Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

TEST - 6 (Paper-I)

ANSWERS
PHYSICS CHEMISTRY MATHEMATICS
1. (4) 31. (1) 61. (2)
2. (1) 32. (3) 62. (2)
3. (2) 33. (2) 63. (3)
4. (2) 34. (4) 64. (4)
5. (1) 35. (3) 65. (3)
6. (3) 36. (2) 66. (1)

n
7. (4) 37. (3) 67. (3)
8. (3) 38. (3)
.i 68. (2)
al
9. (2) 39. (3) 69. (1)
10. (4) 40. (1) 70. (3)
n
11. (3) 41. (2) 71. (2)
ur

12. (3) 42. (4) 72. (3)

o

13. (3) 43. (1) 73. (2)

14. (1) 44. (3) 74. (4)
uj

15. (2) 45. (2) 75. (4)

Ed

16. (1) 46. (3) 76. (2)

17. (1) 47. (4) 77. (3)
18. (4) 48. (4) 78. (4)
19. (1) 49. (3) 79. (3)
20. (1) 50. (2) 80. (2)
21. (1) 51. (3) 81. (3)
22. (2) 52. (2) 82. (3)
23. (1) 53. (1) 83. (4)
24. (3) 54. (4) 84. (4)
25. (1) 55. (2) 85. (2)
26. (2) 56. (2) 86. (1)
27. (4) 57. (4) 87. (1)
28. (2) 58. (3) 88. (3)
29. (1) 59. (4) 89. (4)
30. (2) 60. (3) 90. (3)

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All India Aakash Test Series for JEE (Main)-2014 Test - 6 (Paper-I) (Answers & Hints)

PART - A (PHYSICS)
1. Answer (4) 10. Answer (4)
G JJG
2. Answer (1)
v∫ B.dl = μ0i
In both the cases magnetic field at O will be zero.
a /2
a r
3. Answer (2)
⇒ B × 2π 2 = μ 0 ∫ J0
a
× 2πrdr
μ0 μ0
0
(2R )2 iR 2
B1 = i ; B2 =
2 [(2R )2 + (2R )2 ]3/2 2 [R 2 + R 2 ]3/2
2μ0 πJ0 a /2
⇒ B × πa = ∫r
2
B1 4 / 8 4×2 2
3/2 dr
= = = 1: 2 a 0
B2 1/ 2 2 8 × 2 2
4. Answer (2) a /2
2μ0 πJ0 ⎡ r 3 ⎤
Emf induced in the disc will be same as that = ×⎢ ⎥
a ⎢⎣ 3 ⎥⎦0
induced in a rod rotating in a magnetic field.

1 2 μ0 πJ0 a
3
∴ e= Bωa 2 ⇒ B πa =
2 24 a

n
Bωa2
∴ i= μ0 J 0a
2R
.i⇒ B = 12
al
5. Answer (1)
E × 2 πr = πr2.α 11. Answer (3)
n
r 100 100
∴ E= α, XL = ,R= = 10 Ω
ur

2 8 10
At P, E will be along iˆ 1
∴ L=
8π
o

r
uj

∴ F= αe(− iˆ) ⎛ 1 ⎞
2
2 z= ⎜ × 2π × 40 ⎟ + (10)2 = 10 2
⎝ 8 π ⎠
Ed

6. Answer (3)
E 100
−t / τ ∴ i= = =5 2A
i = i0 (1 − e ) z 10 2
12. Answer (3)
i0 ⎛ − R⎞
t
= i0 ⎜ 1 − e L ⎟ i = a + bsinωt
2 ⎜ ⎟
⎝ ⎠ ∴ <i2> = a2 + b2 <sin2ωt> + 2ab<sinωt>

L b2
∴ t= ln 2 = a +
2
R 2
7. Answer (4)
b2
8. Answer (3) ∴ Irms = a 2 +
2
For an element at a distance x of length dx
13. Answer (3)
3
μ 5 5μ0 14. Answer (1)
∫ de = ∫ 2π0 x × 2 × dx = π
ln3
1 15. Answer (2)
9. Answer (2) e = 2 × 2 × 1 = 4 volt, Req = 4Ω
E – (3 + 5t)4 – 6 × (5) = 0 ∴ i through slider is 1 A
⇒ E = 20t + 42 ∴ Fm = i BA = 1 × 2 × 1 = 2N = Fext

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Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

