1505 lts answerkey olutions aiot dlp

7/25/2019 1505 LTS Answerkey Olutions AIOT Dlp

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SOLUTION

PAPER-1

HS-1/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

PART-1 : PHYSICS ANSWER KEY

SECTION–I

1. Ans. (A,D)

Sol. By conservation of linear momentum both will get

same velocity of centre of mass

2

cm

1mv mgh

2

h1 = h

2

But by conservation of angular momentum about

centre of mass

System 2 have angular velocity also

K 1 < K

2

[where k =2 2

cm1 1mv I2 2

]

2. Ans. (C,D)

Sol. 3

2 0.085 4h 1mm

r 13.6 10 10 500

3. Ans. (A,C,D)

Sol. (A) Sx = –4 cos kx sin t

= –4 cos2 20.05 sin 0.05

0.40 0.2

= 2 2

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced

TEST # 12 TEST DATE : 15 - 05 - 2016

TARGET : JEE (ADVANCED) 2016

LEADER TEST SERIES / JOINT PACKAGE COURSE

Paper Code : 0000CT103115008

DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)

Q. 1 2 3 4 5 6 7 8 9 10

A. A,D C,D A,C,D A,D A,C,D B B A,B,C,D B,C A,C

A B C D A B C D

R R S P Q,S,T P,R,T P,R,T Q,S,T

Q. 1 2 3 4 5 6 7 8

A. 0 5 5 2 5 1 0 2

SECTION-I

SECTION-II Q.1 Q.2

SECTION-IV

(C)0.4

v 2m / sT 0.2

p

2 1 2 2v 4cos cos 0.1

0.4 15 0.2 0.2

= +40 ×1

2 = 20

4. Ans. (A, D)

Sol.

2

0 12

2 12 0 2

1k E

k 21 k k E2

=

5

3

1

1 3

,

1

1 5

=

2 4

3 5

=

2

15

5. Ans. (A,C,D)

Sol. Parallel ray will pass through focus after passing

through lens.

Slab will shift the image by an amount

of =1

t 1

=

23 1

3

= 1 cm

Final image will form at 11 cm

If converging rays meet at the point of image, image

is known as real. If diverging rays meet at thepoint

of image, image is known as virtual.



LTS-2/16

ALL INDIA OPEN TEST/JEE (Advanced)/15-05-2016/PAPER-1

0000CT103115008

6. Ans. (B)

Sol. As tension is always perpendicular to the velocity

its speed remain same.

7. Ans. (B)

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

Sol. x 2

2cxF

a

y 2

2cyma F

b

9. Ans. (B,C)

10. Ans. (A,C)

Sol.2 2 2

2 1 2

1 1 1kx m 4 m x

2 2 2

2

1

m 4 1

m 16 4

m1v

1 = m

2v

2

1

2

v 1

v 4

2 2 2

1 1 2 2

1 1 1kx m v m v

2 2 2

22 1

1 1 1

1 1 mm v 4v

2 2 4

2 2

1 1 1

1 1m 2 m 5v

2 2

1

4 2v

5 5 2

8v

5

2 mT

4 R

SECTION-II

1. Ans. (A)-(R); (B)-(R); (C)-(S); (D)-(P)

Sol. Path diffrence due to slab-1 = x = t(u –1)

=1 5

5 m2 2

Phase diffrence2 x

=6

9

2 510 m

500 10 2

= 10

Path diffrence due to slab (2) = x = t[µ – 1]

= 3 3 9 m2 3 4

62 910 9

4

Path diffrence due to slab (3) = x = t[µ – 1]

=1 1

[1] m4 4

6

92 1 10

500 10 4

(A) 10 fifth Maxima

(B) (10 – ) = 9 fifth minima

(C) (10 – (9 + ) = 0 Central Maxima

(D) [[10 + ] – 9] = 2 first Maxima

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

(D)-(Q,S,T)

SECTION-IV

1. Ans. 0

2. Ans. 5

3. Ans. 5

Sol.

