



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7. Ans. (B,D)

2

2

2 ( 1)()

2 ( 1)

p - -=

+ -

\

0 1

( 1)() 1

( 1)

1 1

p < -

- =

\ 1 () 1

=

() = 1.

8. Ans. (A,C,D)

0

( ) () ()

+ -=

0

() () 3( ) 1 ()

+ + + - -=

0

() 1 3( )

-= + +

= (0) + 32 0

(0 ) (0)( (0)) + - =

Q

() = 1 + 32

() = + 3+ CQ (0) = 1 C = 1\ () = 3+ + 1Q () 0 " \ -

9. Ans. (A,D)

F

.\ D = 0 ( 1)2 4( 1) = 0 = 1 = 5

10. Ans. (C,D)

.

G () :

1

p/4-p/4

C "

- , 0,4 4

p p =

Paragraph for Question 11 to 13

() =()

()

Q () = 1 & 1 .

() .

\ () =3 2( 1)( 2 2 2)

( 1)( 1)

+ + - +- +

.

\ 1

5 5 ()

2 4-

-= = - = 2

\ () = 2(2 1).11. Ans. (B)

() : 2, )12. Ans. (C)

D3 2

1

4 9 13

2(2) 4=

+= =

13. Ans. (B)

( ) 0( ) () 0

- >- =


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HS-11/11Kota/01CT214002

(D) G

22

p=

2

2

0 1 3

2

4 5

1 0 1

0 1

1 1 3()

0 3

1 3 5

0 5


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PAPER-1

SECTION-I

1. Ans. (D)

2. Ans. (D)

Sol. On two sides of wire force is equal and opposite. So, net force on wire is zero.

3. Ans. (D)

Sol. Let a charge flows into earth( )

2

QK a

KQ K 2Q302R 2R 3R

- - = + + = , a = 0

4. Ans. (C)

Sol. uniform Er

with in cavity

5. Ans. (A)

Sol.2K

Q maR

l=

6. Ans. (B)

Sol. Only half of field lines would pass.

7. Ans. (A,B)

8. Ans. (A,C)

Sol.

M

M

K1

K2

60x

x N

1

N2

30

N2cos30 = N

1...(i) & N

2sin30 = Mg...(ii)

M

60

N2

K x2K x

1

(K1+ K

2)x = N

2cos 30...(iii)

3

Mgx

K

9. Ans. (A, B, C)

Sol. V1+ V2= 2V

V' = 2V V2V" = 2V V1V' + V" = 2V

V < 2V 2V V" < 2V

Path to Success


TM


TEST # 02 (Pattern : JEE Advanced)



HS-1/12

PAPER CODE 0 1 C T 2 1 4 0 0 5


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Kota/01CT214005HS-2/12


V' = 2V 2V < V" 2V < V

0 < V" < V

10. Ans. (A, D)

Sol. 10N would form action reaction pair

11. Ans. (B,C,D)

Sol. At equatorial position direction ofE

ris opposite to direction of p

r

12. Ans. (C)

Sol. E = ( ) ( )0

20

K d1 cos

2 r

ls- q -

l

l

= 25 104 104= 24 104N/C

13. Ans. (A)

Sol. F =( )( )2 0 7

2

K R d25

10 Nr d

-sp l

=

l

l

14. Ans. (B)

Sol. Ft= mg sin q kx cos q; q= 60

at= g sin 60 ( )kx 3g kR

cos60 3 1m 2 4m

= - -

ac= v2/r = 0 released from rest

15. Ans. (B)

Sol. S.E. =

22

2 21 1 1 3 1kR sin kR2 2 2 2

- q - =

= ( )

2 2kR

3 18 -SECTION-IV

1. Ans. 1

Sol. Let charge be present at (x, 0)

9

2 4

9 10 q100

(2 x) 10-

=-

...(i)

3 3tan , sin

4 5q = \ q =Q ...(ii)

q=

q

qand

2

3sin

(3 x) 9q =

- +...(iii)

By (i) (ii) and (iii) x = 1 q = 1 1011C n=1

2. Ans. 8

Sol. Bead will complete the circle if it reaches at B.

3R/2

R / 2

1 1W F.dr km

3R / 2 R / 2

= = - ;21 4k 2W 0 mv v

2 R 3

= - = ;

8kv 8m / s

3R= =


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HS-3/12Kota/01CT214005

3. Ans. 8

4. Ans. 1

Sol.dE

F Pdx

= here0

E2 x

l=

pe; 2

0

dE

dx 2 x

l= -

pe

( )2

0

F q 2a2 x

l \ = - pe ; 2

0

qaF

x

l= -

pe5. Ans. 5

Sol.

2 30 62

40

10 2 10

2 10

-

-

p = w

6. Ans. 2

Sol. P

Q

QV

r=

K2Q 3

R 2

\forP'

Q

1 K2Q 3V'

2 R 2r

=

=3 KQ

2 R

By comparing x = 2


SECTION-I

1. Ans. (D)

2. Ans. (C)

3. Ans. (C)

4. Ans. (C)

5. Ans. (B)

6. Ans. (A)

7. Ans. (A, D)

8. Ans. (A, C)

9. Ans. (A, B, C, D)

10. Ans. (A, B, C, D)

11. Ans. (C, D)

12. Ans. (B)

13. Ans. (D)

14. Ans. (C)

15. Ans. (B)

SECTIION-IV

1. Ans. (44.5) [OMR Ans. 8]

2. Ans. (0) [OMR Ans. 0]

3. Ans. (36) [OMR Ans. 9]

4. Ans. 4

5. Ans. 1

6. Ans. 4


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SECTION-I

1. Ans. (D)

Put x = 3 + t

t 0

9 t sin t 3lim

tcost+ - -

( )t 01 9 t 9 sin t

limcos t tt 9 t 3

+ - -

+ +

1 51 1

6 6

= - = -

2. Ans. (A)

( ) 1x cos cos x2

-p= -

( ) ( ) 7 10 4 10 102 2p p= - p - = -

( ) ( )2

1' x sin x

1 cos x= + -

-'(10) = 1

\(10) + '(10) =7

112

p-

3. Ans. (C)

g"(y) =( )

( )( )3

" x

' x-

Now, if y = 1 x = 0

\g"(1)( )( )

3

"(0)

' 0

-=

Now, '(x) = ex

"(x) = ex

\'(0) = 1 & "(0) = 1\g"(1) = 1

4. Ans. (C)

( )( ) ( )3 3 2d x ' x .3xdx

=

Now, ( )( ) ( ) ( )2

3 3 4 5

2

dx " x .9x ' x .6x

dx= +

( ) ( )6 4 39 x .x 9 x .6x= +\k = 9, a = 4, b = 6, ; p = 6 & c = 3\(a + b + c + p + k) = 28

5. Ans. (C)

( )1sin x

2

1 x e .

1 x

-

=-

\x = 0 y = 1&

( ) ( ) ( ) ( )

1 13

sin x sin x 2 22

1 1' x e . e . 1 x . 2x

21 x

- - - = + - - -

-

= 1 + 0 = 1\Equation of tangent is y 1 = x 0y = x + 1

6. Ans. (A)

Given curve is a homogenous equation

Hence

( )

2

2

1,1

d y0

dx

=

7. Ans. (A,B)

Put x3/2

= t

\3/ 2

1

3 / 23 2

2 dt 2 xI sin c

3 3 aa t

- = = + -

8. Ans. (B,C)

Put 1 x z+ = dx = 2(z1)dz

\

( )8 6 7

2 z 1 1 1

dz 2 cz 6.z 7z

-

= + + -

( ) ( )7 6

2 1c

7 1 x 3 1 x= - +

+ +

\k1= 7 & k2= 69. Ans. (A,C,D)

( ) ( )1 ; x n

x where n I0 ; x n

= p=

p

10. Ans. (B,C)(1 + 3x + 2x2)6= a0+ a1x + a2x

2+ ......+a12x

12

\coefficient of x12= 26

coefficient of x11

=6C5.2

5.3 = 9.2

6

\(B) is correct.Now, a0+ a1+ a2+......+a12= 6

6

a0 a1+ a2......+a12= 0

\a0+ a2+ ....+a12=66

2& a1+ a3+...+a11=

66

2


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HS-5/12Kota/01CT214005

Now,

6

2r 6 6r 0

612

a6 3

a 2.2 2= = =

\(C) is correct.

&

6

6 62r 1r 1

612

a6 3

a 2.2 2

-= = =

11. Ans. (A,B,C)

15C2r>

1

215

C2r+1

( ) ( ) ( ) ( )15! 1 15!

