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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

Paper Code : 0000CT103115002

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

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

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

A B C D A B C D

S R S T T R S P

Q. 1 2 3 4 5 6 7 8

A. 6 1 4 6 2 6 2 6

Q.2

SECTION-IV

SECTION-I

SECTION-II Q.1

SECTION-I

1. Ans. (B,D)

2. Ans. (A,B)

3. Ans. (B,C,D)

Sol.0

QV d

2A

2Q Q

3Q—2

3Q—2

Q—2

–Q—2

If left plate is earthed,

0

QV d

A

–Q Q–––––––

+++++++

If right plate is earthed,

0

2QV d

A

2Q –2Q–––––––

+++++++

4. Ans. (A,B,C)

Sol. 0 = –nmv + m(v0 – v) 0v

vn 1

& vm = v

0 – v = v

0 –

0 0v nv

n 1 n 1

2 2

0 0v nv1 1KE nm m

2 n 1 2 n 1

20

1 nmv

2 n 1

5. Ans. (A,C)

Sol. E1 = 0 & V

1 = 4V

0

E2 = E; V

2 = 3V

0

E3 = 2E ; V

3 = 2 V

0

E4 = E; V

4 = V

0

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

Sol. In all cases

A P100 2 100

A P

7. Ans. (B,C)

Sol.

22 2 02 2

v1 1 2mg R R cos mv mR

2 2 2 5 R

Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced

TEST DATE : 14 - 02 - 2016

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE

HS-2/16

ALL INDIA OPEN TEST/JEE (Advanced)/14-02-2016/PAPER-1

0000CT103115002

= 22

7mv

10

2 2

10v gR 1 cos

7

22

2

mvmgcos

R 2

10g1 cos

7

17 cos 2 = 10

2

10cos

17

cos 2 < cos

1

2 >

1

mg (R – R cos 1) =

21

1mv

2

1 1v 2gR 1 cos

21

1 1

mvmgcos 2mg 1 cos

R

1

2cos

3

1

2 2gRv 2gR 1

3 3

2

10gR 7v

7 17

1

2gRv

3

8. Ans. (B,C)

Sol. Here NA = N

0e–t

NB = N

0(1 – e–t)

Now tB

A

Ne 1

N

If t << T; then

B

A

N1 t 1

N (ex = 1 + x for small x)

B

A

Nt

N

9. Ans. (A,B,C)

Sol. 2.5A (37 – T1) = 1 2 2

KAT T 2.5A T 7

Solving we getT

1 = 25 and T

2 = 19

Also heat lost by body with fur coat

= dH 5

1 37 25dt 2

= 30

10. Ans. (C,D)

Sol. It is clear that, I1 = I'

1 & I

2 = I'

2

But I1 > I

2 & I'

1 > I'

2

SECTION-II

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

Sol. (A)

41

4 1 83 V1 3V 2 5

We know, 1

2

v 3m

u 5

Since v = –ve & |m| < 1

Hence image is virtual & diminished.

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

Sol. (A) U = 3gh1 + 2gh

2

1 2dU dh dh3g 2g

dt dt dt

= –3gU + 2gU = –gU = 2g

t5

2 2

2

d U g

dt 5

(B)21

U Li2

= 2R

2 L0

1Li e

2

2R2 L0

dU 1 2RLi e

dt 2 L

2R2 22 L02 2

d U 1 4RLi e

dt 2 L

(C) U = mgy

y

dU dymg mg a 0

dt dt

2

2

d U

dt= mg a

y

(D)21

U kx2

= 2 20

1kx cos t

2

20

dU 1kx 2sin t cos t

dt 2

20

1kx sin 2 t

2

at t = 0 du

0dt

22 202

d U 1kx 2 cos 2 t

dt 2

at t = 0 2

2 202

d ukx

dt

2

2

d u0

dt


HS-3/160000CT103115002

SECTION-IV1. Ans. 6

Sol. 21

x 10 0.2 0.2m2

2 2D x

f 24cm4D

f

6cm4

2. Ans. 1

Sol. mvd

dx

= µqvB

m dv qB dx 15

6

0.2 1.5 10x

0.3 50 10 0.2

3. Ans. 4

Sol.

