A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx =...

12
Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005 PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.in KOTA / HS - 1/12 SOLUTION PART-1 : MATHEMATICS ANSWER KEY PAPER-1 PAPER CODE TM Path to success KOTA (RAJASTHAN) 0 1 C T 3 1 3 0 8 8 PATTERN : JEE (Advanced) LEADER & ENTHUSIAST COURSE Date : 09 - 05 - 2014 TARGET : JEE 2014 SCORE-II : TEST # 06 SECTION-I 1. Ans. (C) Total number of desired paths = (A ® C, C ® B) – (A ® C, C ® D, D ® B) Þ 6 C 3 . 9 C 4 6 C 3 . 4 C 2 . 5 C 2 = 1320 2. Ans. (B) 2012 2 2014 x x 1 x 1 + + + is always positive at x = 0 2012 2 2014 x x 1 1 x 1 + + = + as x ® ¥ 2012 2 2014 x x 1 0 x 1 + + ® + Þ (0,1] hence range is 0, 2 p æ ù ç ú è û integral element is 1. 3. Ans. (C) –1 1 2 1 2 1 1 1 x 1, ,0, ,1 2 2 =- - does not lies in the domain of ƒ'(x) 4. Ans. (D) P y = x–2 2 x = y–2 2 Q. 1 2 3 4 5 6 7 8 9 10 A. C B C D C B C A B B Q. 11 12 13 14 15 16 A. A C D D C C Q. 1 2 3 4 5 A. 3 5 2 0 3 SECTION-IV SECTION-I Let point be P(a 2 + 2,a) p dy 1 91 1 P , dx 2 42 æ ö æ ö = Þa= Þ ç ÷ ç ÷ è ø è ø 5. Ans. (C) April has 30 days. For minimum probability, Saturday & Sundays should be maximum = 5 Sundays & 5 Saturdays Þ Sunday's dates Þ 2,9,16,23,30 Þ Saturday's dates Þ 1,8,15,22,29 Þ Prime dates Þ 2,3,5,7,11,13,17,19,23,29 Maximum number of days he does'nt hunt Þ 5 + 5 + 10 – 3 = 17 hence minimum probability is 17 13 1 30 30 - = 6. Ans. (B) Domain of g(x) 8x – x 2 – 15 > 0 Þ [3,5] ƒ'(x) = 3x– x 2 – 4 ƒ'(x) < 0 " x Î [3,5] ƒ(x) max = ƒ(3), ƒ(x) min = ƒ(5) () 5 2 3 0 3z z 145 ƒ5 4z 2 3 6 = - - =- ƒ(3) = 45 6 - Þ ƒ(3) – ƒ(5) = 50 3

Transcript of A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx =...

Page 1: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.in KOTA / HS - 1/12

SOLUTION

PART-1 : MATHEMATICS ANSWER KEYPAPER-1

PAPER CODETM

Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 8

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

Date : 09 - 05 - 2014TARGET : JEE 2014

SCORE-II : TEST # 06

SECTION-I1. Ans. (C)

Total number of desired paths =(A ® C, C ® B) – (A ® C, C ® D, D ® B)Þ 6C3.9C4 – 6C3.4C2.5C2 = 1320

2. Ans. (B)2012 2

2014

x x 1x 1

+ ++

is always positive

at x = 02012 2

2014

x x 11

x 1+ +

=+

as x ® ¥ 2012 2

2014

x x 10

x 1+ +

®+

Þ (0,1]

hence range is 0,2pæ ù

ç úè ûintegral element is 1.