2r ! 15 2r ! 2 2r 1 ! 19 2r !>

- + -

4r + 2 > 15 2r 13

r r 2

6

> >

\r =5,6,7 but will not hold when r = 8Paragraph for Question 12 & 13

P(x) = (x3 x) Q(x) + (ax

2+ bx + c)

where R(x) = ax2+ bx + c

\P(0) = c = 2P(1) = a + b + c = 4 a + b = 2P(1) = a b + c = 2 a b = 0\a = 1 & b = 1\R(x) = x2+ x + 2

12. Ans. (C)

Now, ( ) ( )

2 2

3

x x 2 x x 2dx dx

x x 1 x 1 x x

+ + + +=

- + -

3

2

dx dx2

1x 1x 1

x

= +- -

( ) 21

n x 1 n 1 cx

= - + - +

l l

( ) ( )2

2

x 1 x 1n c

x

- + = +

l

\ ( ) ( ) ( )

2

2

x 1 . x 1 x

x

- +=

13. Ans. (A)

Now, (R(x) x)10

= (2 + x2)10

\Coefficient of x3in (2 + x2)10is 0

Paragraph for Question 14 & 15

(x) = (1).x2+ x '(x) + "(x)

& (x) = x2+ xg'(1) + g"(2)

'(x) = 2x + g'(1) & "(x) = 2\g(x) = (1 + g'(1) + g"(2)). x2+ x(2x + g'(1))+2Now, g'(x) = (1+ g'(1) + g"(2)). 2x + 4x + g'(1)Now, g'(1) = 2 + 2g'(1) + 2g"(2) + 4 + g'(1)

g"(2) + g'(1) = 3Also, g"(x) = (1 + g'(1) + g"(2))2 + 4g"(2) + 2g'(1) + 6 = 0\g'(1) = 3 & g"(2) = 0\g(x) = 2 3x & (x) = x2 3x

14. Ans. (A)

3 2

6 4 5

2

2 32 3x x xdx dx

1 34x x 3x 4x x

--=

- + - +

22 4x 3x 1 c

x+ - +

15. Ans. (C)

Now, h(x) = (sinx)|x(x3)| + sin 5| x | x 7+ -\h(x) is non differentiable at x = 0,3,7\3 points

SECTION IV

1. Ans. 1

( )2

24

2 2

1 cos xsin xdx dxcos x cos x

-= 2 2sec xdx cos xdx 2 dx= + -

1 sin 2xtan x x 2x c

2 2

= + + - +

\ ( )sin 2x 3x

g x4 2

= -

\1

g 12 8 8

p p - -

\1

8 18 8 8

p p + - =

2. Ans. 7

R.H.D of |3 x| at x = 3 is 1.

Now, ( )

3

x 0

xlim 1

a x tan x bx=

+ -


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Kota/01CT214005HS-6/12


3

3x 05

xlim 1

x 2a x x x bx

3 15

=

+ + + -

Now, if limit is existent\b = 1

& 3 1 a 9a

= =

\a 2b = 73. Ans. 9

( ) ( ) sin x sin xx x cos x 1 e dx xe c= + = +Now, (0) = 0 = 0 + c c = 0\(x) = xesinx

Now,

g(x) = ( )cos x cosxe x sin x 1 dx xe c- = - +Now, g(0) = 0 = c c = 0\g(x) = xecosx

Now, (x) + g(x) = 0 esinx= ecosx& x=0tanx = 1 & x = 0\9 solution

4. Ans. 2

( ) ( ) x

e . ' x dx 2xdx= e(x) = x2 + c (x) = ln(x2+ c)

Now, '(x) = 2

1

x c+ .2x

( )

2' 1 1

1 c= =

+c = 1(x) = ln(x2+ 1)

Now, (x) = '(x) ln(x2+ 1) = 22x

x 1+

ln(x +1)2

Hence 2 solutions.5. Ans. 5

(0, /2)p

1 p 2 p 3p 4pp-2 p1

(x) is non derivable at x =1,0,p1 or 2p,3p6. Ans. 9

n10 10r

nr 1 r 1r 1

C n r 1r. r.

C r= =-

- +=

\ ( )10

r 1

n r 1 5n=

- + == (n + 1)10 55 = 5n

5n = 45 n = 9


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1. Ans. (1)

Sol.kq

10R

=

So, For symmetrical point P.

P

kqV

2 2R=

P2RP 2.5 volt=

2. Ans. (1)

Sol. The potential at centre of sphere in which qcharge is uniformly distributed throughout thevolume is -

c0

1 3qV

4 2R=

pe

By smmetry the potential at centre due to halfsphere will be half of the complete sphere.

\ c 0 0

1 3q / 2 1 3Q q

Q4 2R 4 2R 2

= = = pe pe Q

3. Ans. (2)

Sol. xv

E 4xx

= - = -

; y

vE 3

y

= - = +

x y E E i E j 4xi 3j 3jN / C= + = - + =

r

4. Ans. (2)

5. Ans. (2)

Sol. Electric field at each point on the surface of ring

due to dipole is 3kpE R=

in direction opposite to the dipole moment.(figure below)

R

E

p

Hence net force on ring is3

kpQF QE

R= =

SOLUTION

Path to Success


TM


TEST # 02 (Pattern : JEE-Main)



PAPER CODE 0 1 C T 2 1 4 0 0 6

PAPER-2

6. Ans. (4)

7. Ans. (1)

Sol. Velocity of A w.r.t. plank = V

Velocity of B w.r.t. plank = 3V.

L LT

3V 4V= =

+

Their relative displacements w.r.t. plank are

3L L&4 4

.

8. Ans. (2)

Sol.

- ...(i)

..(ii)

Solving equation (i) & (ii) h = 4R9. Ans. (3)

Sol.

9 6kq 9 10 9 10V

CB 27

- = =

81 10003000Volt

27

= =

10. Ans. (2)

Sol.KQe KQe

R 9R

+- 2

KQe KQe 1mV

R 9R 2

-= + +

22KQe 2KQe 1 mV9R R 2

- + =

2

0

16 eQ 1. mV

4 9R 2=

pe

0

8eQV

9 mR =

pe


19/132

Kota/01CT214006HS-8/12


11. Ans. (4)

Sol. x y E E i E j= +

r, DV = E

xDx E

yDy

for A and B

16 4 = Ex(2 2) E

y(2 2)

Ex= 3V/m

for B and C

12 16 = Ex(2 (2)} E

y(4 2)

Ey= 4V/m.

( ) E 3i 4 j V / m\ = -r

12. Ans. (1)

Sol. Work done by field

W = DU = UAUB= (e)VA(e)VB

1.28 10

18

= 1.6 10

19

(VBVA)

VBVA= 8 volt VEVD= VAVB= 8 volt

13. Ans. (2)

14. Ans. (2)

Sol. q =2 2

0 00

rE.ds 4 r 4 r 6

a = p = p a =

15. Ans. (1)

Sol.

( )

( ) ( )

9 99 10 1 2 103V.

3 3 3

- + =

16. Ans. (4)

Sol.45452p 2p

= 4p

2p

17. Ans. (1)

Sol. Direction of electric field is perpendicular tosurface of circular base, so work done byexternal agent from B to C is zero.

18. Ans. (1)

Sol. For r < a, E r For b < r < c, E = 0

For a < r < b,2

1E

r For 2

1c r, E

r<

19. Ans. (3)

Sol. Vball = 0

KX Q0

r R

K+ =

Qrx

R = -

Potential difference Vs Vb=KQ Kx

R R+

because potential difference depend only on

charge on inner surface after electrostatic

condition is reached after grounding.

s b KQ rV V 1R R

- = -

20. Ans. (2)

Sol.2K

Er

l=

20

1

RV 2k n

R

= l

l

21. Ans. (3)

Sol.1 1 1

KQ ......r 2r 4r

= + + +

KQ 11r 12

=-

= 2KQr

22. Ans. (4)

Sol. Flux =inside

0

q

23. Ans. (1)

24. Ans. (3)

25. Ans. (2)

26. Ans. (4)

Sol. dV E.dr= -

0

V 2

0V 0

dV dZ2

s= -

V V0= 0

Z

2

s

V = V00

Z

2

s

27. Ans. (4)

Sol.dV

Edx

= -

28. Ans. (1)

Sol. Total charge on ring Q = l = l p l

Then VP=


20/132


HS-9/12Kota/01CT214006

29. Ans. (4)

Sol. V0 VA= 400 1

2 2 102= 4 2

V0 VB= 400 1

2 3 102= 6 2

VA VB= 2 2

30. Ans. (1)Sol. W = q DV = q0(VQ VP)

= q0(VP VQ) = q02Klln2r

r

= 2Klq0ln1

2

31. Ans. (3)

32. Ans. (3)

33. Ans. (4)

34. Ans. (3)35. Ans. (2)

36. Ans. (4)

37. Ans. (3)

38. Ans. (2)

39. Ans. (1)

40. Ans. (2)

41. Ans. (4)

42. Ans. (4)

43. Ans. (4)

44. Ans. (3)

45. Ans. (2)

46. Ans. (1)

47. Ans. (3)

48. Ans. (1)

POCl3+ 3H

2O H

3PO

4so oxidation state of

P = +5

49. Ans. (1)

50. Ans. (1)

51. Ans. (3)

52. Ans. (3)

53. Ans. (4)

54. Ans. (4)

55. Ans. (2)

56. Ans. (4)

57. Ans. (3)

58. Ans. (4)

59. Ans. (4)

60. Ans. (3)

61. Ans. (2)

3 3

3(x 2) 1dx dx

(x 2) (x 2)

+-

+ +

2 2

3 1 (6x 11)C

x 2 2(x 2) 2(x 2)

- += + = - +

+ + +

62. Ans. (3)

2 2

(2x 2) dx2

x 2x 2 x 2x 2

+-

+ + + +

= ln(x2+ 2x + 2)2

dx2

(x 1) 1-

+ += ln(x2+ 2x + 2) 2tan1(x + 1) + C

63. Ans. (2)(x) is continuous

x 0 x 0lim (x) lim (x) (0)

- + = = 0 = p

(x) is differentiable

x 0 x 0lim '(x) lim '(x)

+ - = k = 2

64. Ans. (3)

33 x 0x 0

5(sinx x)1 5sinx 1 5limlim 1 .1 5x (1 5x).xx 6e e e

-+ - - + + = =

65. Ans. (2)

At t = 1, x = 3t2 2t dx

6t 2dt

= -

y = t3 12t 2dy 3t 12

dt= -

\2dy 3t 12

dx 6t 2

-=

-

dy

dxat x = 1 is

9

4-


21/132

Kota/01CT214006HS-10/12


66. Ans. (4)

1 1

2

tan x.cot xdx

1 x

- -

+put tan1x = t

2

1dt dt

1 x=

+

32 tt. t dt t C

2 4 3

p p - = - +

1 31 2 (tan x)(tan x) C

4 3

--p= - +

67. Ans. (1)

xe (cos(sin x) cosx sin(sin x))dx-use x xe ((x) '(x))dx e (x) C+ = += excos(sinx) + C\ (x) = excos(sinx)(p) = ep