22 8

0

R 8 10 2 RT2

2

8

20

R Q8 10

4 t 4 R

Q = 8 2 38 10 64 R t

=

3

3

9

18 10 16

8 0.09 109 10

= 34 10

4. Ans. 6

Sol.0330 v320 320

53 330 10 3

v0 = 5 m/s

90

t15

= 6 sec

5. Ans. 2

Sol.8

e B v 3 510

= 12 Volt

q = CE (1 – e–t/)

24 = 6 × 12 (1 – e–t/) e–t/ = 2

3

i = t /E 12 2

e 2R 4 3

6. Ans. 6

Sol.2

max

muT mg

2

2min

mu mT mg u 4g

2

max

muT mg

2

min

muT 5mg

difference = 6 mg

7. Ans. 2

Sol. I0W

0 = IW

I0 × w

0 × q

0 = Iwq

2

2

mgm 402

m3 3

3

22

0

mmg Mg M

2 3

02

2

40 1 1.5 11.5 10

3 2 3

1 1.5 11.5 10 0.5 10 1 0.5 1

2 3

40 1550 23 4 2

25155 1

2

8. Ans. 6

Sol. C

C

q =12CV0

q =12CV0

3C

C

In second case

totalC

total

QV

C

C

24CVV 6Volt

4C

HS-4/16


0000CT103115002

SOLUTION

PART-2 : CHEMISTRY ANSWER KEY

SECTION - I

1. Ans. (A,B,C)

(A) For solution Z2 at P

1 pressure

XA = 0.25 X

B = 0.75

YA = 0.5 Y

B = 0.5

(B) For solution Z2 at P

3 pressure solution will

not vapourise so

XA = 0.4 ; X

B = 0.6

(C) For sulution Z1 at P

2 pressure solution will

not vapourise so

XA = 0.2 ; X

B = 0.8

2. Ans. (B)

For a reaction to occur Gº < 0

2MnO + 2C 2Mn + 2CO ; G = 50 kJ/mol

(not feasible)

2/3 Ca2O

3 + 2C 4/3 Cr + 2CO ;

G = – 50 kJ/mol (feasible)

3. Ans.(A,D)

(A) Fact

(B) Probability of finding an electron is nearly 90%

in an orbital

(C) No of angular nodes are l

(D) For 1s ||2 is maximum at nucleus

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

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

6. Ans. (A, C, D)

7. Ans. (A,B,C)

8. Ans. (A,C)

9. Ans. (A,D)

10. Ans. (A, B)

SECTION - II

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

(D)(P, T)

(A) 10–6 =2x

1 xx = 10–3 ; pOH = 3; pH = 11

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

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

A B C D A B C D

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

Q. 1 2 3 4 5 6 7 8

A. 2 4 4 9 1 2 6 5

Q.2

SECTION-IV

SECTION-I

SECTION-II Q.1

(B) 10–7=2x

0.1 xx=10–4 ; pOH = 4, pH = 10

(C) BOH + HCl BCl + H2O

20 × 0.5 2 × V 20 × 0.5

[BCl] = 20 0.5

0.420 5

2x

0.4 x

4 × 10–5 x = 4 × 10–3

pH = 2.4

(D) 3 14

10

(10 x)(x) 10

0.1 x 5 10

x = 10–3

[H+] = 10–3 + 10–3 = 2 × 10–3 pH = 2.7

2. Ans. (A) (P,R,T); (B) (P,R);

(C) (P,Q,R,S,T); (D) (P,Q,R,S)

SECTION-IV

1. Ans. 22

m

G. K.V

n n

l

2 11 4002Scm mol

100 2

2. Ans. 4

(i) It is enthalpy of hydration

(ii) For atomisation coefficient of NH3 should be

one

(iii) Products should be in gaseous ionic form

(iv) Hºr = 3/2

P–PHº

3. Ans. 4

4. Ans. 9

5. Ans. 1

6. Ans. 2

7. Ans. 6

8. Ans. 5


HS-5/160000CT103115002

SOLUTION

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

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

A B C D A B C D

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

Q. 1 2 3 4 5 6 7 8

A. 9 6 4 4 6 3 3 4

SECTION-I

SECTION-II Q.1 Q.2

SECTION-IV

SECTION-I1. Ans. (A,B)