3. Ans. (C)

–1 – 12

12

1

1 1x 1, ,0, ,1

2 2= - - does not lies in the domain

of ƒ'(x)4. Ans. (D)

Py = x–22

x = y–22

Q. 1 2 3 4 5 6 7 8 9 10A. C B C D C B C A B BQ. 11 12 13 14 15 16A. A C D D C CQ. 1 2 3 4 5A. 3 5 2 0 3

SECTION-IV

SECTION-I

Let point be P(a2 + 2,a)

p

dy 1 9 11 P ,

dx 2 4 2æ ö æ ö= Þ a = Þç ÷ ç ÷è ø è ø

5. Ans. (C)

April has 30 days. For minimum probability,Saturday & Sundays should be maximum = 5Sundays & 5 Saturdays

Þ Sunday's dates Þ 2,9,16,23,30

Þ Saturday's dates Þ 1,8,15,22,29

Þ Prime dates Þ 2,3,5,7,11,13,17,19,23,29

Maximum number of days he does'nt hunt

Þ 5 + 5 + 10 – 3 = 17

hence minimum probability is 17 13

130 30

- =

6. Ans. (B)

Domain of g(x)

8x – x2 – 15 > 0 Þ [3,5]

ƒ'(x) = 3x– x2 – 4

ƒ'(x) < 0 " x Î [3,5]

ƒ(x)max = ƒ(3), ƒ(x)min = ƒ(5)

( )52 3

0

3z z 145ƒ 5 4z

2 3 6= - - = -

ƒ(3) = 456

-

Þ ƒ(3) – ƒ(5) = 503

Page 2: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

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7. Ans. (C)

( )42

23

1

1 x 1x x x 1

x x

æ ö+æ ö+ + +ç ÷ç ÷è øè ø

( ) ( )5 42 4

13

x 1 x 1

x

+ +

Þ cofficient of x22 in (x2 + 1)5(x4+1)4

Þ 5Cr.x2r.4Crx4k

r = 3 & k = 4 r = 5 & k = 35C3.4C4

5C5.4C3

Þ 10 + 4 = 14

8. Ans. (A)ƒ(x) – x2 = (x – 1) (x – 2) (x – 3) (x – a)

ƒ(5) = 25 + 4.3.2 (5 – a)

ƒ(–1) = 1 + 2.3.4(1 + a)

ƒ(5) + ƒ(–1) = 26 + 24.6 = 170

9. Ans. (B)

9 2 8 3 3 3dy dy3x y 9x y y x 0

dx dx+ + - =

3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 03(x3y)2d(x3y) + y3dy – x3dx = 0

( )4 433 y x

x y C4 4

+ - =

y(0) = 0 Þ C = 0

4(x3y)3 + y4 – x4 = 0

10. Ans. (B)For infinite solutions D = Dx = Dy = Dz = 0

& No two planes are parallel

1 2 31

D 2 3 2sin 0 sin2

3 5 2

= - q = Þ q =

two possibilities q = 30º, 150º

So secq is either 2

3 or

2

3-

For 2

sec3

q = -

x

2 2 3

D 9 3 1

7 5 2

= - (Dx ¹ 0)

for secq = 2

3, Dx = Dy = Dz = 0

& No two planes are parallelhence only 1 value.

Paragraph for Question 11 & 12

Let ˆ ˆ ˆr xi yj zk= + +r

( )ˆ ˆ ˆr. 8i 10 j 6k r 8- + - - =r r

Þ x2 + y2 + z2 + 8x – 10y + 6z + 8 = 0

Þ Sphere having centre (–4,5,–3)

and radius 42

11. Ans. (A)

(3,4,–5)(1,1,–2)r

r

a 54 42, b 54 42= - = +

Þ (a2 + b2) = 19212. Ans. (C)

Normal vectors of plane

ˆ ˆ ˆ(5i 4 j k)- + ; through (1,1,–2)

Þ (x – 1)5 + (y – 1)(–4) + (z + 2)1 = 05x – 4y + z + 1 = 0

Paragraph for Question 13 & 14

13. Ans. (D)Number of H > No. of TÞ H are Þ 1008, 1009, ....... 2014

2014 2014 20141008 1009 2014

2014 2014 2014

C C CP ......

