solutions aiits pt-3-jee adv paper 1

AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16

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1

ANSWERS, HINTS & SOLUTIONS

PART TEST-III (Paper - 1)

Q. No. PHYSICS CHEMISTRY MATHEMATICS

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

2. B B, C A, C, D

3. A A, B, D B, D

4. B B, C, D B, D

5. A, B A, B, D A, B, D

6. A, C, D A, C C, D

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

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

9. B A, B A, D

10. A A, D A, B

1. 3 3 3

2. 6 4 3

3. 6 4 4

4. 2 9 7

5. 4 3 2

6. 8 3 2

7. 7 4 6

8. 1 6 0

9. 3 3 1

10. 2 3 5

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2

PPhhyyssiiccss PART I

SECTION A

1. B, C Equivalent circuit is

4

28 V

I1

12

2 6

4 12

S

I

2. B

Loss of energy is 2e1C E2

3. A VA=VC c=a+b 4. B

Current through Battery, t .

2RCV V eR 2R

5. A, B

Final common potential across the capacitors is o o2E C 2E3C 3

after a long time.

6. A, C, D mF q v B

and apply Newtons second law of motion.

7. A, C As four resistance forms a balanced Wheatstone bridge so potential difference across capacitor C1 is

zero. Charge on C1 = 0 Charge on C2 = 4x106 x 8 = 32 C Charge on C3 = 8x106 x 4 = 32 C i = 12/6 = 2 A When k2 is closed. Potential difference across C2= 4V So change on it is 4x106 x 4 = 16C Potential difference across C3 = 8V So change on it is 8x106x8 = 64 C i = 12/6 = 2A 8. B, D mF q(v B).

9. B E.A

10. A B = oni

n B

iturns m

o

1

4 10 11

4107

7 /

B = oni



3

SECTION C

1. 3 Acceleration and magnetic field are perpendicular to each other thus B.a 0

2. 6

Current through 30 is 1240

and apply loop rule then charge on the capacitor C2 is 60C.

3. 6 M = A 4. 2 dt NAB idt 0 NABQ (i)

2 20 max1 1C2 2 .(ii)

From equation (i) and (ii)

maxNABQ

C

5. 4

0net j 2

B2

6. 8 Apply KVL & symmetry charge on the capacitor 1F is zero. 7. 7

B0 = 0 02 : . .2sin 45r4r 42

= 7.14 106T

8. 1

R 2 0

20

T D2 rB Q dr r2 2R

0 2T D .

QR B

9. 3 According to Kirchoffs current rule. Current passing

through A towards left = current passing through B toward right.

2

2

1

1

1 V

2

1

3

1 V

1 V 1 V

1 V

1 V

4 1

1

2

1

A

B

10. 2 Potential difference across 3 resistance is 20V 5V = 15V

Current through 3 resistance = 5 Ampere. 5 Volt = (5 i)r = 2i r = 2

3 2

r

i 5A

5i



4

CChheemmiissttrryy PART II

SECTION A

4. NaBH4 does not reduce ester. 5. Galactose is empimer of D-(+)-glucose. It is monomer. 7.

8.

9. OH

3 3 2CHCl CCl H O

3 2CCl : CCl Cl

OH

OH

O

C Cl2

O

CCl2

(B)

O

CCl2

HOH

O

CHCl2

(A)

OH 2 eq.

O

CHO

(C) cannot exist in presence of OH.

SECTION C

1. One with keto and two with ester group (lactone). 2. All are stereoisomers. 3. COOH OH OH

NCH3 CH3

COOH

are soluble in alkalies.



5

4.