/ 2 / 2

n n / 2 nn

0 0

I x sin x cosx dx 2 x cos x dx4

Put x t4

/ 4n / 2 n

n

/ 4

I 2 t cos tdt4

/ 4n

1 n2

n

0

I 2 cos t dt

/ 4n / 41 n 1 n 2 22

n0

0

I 2 cos t sin t n 1 cos t 1 cos t dt

n

n 2

nI / 22 n 1

I

{After simplification}

2. Ans. (A,C)

Let g(x) = ax2 + bx + c

[ƒ(x)] = [g(x)] x R

–1 < ƒ(x) – g(x) < 1 x R

–1 < (1– a)x2 – x(1 + b) + (1 – c) < 1 x R

a = 1, b = –1

[x2 – x + 1] = [x2 – x + c] [1] = [c]

Let x2 – x + 1 = I + ƒ c = 1 + ƒ'

[I + ƒ] = [I – 1 + ƒ + c]

0 = [ƒ + c] – 1

1 = [c + ƒ]

1 = [1 + ƒ' + ƒ]

[ƒ' + ƒ] = 0 ƒ (0,1)

ƒ' = 0

c = 1

ƒ(x) = g(x) x R

3. Ans. (A,B,D)

ƒ(x) = (x – 1) |x – 3| – 4x + 12

2

2

9 x x 3ƒ x

x 8x 15 x 3

2x x 3

ƒ ' x2x 8, x 3

ƒ'(3–) = –6

ƒ'(3+) = –2

ƒ'(0) = 0

ƒ'(4) = 0

(1,ƒ(1))

4

5310

4. Ans. (A,C)z

P(x,y,z)

x

y

P'(x, ,0)

Obviously 'P' lies on plane y = zTo see the curve rotate y = z about x-axis tobecome z = 0. New co-ordiante of pointcorresponding to P in new plane is

4cos t,4 2 sin t,0

Locus is 2 2x y

1;z 016 32

5. Ans. (A,B,C)

P(22.52) = 22.521 1

1 1 2.5.42 5

no. of numbers multiple of 2 = 2.52

no. of numbers multiple of 5 = 22.5.no. of numbers multiple of both = 2.5 P(22.52) = 22.52 – 2.52 – 22.5+ 2 .5

HS-6/16


0000CT103115002

2 2 1 12 5 1 1

2 5

similarly check other options.

6. Ans. (A,B)

E : Both die shows face 6

A1 : Both are biased

A2 : Both are fair

A3 : One is biased & other is fair.

1

1 1 1A 6 8 8P

1 1 1 1 1 1 4 1 1E . . .6 8 8 6 6 6 6 6 8

9

73

2A 16P

E 73

7. Ans. (A,C)

2 3 1 1 1 1 1

P X3 5 3 2 5 2 5

6 1 7

15 15 15

Y 6 /15 6P

6 1X 7

15 15

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

No. of roots common to

1nz 1 & 2nz 1 is HCF(n1,n

2)

9. Ans. (A,B)

Do yourself

10. Ans. (A,C,D)

1003

1

100

I t dt

100

1

100

2I 1 dt

{king & odd}

I1 = –100

SECTION – II

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

L1

S'(1,1)

y=x

L1

'

S

L2

'

(1,0)L2

y=0

Directrix

O

axis of parabola

M

In the figure ; line MO is parallel to axis of parabola.

Slope of axis 1

2 . L

1 & L

2 shown in figure

are parallel to axis of parabola and image of L1

& L2 in corresponding tangents pass through focus.

L1 x – 2y + 1 = 0

L1' –2x + y + 1 = 0

L2 x – 2y – 1 = 0

& L2' x + 2y – 1 = 0

3 1S ,

5 5

Now image of S lies on directrix

1 3

S' ,5 5

Equation of directrix 2x + y = 1now check options.

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

R

P

A( 0)A'(– ,0)

2 2x y1

2 1

Equation of AP a

y cot x ab 2

Equation of A'P a

y tan x ab 2

2 2 2 4

2 2 2

2 2 2 2

a y x ay x a

b a b b

2 2 4

2 2 2

x y a

b a b


HS-7/160000CT103115002

22 y

x 42

2 2x y1

4 8

Now check options.SECTION – IV

1. Ans. 9

(0,16)

Q( n3,24–18 n3)

(n3,0)0

P

A

Area of trap OAQP = 1

40 18 n3 n32

Req. area n3

2x

0

140 18 n3 e 18x 15 dx

2

5 n3 4 2. Ans. 6

ƒ(x) = x4 + px2 + qx + r

ƒ'(x) = 4x3 + 2px + q

2ƒ" x 12x 2p

ƒ"(x) = 0

2 p

6

Now

(1) 4 + p2 + q + r = 0

(2) 43 + 2p + q = 0

(1) - (2) ×

–34 – p2 + r = 02 23p p

r 036 6

p2 + 12r = 03. Ans. 4

16 3

8 0

1 82x u I ƒ u du I . ƒ u du 4

2 2

4. Ans. 4

ƒ(x) = 0 has 3 distinct real roots ƒ'(x) = 0 has atleast 2 distinct real roots. (ƒ(x).ƒ'(x))' = 0 has atleast 4 distinct real roots.