2 2 2= + +

2014 2014 20141008 1009 2014

2014

C C ..... CP

2+ +

=

2014 20141007

2014

2 CP

2.2-

=

0141007

2015

C1P

2 2

2

= -

Page 3: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

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14. Ans. (D)

990 1024Þ 990C0.1024C0 +

990C1.1024C1+ ......+ ....... 990C990.1024C990

—————————————— 22014

Þ2014

9902014

C2

Paragraph for Question 15 & 1615. Ans. (C)

O

P

Q R(h,k)

Equation of PQ : xh 9yk1

9+

=

Homoginise C2 with PQ2

2 2 xh 9ykx 9y 9 0

9+æ ö+ - =ç ÷

è ø2

2 2 2hx 1 y (9 9k ) 18hk 0

9

æ ö- + - - =ç ÷

è øabove equation is same as C1

2 29 h 9(1 k )9 9

- -= Þ x2 = 9y2

Þ (x + 3y)(x – 3y) = 016. Ans. (C)

If OPRQ is a cyclic quadrilateralÞ ÐPOQ = 45°

Þ

2

12 9

4tan 4510

-

l-

= ° Þ 2 136l =

SECTION-IV1. Ans. 3

aeee = ah, aheh = ae

2 2e he e+ =

2 2e h2 2h e

a a2

a a+ ³

but can't be 2(ae ¹ ah) Þ 32. Ans. 5 A

4A5

A3

A2

A1

A12

A11A

10

A9

A8

A7

A6

both the diagonals of rectangle are equal.

There are six diameters of polygon all are equal

So to form rectangle we take any of the two.

Þ 6C2 = 15 = k

k5

3=

3. Ans. 2Let circle be (x – a)2 + (y – a)2 = 2(a – 1)2

eq. of common chord

Þ S1 – S2 = 0

(4x – 6y + 7) + a(2x + 2y – 2) = 0

that fixed point lies on 4x – 6y + 7 = 0

x + y = 2 Þ l + m = 2

4. Ans. 0

( ) ( ) ( )3 3 31 1 2 2 3 3p , p , p , p , p , p

O, G, H are collinear

Þ

31 1

32 2

33 3

1 p p

1 p p 0

1 p p

=

Þ (p1 – p2)(p2 – p3)(p3 – p1)(p1 + p2 + p3)

Þ p1 + p2 + p3 = 0 (p1 ¹ p2 ¹ p3)

m1 = 21p

3k 1

k 1 2 3k 1

( 1) m p | p | p-

=

- = - +å = 0

5. Ans. 3

O

y=ƒ(x)

y=xy=x2

1 2

1/2

3/2

1

1ƒ(x) {x}

2= +

1 22

0 1

1A x dx x dx

2æ ö= + +ç ÷è øò ò

1 22 3

0 1

x x x 8 1 101

2 2 3 3 3 3= + + = + - =

[A] = 3

Page 4: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

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SOLUTION

PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. B A C D B A D A C AQ. 11 12 13 14 15 16A. A A A D A CQ. 1 2 3 4 5A. 5 8 1 8 8

SECTION-I

SECTION-IV

SECTION–I

1. Ans. (B)

Sol. 00

2E

2 dl

=pÎ

0 0del = pÎ

So ep = 0 2

dE

4rl

2. Ans. (A)

3. Ans. (C)

Sol. At terminal stage, torque applied on smaller discby rope = mga

current to the disc = 2B r

2Rw

(where w is terminal

angular velcity)

torque applied by magnetic field = 2 4B r4Rw

So, 2 4B r

mga4Rw

=

w = 100 rad/sec

4. Ans. (D)

Sol. Induced current in ring will create such amagnetic field which will repel bar magnet.

