Crt i (Advance) Answer

ANSWERS, HINTS & SOLUTIONS

CRT-I(Set-I) PAPER -2 ANSWERS KEY

ALL INDIA TEST SERIES

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FIITJEE JEE(Advanced)-2013

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

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 (A) → (p, q, r, s) (B) → (p, q, r, s, t) (C) → (q, r) (D) → (q, r, s, t) (A) → (r) (B) → (p) (C) → (q) (D) → (q) (A) → (s) (B) → (s) (C) →(t) (D) → (s)

B A→q B→r C → q, r, t D → p, s, t A → q, s B → p, q, r, s C → p, r, s D→r A → (r) B → (p) C→ (s) D→(q, t)

A (A) → (p) (B) → (q), (r) (C) → (p) (D) → (q), (r) (A) → (r) (B) → (s) (C) → (s) (D) → (q) (A) → (p) (B) → (r) (C) → (q) (D) → (s)

1.

2.

3.

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2

Physics

AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

PART – I SECTION – A

1.

V = Es ; 2sm E= qt 2 ⇒ V=

2.

3.

2s2m = 11.375 Volt qt 2

Static friction depends on tendency.

vA =

vB =

λL g 4 λ 3L g 4 λ

vB = 3v A

4. 5

When impulse of external force is zero.

sin30°

=

' sin 60°

v

⇒ '= 3 ' ω 'cos 30° = v cos 30° ω = cos 60°(v + ' ω ') = cos 60°(2v) = v ω = v/ =

6.

40 3 0.1 3

′ ω′

30° 120°

ω

= 400 rad/s

for x0 It will perform periodic motion under constant force 2E T2 = 2 mg2 Hence time period T = T1 + T2

T=π

7.

m 2E +2 k mg2

mv 02 R mv 0 B= eR

ev 0B =

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AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

3

a2 + R 2 = b − R b2 − a2 R= 2b 2bmv 0 B= 2 (b − a2 )e 8.

Here the total time is given T Time spent m the region is ( π + 2θ) Hence total time spent  π + 2θ  = T   2π 

θ π + 2θ

π − 2θ θ

θ

9.

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

17.

µ I  2kλ  FB = FE ⇒ 9  0  u =  q  r   2πr  λ ⇒ I= µ0 ε0u If the current a half then to maintain the force velocity should be increased by two times.

u quµ0I 2πr

18.

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

19.

If this case the only force is magnetic force u2 quIµ0 m = R 2πr 2πrum R= qIµ0

1.

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

λq 2πε0r

SECTION - B

Chemistry

PART – II SECTION – A

1.

OH

O

+CH3 − COOH →

-

OH ||

O

CH3 − CH OH →

H3C NH2

NH2

NH2

OH OH

-H2O H N

N −H O

2 CH3 ←

O

O

H2N O OH CH3

H3C

O

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AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

4

2.

O

O

O

O H3C

O

→

O

→

CH3 O Ph

O H

+ CN−

CN Ph

H

− CO

2   →

CH3

Ph OH

Ph

−CN−

Ph

4.

-

Ph

3 PhCHO + 2NH3  → Ph

-

Ph

Ph

Ph O

H

CN O

CN OH

OH

O

C

CN

O

OH

CH3

H

-

Ph

Ph

CH3

O

O

CH3

3.

O

OH H

H

N N

+3H2O Ph

Hydrobenzamide

As there in no asymmetric carbon in hydrobenzamide, it is optically inactive 5.

The rate equation of 3rd order reaction is

dx = k(a − x)3 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 2O4 Moles at t = 0 Moles at eqm.

2 NO2

1 1–x

0 2x

2

' ∆n  PNO  4x2  P  kp =  ' 2  = PN2O4 (1 − x )  Σn 

.... ( i )

Σn = 1 − x + 2 x = 1 + x On putting values in eqn. (i), we get x = 70.7% For 50% dissociation x = 0.5 From (i) P = 780 mm Hg 10. 17.

Spontaneous change must have +ve sign for ∆S total. NH2

NH3

NH3 H2SO 4 / HNO 3

NO2 NO2

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AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

18.