5. Ans. 6

8k 8k22 22 22

k 2k 7k

24k kk 1 k 1 k 1

z 1 z 1N 1 z z .... z

z 1 z 1

= 22 + (0–1)× 7 = 15

6. Ans. 3

ˆ ˆ ˆ ˆˆ ˆi j k . ai bj ck 3

3cos

2

/6 2

i–j+k

1

P

3.2cos 3 { a2 + b2 + c2 = 4}

locus of 'P' is circle with radius A = [A] = 3

7. Ans. 3D(origin)

A(a)C(c)

B(b)

a b c

2 2 2a b c 2a.b 0

2 2 2b c a 2b.c 0

2 2 2c a b 2a.c 0

2 2 2a b c 2a.b 2b.c 2c.a 0 ...(1)

Also 2

a b c 0

2 2 2a b c 2a.b 2a.c 2b.c 0 ....(2)

(1) & (2) similarly

a.b 0

b.c 0

c.a 0

equality cannot hold as points becomes collinear.8. Ans. 4

Let a b c d x

Consider

a c . b d c.d a.b c.b a.d

= 0

2xa.b c.d b.c a.d

2

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

Paper Code : 0000CT103115003

CLASSROOM CONTACT PROGRAMME(Academic Session : 2015 - 2016)

PART-1 : PHYSICS ANSWER KEY

SECTION-I1. Ans. (A,D)

Sol.21

nR 10 425

m 4

332 25

96 4m 15.36gm

25

nRTV

p

Vmax

at 3

16 40nR

400

1

1 300V

25 1

2

1 400V

25 2

2. Ans. (A,D)

Sol.

2 2R2h h

2g

2

2gh

R

2 21dK dm x

2

22 21 x

2 xdx h x2 2g

4 2 62 R R

K h2 2g 6

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

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

Q. 11 12

A. A A,D

Q. 1 2 3 4 5 6 7 8

A. 6 8 5 4 2 7 4 2

SECTION-I

SECTION-IV

4 42 R 2gh R

h2 12g =

2 45 R h

6

3. Ans. (B,C)

Sol. 1

100

1 = 10 rad/s

2

400

1 = 20 rad/s

20t10t

30 t =

t = 30

sec

4. Ans. (A,B,C)

Sol. (A) t = 0

iL = 0

VV

2

(B) t =

2Vi

3R 2

Vi

3R A

VV

3

(C) after opening

A

Vi

3R

VL = V

A + V

R

V2R

3R

2V

3

A

VV

3

Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Advanced

TEST DATE : 14 - 02 - 2016

TARGET : JEE (ADVANCED) 2016

JEE (Main + Advanced) : ENTHUSIAST & LEADER COURSE


HS-9/160000CT103115003

5. Ans. (B,C)

Sol. Pmax

at resonance 1

LC

61f 10 40 10

2

25Hz

Qfactor

= L

2 R

R 1005

2L 20

45f

2

6. Ans. (A, D)

Sol. mgˆ ˆ0.5g 1i 5i

Bˆ0.5 4i B

Option A Bˆ ˆ5i 6k those B

for which

has – ˆ5i are useful.

7. Ans. (A,C)

Sol. (B) unstable equilibrium

(C)

x

2 3

GMm GMmxF cos

R R so SHM.

(D) F proportional to x3

8. Ans. (C)

Sol. Pipe fixed

1 1S pdt m v

2 2 2S pdt m v1 2

2

1 2

m v Sv

S m

Pipe free same effect9. Ans. (A, D)

Sol.