5. Ans. (B)

Sol. Electric field lines originate from positive chargeand terminate at negative charge

6. Ans. (A)

Sol. For the electric field, 2 2

0 0E x E yV K

2 2= - +

l l

from energy conservation2 2

20 0QE QE1k Mv 2 K

2 2 8- + + = - ´ +

l l

l l

v = 2 m/s

7. Ans. (D)

Sol.dm

2a v 5 1dt

= = ´

8. Ans. (A)

9. Ans. (C)

Sol.t

RCq Qe-

=

where C = 4pÎ0r

10. Ans. (A)

Sol. J – J' = mv

J' = 2mv

Jv

3mÞ =

11. Ans. (A)

12. Ans. (A)

Sol. (11 and 12)

In the case of water, (5 × 103) × 10 × 15 × 3600

= V × rw × S

w × (100 – 20)

In case of oil, (5 × 103) × 10 × 3.15 × 3600

= V × rO × S

O × (60 – 20)

After solving SO/S

w = 0.6 and

V × r0 = 5.7 × 103 kg

13. Ans. (A)

Sol. Using calorimetery principal

150 × 540 + 150 × 1 × (100 – T)

= (250 + 50) × 1 × (T – (–10))

+ 800 × 12

× (0– (–10)) + 800 × 80

+ 800 × 1 × (T – 0°)

Page 5: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

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By solving it t = 20°C hence steam and ice bothbecome water so amount of water = 150 + 800

14. Ans. (D)

Sol. rB (20°C) = r

B (0°C)

11

10æ ö-ç ÷è ø

Hence density is decreased to 1

10 of its density

at 0°C

15. Ans. (A)16. Ans. (C)

SECTION–IV

1. Ans. 5

Sol. 2 21 2

1 1 1R

n n

é ù= -ê úl ë û

n1 = 3 for Paschen series

2. Ans. 8

Sol. 42 1Q T T 2Tµ Þ =

l1T

1 = l

2T

2

12 3000Å

2l

Þ l = =

K.Emax

= 2

hcW-

l

where K.Emax

= 13.6 2 2

1 12 4

é ù-ê úë û eV

3. Ans. 1

Sol. p(0.02)2 (1) = ( )2 4x0.01 2g

5æ öp ç ÷è ø

4. Ans. 85. Ans. 8

Sol. P = 700 × 103 × 1.6 × 10–19 × dtdN

= 10 × 10–3;

dtdN

= 167

11010

14

2

´´-

-

= 2.11

1012 = l N

0

l = 86400142n

´l

Þ N0 = 2n2.11

108640014 12

l´´

= 160 × 1015

SOLUTION

PART-3 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. B A C C C A B A C DQ. 11 12 13 14 15 16A. B A C B C CQ. 1 2 3 4 5A. 3 7 4 5 3

SECTION-I

SECTION-IV

SECTION-I

1. Ans. (B)2. Ans. (A)3. Ans. (C)

4. Ans. (C)

5. Ans. (C)

6. Ans. (A)

7. Ans. (B)

8. Ans. (A)

9. Ans. (C)

10. Ans. (D)

11. Ans. (B)

12. Ans. (A)

13. Ans. (C)

14. Ans. (B)

15. Ans. (C)

16. Ans. (C)

SECTION-IV1. Ans. 31. Ans. (25) [OMR Ans. 7]

3. Ans. 4

4. Ans. 5

5. Ans. 3

Page 6: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

PAPER-2

PAPER CODETM

Path to success KOTA (RAJASTHAN)0 1 C T 3 1 3 0 8 9

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

Date : 09 - 05 - 2014TARGET : JEE 2014

SCORE-II : TEST # 06

Corporate Office : ALLEN CAREER INSTITUTE, “SANKALP”, CP-6, INDRA VIHAR, KOTA-324005PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected] Website: www.allen.ac.inKOTA / HS - 6/12

SOLUTION

PART-1 : MATHEMATICS ANSWER KEY

SECTION-I1. Ans. (A,B)

–ae = –4& 2b = 4A

(–4,0)