CH3 C

CH3

CH2

OCH MgBr3

Attack at lesshindered carbon

CH3 C

CH3

CH2

O MgBr

CH3

H

CH3 C

CH3

CH CH3

5. Pinacol-Pinacolone rearrangement reaction. The migratory aptitude is H > Ar > R.

HC

CH3

OH

C

CH3

OH

C

CH3

C

OH

Ph

CH3 C

CH3

C

O

CH3phenyl shift

H



6

MMaatthheemmaattiiccss PART III

SECTION A

2. f(x) = (x 1)4 (x 2)n; n N f(x) = 4(x 1)3 [x 2)n + (x 1)4 n(x 2)n1 = (x 1)3 (x 2)n 1 [(n + 4)x (n + 8)] If n is odd then f(x) > 0 and if x < 1 and sufficiently close to 1 f(x) < 0 if x > 1 and sufficiently close to 1 x = 1 is point of local maxima Similarly if n is even then x = 1 is a point of local minima further if n is even then f(x) < 0 for

x < 2 and sufficiently close to 2 and f(x) > 0 for x > 2 and sufficiently close to 2 x = 2 local minima.

3. f(x) = 2x3 3(a + 1)x2 + 6ax 12 f(x) = 6{x2 (a + 1)x + a} = 0 6{(x 1) (x a)} = 0 f (x) = 12x 6(a + 1) f(1) = 6 6a > 0 f(a) = 12a 6a 6 = 6a 6 if a < 1 then x = a is a local max

2a = 1 a = 12

if a > 1 then x = a is a local minima a = 2.

5.

1 0 x 1x 1

2f x 1 x 2x3 52 x

x 1 2

8. 2 2

1x lnx 1 lnxxf xx x

for horizontal tangent ln x = 1 x = 1 if cut the x axis

lnx 0x

x = 1

f(x) is +ve x (0, e) and f(x) is ve if x (0, ) f(x) is not monotonic for vertical tangent f(x) =

21 lnx

x

2x 0

1 lnx

x = 0 which is not in domain.

9. x 2 23 202 1F x 4t 2F t dt 4x 2F x

x x put x = 4

1F 4 0 4 16 2F 416



7

32F 49

for option D

x3k 2

2 204

1 1 4tf x 4t F t dt 2F t3x x

3 3

21 4.8 4.4F 8 2F 8 2F 4

3 38

32 28F 833 3

.

10. (A) x 0

1lim x 02

x 0

1lim 2x 03

x 0lim f x 0

(B) x 0 x 0

1 1 1 1lim x ; lim 2x2 2 3 3

f(0) f(0+)

(C) x 0lim f x

does not exist so x 0lim f x

no exist

(D) x 0

1x2lim 0

x

;

x 0

12x3lim 0

x

x 0

f xlim

x .

SECTION C

1. x = cos y = sin x + y = sin + cos

2. 1 1f x f x f x2 2

1f x 1 f x f xx

1f x 1 f x 02

3f x f x2

3f x 3 f x f x2



8

3. Function is not differentiable in (,

) at points ,4 4

and ,2 2

.

0 - -/4 /2

4. Point of intersection are 2,1 y = |x2 1| and y = |x2 3|

dy 2xdx

dy 2xdx

12,1

dym 2 2dx

and

22,1

dym 2 2dx

4 2tan7

.

5.

4 4

3/5 3/53 5 5 6 5

cos x dx cos x dxIsin x sin x cos x sin x 1 cot x

2

5 53/5

1 dt 1 1 cot x c5 2t

.

6. 2 2 20cospx sinqx dx cos px sin qx 2sinqxcospx dx

0

1 cos2px 1 cos2qx2 dx2 2

7. n 1 n 1

2 2 2n nr 0 r 0

1 1 1lim limn4n r r4

n

1

1 10 2

0

dx xsin 02 6 64 x

.

9. Volume 13

(area of base x height)

21 3 a 6 h3 4

2a 1 h

23v h 1 h2

3v2

(1 h2 + h(2h)] = 0



9

1h3

v < 0 at 1h3

1h3

.

10. F(x) = x x x xf f g g2 2 2 2

here g(x) = f(x)

F (x) = x x x xf g f g 02 2 2 2

F(x) is a constant function.

solutions aiits pt-3-jee adv paper 1

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