100V200

50

100 – x

100

80V

0

100

i = 100

0.4A250

Vs = 1.5 V

no light

for current to flow initially

80 > 100 – xlight x > 20V

200

50

100 – x

100

y

0

100

1A

100

1 + y 0 y 100

0200 50

y y1

200 50

y 4y

1200

y = 4040 > 100 – x

x > 60V10. Ans. (C,D)

Sol.h

=

hC

2.5 =

12402 496nm

5

11. Ans. (A)

Sol. S

E

x1

L1

L2

x2

2 s3

GM

R

2s e

12 211

GM GMR x

xR x

s e s

12 2 311

M M MR x

x RR x

33 e1

s

M Rx

3M

s e s

22 2 322

GM GM GMR x

x RR x

s e22 2

2

M M3x

R R x

12. Ans. (A, D)

HS-10/16


0000CT103115003

SECTION-IV

1. Ans. 6

Sol. 14 10 47 5 2N B He

Q = (0.00307 – 0.01294 – 0.00260) 931

= 0.01247 × 931

2. Ans. 8

Sol.53°

37°

x

37° 53°

T2T1

T2 sin 37 = T

1 sin 53

3T2 = 4T

1

2

3xx cos53

5

1

4xx cos37

5

32

2

3x

5t 3 3 10T

3

1

3x 33 10

5 4T

31

11

4x4 45t x 80 105 T 5T

x

= 8 × 10–3 sec

3. Ans. 5

Sol.

120°

h

x

15°45°

2h

2 2 2 1.05 2x

3 1 3

2.1 3x

3 1

2.1 2.10x

1 0.577 0.42

4. Ans. 4

Sol.2 2

2 2

kQ kQ mv

r 4r r

22 3kQ

mv4r

2 221 kQ 2kQ

E 2 mv2 2r r

2 2 23kQ kQ 2kQ

4r 2r r

23kQ

4r

5. Ans. 2

Sol.

2r × S × 2 = 0.56 × 10–3

2r × 0.07 = 0.56 × 10–3

r = 2mm

6. Ans. 7

Sol.L

L g

g

1 Q 41 Q Q

5 Q 5

L

L

1 Q '1

10 Q

L L g

9 36Q ' Q Q

10 50

L

g

Q ' 36 141 1

Q 50 15

28%

7. Ans. 4

Sol. Rapprox

= 120

602

x v

120 15i i i 2 A

960 8

120R 64 R 4

15

8

8. Ans. 2

Sol.

R – x2 2

Rx

2 202 R x µ ni i = ma

= mvdv

dx


HS-11/160000CT103115003

R22 20

0

2 nivdv R x dx

m

2v

2 =

7 22 4 10 4000 i

m

R2 2

0

R x dx

x = R sin

dx = R cos d

/ 2 2

2 2

0

RR cos d 1 cos 2 d

2

= 2R 1

sin sin 02 2 2

= 2R

4

2 7 2v 8 10 4000 i2

2 0.02 4

2 7 22

3

v 8 10 4000 i1

2 4 10 4

v = 2m/s

SOLUTION

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

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

Q. 11 12

A. A,B,D B

Q. 1 2 3 4 5 6 7 8

A. 2 1 2 0 6 4 6 3

SECTION-I

SECTION-IV

SECTION-I

1. Ans. (A,C,D)

2P(atm)

1 10 20

T=300K

V(Litre)

1

QP

(A) A real gas condense if external pressure is

just greater than vapour pressure

(B,C,D)

At point 'P' whole real gas is in gaseous phase

so density of gas = 1000

0.05gm / ml20 1000

At point 'Q' whole substance is in liquid phase

so density of liquid = 1000

1gm / ml1 1000

Note : during condensation at 300K density of

gas and liquid remains constant, only amount

varies.

2. Ans.(D)

In terms of edge length 'a'

X = 2

2a

Y = a

2

Z = 3a

4

3. Ans. (B, D)

4. Ans. (C, D)

5. Ans. (A,C,D)

6. Ans. (A,B,D)

7. Ans. (B)

8. Ans. (A,C,D)

9 Ans. (B, C)

If n/p ratio is smaller than required possibleemission is

53P135

54Q135 +

–1e0

HS-12/16


0000CT103115003

If n/p is smaller than required possible emission

are

54Q135

52R131 +

2He4

or

54Q135

53R135 +

+1e0

10. Ans.(D)

For 'Q' using steady state concept

Rate of appereance of Q = Rate of

disappeareance of Q

K1[P] = K

2[Q]

Number of nuclei of Q

= 1

2

k

k no. of nuclei of P

= 23 2010 / 60

6 10 101000

For 'R'

Since rate constant is very high for second step

than first step

Number of nuclei of R = Number of nuclei of

P disintegrated = 6 × 1023

11. Ans. (A,B,D)

12. Ans. (B)

SECTION-IV

1. Ans. 200 [OMR Ans. 2]

At freezing point vapour pressure of solid

= vapour pressure if liquid

10 – 3000 2000

5T T

5 = 1000

T

T = 200 K

2. Ans. (T5 = 100) OMR ANS (1)

Let heat involved in step 5 is Q5

qtotal

= – Wtotal

500 + 800 + Q5 = 700

Q5 = – 600 J

Since S total = 0

5

500 800 6000

200 200 T

2 + 4 – 6 = 0.