68. Ans. (1)

1y sin 3x

4=

dy 3cos3x

dx 4=

x3

dy 3

dx 4p= = -

69. Ans. (3)

2

1[x 1], 1 x

2(x)

1x 1, x 3

2

- -

= + <

\ (x) is discontinuous at x = 0,1

2

Sum of values =1

2

70. Ans. (4)

put log (x) = t

1. '(x)dx dt

(x)=

2 2t log (x)tdt C C

2 2= + = +

71. Ans. (3)

1/ 2

2 1/ 2

3 2

11

(x x) xdx dx

x x

- - =

put1

1 t

x

- = 21

dx dtx

=

3 / 2

3 / 21/ 2

x 12

t xt dt C C

3 32

- = + = +

72. Ans. (2)

xx e [cos 2x 2 sin 2x]e cos2xdx C

5

+= +

\ (x) = cos 2x 2 sin 2x

5

+

\ range is1 1

,5 5

-

73. Ans. (2)

2 2

2 2x 1 x 1

x( x 1) (x x 1)lim (x) lim

( x 1) ( x 1) - + +

=- +

=9

4

(x) is continuous k =9

4

74. Ans. (1)

3x yy e +=

\dy

dx=

3x y 2 dye 3xdx

+ +

2dy(1 y) 3yxdx

- = 2dy 3x y

dx 1 y=

-75. Ans. (1)

We get only even power of x\ a

1= a

3= a

5= ....... = a

99= 0

\50

2r 1r 1

a 0-=

=76. Ans. (3)

dy

dx=

2

1 1 ' .

x x

-

2

2

1 1sin

x x

- =

3x

dy 3.

dx 3 2 2 3=p

p p= - = -


22/132


HS-11/12Kota/01CT214006

77. Ans. (4)

Middle term has greatest binomial coefficient

\m

40 m m 40m i m i

20i 02

C C . C+ -+

=

= 40 m 40 m

m m20

2

C C+ ++

=

\m

20 m2

+ =

m = 4078. Ans. (4)

Put

2xe

e t =

2x2e xe .e .2xdx dt

=

\2xe1 1dt e C

2 2

= +79. Ans. (4)

'(x) = log(1 + x) x

2x 0

(x)lim

x0

0

form

x 0

'(x)lim

2xL-Hpital

x 0

log(1 x) xlim 0

2x+ -

=

80. Ans. (2)

(x) is non-differentiable

where sinx + cosx = 0

3 7

x ,4 4

p p=

81. Ans. (3)

1 2

tan x cot xI I dx

1 cot x 1 tan x

+ = + - -

2

tan x 1 dxtan x 1 tan x(tan x 1)

= - - -

2tan x tan x 1dx

tanx

+ +=

11 dx

sinxcosx

= +

2 cosec2xdx dx= +

= x n cosec2x cot 2x C+ - +l

82. Ans. (3)

'(x) = ex(cosx sinx)

''(x) = ex(2sinx)

'''(x) = ex(2sinx 2cosx)

'''(x) = 4excosx = 4(x)

\ 8(x) = 16excosx = 16(x)

83. Ans. (1)

2 0 1/2 1 3

\ non differentiable at 5 points.84. Ans. (2)

2

4

x 2dx

x 4

-

+

2

2

2

21

x dx4

xx

-

+

2

2

21

x dx2

x 4x

-

+ -

put2

xx

+

= t

2

dt

t 4-

1 t 2log C

4 t 2- ++

=2

2

1 x 2x 2log C

4 x 2x 2- + ++ +

85. Ans. (3)

I1= 49, I

2= 50, I

3= 50, I

4= 50

\ I1+ I

2+ I

3+ I

4= 199

86. Ans. (4)

(x) = 1 + xnor (x) = 1 xn

QP(3) = 10 (x) = 1 + x

2

2(x)dx (1 x )dx= + 3x

x C3

= + +

g(x) =3

x C3

+ + Q4

g(1) C 03

= =

g(3) =33

3 123

+ =


23/132

Kota/01CT214006HS-12/12


87. Ans. (1)

P =

40 n

40n

1C .

2

-

......(i)

( )n

39 n10 9 9

n 9

12 P C . 2 .

2

- =

......(ii)

\ 210

40

Cn2n40

=39

Cn

235118n

40381 19nn

39n

C2

C-=

\ n = 20 m = 20, q = 1988. Ans. (4)

n

1 1( n) 2x 2 | x | 2x 2 | x |

n nlim

12x 2 | x |

n

- - - - +

- +

2

n n

4x 1( n)

| x |n nlim lim , x 0

1 x2x 2 | x |

n

- - +

=

- +

at x = 0 value is 089. Ans. (2)

Let1

(x 1)(x 2).....(x 10)+ + +

3 101 2 a aa a

.....x 1 x 2 x 3 x 10= + + + +

+ + + +

put x = 1 \9

01

C1a

9! 9!= =

x = 29

12

1 Ca

8!1! 9!

-= = -

M M

x = 10

9

910 C1a

9! 9!= - = -

\10

i 1i

i 1

( 1) a+

=

- = a1 a2+ a3...... a109 9 9 9

0 1 9C C ..... C 2 4

9! 9! 2835

+ + += = =

90. Ans. (3)

(sin 50x) (sin 20x) dx= sin(30x 20x) sin20xdx+

= sin 20xcos30x sin 20xdxsin30xcos20x sin 20xdx+

=3 / 2sin30x (sin20x)

30

1/ 2sin 30x 3. (sin20x) .20cos20xdx

30 2-

(sin30x)(cos2x) sin 20xdx+

\ a = 30, b = 20,

3p

2=


24/132




PAPER-1

Path to Success


TM




HS-1/11

PAPER CODE 0 1 C T 2 1 4 0 0 9

SECTION-I

1. Ans. (C)

Sol. 10

E 2

s=

e

r

1

2

3

4

x

y

x

x

s

2 3

0

E 0, E 2

s= = +

e

r r

4

0

E

2

s= -

e

r

2. Ans. (D)

Sol.

2QJ

t= ( )

221 1 1 1

2

2 2 2 2

J Q2 2 4 32

J Q

t= = =

t

3. Ans. (B)

Sol. B

3v0V0

Vcm

t

4. Ans. (D)

Sol. B

N C2

2

= P ; =

5. Ans. (A)

Sol.

O

1 /2 2

( )2 2 2 20 1 1 1 2 2 2L mu gt m gt ut mugt mugt mugt k= - = - = -

6. Ans. (A, B, C)

Sol.

2M

2R/3R/3

M

2

2

G2M

R= 2M w2

R

3

w= 33GM

R

KM

=1

2M(w 2R/3)2

7. Ans. (B,C)

8. Ans. (A, C, D)

Sol. D

S

9. Ans. (A, B, C, D or C,D)

Sol. C

0

Q =

. F

.

I

.

10. Ans. (A, C, D)

Sol. ll

3l

TEST # 03 (Pattern : JEE-Advanced)


25/132

Kota/01CT214009HS-2/11


2KE 2 ( R)

l =

r

11. Ans. (C,D)

12. Ans. (C)

13. Ans. (D)

Sol. (12 & 13)

L &

w A

L

= L

0 = R 2

MR

2

w

w=2

MR... ()

E

1

22=

1

22+

1

2Iw2

... ()

=M

M 2+

, w= M

2MR M 2+

14. Ans. (A)

Sol. 0= 2 ... (1)

2

00

12

= +

l...(2)

15. Ans. (A)

Sol. V

1

4

=0

2

0

2

p

SECTION-IV

1. Ans. 4

Sol. ( )1/ 2

2 2

0

R 8

s + - e

2. Ans. 5

Sol.

2 F ( ) 2 ( ) (2 )

= + l + + l + + l

l

l

rl l l l l l l

l l l

3. Ans. 3 or 4

Sol. ( ) ( )2 21 1 2 22 2

MR MR5 5

w = w

2 2

1 2

1 2

R R

T T= R2=

1/ 2

221

1

TR

T

4. Ans. 6

Sol.q

R

a a qq

F2F1 qq0

F1= F2= ( )

0

2 2

R+F = 2F

1q

F 0 R

R 2= =

2

2

R2

F0

R=

2

5. Ans. (D)

I = + = + = -11

3 / 2 3/ 2

00

2 21 (1 ) (2 1)

3 3

2(2 2 1)

3= -


27/132

Kota/01CT214009HS-4/11


6. Ans. (A,B,C)

(A) ()=

+

x sin x x [0,1]

x sin x x [ 1, 0)

+ =

- - -() (1) = (1)

() - = 0

(B) ( ) -

+ = - -

=

l2

2

( 1) ( ) 1,0

1 1

0 1,0

() = 0(C) () = + = + ()

0 x I([ x] [ x])

1 x J

= - + - -

/\ D = 1, 0

(D) () R .7. Ans. (A,D)

(A)cosx

0

dxI

1 (sin x)

p

=+

/ 2

cos x cos x

0

1 1dx

1 (sin x) 1 (sin x)

p

-

= + + +

(B)p p

= =+

/ 2 5

5 5

0

I

4

(C)

+

- +

2

2

0

11

1

1

=1

0

1

- -

2 2

p p= + = p

(D) ( )p p

=l l/ 2

0

2 2

8. Ans. (B,C)T () = + ( 0)

1() =x b

a

-

9. Ans . (Bonus)

(A) {

3 2

I II

. = 142432 2tan x tan x

x. dx2 2

= -

21 (x tan x (tan x x)) C2

= - - +

21 (x sec x tan x ) C2

= - +

\ () = 2 & () = \ A

(B) = = 0 & = 1

\ = p, 0, p, 2p\ 4 .