16. Answer (1) 21. Answer (1)

IC
1
φ = B πa2 cos ωt
2

1
⇒ e= Bπa 2ω sin ωt v
2
1
∴ Q= Bπa 2c ω sin ωt
2
IL
dQ ⎛ 1 ⎞
∴ i= = Bπa 2cω2 ⎟ cos ωt
dt ⎜⎝ 2
Drawing phasor diagram
⎠
∴ Inet = | IL − IC | = 0.2 A
17. Answer (1)
22. Answer (2)
1
qE0 d = mv 2 If charge on capacitor at time t is θ
2
Q
2qE0 d E=
⇒ v= ε0 A
m
Q A Q
∴ φE = × =

n
18. Answer (4) ε0 A 2 2ε0
19. Answer (1) dφ
20. Answer (1) .i∴ i d = ε0
dt
= ε0 ×
1 ⎛ dQ ⎞ i
⎜ ⎟=
2ε0 ⎝ dt ⎠ 2
al
23. Answer (1)
Q 8CV0 + CV0
n
VC = net = = 3V0 24. Answer (3)
Ceq 3C
ur

25. Answer (1)

di 400
Current will be maximum when L =0 n=
2πR
o

dt
H =ni = 2000 A/m
uj

(∴ PD across capacitors will be same)

I 4.8 × 10−2
Using energy conservation Susceptibility(χ) = = = 2.4 × 10−5
Ed

H 2000
1 1 1 1 2 26. Answer (2)
(2C ) (4V0 )2 + C(V0 )2 = (3C )(3V0 )2 + Limax
2 2 2 2 27. Answer (4)
28. Answer (2)
6C
⇒ Imax = V0 29. Answer (1)
L
30. Answer (2)

PART - B (CHEMISTRY)
31. Answer (1) 32. Answer (3)
p-Nitrochlorobenzene undergoes nucleophilic
aromatic substitution involving carbanion as CH3 – CH = CH – CH = O
intermediate which is stabilished by resonance.
–
Cl O OH +
CH3 – CH – CH = CH – O–
Δ
+
H3O
+ OH
– In addition to resonance shown above, the methyl
group shows + I effect and hyperconjugation.
NO2 NO2 NO2 33. Answer (2)
(X) (Y)
Addition of HOCl to alkenes follows Markownikoff's
rule.
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All India Aakash Test Series for JEE (Main)-2014 Test - 6 (Paper-I) (Answers & Hints)

36. Answer (2)

Cl 4-Bromobutan-2-one (Y) more reactive than

CH3 – CH = CH2 + HOCl CH3 – CH – CH2 3-bromobutan-2-one (X) because its α-hydrogen is
more acidic.

KOH
OH
(A) 37. Answer (3)

CH3 – CH – CH2
O 18
(B) HOCH2 – C ≡ C – CH2 – CH2OH + H2

Reagent
Lindlar's
18
– HOCH2 CH2 – CH2OH
– O
C2H5O C=C
CH3 – CH – CH2 H H
(SN2)
OC2H5

H+/Δ
C2H5OH 18

n
18
OH HO – CH2 O – CH2
+ +
–H3O
CH3 – CH – CH2
.i H2O – CH2 CH2 CH2 CH2
al
OC2H5 C=C C=C
(C) H H H H
n
ur

34. Answer (4) 38. Answer (3)

1-Methylcyclohexene is oxidised by Baeyer's reagent to
o

1-Methylcyclohexane-1, 2-diol(A). The secondary

uj

CH3 alcohol part of (A) is oxidised by K2Cr2O7/H+ and at

(i) O3 the same time the tertiary alcohol part of (A) is not
CH3 – C = C – CH = CH – CH3
Ed

(ii) Zn/H2O oxidised.

CH3
Me Me Me
K2Cr2O7
(i) O3 Baeyer's OH OH
CH3 – CH = C – CH = C – CH3 Reagent H
+

(ii) Zn/H2O OH O
CH3 CH3 (A)

CH3 O 39. Answer (3)

C = O + CH3 – C – CHO + CH3 – CHO

CH3 Cl Cl
2–
CH3 – CH – CH2 – CH2 – CH – CH3 + S

35. Answer (3)

Cl
Cyclic secondary amine is more basic than open
chain secondary amine. Pyridine is weaker base CH3 – CH – CH2 – CH2 – CH – CH3
than both of them because it has sp2 hybridised –
S
N-atom which is more electronegative. Aniline is the
weakest base because lone pair of electrons at CH2 – CH2
N-atom is delocalised with the benzene ring and CH3 – CH
CH – CH3
hence partially available. S

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Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