T5 = 100 K

3. Ans. 2

4. Ans. 0

Sol. FeO.Cr2O

3 + Na

2O

2 2NaFeO

2 + 4Na

2CrO

4

+ 2Na2O

(A) (B)

2NaFeO2 + H

2O NaOH + Fe

2O

3

(A) (C)

red brown ppt.

Na2CrO

4 + H

2O CrO

4–2 + 2Na+

(B) (D)

CrO4

–2 + 2H+ Cr2O

72– + H

2O

(D) (E)

(orange)

5. Ans. 6

6. Ans. 4

7. Ans. 6

8. Ans. 3


HS-13/160000CT103115003

SOLUTION

PART-3 : MATHEMATICS ANSWER KEY

SECTION-I1. Ans. (A,B,C)

1 x

0 0

ƒ(xt)dt 0 ƒ(t)dt 0 ƒ(x) 0

(A) y = 0 & y = x3 – 3x2 + P

intersect at 3 distinct points.

x3 – 3x2 + P = 0 P = 3x2 – x3

2

34

P (–4,0) P = –3,–2,–1

(B) ƒ(x) = 0 is a periodic function

(C) for minimum area ƒ(3) = 0

81 – 12 – a = 0 a = 69

(D) ƒ(x) = 0 is even as well as odd.

2. Ans. (B,D)

For k1 > k

2 k

1cos2x > k

2cos2x

–k1cos2x < – k

2cos2x

2 21 21 k cos x 1 k cos x

2 21 2

1 1

1 k cos x 1 k cos x

ƒ(k1) > ƒ(k

2)

Hence ƒ(x) is increasing function.

3. Ans. (A,B)

AB = BT

A = BTB–1

|A| = |BT|. |B–1| = 1 ...(1)

Also Adj A = Adj(BT B–1)

= Adj(B–1) Adj(BT)

AdjA = (AdjB)–1 Adj(BT)

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

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

Q. 11 12

A. C C

Q. 1 2 3 4 5 6 7 8

A. 0 0 4 9 2 3 5 5

SECTION-I

SECTION-IV

(AdjB)T = (AdjB) AdjA

= (AdjB) B

(AdjB)T = |B|.I

AdjB = |B|.I

Adj(AdjA) = |AdjA| I

|A|n–2A = |A|n–1I (order n)

A = |A|I A = I (By (1))

B = I

4. Ans. (A,C)

If ƒ(x) touches x-axis at 2,0

it also touches at 2,0

Roots are 2, 2, 2, 2

5. Ans. (B,C)

1

n2

0

I n 2016 x cos x dx

2

0

1I 2016 x cos xdx

2

0

1I 2016 x cos xdx

2

0

12I 2016 cos xdx

40321I

4

1/ n2

0

n n

x 2016 cos xdx

lim nI n lim1/ n

2

2

2n

1 1 12016 cos

n n nlim 2016

1/ n

HS-14/16


0000CT103115003

6. Ans. (A,D)

z1 = , z

2 = 2

n ni2 i2n n 3 31 2z z e e

2n2cos

3

2016 20161 2z z 2

2015 20151 2

4030z z 2cos

3

2cos 13433

12 1

2

7. Ans. (A,C)

Q( )P(t)

A

B

Equation of chord of contact

x + y = 4 ...(1)

Also, Equation of tangent at P(t)

is ty = x + 2t2

Also t = + 2t2 ...(2)

from (1) & (2)

(t – 2t2)x + y = 4

(tx + y) – (2t2x + 4) = 0

tx + y = 0 & 2t2x + 4 = 0

yt

x

2

2

y2. x 4 0

x

y2 + 2x = 0

(–1/2,0)

Area bounded by S = 0 & the line 2x + 1 = 0

2 1 22

3 2 3

2x – y + 1 = 0 is a focal chord.

Hence angle is 90°.