2B ai

2R

21

0.1 4020

2 1

= 2 ×

1

400

4. Ans. 2

Sol.dm / 0

0.08 0.081dt 0 1 r T

0.8 + 0.8 r T = 0.81

0.01r T

0.80 × 103




LTS-3/160000CT103115008

1 100T 2

80 16

dmsT

dt

33 0.8 10 25

10 400010 4

= 2 × 103 W

5. Ans. 5

Sol. Potential at centre,

kQ kQ kQ kQ

12 12 9 9 =

1 2 1kQ kQ

6 9 18

Q –Q

Q –Q

12m9m

37°

15m

Potential at centre

=

9 69 10 0.01 10 90V 5V

18 18

6. Ans. 1

Sol.H

n T

QnC T W

n T

= 2R – R = R

7. Ans. 0

8. Ans. 2

Sol. Momentum of electron pe = 2meV

Momentum of particle, p

m p 2 eV

4

p p

e e

h / p m2

h / p m / 4

SOLUTION

PART-2 : CHEMISTRY ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. C,D C,D A,C,D B,C,D A,B,C A,B B,D B,C,D B,C A,C,D

A B C D A B C D

P S R,T Q P,Q,S P,Q,S P,T Q,R

Q. 1 2 3 4 5 6 7 8

A. 8 8 4 2 4 3 2 4

SECTION-I

SECTION-II Q.1 Q.2

SECTION-IV

SECTION - I

1. Ans. (C, D)

All molecules move with different speeds and due

to molecular collision, kinetic energy of molecules

will change.

2. Ans. (C,D)

4 2 3

1 3P (s) Cl (g) PCl (g)32

4 2 ; H = 140

140 = P Cl

1 33 (B E)320 120

4 2

(B–E)P–Cl

=120

40kJ / mole3

4 2 3

1 3P (s) H (g) PH (g)4 2 ; H = 8

8 = P H

1 33 (B E)320 216

4 2

(B–E)P–H

= 132 kJ/mole

H2 2H ; H = 216 (H

f )

H =

216108

2

Cl2 2Cl ; H = 120 (H

f )

Cl =

12060

2



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3. Ans. (A,C,D)

Anode : 2Br – Br 2 + 2e –

Cathode : 2H2O + 2e – H2 + 2OH –

K + combines with OH – so KOH will also form.

4. Ans. (B,C,D)

5. Ans. (A,B,C)

6. Ans. (A,B)

7. Ans. (B,D)

8. Ans. (B,C,D)

9. Ans. (B,C)

10. Ans. (A,C,D)

SECTION - II

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

(A) pH = 10 + log0.1

100.1

(P)

(B) pOH = 6 + log0.1

60.1

pH = 8 (S)

(C) pH =1

2 [14 + 5 – 7] = 6 (R), (T)

(D) [H+] =500 0.02

0.011000

,

[OH – ] =500 0.02

0.011000

so solution is neutral (Q).

2. Ans. (A)-(P,Q,S); (B)-(P,Q,S); (C)-(P,T);

(D)-(Q,R)

SECTION - IV

1. Ans. 8

10 H2O + SCN – SO

4

2– +CO3

2– + NO3

– +2OH+ +16 e –

'n' factor or Ba(SCN)2 = 2 × 16 = 32

n 32

84 4

2. Ans. 125 [OMR Ans. 8]

2A(g) + B(g) 2C(g)

ng = 2 – (1 + 2) = – 1

Hr = 25.6 +

( 1) 2 30025kcal

1000

Sr = 2 × 500 – (2 × 200 + 100) = 500 cal.