(C) () = ()

(D) = 2 2 + 1 = 1

.

10. Ans. (C,D)

() =

+ = 32 + 3 = 32 32 = .2(3 )

g ' 3g 3 x

=-

I n | g(x) | 3 n | 3 x | C= - - +l lQ () = 1 = 0

3

1g(x)

| x 3 |=

-

11. Ans. (C,D)

= 2 5 7 13

\ N = 3 3 3 3 = 81Paragraph for Question 12 & 13

() = ( ) - -

0 0

( ) .

x x

0 0

(x) (t ) cos tdt cos t dt= - () = ()

'(x)

cosx

(x ) 1

=

-I ... ,

l()1 = + C

N, (0) = 0

= 0 () 1 =

() = 1

.

B (0) = 0 () = 1 .12. Ans. (D)

N () .

A () = 0

= 1 = 0 = 0, p, 2p


28/132


HS-5/11Kota/01CT214009

A () = 0 a & b\ (a) = (b)\ U R () = 0

(a, b)B () -.

\ (D) .13. Ans. (B)

\ R () 1

1 e, 1e

- - \ 2 .

Paragraph for Question 14 & 15

"() = 2

\ () = ( 2)2+ \ () = 2( 2)Q (3) = 24 \ = 12\ () = 12( 2)2 4(Q = 4)\ () = 43 242+ 44 24\ () = 4 83+ 222 24 ((0) = 0)

14. Ans. (D)

( ) ( )

1

0 - -1

3

0

16 48 28= + =15. Ans. (B)

= 2 0

= 2 : = 8

\E : + 8 = 0SECTION-IV

1. Ans. 8I

13

(12 2) .. 2122

A 13 > < 7N = 210+ 28+ 26+ ..... + 20

= 1365

\ N = 1365 = 3 5 7 13\ D 5 8

2. Ans. 7

( ) ( )4 3

1

1 5

-+ = 12 5 = 7

Q .

\ (1) = 5 & (4) = 33. Ans. 2

S = 12C2= 66

T = 13C3 6 = 286 6 = 280

D = 12C2 12 = 54\ T + D + S = 400T D S = 160

\ T D S

2T D S

+ + = - - 4. Ans. 6

( )1

0

2= =

1

2

N, () = xba

\a

0

(x)dx 2ab=

1 a

0 1

(x)dx (x)dx+ = 2

= 6

5. Ans. 2

1

3

1

( ) ()-

-

= - + - - -1

3 3

0

(( ) () ( ) ( )

1

3 3 3 3

0

(x)(x k 3xk(x k) x k 3xk(x k))dx= + + + + - - -

1

3 2

0

(x )(2 x 6xk )dx 64= + = 2.3+ 6.2. = 643= 8 = 2

6. Ans. 1

( )2

1

-

0 1 2

3 4

1 0 1

x dx x dx 1.dx-

= + + 0 1

4 5

1 0

x x1

4 5-

= + + 1 1 1914 5 20

= - + + =

\ 20 19 = 1


29/132




PAPER-2

SECTIONI

1. Ans. (A,B,C,D)

2 . Ans. (A,B,C,D)

Sol.

=0 02 2

3

=

+ +

2 2

1 1

2 2m = =

22 00

1

2 2 4 =

(C)0 0

2 2=

(D) A = 0

3. Ans. (A, B, C)4. Ans. (A, B, C)

5. Ans. (B, C, D)

6. Ans. (B, C)

7. Ans. (A,B,C)

Sol. I

8. Ans. (A,C,D)

9. Ans. (D)

10. Ans. (B)

Sol. K

=2

C

12

2 = ( )

2

0 0 2 1 22 2

-

=

2 2

0 0 1 22 2 4

=

11. Ans. (B)

Sol.

a- .

SECTIONII

1. Ans. (A) (P,Q, T) ; (B) (Q,R,T) ; (C) (Q) ; (D) (Q, T)





PAPER CODE 0 1 C T 2 1 4 0 1 0

Path to Success


TM


30/132


HS-7/11Kota/01CT214010

SECTIONIII

1. Ans. 002

Sol.2

l =

mg

2kl

hq

hF =2

l -

=

2

p

2. Ans. 020

3. Ans. 000

Sol. F & . 0E r E r = ^r rr r . 0E Af= =

rr

4. Ans. 004Sol. F LC

= 1+

2

& = 1

= 2

1

1

2

1

4=

= 1 =

2

4

=

SECTIONIV1. Ans. 3

Sol.

3 / 2

2

1

10500 1

42000 8

= =

2= 3

2. Ans. 7

Sol. 0 322

4 1 1G

3 18

2

pr - - 3. Ans. 6

Sol. =

4. Ans. 5

Sol. 0R

kQ

x

kq=+

= x

qR

.

q

Q

R

x

++

+

dt

dQ=

dt

dx

x

qR2 =

3 310 2 10

2 2

- -

= 0.5 106


31/132

HS-8/11


Kota/01CT214010


SECTION-I

1. Ans.(A, B, C)

2. Ans. (B,C)

3. Ans. (A)

4. Ans. (A, B, D)

5. Ans. (B, C)

6. Ans. (B)

7. Ans. (A,B,C,D)

8. Ans. (A,B,C)

9. Ans. (D)

10. Ans. (D)

11. Ans. (B)

SECTION-II

1. Ans. (A)

(Q,S);(B)

(R,S);(C)

(P,T);

(D)(T)


SECTION-I

1. Ans. (A,C)

( ) . 1

1 1

1

+= =

+ +++

+ 2+ = (1 + ) ...(1) = 0, 2+ 2 = 0 2\ = 0, (1) ,

( )

2 . 1 . 0

+ - + - - + =

= = 0 ( )1

02

=

2. Ans. (A,B,D)

B I.. (C)

A,B,D .

3. Ans. (A,D) & ,

22 3 13 3 4

2 1 2 2 2 9 9

C C . C C . C

= - -

= 231 234 18 = 15

A : B ,

( ) ( ) ( ) ( )3 3 3 6

8,8, 4 , 7, 7, 6 , 6, 6, 8 , 7,8, 5 ,

\ = 15.

SECTION-III

1. Ans. 075

2. Ans. 9 1022

OMR Ans. 922 or 036 or 1983. Ans. 002

4. Ans. 004

SECTION-IV

1. Ans. (2400)

OMR Ans. 6

2. Ans.(16)

OMR Ans. 7

3. Ans. 6

4. Ans. 1

4. Ans. (B,C,D)

G, ( + ) = (). () ...(1)

= = 0 (0) = 1 (Q(0) 0) = (1)

\(1) (). () = (0) = 1

( )( )1

- = ...(2)

B (2), F() .

(B)

2014 0 2014

2013 2013 0

F() F() F()- -

= +

2014 2013 2014

2013 0 0

F() F() F()-

- = 0 2013

2013 0

F() F()-

=

Q

(C) L

2013

2013

F()I ()-

= ...(1)

( )

2013

2013

F()I

-=

- (QF() = F())

( )( )

22013

2

2013

I

1 -=

+ ...(2)

(1) + (2)

2013

2013

2I 1.

-

= I = 2013

(D) F() F() = 0


32/132


HS-9/11Kota/01CT214010

5. Ans.(C,D)

G

( )( )1 1

2

2 2 0 1 0

1 / 1 1

1 21

=

= = = ++ +

l

12 2

2= =l l

6. Ans. (A,B,C,D)

( ) ( )/ 2 1

0

2 1 2 1

p

+

+ - -- =

/ 2

0

22.

p

= / 2

0

22 0

2

p = =

\+1= 1= 2= 3=......../ 2

0

2

p p= =,

( )2 2/ 2 1 2

0

1

p

+

+ -- =

( )/ 22

0

2 1 .

p +=

( )/ 2 1

0

2 1

p

++= =

7. Ans. (A,B,D)

(A) () 0,2

,

.

(B) Q() = ( 1)2+ 2

\ () = (1,2), () (1,3).

(C) L ( ) ( ) 1 5 4

p= + - + p

0,1 () 0,1.A, (0) = 5 (1) = 4

\(0). (1) < 0, I... () = 0 (0,1).

(D) C ( )1

2 3

0

I = + + +

12 3 4

0

I 2 3 4 2 3 4

= + + + = + + +

12 6 4 30

12

+ + += =

\() (0,1).8. Ans. (A,B,D)

P

L1

L2

I L1

L2

L1 L

2.

\

=C

2.C

2=

( )2

1

2

-

A,( )

2

1100,225, 784

2

-=

= 5,6,8

.

Paragraph for Question 9 to 11

2() = (())+ (())",

= 1

\ 2() = () + 5() = 5

9. Ans. (D)

Q = 5 =

2

\1=

2 5l5 = .2

2.l5 = .2 2

5

=l

1

5= .

2

22

52

5 .5

=

l

l

1

22

2 52

5 . 5 .5

= =

l

l

5

2

2 25 . 5

= l

2 5

2

=

l


33/132

HS-10/11


Kota/01CT214010

10. Ans. (D)

() = (5) () = (5).5l5.

Q 5 (0,)

(0,), () -

5= (2 + 1) 2

p( I)

. C () 1,1.

