Download - Solutions AIITS PT-3-Jee Adv Paper 1
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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.
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com
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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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
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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
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AIITS-PT-III (Paper-1)-PCM (Sol)-JEE(Advanced)/16
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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.