O

Focus of parabolais (–4 + A,0)Þ A = 4Þ b = 2A = 8 & ae = 4

b2 = a2(1 – e2) Þ a2 = 80 & 1

e5

=

Directrix of parabola is x + 4 + A = 0or x = –8

2. Ans. (A,B,C)

30º30º

60º

A

B CDÐB = 60ºA + C = 120º Þ A = 90º & C = 30º

( ) A A B Cr s a tan 4Rsin sin sin

2 s 2 2 2D

= - = =

Þ b c a bc 3 1

r R2 a b c 2

æ ö+ - -= = = ç ÷ç ÷+ + è ø

3. Ans. (B,C)Apply L'Hôpitals rule on L.H.S.

Þ 2

a x

ƒ(x) xƒ '(a)lim x

2a®

-=

-Þ ƒ(x) – xƒ'(x) = –2x3

Þ ƒ'(x) – 2ƒ(x)

2xx

=

Q. 1 2 3 4 5 6 7 8 9 10A. A,B A,B,C B,C A,B,C A,C A,B,C,D A,B A,C A,B,C,D A,B,D

A B C D A B C DP,Q,R,S,T R,S P,Q,R,T P P Q Q,T R,S

Q. 1 2 3 4A. 006 000 032 003

SECTION-I

SECTION-II Q.1

SECTION-III

Q.2

I.F = 1

dxx 1

ex

-ò =

Þ 21 1ƒ(x). 2x . dx

x x= ò

ƒ(x) = x3 + cxQ ƒ(1) = 0 Þ ƒ(x) = x3 – x

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

1 1

sin x cos x ; 1 x 0y

sin x cos x ; 0 x 1

- -

- -

ì- + - £ £ï= í+ £ £ïî

12sin x ; 1 x 02y

; 0 x 12

-pì - - £ £ïï= ípï £ £ïî

3y ,

2 2p pé ùÎ ê úë û

5. Ans. (A,C)Possible when two circles touch each other &having same radii

r1 = r2 Þ 22 2a b= - ...(i)c1c2 = r1 + r2

Þ ( ) ( )2 2a 1 a 1 2 2+ + + = ...(ii)

Þ a = –3 & b = 16

(2,2)(3,3)(1,1)

L

Line L is x + y = 4

Page 7: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

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PAPER – 2

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

y | x 1 |= - and y | x 1 |= -

0 1 2

1

y =(x–1)12

y = x–12 Ö

y =(x–1)32

y = –x4 Ö1

both curves intersect at (0, 1), (1, 0) and (2, 1)

required area =2

2 1

1

2 (y y )dx-ò

Þ A = 2

2

1

2 [ x 1 (x 1) ]dx- - -ò .......(1)

233/ 2

1

2 (x 1) 22 (x 1)

3 3 3

é ù-= - - =ê ú

ë û

Apply b b

a a

ƒ(x)dx ƒ(a b x)dx= + -ò ò in (1)

Þ2

2

1

A 2 [ 2 x (2 x) ]dx= - - -òRequired area can also be calculated as

1

4 3

0

2 (y y )dx-ò1

2

0

2 1 x (x 1) dxé ù= - - -ë ûò

Apply b b

a a

ƒ(x)dx ƒ(a b x)dx= + -ò ò

Þ1

2

0

A 2 x x dxé ù= -ë ûò7. Ans. (A,B)

Q opp

dym

dx=

Þ4 3

3 2h h h4h 3 h

h+ a + b

= + a + b

Þ h4 + ah3 + bh = 4h4 + 3ah3 + bh3h4 + 2ah3 = 0

Þ h3(3h + 2a) = 0 Þ h = 0, 2

h3a

= -

Þ a ¹ 0 and b Î Rfor two different values of h.