11. Ans. (A)

2

1 2

() 5 5

() (1 ) 5 5 5 5-= =

+ - + +

( )21

. 5 5 C2 5

= + +ll

( )( )

1. 5 2 C

2 5= + +l

l

SECTION II

1. Ans. (A)(T); (B)(P); (C)(Bonus);

(D)

(R)

(A) G = 5 ...(1)

= - =

, + + = 0

= - .

() . = ()

(B) 2 = 3

2 + ...(1)

1 3

2 6

= = =

2= 4 ...(2)

2

2

2 4 = = =

\1

2= 1

2

2

61 6

= - = -

(2) 4 = 6

= 0 \ = 0 = 0

1

2 .

.

(C) () = (

2

1).

2

> 0 " 0,p (2 1) > 0 > 1 < 1.

\ = 2,3( = 1 )

(D) (x) = min.{ 2 x2,2 ,

2+ 1

\\\\\\\\\\\\\\\\\\\\\\\

\\\\\\\\\\\\\\\

\

\\\\\\\\\\\\\

\

y = 2x2 (0,1)

\\\\\\\\

(2,0)( 2,0)

R(1,1)

y=1+x2(0,2)

P

y=2x

() - ,, ..

3 .

SECTION-III

1. Ans. 078

() = 9 2 l> 0 ,

6 3

p p "

29 ,6 3

p p l "

29.

6

pl .. l< 12

\ l

12 1378

2

= = .

2. Ans. 010D .

( ) ( )

( )

( )

4 24

2 2 22 2 2

2 2 1 2

1 1 . 1

+ - ++= - =

+ + +

, = = 1

\G 0

10 1

+ -=

2 0

10 1

- = +

= 10 + 0 = 10.

3. Ans. 035

(0,0). I

, , . , . , (0,0) (3,4) . =

7C

3= 35


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HS-11/11Kota/01CT214010

4. Ans. 009

( ) ( )

( )

1

1 1

+ +=

+ +&

( ) ( )( ) ( )

1 2 , 0,

,0 ,2

1 2 , ,0, ,2

2

- p p p

= -p p p + = - p -p p p

() = 2p,p,p,2p () = 0 = 0\ = 4, = 5 + = 9

SECTION-IV

1. Ans. 7

Q ( ) ( )( ) ( )4 4

4

0II I0 0

A . .= = - 4

2

0

40 . 9 = - + L 9 + 2= 2 = 5

2

3

40 = -

( )3 31 98 22

40 5 3 403 3 3

= - - = - =

\A = 72. Ans. 4

Q3 24 4 . 10

3 = p = p =

2 10

4 =

p...(1)

A, = 4p2

58 . 5 8 .

8 = p = p =

p...(2)

(1) (2)

2

10 5 4

4 8 = =

p p3. Ans. 6

( ) ( ) 1 1

1 1 1 1

5 1 5 4 3 5 4 5 1

= =

= - - - - -

1 1

5 5 5 2

1 0 0

1

3

- -

=

= -

1 1

5 5 5 2

1 10 0

1

3

- -

= =

= -

( ) ( )1 1

5 10 3 8 13

0 0

11 .... 10 ....

3

= + + + + - + + +

1 1 13 3

5 5 5

0 0 0

1 1 1 1

3 1 1 3 1

-= - = - - -

1 2 2

4 3 2 4 3 2

0 0

1 1 1 1

3 1 6 1

+ + + += =

+ + + + + + + +

Q1 2 2

4 3 2 4 3 3

0 1

1 1

1 1

+ + + +=

+ + + + + + + +

1

=

\ = 64. Ans. 8

C 1=2,

2= 3 (00,01,10)

.

I

0 (1)

1

.

1 0

\1 0 (2)

2.

=

1+

2...(1)

1

2

1

-

-

- = A,

2

5 6 5 6 55 5

4 4

- -= =

,

5=

4+

3= 2

3+

2= 2(

2+

1) +

2

= 32+ 2

1= 3 3 + 2 2 = 13

\G = 13 5 1 = 8


35/132



1. Ans. (2)

11.8

6 83 2

6 8

+ +

+

8I I

6 8

= + 2. Ans. (1)

/.

3. Ans. (3)

4

( )

4

5 4

427 27310

10

+= = 7000 K

4. Ans. (4)

80 90 40- q - q q -+ =

l l l

5. Ans. (1)

L= 24 10 W

=

12 102

= W

6. Ans. (3)

(100 + 10)(q 10) = 330

(90 q)

4q= 280 q= 70C7. Ans. (4)

BF

= DE

BF DE2

Dq Dq = D D l l

HBF

=DE

1H

2

BF

= 2DE

.

8. Ans. (4)

0

2

1

2

=0

GM

SOLUTION

Path to Success


TM


TEST # 02 (Pattern : JEE-Main)



PAPER CODE 0 1 C T 2 1 4 0 1 6

9. Ans. (3)

= I

45 = 3( + 1 + 8)

= 6W10. Ans. (2)

(e I).

11. Ans. (3)

I=

20

11000

20I

100010000

1000

=+

+

II I

1000 2

= = +

12. Ans. (1)

( )KA 30H

q -=

l

( )480 4 10 302

0.8

- q -=

13. Ans. (2)

( )10

I 2A C1 1 3

= =+ +

1I 2 =

= 4

14. Ans. (3)

F

As =

F

= 3 F

F

= F

(2 )

= 0.5 15. Ans. (1, 3)

B D .

I A B .

16. Ans. (3)


36/132

Kota/01CT214016HS-2/5

Target : JEE (Main + Advanced) 2015/29-06-2014

17. Ans. (1)

C

4W& 6W

=( )

230

3

= 300

I =30

10A3

=

18. Ans. (4)

19. Ans. (4)

Dl= laDq0.06 = 60(12 106) (q 0)

q=1000

12= 83.33 C

20. Ans. (1)

1

1 2

31

h = = +

()

( )

2

1

1

/10

h =+

, 21

6h =

21. Ans. (4)I = 2

e=

+

=B .

C22. Ans. (3)

23. Ans. (2)

24. Ans. (2)

I F =1 2

2

G

F =( ) ( )

( )1 2

2

G 2 / 2 F

93=

25. Ans. (3)

26. Ans. (3)

.D. = 2

1= E ( q

1 q

2)

27. Ans. (1)

=C

Q

= 800 10 106

= 66

1050

1010800

A.

28. Ans. (4)

.D. = (2

1)

2= 0

1=

0

1

4

-

p 29. Ans. (1)2

2

0

1 1

4 2

= p

9 10912

36 10

2

-

= ( ) 21

0.1 2

30. Ans. (3)

= 201

E 2

= ( ) ( )2 2

0

13

2

=

0

=

= = 50

3

2

31. Ans. (1)

32. Ans. (3)

33. Ans. (3)

34. Ans. (3)

35. Ans. (4)

36. Ans. (3)

37. Ans. (1)

38. Ans. (1)39. Ans. (2)

40. Ans. (4)

41. Ans. (2)

42. Ans. (3)

43. Ans. (2)

44. Ans. (3)

45. Ans. (2)

46. Ans. (1, 2)

47. Ans. (3)

48. Ans. (2)49. Ans. (1)

50. Ans. (1)

51. Ans. (3)

52. Ans. (4)

53. Ans. (1)

54. Ans. (2)

55. Ans. (4)

56. Ans. (3)

57. Ans. (1)

58. Ans. (1)

59. Ans. (2)

60. Ans. (2)


37/132

Enthusiast Course/Phase-I/29-06-2014

HS-3/5Kota/01CT214016

61. Ans. (4)

( )

3

22 3 0

3 3

1 3 .

- - -

( ) 2 22 2 03 1 3

1 3 1 3 = - ---

( )

2 2 4

2 2 2 0

3 3 1

1 3 .

- + = - -

( ) ( )

( )2 2 2 03

9 11 3 . .

- += + -

-

13 2 8 6

3

= - + =

62. Ans. (3)

( )3

0

I 6 = - = 62q/ 4

2 2

0

6 .6

p

= q q 12 qqq/ 4 / 2

2 2

0 0

918 2 9

4

p p p= q q = q q =

63. Ans. (3)

() = 3( 2)2/3 ( 2)

( )( )

1

3

2 1 0

2

= - =-

( 2) = 8 = 10\

( ) 1/3 2 / 3 0 2 10 20

3.4 2 0 4 3.18 18+ -

0

64. Ans. (2)

( )( )1 1 2 3

0 0

1 1 1 ....

2 3

+ = - + -

l

2 2 2

1 1 11 ....

2 3 4= - + - +

, 2 2 2 2 2 21 1 1 1 1 1

1 ... 1 ...2 3 4 2 3 4

- + - + = + + + +

2 2 2

1 1 12 ...

2 4 6

- + + + 2 2 2 2

22 3.

6 2 6 12 36p p p= - = =

\ = 36

65. Ans. (1)

() ( - a) () ( a) ( 1) ( & > 1)

66. Ans. (2)

( )2

3 3

1 2 2 A

-= - =

\ ( )2

3

2

-=

& ( ) ( )3 2 2

6

.2 2 .3

- -=

( )2 2 24 4

2 3 2 6

- - -

= =

( ) ( )4

6 6

+ -= -

\L 6=

& ( )

4 6

96 6= = \( + )

= 15

67. Ans. (4)

( )( )

1/

1

0

0 0

1 3 1 3

1

++

=

1.3

3

0

= =

68. Ans. (2)

( ) ( )

( )

( ) ( )( )

2

0

() ()

.

. . () . ()

=-

( )( ) ( ) ( )

2

. =

-

( )

( )

( )

( )

2

1 1

2014

= =

-


38/132

Kota/01CT214016HS-4/5

Target : JEE (Main + Advanced) 2015/29-06-2014

69. Ans. (1)

P(1,4)Q(4,5)

R(m,m)

= . L

(5,4) , = 4,

& = .