40. Answer (1) 43. Answer (1)

HCOOH is most acidic because its conjugate base Allyl halide is more reactive than primary alkyl halide
is stabilised by resonance. towards SN2 reaction because its transition state is
– stabilised by delocalisation. Vinyl halide is least
O O O
reactive towards SN2 because halogen is attached to
+ –
H – C – OH H +H–C–O H–C=O sp2 hybridised C-atom.
Benzoic acid is weaker than formic acid because 44. Answer (3)
carbonyl group is partially available.
Squaric acid is most acidic because its conjugate
base is aromatic and stabilised by resonance.
– –
O OH O O O O
–
C C C O OH O O
+
+ + 2H
H +
O OH O O–

– – – –
O O O– O O– O O O O O

n
C C C
+
.i O O
–
O– O
–
al
+
n
– – –
– – – O O O O
O O
ur

O O
C C
o

–
+ O O O O
uj

45. Answer (2)

Ed

Acetic acid is the weakest acid because the +I 2, 4, 6-trinitrophenol (Picric acid) and phenol both
effect of CH3 group destabilises its conjugate base. give violet colour with FeCl3.
O O 46. Answer (3)
–
CH3 – C – OH H+ + CH3 → C – O Aliphatic and aromatic primary amines react with
O – CHCl3 and KOH to give a foul smelling isocyanides.

CH3 → C = O 47. Answer (4)

41. Answer (2) Phenol is oxidised by K 2S2O 8 in presence of a

base to quinol.
Specific rotation of (+) mandelic acid = +158°
Specific rotation of a solution containing 35% of (–)
OH OH
mandelic acid and 65% of (+) mandelic acid
–
= –0.35 × 158 + 0.65 × 158° = +47.4° OH
+ K2S2O8
H2O
42. Answer (4)
PBr 3 and SOBr 2 react with alcohols to give OH
bromides without rearrangement by SN2 mechanism
whereas HBr reacts with alcohols to give bromides 48. Answer (4)
with the possibility of rearrangement.
Pesticides are compounds used to kill insects,
weeds and fungi.
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All India Aakash Test Series for JEE (Main)-2014 Test - 6 (Paper-I) (Answers & Hints)

49. Answer (3) 54. Answer (4)

Carbon monoxide is not a greenhouse gas. Tropylium chloride gives instant precipitate of AgCl
with aq. AgNO3.
50. Answer (2)
Cl
CH3 +
is a 2° allyl radical, it is stabilised by + aq.AgNO3
–
+ NO3 + AgCl

Aromatic
resonance as well as hyperconjugation due to two
α-hydrogen atoms. Other compounds do not give Cl– ions because the
51. Answer (3) carbocation formed after loosing Cl– ion is not stable.

Conjugated product is most stable. 55. Answer (2)

Hence, (3) is the major product.

OH OH
52. Answer (2) + +
H
The given compound has 6 stereoisomers, two of O O

n
which are meso and the remaining 4 are optically
active.
CH3 CH3 .i OH
OH
al
H H CH3 H Cl CH3
C C C C + +
O
Cl C C Cl Cl C C H
n
H H H H O
(Meso) (Dextro)
ur

+
–H
CH3 H CH3 O
Cl CH3 H
o

C C C H
H C C Cl Cl C C Cl
uj

H H H C O
(Levo) (Meso) H CH3
Ed

CH3 H CH3 Cl
C H C H 56. Answer (2)
Cl C C H H C C Cl 57. Answer (4)
H C H C
Cl CH3 H CH3 Dipole moment of CH3 – F is lower than that of
(Dextro) (Levo) CH3 – Cl because C – F bond length is shorter than
53. Answer (1) that of C – Cl bond even though the polarity of
C – F bond is higher.
+ + 58. Answer (3)
CH3 − O − CH2 and CH3 − S − CH2 are stabilised
Butan-2-ol gives positive iodoform test while butan-1-
+ ol does not. Acidic strength of alcohols follows the
by resonance whereas CH3 − CH− CH3 is stabilised
order CH3OH > 1º ROH > 2º ROH > 3º ROH
by +I effect of two CH3 groups and hyperconjugation 59. Answer (4)
due to 6 α-hydrogen atoms.
If BOD level of water is greater than 5 ppm, it is
+ + highly pure water.
CH3 – S – CH2 CH 3 – S = CH2 is more stable
60. Answer (3)
+ +
than CH3 – O – CH2 CH3 – O = CH2 because Aromatic compounds containing strongly deactivating
the latter has +ve charge on more electronegative group like nitrobenzene do not undergo Friedel
Craft's reaction.
O-atom.