NH2

5

N2Cl H3PO2

NaNO2 / HCl Br

Br

Br

Br

Br

Br

Br Br

Br

SECTION - B 3.

The reaction at high temperature in the blast furnace is 2CuFeS2 + O2 → Cu2S + 2FeS + SO2

Mathematics

PART – III SECTION – A

1.

2.

Put x – [x] = y The equation becomes f(y) = (a –2)y2 + 2y + a2 = 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 + a2) < 0 ⇒ a2 (a + a2) < 0 ⇒ a3 (1 + a) < 0 ⇒ a ∈ (–1, 0) If a = 2, 2y + 4 ⇒ y = –2 (not possible)

I=

∫

= −2

3.

4.

5.

 2 1   2+  dx  1 1 1 1 x x x  put 1 + + = t; −  + 2  dx = dt 2 x x  2x x x   1 1 +  1+ x x  dt

∫t

(

2

=

2 2x +c⇒I= + c. t x + x +1

)

(

)

d d g′ ( x ) e−3x > 3e−3x ⇒ g′ ( x ) e−3x + e−3x > 0 ⇒ (g′(x) + 1)e–3x is increasing function dx dx Now (g′(x) + 1) e-3x > (g′(0) + 1) ∀ x > 0 ⇒ (g′(x) + 1) > 0 ⇒ g(x) + x is an increasing function. 2

 x cos α y sin α  cos2 α  sin2 α  2 1 2 1 − = + ⇒ − − + x y     + ........ = 0    a2  b2 p  p2  p2  a2 b2  p   These lines are perpendicular 1 cos2 α 1 sin2 α b2 − a 2 1 ab ∴ 2− ⇒ . = 2 ⇒p= − − = 0 2 2 2 2 a p b p p ( ab ) b 2 − a2 x2

y2

Since |z – 6| = |z – 8| ⇒ z is locus of perpendicular bisector of points joining (6, 0) and (8, 0) ⇒ If z = x + iy ⇒ x = 7

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6

6.

AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

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

y A

π

1

B 0

C

0

45

D

45

1 π/2

π

3π/2

2π

x

2π–1

7.

⇒m=

8.

2 cos x + 2 sin x + 7 = 2 sin ( x + φ ) + 7 ⇒ −2 + 7 ≤ α ≤ 2 + 7

Suppose α =

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

(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 ( x ln sin 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. 4 Required probability = . 5 2

11.

( 8 − e )2 16 − e2 a2 + b2 + c 2 + d2 a +b + c + d ⇒ ≤ ≤   4 16 4  4  2 2 64 + e – 16e ≤ 64 – 4e 5e2 – 16e ≤ 0 16 . 0≤e≤ 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 3 – 2 – 2 –1 –1 3 – 2 –1 3 – 2 – 2 3 – 2 – 2 – 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

AITS-CRT(Set-I)-Paper-2-PCM(S)-JEE(Adv)/13

17.

7

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 ( x1′ ) = x1 and g ( x ′2 ) = x 2 ⇒ f ( g ( x1′ ) ) = f ( g ( x ′2 ) )

⇒ f(g(x)) can’t be one-one. 2.

x π + tan x = 2 4 from graph, the equation has 3 solutions in [−π, π]. π (B) sin−1 | x 2 − 1| + cos−1 | 2x 2 − 5 | = ⇒ |x2 − 1| = |2x2 − 5| 2 ⇒ x2 = 2 ⇒ x = ± 2 (Two solutions)

(A)

2

πx  πx  4 πx + 1 = 0 ⇒  x 2 − sin2  + 1 − sin 2 = 0 2 2   π x 2 =1 ⇒ x = (2n + 1), n ∈ I and x = sin2 2 ⇒ x = ± 1 is the solution. (D) Let y = x2 + 2x + 2 sec2 πx + tan2 πx ⇒ y = (x + 1)2 + (2 sec2 πx − 1) + tan2 πx > 0 ⇒ no solution.

(C) x4 − 2x 2 sin2

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