Crt i (Advance) Answer

ANSWERS, HINTS & SOLUTIONS

CRT-I(Set-I)

PAPER -2

ANSWERS KEY

PHYSICS CHEMISTRY MATHEMATICS Q. No. ANSWER ANSWER ANSWER

1. A B D 2. D A B 3. A C A 4. C A C 5. A D B 6. C A B 7. A D A 8. C B B 9. C B C 10. B C A 11. D C A 12. D D B 13. A C A 14. A D C 15. B B A 16. C D D 17. C D A 18. C B C 19. A B A 1. (A) → (p, q, r, s) (B) → (p, q, r, s, t) (C) → (q, r) (D) → (q, r, s, t) A → q B → r C → q, r, t D → p, s, t (A) → (p) (B) → (q), (r) (C) → (p) (D) → (q), (r) 2. (A) → (r) (B) → (p) (C) → (q) (D) → (q) A → q, s B → p, q, r, s C → p, r, s D → r (A) → (r) (B) → (s) (C) → (s) (D) → (q) 3. (A) → (s) (B) → (s) (C) →(t) (D) → (s) A → (r) B → (p) C→ (s) D→(q, t) (A) → (p) (B) → (r) (C) → (q) (D) → (s)

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FIITJEE

JEE(Advanced)-2013

From Long Term Classroom Programs and Medium

/ Short Classroom Program 4 in Top 10

, 10 in Top 20, 43 in Top 100,

75 in Top 200, 159 in Top 500 Ranks & 3542 t

o

ta

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e

le

c

ti

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IT

-J

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E

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0

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P

P

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PART – I

SECTION – A

1. V = Es ; E = 2sm₂ qt ⇒ V 2s m2₂ qt = = 11.375 Volt

2. Static friction depends on tendency.

3. _A L g 4 v λ = λ B 3L_g 4 v = λ B A v = 3v

4. When impulse of external force is zero.

5 ' sin30° =sin60° ⇒ '= 3 ' 'cos30ω ° =v cos30° cos60 (v ' ') cos 60 (2v) v ω = ° + ω = ° = ω = v/ = 40 3 400 0.1 3 = rad/s ω 30° 120° ′ ω′ v 6. for x<0

It will oscillate like SHM

Time period 1 m T k = π For x>0

It will perform periodic motion under constant force

2 2

2E

T 2

mg =

Hence time period T=T₁+T₂

2 m 2E T 2 k mg = π + 7. 2 0 0 mv ev B R = 0 mv B eR =

3 2 2 a +R = − b R 2 2 b a R 2b − = 0 2 2 2bmv B (b a )e = −

8. Here the total time is given T Time spent m the region is (π + θ2 )

Hence total time spent

2 T 2 π + θ   = _{ π} _ θ θ θ 2 π + θ 2 π − θ

9. Torque of Internal force is always zero. All other quantities will vary.

17. F_B =F_E⇒ ₉ 0I _u 2k _q 2 r r µ λ   ₌   _π  _ _   ⇒ 0 0 I u λ = µ ε

If the current a half then to maintain the force velocity should be increased by two times.

18. If we increase the radius by the times then there is no need to change m velocity

0 q 2 r λ πε 0 qu I 2 r µ π u

19. If this case the only force is magnetic force

2 0 quI u m R 2 r µ = π 0 2 rum R qI π = µ

SECTION - B

1. Adiabatic dQ = 0 Isochoric dW = 0 Isothermal dT = 0 ⇒ dU = 0

C

C

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e

m

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s

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t

r

r

y

y

PART – II

SECTION – A

1. OH NH₂ 3 CH COOH + − → O -NH₂ 3 OH || CH −CH OH→ O NH₂ OH C H₃ OH -H₂O O N H₂ O C H N H OH 2 H O − ← N CH₃

4 2. → O O CH₃ O H O CH₃ O OH → _−CO2_→ CH₃ O CH₃ CH₃ O O C H₃ O 3. CN− + O Ph H O -Ph H CN CN− − C OH Ph CN H Ph O CN Ph OH H O -Ph CN Ph O- H OH Ph Ph O OH Ph 4. 3 PhCHO 2NH+ → +3H O₂ Ph N N Ph Ph 3 Hydrobenzamide