\ = 470. Ans. (4)

62 + 42 (3 + 4) (2 3)\4 = 12

= 371. Ans. (3)

2= I

3= 2= I + = 2 I

4= 22 = 2(I ) = 2I 3

5= 2 32= 2 3(I )

= 5 3I

6= 52 3 = 5(I ) 3 = 5I 8

\ = 672. Ans. (3)

1 1

1 2

1 2

+ -D = + +

- +

2

2

1&

3

3

1

1 1

1 1 3

1 1 1

+ -

= -

-

= (4) ( + 1) (4) + ( 1) 0 = 04( 1) = 01

2

= -

73. Ans. (4)/ 2 / 22

2

/ 2 0

I

1 5 4

p p

-p

p= = =

+ 74. Ans. (4)

L-H :

12 8

5 8 2

1 5 8 2 1 26 2 7 6 2 6

+ + = =+ +

75. Ans. (1)

( ) ( )( )5

4

1I 3 3

3= + =

( ) ( )( )15

12

17

12

+

76. Ans. (3)2 4 8 16

= p = p =

2

=

p

,34

3= p

2 2 24 4 8

= p = p =

p77. Ans. (3)2

1

I

=

l; l = =

2

2

1

I

= =

78. Ans. (1)

L () = A4+ B3+ 2

A ( )2 0 1 =

, ()= 4A3+ 3B2+ 2

= (4A2+ 3B + 2)

Q = 1 & = 2 () = 0

\2

2 4A 14A

= =

&3B

3 B 14A

-= = -

\() = 4 3 21 4 - +

\(2) = 48 + 4 = 079. Ans. (1)

C3C

3 C

1

1 1 0

1 1 0 0

1 1

+ q =

q (3),

q(1 + q 1) = 0

q = 0 3

,2 2

p pq =


39/132

Enthusiast Course/Phase-I/29-06-2014

HS-5/5Kota/01CT214016

80. Ans. (4)

3 3

q=

- q

\ 3 3

q=

q\

( q) ( )3 3 3 3

q= - q

q

\ 3 3

1 27

- = -q q

\3 3

126 26

- =q q

\A D 1 26 262 3 3 q= q

1692

3 3= q

\M 4

pq =

81. Ans. (4)

() = 12 9+ 4 + 1;

(x) i s (+)ve for x < 0

> 1 () = 9(3 1) + (31) +1

() > 0A 0 < < 1 () = 12+ 4(1 5) +1

\() > 0.82. Ans. (2)

L C (,)

\

11

4 0 1 182

0 12 1

=

12 4 + 48 = 36

3 + 12 = m 9(3 + 12)2= 81

83. Ans. (3)

F 1 ,2 2. 4

0 4

+ = = -

2 ,2

2

163 16

3= =

Q

2 3

4 16 641 1

3 3

- = - =

64 4

1 3.16 3

= =

84. Ans. (1)

1

0

1

0

1 2 ... 1

1 2 ... 1

aa a a

b-ab b b

b

+ + + b += = + + + a +

85. Ans. (2)

( )2 1

- +=

Q > 0 2 + 1 > 0 " +

D < 0

1 4 < 0

1

4

.

II A.M & G.M

86. Ans. (3)

L

L &

L

= 83

2

=

\ 2

18 3

+ =

\1 3 3

.8 2 16= - = -87. Ans. (3)

( )2

1

+ =

( )2

-=

88. Ans. (1)

.

0 3 3

= =

= l389. Ans. (3)

(a) = a& 90. Ans. (3)

22 4 1 24. 4

2 4 2=

( )81

2 8

-= +

\1

A16

= -


40/132




PAPER-1

Path to Success


TM




HS-1/7

PAPER CODE 0 1 C T 2 1 4 0 2 2

SECTION-I

1. Ans. (B,C)

2. Ans. (A,C,D)

3. Ans. (A,B)

Sol. L : . ,

: ,

4. Ans. (A, B,C)

Sol. H 2

ratedV

RP

=

2

V

R

5. Ans. (A,B,D)

Sol. A = 0, & = 0.

A = , =2 A;

C= 0.

6. Ans. (A,C,D)

Sol. =( )1 1 2 2 1

1 2

a - a D

+

l

7. Ans. (B,C)

Sol.

2T05R/6 R

Tx T0

( )0 6 2 5

-=

( )0 06 2

11

-=

8. Ans. (A)

Sol. 200 = 1 200 =

2 100

1= 400 J,

2= 300 J

9. Ans. (D)

Sol. 2 K

K E 22

1E 77 K2

1

= =

10. Ans. (B,C)



41/132

Kota/01CT214022HS-2/7


SECTION-IV

1. Ans. 5

Sol. I 0 0

050

4.45.5

- +

0= 40

\ 0= 0\ 1.1 2

050

1.15.5

- +

0= 2

10 = 2

\ = 52. Ans. 9

Sol.0

1 1 1

4

= + - p

3. Ans. 4

4. Ans. 8

Sol. J :

I2 > (

).

B , 10 A. ,

( )21 1 1I p l ( ) l ,

, I

1

1

.

2=

1/16. , I

2

2

2 2 2I p l ( )F,

I28I

1= 8.0 A.

5. Ans. 2

Sol. 20

E

4 =

p

0

E 2

4

= - - p

,

11 1

0

E 2

4

= - - p

22 1

0

E 2

4

= - - p

,

1

2

2 1

E E

- =

1

2

0.3 = 1

2

2 =


42/132


HS-3/7Kota/01CT214022

6. Ans. 2

Sol. D .q v

rv0

dQ

, 1 & .

0 =

&2 2

0

1 K 10

2 2+ = +

2

2 00

1 K 1

2 2

= +

22

0 2

1 K 1

2

- =

( )2 2 201

K2

- =

( )2 2 2 20 01 1

K 02 2

- - =

( )2 2 4 2

0

2

0

K K

+ += = 2

K = 9

2

0 12= = 1

7. Ans. 3

Sol. C & D , .

( ) ( )1

2I

2

e e= =

+ +

C & D = ( )1I

2

-

(0) =

( )( )

e -

+

140 = 420

- +

1 2

3

- = = +

, C D ,

2I 22 2

3

e e

= = +

3 3 420 3154 4 e = = =


43/132

Kota/01CT214022HS-4/7


C C & D

0 2 2

I I I

= -

+ +

= I2

-

+

315 121

3

=

= 5W, = 10 WF,

D

C

BA

r=5W

420V

R=10Wr=5W

R=10W

5W

, KL & KCL, C & D = 12 A

8. Ans.2

Sol. L & , .

= ( )rx

R xr x

+ -+

,

( ) ( )

0 0A

I r V r I

r x rx R x r x

= =

+ + - + ,

AdI 0dx

=

=R

2

, ( ) ( ) ( )

( )0

A min

12 2V rI

R 4 2 1R r

4

= =+ +

= 2A

9. Ans. 2

Sol. F 1 2

r R R=

= ( ) ( )xR 1 x R - = ( )R 1 x-

( )

R 1

r x 1 x-

-

R

r

1x

2=

R

2r =10. Ans. 8


44/132


HS-5/7Kota/01CT214022


SECTION-I

1. Ans. (A, C)

2. Ans.(A,C)

3. Ans.(B, C)

4. Ans. (B, C)

5. Ans. (C, D)

6. Ans. (C, D)

7. Ans. (B, C, D)

8. Ans. (C, D)

9. Ans. (B,C)

10. Ans. (C, D)

SECTION-IV

1. Ans. 6

2. Ans. 6

3. Ans. 1

1 2 A B C

, 1>>

2

\A 2k C

C = A

0 2

k t[1 e ]--

210 1010[1 e ]

-- = - 0.110[1 e ]-= -

10[1 0.9] 1= - =

4. Ans. 4

5. Ans. 2

6. Ans. 2

7. Ans. 4

8. Ans. 5

9. Ans. 6

10. Ans. 6


SECTION-I

1. Ans. (B,D)

6 2 5 > 0 (1,5)

( ) ( )

2 3 2

0

3 2 2 2= + + = + +() (1,5)

\ = 4, M = 1602. Ans. (D)

. 2

q = =

rr

rr

4 4

22 2

4 4

+ +p p

p-q

= =p p

- -

4 4

22 2

4 4

- -p p

p-q

= = -p p - -

\L .

3. Ans. (B,C)

(+1

) + 2

( + 1)

+1= 0

+1

= 2(+1)

\( )

1 1

1

1

2 4

-- +

=

p= + =

4. Ans. (A,C,D)

( )( )2

1

=

+Q () 1,) () > 0 () () > 1

( )( )

( ) 12 21 1

1

-= ++ +

Q(1) = 1 ( ) 14

p +

5. Ans. (A,C)

1 3 1 2

1 4 2 3 0 0

2 3 1 3 3

- +

- - + = =

+ -

& = 3

6. Ans. (A,C,D)

B I7. Ans. (A,C,D)

R

EP Q

H O2

1

C &

2 : 1 5 8

,3 3

, E

E & E(2,5) G E 2 : 1 (9,2)


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\E ( 1)

2+ ( 2)

2= 80

8. Ans. (B,C)

() 0,1

() 0,1, (0,1) & (0) = (1)

\b 1 1

,3 4

- -

9. Ans. (A,C)

L L

2 0 55 33 5 33

1 1 0 ,4 4

1 2

+ -- m = m =

m10. Ans. (C)

( ) ( )

2 2 2 21

I

1=

+ + l

...(1)

21 1

,

= = -

( ) ( ) ( ) ( )

1

2 2 2 2 2 2 2 21

I

1 1 = = -

+ + + + l l

(1) & (2) 2I = 0SECTION IV

1. Ans. 9

L () = 4+ + 5

() = 43+ = 0 1/3

4

- =

1/ 3 4

3

256 0

4 27

- >


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HS-7/7Kota/01CT214022

5. Ans. 7

1 2 0 1, 2+ + - = = - =

11

22

03 2 3 2AB

0 4 4

+ + = =

- + - +

\ 113

, 8 282

= = =

22= 7

6. Ans. 5

QP

xy=1P(sinA,cosB)Q(tanC,cotC)

(A, B) 2+ 2= 1

(C, C) = 1

\

(C A)2+ (C B)2= ()2

( )2 2

C C 1= + -

( )

2

2 1 3 2 2= - = -7. Ans. 2

1 1

1 0

0 11

2 1 0

++

+

- - =

r r r

11

1 0 0

0 11

2 1 0

- - =

r r r

22

2 1 0

0 11

2 1 0

- - =

r r r

1 0

0 11

2 1 0-

- - =

r r r

\ 0 0 a b

= + g d

r r r

4 2

15 5

2 4 05 5

- =

4

5

25

=

1 1 1 1

....4 16 64 5

a = - + - + = -

1 1 1 2....