G = 25 –300 500

–125kcal1000

|G| = 125

Ans. 8

3. Ans. (4)

4. Ans. (2)

5. Ans. (4)

i, ii, iv, v

6. Ans. (3)

i, ii, vi

7. Ans. (2)

ii, iv

8. Ans. (4)

i, ii, iv, v




LTS-5/160000CT103115008

SOLUTION

PART-3 : MATHEMATICS ANSWER KEY

SECTION-I

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

x 13 y 11 z 15L

2 2 1

A(13,–11,15)

L

P

A'( )

P 2 13, 2 11, 15

Putting P in plane – 6 F(1,1,9)

+ 13 = 2, –11 = 2, + 15 = 18

A' (–11,13,3)

Perpendicular distance26 22 15 9

183

volume5

3

1 9 9 39

6 2 2 2

2. Ans. (A,B,D)

z4 – z2 + 1 = 0 2z cos

3

2 5z cos

3

5z cos m z cos n

3 6

now Doyourself.

3. Ans. (A,B,C)

2

2 2

dy 2x 1 dy 1 x y 1

dx 1 x 1 x dx

2

xƒ x

x 1

12

–10

–1/2

1

4. Ans. (B,C,D)

Do yourself

5. Ans. (A,B,D)

LR = 4a = 4

AB = 4 cosec2 = 16

1 2

2 1

t t 3

1

1

12 3 t

t

2

1 1t 2 3t 1

6. Ans. (A,D)

a and1

b are the roots of the equation

6x2 + 20x + 15 = 0

1 10a

b 3 and

a 5

b 2

3

3 32

b 1

ab 9 ab 1 1 1a. 9 a

b b

1 6

5 1000 20159.

2 27

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

x

0

I ƒ(3 )dx 1

1 x

0

I ƒ(3 )dx 1

0

2I ƒ(3)dx Also ƒ(a.b) = ƒ(a) + ƒ(b)

ƒ(a2) = 2ƒ(a)using this C & D also correct.

Q. 1 2 3 4 5 6 7 8 9 10

A. A,B,C,D A,B,D A,B,C B,C,D A,B,D A,D A,C,D A,B,C,D B,C,D A,C

A B C D A B C D

Q,R P,R Q,T Q,S P,Q,R T T S

Q. 1 2 3 4 5 6 7 8

A. 3 4 6 2 0 3 2 1

SECTION-I

SECTION-II Q.1 Q.2

SECTION-IV



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

1

0

1 x 1

0 0 x

1

0

t x dt,x 0

ƒ x x t dt x t dt t x dt, 0 x 1

x t dt,x 1

2

1 2x, x 0

2

2x 2x 1x ,0 x 1

2

2x 1, x 1

2

9. Ans. (B,C,D)

33

3

1 1x x

x xy1 1 1x 2 x 3 x 2x x x

3

1 tx t y

x t 3t 2

2

1y 2

t 3t

2 21 1 2 1t 3 t 3 6 y

t t t 6

10. Ans. (A,C)

ƒ'(x) < ƒ'(x) + xƒ"(x) ƒ'(x) < (xƒ'(x))1

x x

1

0 0

' x dx xƒ ' x dx

x xƒ ' x ...(1)

Now at

2

ƒ x xƒ ' x ƒ xh x h ' x 0

x x

x (0,1) (from (1))

h Now g(x) > x {As ƒ is concave up in (0,1)}h(g(x)) > h(x)

ƒ g x x

g x x

ƒ(x) g(x) < x2

x(0,1)

SECTION – II

1. Ans. (A)

(Q,R); (B)

(P,R); (C)

(Q,T);

(D) (Q,S)

6 6 6

4 5 6

6

C C C 11 a

P A 2 32 b

1 p

P B2 q

5 5 5

3 4 5

6

P AB C C CA 1 uP 2

B P B 2 2 v

P BAB 32 16 8 m

P A P A 11 64 11 n

2. Ans. (A)

(P,Q,R); (B)

(T); (C)

(T);

(D)

(S)

(A)

4991000

2k 1

k 0

2k 1 C

100

499999

2k

k 0

10 C

= 10.2998

(B) 4

3 1 I ƒ

4

3 1 ƒ '