8. Ans. (A,C)

( ) 4

3 3 3 1P A

3 3´ ´

= =

( ) 4

3 1P B

3 27= =

A Ç B is 0 0

0 0

é ùê úë û

only

Þ P(A Ç B) = 4

13

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

|z|max = OA 2 1= + p/4p/4

q

(1,2)A(z )1

B(z )2

(1,0)O

(1,1)P

|z|min = OB 2 1 3= + =maximum value of amp(z)

is corresponding to point A i.e. 4p

minimum value of amp(z) is corresponding to

point B i.e. 1 1tan

4 4 2-p p

- q = -

(Q 1

tan2

q = in triangle OPB)

10. Ans. (A,B,D)

1ˆ ˆˆ ˆ ˆ ˆa.b b.c c.a2

= = =

Let ˆ ˆˆ ˆ ˆ ˆa b x, b c y,c a z+ = + = + =r r r

v (x y) z (x.z)y (y.z)x= ´ ´ = -r r r r r r r r r r

{ }{ } { }{ }ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ(a b).(c a) b c (b c).(c a) a b= + + + - + + +

{ } { }5 5ˆ ˆˆ ˆb c b a2 2

= + - + { }5 ˆ ˆc a2

= -

Þ vr

is coplanar with a and cÞ D is correct.

(A)5 ˆ ˆv c a2

= -r 5 11 1 2.

2 2= + -

52

=

(B)5 1 5ˆv.a 12 2 4

æ ö= - = -ç ÷è ø

r,

5 1 1ˆv.b 02 2 2

æ ö= - =ç ÷è ø

r,

5 1 5ˆv.c 12 2 4

æ ö= - =ç ÷è ø

r

Page 8: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

01CT313089KOTA / HS - 8/12

09-05-2014TARGET : JEE 2014TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

Þ ˆˆ ˆv.(a b c) 0+ + =r

Þ vr is perpendicular to ˆˆ ˆa b c+ +

(C)2

1 11

2 21 1ˆˆ ˆa b c 1 02 21 1

12 2

é ù = ¹ë û

SECTION – II

1. Ans. (A)®(P,Q,R,S,T); (B)®(R,S);

(C)®(P,Q,R,T); (D)®(P)

(A) 2x2 + mx + 6 = 0

H.M. of a & b =2 12

0m

ab= - <

a + b

(B) 72n+1 = 7(50 – 1)n

= 7 (nC050n – nC

150n+1+......+(–1)n nC

n)

( )7 50 1 : n even. N

7(50 1) : n odd

ì a += a Îí

a -î

Þ Remainder 7 : n even

18 : n odd

ì= í

î

(C) –10 < 2[x] – 5 < 10

5 15[x]

2 2- £ £ Þ n Î [–2,8)

(D) 2

1 1/ xI dx

x+

= ò put

22

1 11 t dx 2tdt

x x+ = Þ = -

Þ 3/ 2

2 2 1I 2 t dt 1 c

3 xæ ö= - = - + +ç ÷è øò

Þ a = –2; b = 3 : g = –1

2. Ans. (A)®(P); (B)®(Q); (C)®(Q,T);(D)®(R,S)

(A) Apply L o pital rule

( )( )ax

axx 0

ae blim 0

e bx c.coscx®

-=

-

Þ a b

0c-

= Þ a = b

(B) TEETWENTY

Þ TETET, W,N,E,Y

Þ Number of reqd. words =5! = 120

(C)2 |sin x|

|sinx| |cos |0

eI dx

e e

p

a=+ò

|sin x|

|sin x| |cosx|0

e2 dx

e e

p

=+ò

/ 2 sin x

sin x cosx0

e2.2 dx

e e

p

=+ò

Þ / 2 sinx cosx

sinx cosx0

e e2I 4 dx

e e

p +=

+òÞ I = p

(D) sinq(3 – 2cos2q) + 1 = 6sin2qÞ sinq(4 sin2q + 1) + 1 – 6sin2q = 0Þ 4sin3q – 6sin2q + sinq + 1 = 0

(sinq – 1) (4 sin2q – 2sinq – 1) = 0

Þ 2

sin 1 cos2 sin 0

14sin 2sin 1 0 cos2 sin

2

q = Þ q + q =ìïí

q - q - = Þ q + q =ïî

SECTION-III1. Ans. 006

AAT = IÞ a2 + b2 + c2 = 1 & ab + bc + ca = 0possible when a = ±1, b = 0, c = 0 ora = 0, b = ±1, c = 0 ora = 0, b = 0, c = ±1