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Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

PART - C (MATHEMATICS)
61. Answer (2) 64. Answer (4)
Let x = 2secθ Differentiating w.r.t. x, we get
⇒ dx = 2secθ.tanθdθ
dy
3 − cos 2 x + sin( y 2 ).2y . =0
π /3 dx
2 tan θ(2 sec θ.tan θd θ)
Now, I =
∫ 4 sec 2 θ
0
dy − 3 − cos 2 x
⇒ =
dx 2y .sin( y 2 )
π /3 π /3
sin2 θ
= ∫ cos θ
dθ = ∫ (sec θ − cos θ)d θ 65. Answer (3)
0 0

⎛ a⎞
π /3 Equation of tangent at ⎜ at , ⎟ is
= {log(sec θ + tan θ) − sin θ} 0 ⎝ t⎠

a 1
3 y− = − 2 ( x − at )
= log(2 + 3) − t t
2
⇒ x + t2y = 2at

n
− 3
i.e., α = 2 + 3, β =
2 .iSo that A(2at, 0) and B ⎛⎜ 0,
⎝
2a ⎞
t ⎟⎠
al
62. Answer (2)
⎛ a⎞
n
Local maxima at x = 2, Mid point of AB = ⎜ at , ⎟ ≡ P
⎝ t⎠
ur

So, hlim f (2 + h) ≤ f (2)

→0 So, AP : PB = 1 : 1
o

66. Answer (1)

⇒ a + 14 − | 46 | ≤ 3 − 22
uj

dy 2ax
As, = 2
⇒ a + 14 ≤ 45 dx x − a2
Ed

So, length of tangent at (x, y) is

⇒ a ≤ 2011
2
So, greatest value of a may be 2011. ⎛ dx ⎞ 4a 2 x 2 + ( x 2 − a 2 )2
y 1+ ⎜ ⎟ =y
63. Answer (3) ⎝ dy ⎠ 2ax

log(e x ) − log(1 + e x ) y ( x 2 + a2 )
f (x) + c = ∫ dx =
2ax
1+ ex
⎛ dx ⎞ ⎛ x 2 − a2 ⎞
⎛ e x ⎞ Length of subtangent = y ⎜ ⎟ = y ⎜⎜ ⎟⎟
log ⎜⎜ ⎟⎟ ⎝ dy ⎠ ⎝ 2ax ⎠
⎝ 1+ e
x
⎠ dx
∫ 1+ e x
= Therefore required sum

y (2.x 2 ) xy
= =

= −∫
(
log 1 + e − x ). e −x
dx
2ax
67. Answer (3)
a
−x
1+ e
As, f(x) is a polynomial function, is continuous and
differentiable for all real x.
1
{ }
2
= log(1 + e − x ) +c
2 Also,

{taking log (1 + e–x) = t} f(x) = (x2 – 2) (2x + 1)

and f (− 2) = 0 = f ( 2)
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All India Aakash Test Series for JEE (Main)-2014 Test - 6 (Paper-I) (Answers & Hints)

Hence, by Rolle's theorem for at least one 70. Answer (3)

x ∈ [− 2, 2]
, Let x = πtanθ

f′(x) = 0 ⇒ dx = πsec2θdθ
⇒ 6x2 + 2x – 4 = 0
1⎛ ⎞
∞ π /2 π /2
log x
2 ∴ ∫ x 2 + π2 dx = π ⎜⎜ ∫ log πd θ + ∫ logtan θd θ ⎟⎟
⇒ x= , −1 0 ⎝ 0 0 ⎠
3
68. Answer (2) 1 1 1
= log π + × 0 = log π
f′(x) = 6x – 1 = 0 2 π 2

1 71. Answer (2)

⇒ x=
6
as f′′(x) = 6
Y ⎛ 3⎞
B ⎜ 1, ⎟
1 ⎝ 2⎠
⇒ f is minimum at x = and least value of f is
6

⎛ 1 ⎞ 3 1 25 ⎛ 1⎞
f⎜ ⎟= − + =2 A ⎜ 1, ⎟

n
⎝ 6 ⎠ 36 6 12 ⎝ 2⎠

Let r be the common ratio

.i x=1
X
al
r2
Given =2
n
1− r
ur

⇒ r = 3 −1 We can write
69. Answer (1)
o

x 3/2
y =x±
uj

⎧ x 2 − 5 x + 6, 0 ≤ x ≤ 2 2
⎪
As, f ( x ) = ⎨ 12 From graph Required area
⎪ −( x − 5 x + 6), 2 < x ≤
2
Ed