4 2

4 4 4

0 2 4I ƒ ƒ ' 2 C 3 C 3 C

I + 1 = 2{9 + 18 + 1}

= 56

(C) Do yourself

(D) No. of 4 digit numbers

= Dearrangment of 4 objects




LTS-7/160000CT103115008

SECTION – IV

1. Ans. 3Do yourself

2. Ans. 4

Do yourself

3. Ans. 6Use |PS – PS'| = 2b

4. Ans. 2

2 x 2x

x 2x

1 1log y 2 2 log y 2 2 0

2 2

2

x 2x 2x

x 2x 2x

1 1 1log y 2 2 2 2 2

2 2 2

2

x 2x

x 2x

0 0

1 1log y 2 2 2

2 2

x = 0; y = e –2

5. Ans. 0

c a c a b

...(1)

a.c 0

...(2)

Also

a c a a c a a b

{Taking cross with a

}

2 2c a b a c a.b a a b

21 a c b a a b a 0

6. Ans. 3

x

1 x 1lim cot

x 1 2 2x 2

x

1 2 2lim cot

x 1 4 x 1

x

1

2 x 1 2lim

1 1tan2 x 1

7. Ans. 2

Do yourself

8. Ans. 1

ƒ(1 – ) > ƒ(1)

2 > 1 + k

k < 1



SOLUTION

PAPER-2

HS-8/16Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Test Type : ALL INDIA OPEN TEST (MAJOR) Test Pattern : JEE-Advanced

TEST # 12 TEST DATE : 15 - 05 - 2016

TARGET : JEE (ADVANCED) 2016

LEADER TEST SERIES / JOINT PACKAGE COURSE

Paper Code : 0000CT103115009

DISTANCE LEARNING PROGRAMME(Academic Session : 2015 - 2016)

PART-1 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. A,D A,B,D C,D B,C A,B A,C B A,B,D B,D A,D

Q. 11 12A. D A

Q. 1 2 3 4 5 6 7 8

A. 5 5 4 9 4 4 9 6

SECTION-I

SECTION-IV

SECTION-I

1. Ans. (A,D)

Sol. max 1

V 1i Q CV

R LC

V C2V

R L

61 L 1 2R 10

2 C 2 8 = 250

2

0

1H V idt Li

2

Rt

L

0

VV e dtR

–

21 Li2

0

Rt2L 2V L 1

e LiR R 2

22

2

V L 1Li

R 2

2 2 2

2 2 2V L 1 V 1 V LLR 2 R 2 R

2. Ans. (A,B,D)

Sol.CA B8RT3RT 2RT

M M M

C

A B

8T3T 2T

C

280 3T 105

8

& TT

B = 420

B B

A A

P T 3

P T 2

C C

A A

V T 3

V T 8

B C C B

3R T T2

3

R 420 1052

CA

= (280 – 105)R

3. Ans. (C,D)

Sol. (A)

00

R JJ 20

2 2




LTS-9/160000CT103115009

0

22

R I

4 R R

4

02R I

3 R 4

4

(B)R/2

0

R J

2 02

03

R I

3 R 4

4

0I

3 R

4. Ans. (B, C)

Sol. (A)22 2

kQ kQ kQ0

a 9a 3a

5

(B)22 2

4kQ kQ kQ0

a aa5

(C)