2. Ans. 000y = –xey/x

y / x y / x2

xy ' y.1y ' x e e

x

é ù-ì ü= - -í ýê úî þë û

y / x xy ' y

e 1x-é ù= - +ê úë û

Page 9: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

KOTA / HS - 9/1201CT313089

LEADER & ENTHUSIAST COURSE 09-05-2014TM

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PAPER – 2

y yy ' y ' 1

x xé ù= - +ê úë û

Þy y y

y ' 1 1x x x

æ ö æ ö- = -ç ÷ ç ÷è ø è ø

Þy

y 'x

=y

1x

ì ü¹í ýî þQ

Þ 2 2

yx. yxy ' y xy '' 0

x x

--= = =

3. Ans. 032

A(1,–1,2) C

B(0,0,0)D

x–y–2=0

AB 6=uuur

, ˆ ˆ ˆAB i j 2k= - + -uuur

C is foot of perpendicular from (0, 0, 0)

upon y = x – 2

Let C(a, b, g)

Þ0 0 0 0 0 2

1 –1 0 2a - b - g - - -æ ö= = = -ç ÷

è ø

Þ coordinates of C are (1, –1, 0)

Þ | AC | 2 | AD |= =uuur uuur

Þ AC and AD are sides of square which isbase of parallelopiped.

q

B

A DC

i – j = n

  V = Base area × height

= (AC)(AD) ˆAB.nuuur

= ˆ ˆ(i j)ˆ ˆ ˆ2 2 ( i j 2k).

2

-´ ´ - + -

= 4 | 1 1 |

2

- -4 2=

4. Ans. 003

x.ƒ(x) – 0.1 = sinx

ƒ(b)b1

a ƒ(a)

ƒ(x)dx ƒ (x)dx bƒ(b) aƒ(a)-ì üï ï+ = -í ýï ïî þ

ò òQ

Þsin x

ƒ(x)x

=

.1 / 2ƒ

6 / 6p pæ öp =ç ÷ pè ø

= 3

Alter :

Apply Newton - Leibnitz theorem

Þ ƒ(x) + g(ƒ(x)).ƒ'(x) = cosx

Þ ƒ(x) + x.ƒ'(x) = cosx

(Q g(x) is inverse of ƒ(x))

Þ ( )dxƒ(x) cosx

dx=

Þ xƒ(x) = sinx + C

Q ƒ(0) = 1 Þsin x

ƒ(x)x

=

Page 10: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

01CT313089KOTA / HS - 10/12

09-05-2014TARGET : JEE 2014TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

SECTION–I1. Ans. (C, D)Sol. Stress is the property of material.

load = stress × cross-sectional area2. Ans. (A, B)

Sol. ( )mg

am M

=+

and Mg – N = Ma

3. Ans. (A, C)4. Ans. (A, B, C)Sol. Fm = 2 (Bi) (AC)

VB = VO = VD5. Ans. (A, D)6. Ans. (A, B, C, D)Sol. For convex lens

1 1 1f v u

= - m1 – m

2 =

( )( )2 2 u v v uv uuv uv

- +-= ...(ii)

1 u vf uv

-= ...(i) x = v – u ...(iii)

Using (i) & (ii) & (iii)

1 2

xf

m m=

-7. Ans. (A, D)8. Ans. (A, C)

Sol. The force exerted = I

AC

´

= 3

98

0.84 10100 0.28 10 N

3 10

--´

´ = ´´

The no. of photons per unit volume = 6I

7 10C h

= ´´ n

9. Ans. (B)

Sol. En = – 13.6 2

2

Zn

10. Ans. (A, B, C)

Sol.( )( )

2

1 2max2

min 1 2

a aII a a

+=

-

SOLUTION

PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. C,D A,B A,C A,B,C A,D A,B,C,D A,D A,C B A,B,C