2 8 64 5b = - + - + = -

1 1 1 2....2 8 64 5

g = - + - =

1

5d = a = -

8. Ans. 0

( )( )/ 4

3

0

2 1

1 I

2

p + +

+=

l

( )2

/4

1

0

2 1

1 2

p

-

+ ++=

l

( )2 1

4

2

0

1 1

2

p + +

=

l

l

1

4

p = +

l

9. Ans. 7

2 - +rr r

= 2 2 2 4 4. 2. 4.+ + - + -r rr r r r

13 6 2= -

\ + = 710. Ans. 4

L + 4 = =

2 2y 4 4 y

2

y 2 y 4y 4I e dy e e dy

y y

- - - - += =

y 4 4e 1 Cy

- = - +

xx e Cx 4

= ++

a= 0 & b= 4


47/132




PAPER-2

SECTION-I

1. Ans. (D)

Sol. 2B q= 20I

2

m p

2. Ans. (B)

3. Ans. (B)

Sol. C ( )

E

= G

E

= G

E

= G

E

= G

E= E

E

= 0

E

= E

E

= E

+ E

= 2GM

R

Eext

Ein

Eself

Eext

E

= 2GM

2R

A E

E

.

E

= E

= 2GM

2R4. Ans. (C)

Sol. 60 3 2GM

100 5 R= =

A ( )

2GM 1 GMR 2 R

- + = -+

2 2GM 2GMR R

= -+

9 2GM 1 1

. 2GM25 R R R

= - + ( )

9

25R R R =

+

=9R

16

5. Ans. (B)

Sol. R R

R 2 R4 2

= + =





PAPER CODE 0 1 C T 2 1 4 0 2 3

Path to Success


TM


48/132

HS-2/7


Kota/01CT214023

6. Ans. (B)

Sol. min6t2

3

4

t

)20(k

t

2

)30(k4

2

==

=

=

7. Ans. (B)

8. Ans. (B)

Sol.0

1

IB

2R

m=

( )

2

02 3 /2

22

IRB

2 R 2 2R

m=

+

9. Ans. (C)Sol. F B & A

10. Ans. (C)

Sol.E

F

=r r

11. Ans. (C)

Sol. O .

12. Ans. (B)

13. Ans. (B)

14. Ans. (D)15. Ans. (B)

16. Ans. (D)

Sol. (15 and 16)

Q KS

= -

4 4

= - -

A B

K4, 12

= - = -

A B

Q Q Q

= -

= KS 4 + 12 = 100 0.1 8 = 80

17. Ans. (A)

Sol. 2G

=

< :

= :

> :


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HS-3/7Kota/01CT214023

18. Ans. (D)

Sol. (Moderate) = C= 1 2 mC = 2 mC = C= 1 4 mC = 4 mC\N 4

= 4 2 = 2 mC 4

= ( ) 4 = 8 mJ

D =2 2

1C( )

2- =

1

2 42 22 mJ = 6 mJ

N

DH = D = 8 6 = 2 mJ

19. Ans. (A)

Sol. ( )0 RI

B 5 R

m= qp

( )0 1 2I

B 4

m= q + q

p20. Ans. (C)

Sol.

= -

= -g


SECTION-I

1. Ans. (D)

104=

2x

1 x-, = 102, H = 2

104=

4 2

4

(10 )

C 10

-

--

C = 2 10

4

2 104= 1

= 5000

A = 4999

2. Ans. (B)

C5H

5NH++ H

2O C

5H

5N + H

3O+

0.01 0.002 +

103 =x(0.002 x)

0.01 x

+-

= 2 10

3

H+ = 0.002 +

= 4 103

3. Ans.(B)

4. Ans. (C)

5. Ans. (B)

6. Ans. (A)

7. Ans. (B)

8. Ans. (B)

9. Ans. (A)

10. Ans. (D)

11. Ans. (B)

12. Ans. (D)

13. Ans. (C)

14. Ans. (C)

15. Ans. (C)

16. Ans. (A)

17. Ans. (A)

18. Ans. (A)

19. Ans. (C)

20. Ans. (A)


50/132

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Kota/01CT214023


SECTION-I

1. Ans. (B)

r

( ) . =r r

( ) 3 4 3 4 4 6 .

5 5

+ += +

( )18 3 4 25

= +

2. Ans. (D)

1 1 1 1

1 . . . .

- -

+

1

1

= +

1 1

0

1 1

1

+= + = =

+

3. Ans. (A)

( )

( )

( )

2 40

.10 .2 40 40

= = +

l

( )

2 .10 .2

40

= +l

4. Ans. (B)

C ()

- ()

3, + 3

+ = 2 =

2-

\ + =

-

N, . 2( + ) + .(+)

= .2

.

- + -

= .0 + .0 = 0

(Q

= - () = 0).

5. Ans. (D)

QM. 1 + 2 1

1,2

. + 3 3

0,3\M. 1 + + 2 + 3 4 1,2 0,3.

\1 2

1 50 3

+

6. Ans. (B)

Q . 014243

,6 2

p p

7. Ans. (C)

P = q, = q (q) = 8(q+ q).(3q+3q)= 8(q+ q)2(1 qq)\(q) = 8(1 + 2qq).(1 qq)= 4(1 + 2q)(2 2q)

2q

= \

1,1() = 4(1+ ) (2 )

\()

=9

4.4

1

2

=

\ = 9.8. Ans. (A)

Q 1 1,1,1 1,0,0

\N

= (3 33) 2 = 1629. Ans. (A)

Q( )

( )

0

1

=

+ P =

\

( )( )

1

0

1

=

+

...(1)

0

+

\(1) ( )

( )

1

00

1

+=

+


51/132


HS-5/7Kota/01CT214023

( )( )

1

00

0 . 1

+=

+ ...(2)

( () 0,)

( )

0

+

L.H.S (2) 0

A,( )

0

0

+

L.H.S (2)

,( )

0

+= (0,)

(0) = 0N (2)

( ) ( ) ( )( )( )

11

00

0 1 0 1

0= = + =

+ l

( ) ( )1/ 01

0 1

=-

10. Ans. (B)

() = 2

+ ( 1) 1/4+

4 +

4 > 0 " R " R.

2 ( 1) +1

4

< 4 +

4 "R " R...(1)

Q 4 +

4

1

2

\(1) ( )

2 1 1

1 R4 2- - - + "

2+ ( 1) +1

4> 0 " R

D < 0 ( 1)2 4.1

04

( 1)2< 1 0,2

Paragraph for Question 11 and 12

11. Ans. (B)

QAC A(2,6) S2

\ AC 4 36 4 24 11 9 3= + + - - = =

\AB = 2 3 = 6.

C

B

A

(2,6)D

S2

E

P

S1

12. Ans. (C)

M AP = AE

= AD + DE

= 5 + 4 = 9.


13. Ans. (C)

+ + = 4 ...(1), 2 + + 3 = 6 ...(2) + 2 + = ...(3)

S (1) (2) = 2 2 = 2+ (3), = 6

H, , 0, R\N. = 20 21 = 420.

14. Ans. (A)

F = 0, 6N. = 1 20 = 20.F , = 0 = 6 N.

= 1 1 = 1.

\ = 20 + 1 = 21.


15. Ans. (C)

E ( ) 0 - = = +r r rr r r r

( ) 2 3 2 = + - + - +r

L M ^ P(0,1,2)

\ ( ) ( ) ( ) PM 2 1 1 5 = + + - + -

Q ( ) PM. 2 0- + =2(2 + 1) 1.(1 ) + 1.( 5) = 0

4 + 2 1 + + 5 = 0

2i j + k

M(2t+1, 2t, t3)

6 = 4 2

3

=

\7 4 7

M , ,3 3 3

= -

\S = PM

49 1 169 219

9 9 9 3= + + =


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HS-6/7


Kota/01CT214023

16. Ans. (B)

L R xi yj= +r

C

2+

2 2 + 2 + 1 = 0 = (1,1)

N L

(1,1,0). M. 2 3= .

17. Ans. (B)

(P) G....

= =

,....

. 1

=l .. l = 1

1 1

1

= = =l \ = 1

(Q) [ ] [ ]( ) [ ]( )6 2 6

2 2 2

- -

= +

[ ]( ) [ ]( )6 2

2 6

3 0 3- = + =

(R) Q ( )2 2 22

5 3 2+ + - = + = - =r r rr r r r r

(S) A

( ) ( ) 3 3 9 = + + = - r r r rr r r r

8 8 30 4= = =r rr r

18. Ans. (C)

(P) P - ,32

p=

(Q)2 3 3 0 1

= - = =

N (0) = 2, (1) = 0, (2) = 106=4

\ 4.