2 2 2

kQ 34kQ kQ0

9a 25a 225a

(D)2 2 2

22 2

kQ kQ kQ0

16a 4a 3a

15

5. Ans. (A,B)

Sol.2 21 1 5

MR MR '2 2 4

4

' 5

2

2 2 21 5 4 1k MR MR

2 8 5 5

6. Ans. (A,C)

Sol. z

1C z 1

1

1

1C z 1

1 z

1

z 12

z 1

z1 = 2z – 1

2z 1 1

z 1 2

z 1z

2 2

7. Ans. (B)

Sol. 1 2

1 2

T T 10 30T 7.5days

T T 40

t = 15 days

eliminated =1

5 12

Total driving till then =n2

15105e

=

3

21 55

2 2 2

µg

decayed =1

5 12 2

Remaining in body =5

2

decayed in body =5 1

12 2



LTS-10/16


0000CT103115009

8. Ans. (A,B,D)

Sol. Let magnetic field at centre of square due to

current in square is B1

2

2 213B B B

45°

1B 2B

0ia

42

[sin 45] × 2 × 4 2B

02 2 i2B

a

0

aBi

2

= M × B = ia2 B =

3 2

0

a B

2

(B) U = MB cos 90 = 0

(C) Bnet

= 2B B B 2 1

(D) Bnet

= 2B B B 2 1

9. Ans. (B,D)

Sol. A B,

mg(4) + 2 2B

1 1k 2 0 11 mv

2 2

B C,

mg(3) – 4 =

2 2C B

1m v v

2

10. Ans. (A,D)

Sol.

2B

B

mv N

R

2C

C

mv N mg

R

11. Ans. (D)

Sol.h h h

p 2mE 2mqV where E = kinetic

energy

12. Ans. (A)

Sol.150 150

Å 1ÅV 150

SECTION-IV

1. Ans. 5

Sol. At equilibrium

8g = k × 0.2

8gk

0.2

When mass get turn then new equilibrium

shift by

1g 1g 0.2x

k 8g

x = 2.5 cm

If will be amplitude for SHM maximum height

= 2A = 5 cm

2. Ans. 5

Sol. 3

GM 4G

R 3

t3




LTS-11/160000CT103115009

– 3R

—— 2

30°

30°

R

1t

43 3 G3

=11

11250sec

4 203 10 8003 3

3. Ans. 4

Sol. T = c sxyr z

[T] = [s]x []y [r]z

= [MT –2]x [ML –3]y [L]z

[T] = Mx+y L –3y + z T –2x

x + y = 0

–2x = 1 x = – 12

, y =12

–3y + z = 0

3z

2

3r T c

s

T 1 3 r 1 s

T 2 2 r 2 s

1 13 4

2 2

4. Ans. 9

Sol.d

dT =

36810 m / K

9

Radius of wire r = 2mm

2

2i

r

= e2r (T4 – TT04)

2

2

i

r

e2r (T4)

2 3 42e r Ti

....(i)

As = 368

109

TT

Put = 368

109

T in equation (i) and after

solving we will get i = 0.18 A

5. Ans. 4

Sol. v Rg

2 1R 10

8

24R

5

2

R 4T 4sec2 2 5v

6. Ans. 4

Sol.1

330330 300

300 v

v1 = 30 m/s

2330360 300330 v

2

330330 v 10 275

12

v2 = 55 m/s

V = 25 m/s

100 = V × T

T = 4 sec



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7. Ans. 9

Sol. Using Kirchhoff’s first law at junctions a and

b, we have found the current in other wires of

the circuit on which currents were not shown.

4 F

1A

2A

13V

5

1

4

3

34V

2A

C

1A

2A a

2A

b

3A

6A

3A 2

Now, to calculate the energy stored in the

capacitor we will have to first find the potntial

difference Vab

across it.

Va – 3 × 5 – 3 × 1 + 3 × 2 = V b

Va – V

b = V

ab = 12 volt

U =1

2CV2

ab

=1

2 × (0.125 × 10 –6) (12)2 J

= 9 mJ8. Ans. 6

Sol. From situation, it is clear that it is diverging

lens to left of C.

C B A

x 1/2 1

1

1x

2

+1

3x

2

=1

f

1 1 1

1x f x

2

2 2 1

2x 1 2x 3 f

2

1 9x 2 4x 6 4

f 2x 3 2x 1 4x 8x 3

–f =2 3

x 2x4

1 2 1

x 2x 1 f

2x 2x 1 1

2x 1 x f

–f = 2x2 + x = x2 + 2x +3

4

2 3x x 0

4

1 1 3

2

=

3

2 or

1

2

–f = 2x2 + x

=9 3

24 2

= 6cm




LTS-13/160000CT103115009

SOLUTION

PART-2 : CHEMISTRY ANSWER KEY

SECTION–I

1. Ans. (A, B, D)

(a) Packing fraction of,

=

2

2

13 R

6 0.913 3 3

4R 4

(b) C.N. = 6 ; it is a hexagonal 2D lattice

(c)r

0.155R

so, diameter = 2r = 0.31R

(d)Distance between 2 layers is2 2

R 3

2. Ans. (A, B)

For the equal mix, Y ben

= xtol

& ytol

= xBen

y ben – xBen = minimum, which occurs at

Ptotal

= º ºBen TolP .P

Hence (a), (b) are correct

(C) At this, instant V.P. of equilibrium mixture

corresponds to V'

(D) If external pressure is increased then

condensation occurs and not vaporisation.