A B C D A B C DR P Q S R,T Q,S P R

Q. 1 2 3 4A. 004 003 002 002

SECTION-III

Q.2

SECTION-I

SECTION-II Q.1

Page 11: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

KOTA / HS - 11/1201CT313089

LEADER & ENTHUSIAST COURSE 09-05-2014TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

1 1

2 2

a I 4 2a I 9 3

= = =

max

min

I25

I=

If idential paper is pasted acros second slit shifted C.B.F. will back in central point shift = (µ – 1) tDd

fringe width = n D

dl

n = ( )1 tshift

15fringe width

m -= =

lSECTION–II

1. Ans. (A) ® (R); (B) ® (P); (C) ® (Q); (D) ® (S)2. Ans. (A) ® (R, T); (B) ® (Q, S); (C) ® (P); (D) ® (R)Sol. F

net on the strip r

0vg – rvg = r

0vg – r

0vg (1 – 4y) = –4r

0vgy

\ a = 0

0

4 vgy4gy

vr

- = -r

Since a µ –y Þ SHM

\ Time period = 2 2

1sec4g

p p= =

wWhen the strip is in front of lower slit, central maxima shifts below point C & vice-versa.

SECTION–III1. Ans. 004Sol. mA = mB

rAvB = rBvB2

A B2

B A

rr

r=

rSo, mAsADq = mBsBDqri × vA sA = ri vBsB, depth of pentration of ice

A

B

h 4h 1

=

2. Ans. 003Sol. We have 10.2 = w + Kmax, 1 and ...(i)

10.2 Z2 = W + Kmax, 2 ...(ii)

Also lde-Broglie = h h

2mK=

r

\ 1 2

2 12 1

K2.3 K 5.25K

Kl

= = Þ =l ...(iii)

and 10.2 Z2 = ebergy corresponding to longest wavelength of the Lyman series = 3 × 13.6Þ Z = 2\ From equations (i), (ii) and (iii)

W = 3 eV.

Page 12: A-1 PART-1 : MATHEMATICS ANSWER KEY€¦ · dy dy dx dx + + -= 3x6y2(x3dy + 3x2ydx)+y3dy – x3dx = 0 3(x3y)2d(x3y) + y3dy – x3dx = 0 ()xyC33yx44 44 +-= y(0) = 0 Þ C = 0 4(x3y)3

01CT313089KOTA / HS - 12/12

09-05-2014TARGET : JEE 2014TM

Path to success KOTA (RAJASTHAN)

PAPER – 2

3. Ans. 002

Sol.i

r1r2

e

sin i = 21 21 sin r cos rm - -

since light strike grazingly so i = 90° r1 and r2 = 60°

so, µ = 2

4. Ans. 002

Sol. 22

113.6 Z 1 52.224

né ù- =ê úë û

( )2

2 2

1 113.6 Z 1.224

nn 1

é ùê ú- =ê ú-ë û

SOLUTION

PART-3 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10A. A, D A, B, D C, D A, B, C A,B,C,D B, C, D B, D B, C A, B, D B, D

A B C D A B C DR P, S Q, T P,Q,R,S Q, S P, R, S T Q, T

Q. 1 2 3 4A. 146 007 014 006

SECTION-III

Q.2

SECTION-I

SECTION-II Q.1

SECTION-I

1. Ans. (A,D)

2. Ans. (A, B, D)

3. Ans. (C, D)

4. Ans. (A,B,C)

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

6. Ans. (B, C, D)

7. Ans. (B, D)

8. Ans. (B,C)

9. Ans. (A,B,D)

10. Ans. (B,D)

SECTION-II

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

2. Ans. (A) ® (Q, S); (B) ® (P, R, S);(C) ® (T); (D) ® (Q, T)

SECTION-III

1. Ans. 146

2. Ans. (6.972 kJ) [OMR Ans. 007]

3. Ans. 014

4. Ans. 006