As there in no asymmetric carbon in hydrobenzamide, it is optically inactive 5. The rate equation of 3rd order reaction is

3

dx

k(a x)

dt

=

−

As it is cubical equation, the above graph is cubical parabola 6. Na2Al2SiO3. xH2O + Ca2+ → CaAl2SiO3. xH2O + 2Na+

7. The dissociation equilibrium is

N O

_{2 4}

2

NO

₂ Moles at t = 0 1 0 Moles at eqm. 1 – x 2x

(

)

( )

2 2 4 2 ' ₂ '

4

.... i

1

n NO p N O

P

_x

_P

k

P

x

n

∆





_

_





=

=

_

_

−



Σ



1

2

1

n

x

x

x

Σ = − +

= +

On putting values in eqn. (i), we get x = 70.7% For 50% dissociation x = 0.5

From (i)

P = 780 mm Hg

10. Spontaneous change must have +ve sign for ∆S_total.

17. NH₂ H₂SO₄ / HNO₃ NH₃ NO₂ NH₃ NO₂

5 18. NH₂ Br Br Br NaNO₂ / HCl N₂Cl Br Br Br H₃PO₂ Br Br Br

SECTION - B

3. The reaction at high temperature in the blast furnace is

2 2 2 2 2CuFeS +O →Cu S 2FeS SO+ +

M

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PART – III

SECTION – A

1. Put x – [x] = y

The equation becomes f(y) = (a –2)y2 _{+ 2y + a}2 _{= 0 ….. (1)}

Here 0 ≤ y < 1

Since the given equation has exactly one solution in (2, 3), so equation (1) has exactly one root in (0, 1). So, f(0) . f(1) < 0 ⇒ a2 _{(a –2 + 2 + a}2_{) < 0 ⇒ a}2 _{(a + a}2_{) < 0} ⇒ a3 _{(1 + a) < 0 ⇒ a ∈ (–1, 0)} If a = 2, 2y + 4 ⇒ y = –2 (not possible) 2. I = 2 ₂ 2 1 _dx x x x 1 1 1 x x  ₊       _{+ +}     

∫

put 1 + 1 1 x x + = t; − 2 1 1 dx dt x 2x x  ₊  ₌     = −2 dt₂ 2 t t =

∫

+ c ⇒ I = 2x x+ x 1+ + c. 3. d

(

_{g x e}

( )

3x

)

_3e 3x dx − − ′ > ⇒ d

(

_{g x e}

( )

3x _e 3x

)

₀ dx − −

′ + > ⇒ (g′(x) + 1)e–3x _{is increasing function}

Now (g′(x) + 1) e-3x _{> (g′(0) + 1) ∀ x > 0} ⇒ (g′(x) + 1) > 0 ⇒ g(x) + x is an increasing function. 4. 2 2 2 2 2 x y x cos y sin p p a b  α α − =_ + _   ⇒ 2 2 2 2 2 2 2 2 1 cos 1 sin x y ... 0 a p b p  α  α − − + + =            

These lines are perpendicular ∴ 1₂ cos₂2 1₂ sin2₂ 0 a p b p α α − − − = ⇒

( )

2 2 2 2 _{2 2} b a 1 _p ab p ab b a − = ⇒ = − . 5. Since |z – 6| = |z – 8|

6

6. Tangent at x = π to the curve f (x) = |cos x| will be parallel to x–axis and cuts the curve f (x) = cos–1 (cos x) at B and C. AD = (π – 1) Area of ∆ABC = (π – 1)2 π/2 π 3π/2 _2π 1 2π–1 B A C y x π 1 D 450 450

7. Suppose α = 2 cos x+ 2 sin x+ 7 = 2sin x

(

+ φ +

)

7 ⇒ − +2 7≤ α ≤ +2 7

⇒ m = 1 α ⇒ m ∈ 1 1 , , 2 7 2 7     −∞ ∪ ∞  ₋   ₊     .