3. Ans. (B, D)

As2S

3 sol is negatively charged so the D.M.

moves to cathode while sol particles do not

move in either direction.

4. Ans. (A,D)

5. Ans. (B,C)

6. Ans. (A,C)

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

9. Ans. (A)

H2O

2 decomposes by 1st order kinetics

K × 5 = ln20

15

K = 0.06 min –1 ; t1/2

=ln 2

11.67 min0.06

10. Ans.(A)

N1V

1 = N

2V

2 ;

10

11.35/2 × 11.35

2 = 0.1 × 5 × VV

11. Ans. (C,D)

12. Ans. (B,D)

SECTION–IV

1. Ans. (8)

0 0 01 1 2 2 3 3n E n E n E

(n1 = x – y , n

2 = y – z, n

3 = x – z )

(x – y) × 2 + (y – z) × 3 = (x – z) × 102x – 2y + 3y – 3z = 10 x – 10 zy – 8x + 7z = 0

y 7z8

x

2. Ans. (5)

n – l – 1 = 2 ; l = 3 n = 6

13.62 2

2 2

Z 313.6

6 3 Z = 6 ; oxidation

number = +5.

3. Ans. (3)

4. Ans. (3)

5. Ans. (4)

6. Ans. (2)

7. Ans. (3)

8. Ans. (3)

Q. 1 2 3 4 5 6 7 8 9 10

A. A,B,D A,B B,D A,D B,C A,C C A,B,C,D A A

Q. 11 12

A. C,D B,D

Q. 1 2 3 4 5 6 7 8

A. 8 5 3 3 4 2 3 3

SECTION-I

SECTION-IV



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SOLUTION

PART-3 : MATHEMATICS ANSWER KEY

SECTION-I

1. Ans. (A,B,D)

2

h, k

dy 1 k

dx 3k

2 2

2k 0 1 k 3k 1 k

h 2 3k h 2

...(1)

Also,2

2 2h h1 k 1 k

3 9 ...(2)

From (1) and (2), we get

2h3 1

9 h 9 5h ,k

h 2 3 2 2

9 5A ,

2 2

&

9 5B ,

2 2

2. Ans. (B,D)

We have

a aa b a b 1 1

b b

a

b lies on perpendicular bisector of (–1,0)

and (1,0)

so,a

b lies on imaginary axis.

aarg

b 2

arg a arg. b2

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

Let T(h,k) where h = t1

t2

, k = t1

+t2

.

Also t12 = 16t2

2

Q. 1 2 3 4 5 6 7 8 9 10

A. A,B,D B,D A,B,C,D B,C,D A,C A,B,C,D A,B,C,D B,C B A

Q. 11 12

A. D B

Q. 1 2 3 4 5 6 7 8

A. 0 2 9 3 1 4 2 6

SECTION-I

SECTION-IV

On eliminating t1 & t2, we get locus of T(h,k)

is 2 25y x

4

4. Ans. (B,C,D)

23a a4

r 3as 2 32

,

where a is length of side of triangle

As, r Q a is irrational and multiple of 3 .

Also, R = 2r a

R Q3

.

Also 23a Q

4 and r 1 = r 2 = r 3 =

3a Q

2 .