(R) ( )( )

2

2

2 0

2 0

- >=- +

= + 0 (0,1)\ = 0

(S) Q ( )

0

1

0

1

+= < -

() = 0

\ ( ) ( )

( )

2

2

0

1

1 0

1

> +=

< -

A, (0+) = (0

) = 1 (0) = 1

\ ( )( )

2

1

1 =

+

(1 + )2.() = 1

19. Ans. (C)

(P) E S1 S

2= 0

+ 5 = 0

0

5q

M P

x+y+5=0

N0 0 5 5

OM2 2

+ += =

5 / 2 1

455 2

q = = q =

22

pq = \ = 2

(Q) (C1C

2)

2 =

2 +

2

(5 2)2+ (6 3)2= 22 = 3

P

r r

C2C1(2,3) (5,6)

90


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HS-7/7Kota/01CT214023

(R)

1 2

2 3 0

3 2 1

- -

- =

-

2

3 4 = 0 ( 4) ( + 1) = 0 = 1,4\

1+

2 2 = 1

(S) QA2= O A= O > 2 N

\ (I + A)6= I + 6A = 620. Ans. (D)

(P)

2

6 2

1 .2

1 =

+

( )

( )

5

26

1 6 .2 2.6

1

+ -

= +

2

2

1

4 122

4=

-= = -

(Q)

1

2

1

(x | x x |)dx-

- - = ( )1

2

1

-

- -(Q )

( ) ( )

0 1

2 2

1 0

5 1 1

6 6-

= - - + - = - - = -

(R) L1

= ,

[ ]2

1 2 3 ....

+ + + +=

[ ] [ ]( )2

1 1

2 2

+= =

(S) Q = 6 2

...(1)

1 2

= - 3

1

= + ...(2)

2

31

= -

= 1A = 1, = 4

(1) (2)

(1,4)

1

3

=

= -


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PAPER-1

Path to Success


TM



ENTHUSIAST COURSE

HS-1/8

PAPER CODE 0 0 C T 2 1 4 0 0 1

SECTIONI

1. Ans. (B,C,D)

2. Ans. (A,B)

3. Ans. (C,D)

4. Ans. (A,B,C)

5. Ans. (A,B,C,D)6. Ans. (A, B, C)

Sol. = w + wr

= = w + w + -w w + w wr

r

1 =

2

2 2

1

w=

+ w

r

7. Ans. (A,D)

Sol. O , .

\T

k C12

k C22

V

8. Ans. (A, C)

Sol. E.F. =0

3

re

r

P = R1 R

2

= ( ) ( )2 2 2 21 20 03R 3R 06 6r r- - -e e9. Ans. (A, B, C, D)

Sol. V = E.

V =

2

C 1 1 C

- = - E = 2 2C V 1

1 1

= -

F = E

0

= E

= 2 E.F. = .

COMMON TEST # 01 (Pattern : JEE-Advanced)


55/132

Kota/00CT214001HS-2/8


10. Ans. (A,B,C)

Sol.

2

00

F F

R

=

r , &

F. 0=

r r

W =

2R 2

200

F .

R

r

=

2R3

0 020 0

8F RF

3 3R

=

SECTIONIV

1. Ans. 5

Sol. A (Q0 ) &

\

( )0K Q K

R

--

=

[ ]0

......()

Q 2

R

=-

()

Q =

2

0 RQ 1 2

- -

2. Ans. 6

Sol. F ,

( )2 22 2A A A 1.52 2

A A

r -r r+

= =r r

ll

l G

5

3

6

2 4

7

8

l

( r , A 2, 3, 4)

3. Ans. 9

Sol.A

BCD

I

O (B + C)

N B = 12 6 C

& C = 6 3C

N (B + C) = 54 C

T = 54C


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Enthusiast Course/27-07-2014/Paper-1

HS-3/8Kota/00CT214001

4. Ans. 4

Sol.

48 V

a

b8W

5W

4W

= 12& V

= 17 12 = 204 V

5. Ans. 1

Sol. R = N

=N

= l2p NR

N

2 R

l=

p= = s

F =

( )3/2

2 20

E NPRP

2 R

l =

+

F =( )

03/2

0

160 0.1 1

2 1 3

+= 1

6. Ans. 2Sol. W s s.

W =2 2

0 0

R R2 2

s s + - -

W =0

.2

s

= 2 2

0

R R2

s + -

7. Ans. 1

Sol. a=2

R

aamg

34

R

R

Rhs

j

A -

R(1 a) + =21

2,

2= 2R(1 a) + P N , :

a=( )2R 1 2

R

- a + =

3 1 R

2

a -

\ =

1R

8


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Kota/00CT214001HS-4/8



8. Ans. 5

Sol. 0 B = m r

l

B 2p = 0

2

0

J 2

p 9. Ans. 9

Sol. () L= L

; w

12

m 2l= MV

2

l V =

M6

m lw

m

V1

w

0 VM

() P= P

; 0 = MV V

1 V =

M

mV1 1V

6=

wl

() =1=

02

)V(V 1

-w--

l V + V

1=

2lw ;

6M6m ll w+w =

2lw ;

+1

Mm = 3

Mm = 2 M =

2m = 9 .

10. Ans. 5

Sol. 3 q 3 q= 3 ... ()

3 q R =23R

2a ...... (I =

23R

2) ... ()

= aR ... ()

= 0.5

SECTION-I

1. Ans. (A,B,C)

2. Ans. (B, C)

3. Ans. (A , C)

4. Ans. (C)

5. Ans. (A, C, D)

6. Ans. (B , C, D)

7. Ans. (A , C , D)

8. Ans. (D)

9. Ans. (A , B, D & B, C, D)

10. Ans. (D)

SECTION-IV

1. Ans (12) [OMR Ans. 3]

2. Ans (75) [OMR Ans. 3]

3. Ans. (13.30) [OMR Ans. 4]

4. Ans. 4

5. Ans. 2

6. Ans. 7

7. Ans. 2

8. Ans. 3

9. Ans. 2

10. Ans. 6


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Enthusiast Course/27-07-2014/Paper-1

HS-5/8Kota/00CT214001


SECTION-I

1. Ans. (A,B,C,D)

1 33 0

7

- - p >

1 3

3 7

-p < < p

7 7

3 6

- < < 2.33 < < 1.16

= 3,2,1,0,1

2. Ans. (B,D)

( )1/

3

3 0

2

-

-=

( )3 0

1

2

-= ...(1)

( )3

0

2 3

- = -

l

...(2)

(1)

4 3 3

3 0

A B C D E 1

2

- + - +=

A C,D,E = 0

0

A B 12 2

- = B 12

- = B = 2

() = A4 2

3...(3)

(2)

( )3

0

2 3

- = -

l

0

A 1

A2 3

A 2

2

+ = -

l

A

1. 32 = - A = 6

() = 64 2

3

() = 243 6

2

(1) = 30 A 3B = 0

3. Ans. (A,B,C)

3512

1 = (8 + 1)256

1

= (1 +256

C18 +

256C

28

2.......+

256C

2568

256) 1

=256

C181+.........

3512 1 = 211 INTEGRAL

3512

1 23,2

7,2

11

4. Ans. (A,B,D)

1 2 117 6 2 2

- - p - + + =

...(1)

2 171 6 1

2

- - + =

( )2 1

1 3 12

- - - =

L 1 1

2

= - ,

.

2

17 D 1 6

2 4 2

= - + = - = -

= 1

(1) 1

,12

-

5. Ans. (D)

2= 3

2 + 2

2 =

5 2

2+ ...(1)

2 = 2 " + ()2= 2

23 2 2

2

2 " 2

4

+ = -

4+

3" =

4

22

2 2

4

22 2 1

2 4

= - -

25 1

2 4

= -

= 6

6. Ans. (B,C)

( )3

1

0 0 1

2 1 2- + +

= p + <


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Kota/00CT214001HS-6/8


() (1+) = (1

)

( )22 1 1

12 3

1 ==

- p p + = +

1 10 3

2 6 + = =

(1)13

2 4 4 6

p p= - - = -

7. Ans. (A,B,C,D)

2() +

2() = 2( 1) =

2( + ) ...(1)

", R

= = 0 (0) = 2

(1)

( ) ( )( ) ( )( )

0. 0 2

-+ +

( ) ( )( ) ( )( )

+ -= + +

I 0(0) ((0) + (0)) + 2 = ().2()

2 (0) () ()+ = =

( )22 2 (0)

2 2+ + = ...(2)

= 0 = 1

2= (0) = 0 ...(3)

(2) ( ) 2 2 = +

( ) 3 11 3 = = , ( ) 7 3=8. Ans. (A,B,D)

= 51 + 57 = 254 3

= 50 + 58 = 254 4

= 49 + 59 = 254 5

3 4 5= =

3 > 4 > 5

> > , > & >

+ = 254 (3 + 5)

= 254 (24 1) > 254 4

+ =

9. Ans. (C,D)

( )

2

1

2

1

2

1 2

-

- =

+

(2 + 1)p

() = 1

() 0,p) 1() = () ()

() = 1

() = 2p "(p,2p)() = 1 " (p,2p)

10. Ans. (A,C)

( )

= +

l l

[ ] [ ] 1 1 = - + -l l = l + l

0 I()

1 I

=

-

l

l

R = 1,0

() = 0 l = = ( I), = 0 = 1() = 1 l

SECTION IV

1. Ans. 9

() + () = 2& + = 2

T Enthusiast Course ALL Test Paper Solution

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