8. (x cot y + ln cos x) dy + (ln sin y − y tan x) dx = 0 x cot y dy + ln sin y dx + ln cos x dy − y tan x dx = 0

(

)

(

)

d xlnsin y + d y lncos x =0

∫

∫

⇒ x ln sin y + y ln cos x = ln k or (sin y)x _{(cos x)}y _{= c} 9. Given equation 2x2 + 2mx + m + 1 = 0 D = 4m2 – 8(m + 1) ≥ 0 m2 _{– 2m – 2 ≥ 0} (m – 1)2 – 3 ≥ 0 ⇒ m = 3, 4, 5, 6, 7, 8, 9, 10. Also, number of ways of choosing m is 10. Required probability = 4 5. 11. 2 _a2 _b2 _c2 _d2 a b c d 4 4 + + + + + +  _{ ≤}     ⇒ ( )2 _{16 e}2 8 e 16 4 − − _≤ 64 + e2 _{– 16e ≤ 64 – 4e}2 5e2 – 16e ≤ 0 0 ≤ e ≤ 16 5 . 14–16. f(x) f′(x) ≥ 0 in [a, c] and ≤ 0 in [c, b] ∵ f(c) = – 2 and f(x) f′(x) ≠ 0 in (a, c) ∪ (c, b) ⇒ f(a), f(b) ≤ 0. f(a) f(b) f(c) 0 0 –3 2 0 –1 –3 2 0 _{– 2} –3 2 –1 –1 –3 2 –1 _{– 2} –3 2 – 2 – 2 _–3 2

7

17. If P is inside the triangle and area of ∆PAB = area of ∆PBC ⇒ BP must be the median. Similarly, AP must be the median.

Hence P is the centroid. If ∠ABC > ∠ACB, then AB > AC and PB > PC ⇒ perimeter of ∆PAB > perimeter of ∆PAC

⇒ ∠ABC = ∠ACB. Similarly, other angles are equal and the triangle must be equilateral.

18−19. If P lies outside the triangle assuming A and B lie on the opposite sides of PC. Hence A, C lie on the same side of PB. Since area PBC = area PBA

A and C must be the same distance from PB so AC is parallel to PB and PA is parallel to BC. So PACB is a parallelogram and PC = AB

⇒ ∠ACB = 90° and parallelogram is a rectangle.

SECTION - B

1. Since f(g(x)) is a one-one function ⇒ f(g(x1)) ≠ f(g(x2)) whenever x1 ≠ x2

⇒ g(x1) ≠ g(x2) whenever x1 ≠ x2

⇒ g(x) is one-one.

If f(x) is not one-one then f(x) = y is satisfied by x = x1, x2

⇒ f(x1) = f(x2) = y also if g(x) is onto then

let g

( )

x′₁ =x₁ and g

( )

x′₂ =x₂ ⇒ f g

(

( )

x₁′

)

=f g

(

( )

x₂′

)

⇒ f(g(x)) can’t be one-one. 2. (A) x tan x 2 4 π + =

from graph, the equation has 3 solutions in [−π, π]. (B) _{sin | x}1 2 _{1| cos | 2x}1 2 _{5 |} 2 − _{− +} − _{− =}π _{⇒ |x}2 _{− 1| = |2x}2 _{− 5|} ⇒ x2 _{= 2 ⇒ x = ± 2 (Two solutions)} (C) x4 − _{2x sin}2 2 x _{1 0} 2 π _{+ = ⇒ x}2 _sin2 x 2 _{1 sin}4 x ₀ 2 2 π π  ₋  _{+ − =}     ⇒ x = (2n + 1), n ∈ I and x2 _{= sin}2 x ₁ 2 π = ⇒ x = ± 1 is the solution.

(D) Let y = x2 + 2x + 2 sec2 πx + tan2 _πx