8. Ans. (A,C)

Equation of family of planes

(2x – y + z – 2) + (x + 2y – z – 3) = 0

It must satisfy (3, 2, 1)

(6 – 2 + 1 – 2) + (3 + 4 – 1 – 3) = 0

3 + 3 = 0 = – 1

x – 3y + 2z + 1 = 0

Equation of acute angle bisector

2x y z 2 x 2y z 3

6 6

x – 3y + 2z + 1 = 0

Paragraph for Question 9 to 10

Solving x

y2

& 2

2xy 1

9 , we get.

6x

13 &

3y

13

(6/ 13, 3/ 13)

(3,0)Shaded region

3 2

6 / 13

1 6 3 x1 dx

2 913 13

3

2 1

6 / 13

9 1 x 9 x9 x sin

3 2 2 313

19 1 9 3 9 9 2. sin

3 2 2 213 13 13 13

13 3 2sin

4 2 13

13 3sin

2 13

similarly calculate the other area

as 1

2

3 1sin

2 1 9m


HS-15/160000CT103115003

9. Ans. (C)

10. Ans. (B)

162 = (a + b + c) (a + b – c) (a + c – b) (b + c – a)

Now, b = 2c & a = 3, we get

162 = (3c + 3) (3 + c) (3 – c) (3c – 3)

= 9(9 – c2) (c2 – 1)

Using A.M > G.M, we get

(9 – c2) + (c2 – 1) > 2 ((9 – c2) (c2 –1))1/2

Maximum value will be attained

for 9 – c2 = c2 – 1

c 5

162 = 9.4.4

Dmax

= 3

11. Ans. (C)

12. Ans. (C)

SECTION – IV

1. Ans. 0

A(2,2)

6 – 42 – 2,3

C

6 – 43

B

3y–2x= –2

6 – 23

3y+2x=6

(3,0)

(1,0)

+22

,6

6 – 4 – 4 + 4 = –2

3

2

Also 3.(12 2 )

( 2) 26

12 – 2 – 2 – 4 = –4

4 = 12 = 3

Slope of BC = 0

2. Ans. 0

(1 + x2 + x4 +......+x100 – x (1 + x2+......+x98))

(1 + x2+....+x100 – x(1 + x2 +....+x98))

= (1 + x2 + x4 +....+x100)2 – x2(1 + x2+....+x98)

which implies there is no term having oddpower of x.

3. Ans. 4

E

C

A F D B

/2

/2

/2–

/2–

Using sine law in BDE

BE BD

3cos sin

2 2

...(1)

Also sine law in BDC

BC BD

cos sin2 2

....(2)

Dividing we get

sinBE 2 EF3BC sin2

0

1lim EF

3 p + q = 4

4. Ans. 9

Using Baye's theorem

21

63 1 1 2

0 16 6 2 6

4

5

a + b = 9.

5. Ans. 2We have to calculate

2 2 2a 1 b 1 c 1

a b c

Now x3 – 2x2 + 3x – 4 = 0

a

b

c

abc = 4

HS-16/16


0000CT103115003

Also x3 – 2x2 + 3x – 4 = (x – a) (x – b) (x –c)

Put x = i & –i and multiply, we get

(a2 + 1) (b2 + 1) (c2 + 1) = 8

2 2 2a 1 b 1 c 1

2abc

6. Ans. 3

VDABC 1 1

6 3 × (area of base) × height

Also 1 1 1

AC BC sin 45 AD6 3 2

AC

.BC.AD 12

Now, 3AC AC

BC AD 3. .BC.AD2 2

(using A.M.- G.M.)

AC

.BC.AD 12

Equality must hold

ACBC AD 1

2

AD must be perpendicular

CD2 = AD2 + AC2 = 3

7. Ans. 5

Given |z| = 1 &1

a z 1z

'a' moves on a circle with centre 1

zz

and radius 1.

1z

z can take any value in the interval

[–2,2], so the set S consists of the locus of unit

circles centered at a point on the line segment

with end points at –2 & 2 on the argand plane.

Therefore, the area S is the area of two semi

circles and area of rectangle which is + 8.

8. Ans. 5

0 2015 1 2015 2015 2015

1 1 1S ...

2 2 2 2 2 2

2015 2015 2014 2015 0 2015

1 1 1S ...

2 2 2 2 2 2

2015 2015

1 12S .......2016 times

2 2

2017

2016S

2

k = 5

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