5. Ans. (A,C)

xn 1 xn 2

n 1 nƒ x

ee

x 2x 2x 3x nxn 1 xn

1 2 2 3 n 1 nlim ....

e e e e ee

x

x nxn

1 nlim e

e e

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

we have

(sin2x + cos3y)2 +(cos3y + tan4z)2

+ (tan4z + sin2x)2 < 0

sin2x = cos3y = tan4z = 0

x ;y , ;z ,2 6 2 4 2

Now verify alternatives7. Ans. (A,B,C,D)

2x

2

dy2x e dx

y

2x1e c

y , As ƒ(0) =

1

2




LTS-15/160000CT103115009

c = 1

2x

1y

1 e

(0,1/2)

y

(1,1/1+e)

(1,0)O(0,0)x

1

0

1 11 0 ƒ x dx 1 0

1 e 2

8. Ans. (B,C)

xlim ƒ(x) 1

ƒ'(x) = 0 x = –1

ƒ(–1) =1

2

Also,x 1 x 1lim ƒ(x) , lim ƒ(x)

Range of ƒ(x) =1

, (1, )2

+ +

x = – 1 1

–

sign of ƒ'(x) x = –1 is point of local maximum of ƒ(x).

Paragraph for Question 9 to 10

C1 : x2 + y2 – 10x + 24 = 0 is circle whose centre

is (5,0) and radius is 1. Also , C2 : x2 + y2 –

10x + 21 = 0 is circle whose centre is (5,0) and

radius is 2.

9. Ans. (B)

10. Ans. (A)Paragraph for Question 14 to 16

1 2 3 4

1 2 3 4P B , P B , P B , P B

10 10 10 10

11. Ans. (D)

1

1 2 1 3 1 4 1P E 1

10 10 2 10 3 10 4

4 2

10 5

12. Ans. (B)

3 2

3 2

2

P B EP B E

P E

3 1

10 3

1 2 1 3 1 4 10

10 10 2 10 3 10 4

1110

3 310

SECTION – IV

1. Ans. 0

3 1 0

M 1 3 0

0 0 2

As, MMT = 4I 2MT + adj.M = 0

|2MT + ady.M| = 0

2. Ans. 2

Let P(x) = 12(x–1)(x– 3) (x – 5) + (2x + 1)

3. Ans. 9

n 6 61 1n 3 3

6P C . 3 . 4

6 n 61 1n 3 3

n 6Q C . 3 . 4

n 6

13

Q12 12 12

P

n 6

1 n 93

4. Ans. 3

we have

3 3 3

3 3 3

3 3 3

a a 1 a b a c

1ab b b 1 b c 11

a.b.cac bc c c 1



LTS-16/16


0000CT103115009

3 3 3

3 3 3

3 3 3

a 1 a a

b b 1 b 11

c c c 1

apply C1 C1 + C2 + C3

& solving

a3 + b3 + c3 + 1 = 11

a3 + b3 + c3 = 10

possibilities are

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

Number of triplets = 3

5. Ans. 1

2 2x y1

9 4 ; P(3sec, 2tan)

Equation of chord of contact AB with respect

to P is T = 0

3x sec + 2y tan = 9 ...(1)

Also, equation of chord whose mid-point is

(h,k) is T = S1

hx + ky – 9 = h2 + k 2 – 9 ...(2)

On comparing (1) and (2), we get

2 2

3sec 2 tan 9

h k h k

as, sec2 – tan2 = 1

locus of (h,k) is2

2 2 2 2x y x y1

9 4 9

= 1

6. Ans. 4

x2

– 6x + 12 = 0; Here ( – 6) =

8

24 24

8

62 1 2

= 224.

7. Ans. 2

Normal vector of required plane is parallel to

vector

ˆ ˆ ˆi j k ˆ ˆ ˆ3 1 3 14i 27j 5k

4 3 5

Equation of required plane is

14(x – 1) – 27(y – 2) + 5(z – 3) = 0

14x – 27y + 5z + 25 = 0

8. Ans. 6

dy 1 2y x

dx x x

(Linear differential

equation)

ƒ(x) = (x – 1)2 + 1

Required area

32

0

x 1 1 dx 3 3 6

1505 lts answerkey olutions aiot dlp

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