⎩ 5
1
⎛ x 3/2 ⎞
1
⎛ x 3/2 ⎞
⎧2 x − 5, 0 ≤ x < 2 = ∫ ⎜⎜ x +
2 ⎠
⎟⎟ dx − ∫ ⎜⎜ x −
2 ⎟⎠
⎟ dx
⎪ 0⎝ 0⎝
⇒ f ′( x ) = ⎨ 12
⎪⎩ −2x + 5, 2 < x ≤ 5
1
2
∫x dx =
3/2
Now, f′(x) = 0 = sq. units
0
5
5 ⎡ 12 ⎤ 72. Answer (3)
⇒ x = ∉ ⎢0, ⎥
2 ⎣ 5⎦
Differentiating w.r.t. x, we get
Also, f ′(2) does not exist. (x4 + 1)1/3x7
Therefore x = 2 is the only critical point.
⎧ 4 4 4 ⎫
= k ⎨( x + 1) .16 x + ( x + 1) .4 x (4 x − 3)⎬
4/3 3 1/3 3 4
f′(x) changes sign from negative to positive ⎩ 3 ⎭
So, f is minimum at x = 2 and minimum value
⎪⎧ 16 x 3 ⎪⎫
= f(2) = 0 ⇒ x 7 = k ⎨16 x 3 ( x 4 + 1) + (4 x 4 − 3)⎬
⎪⎩ 3 ⎪⎭
⎛ 12 ⎞ 6
Also, f(0) = 6, f ⎜ ⎟ = , hence maximum value = 6
⎝ 5 ⎠ 5
3
⇒ k=
So, required sum = 0 + 6 = 6. 112
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Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

73. Answer (2) 76. Answer (2)

Put x – y = t a +1
− ( x −1)2
So that y(x – y)2 =x
Let F (a ) =
∫e dx
a −1
Becomes (x – t)t2 = x ⇒ F ′(a ) = e −a − e −(a−2)
2 2

t3 For maximum, F′(a) = 0

⇒ x=
t −1
2
⇒ a=1
t (t − 3)
2 2 2 2
⇒ dx = dt Also, F ′′(a ) = −2ae −a + 2(a − 2)e −(a −2)
(t 2 − 1)2
⇒ F ′′(1) = −2e −1 − 2e −1 < 0
t (t 2 − 3)
Also, x − 3 y = 2 So, F is maximum at a = 1
t −1
77. Answer (3)
dx t 2 − 1 t 2 (t 2 − 3)
Here ∫
x − 3y ∫ t (t 2 − 3) (t 2 − 1)2
= . dt
5 π /12 π /4 tan−1 2 tan−1 3

∫ [tan x ]dx = ∫ 0.dx + ∫π 1dx + ∫−1 2 dx

t 0 0
= ∫ 2
tan 2
dt 4
t −1

n
−1
tan (2+ 3 )

∫ 3dx
=
1
2
loge (t 2 − 1) + c
.i
+
tan −1
3
al
⎛ −1 π⎞ −1 −1
=
1
loge {( x − y )2 − 1} + c = ⎜ tan 2 − ⎟ + 2(tan 3 − tan 2)
⎝ 4⎠
n
2
+ 3(tan−1(2 + 3 ) − tan−1 3)
ur

74. Answer (4)

2cos x + 1 −π 5π
I=∫ dx = − (tan−1 2 + tan−1 3) + 3 ×
o

(2 + cos x )2 4 12
5 π π ⎧ −1 ⎛ 3 + 2 ⎞ ⎫
uj

Dividing numerator and denominator by sin2x, we get − − ⎨ tan ⎜

= ⎟ + π⎬
4 4 ⎩ ⎝ 1− 6 ⎠ ⎭
(2cosec x cot x + cosec 2 x )dx
Ed

I= ∫ (2cosecx + cot x )2 =
π
4
dt
= −∫ , where 2 cosecx + cotx = t 78. Answer (4)
t2
We have,
1
= +c F′(x) = 4sinx + 3cosx
t
⎛ −1 3 ⎞
1 = 5 sin ⎜ x + tan
=
2cosecx + cot x
+c ⎝ 4 ⎟⎠

75. Answer (4) π 3π

Now, ≤x ≤
4 4
Put ex – 1 = t2, we get
π 3
2 2 ⇒ < x + tan−1 < π
2t 2 ⎛ 4 ⎞
I=∫ 2 dt = 2 ∫ ⎜ 1 − 2 ⎟ dt
2 4
0t +4 0⎝ t +4⎠ Hence F′(x) > 0
2
⎧ −1 ⎛ t ⎞ ⎫
= 4 − 4 ⎨tan ⎜ ⎟ ⎬ ⎡ π 3π ⎤
⎩ ⎝ 2 ⎠ ⎭0 So, F is increasing in ⎢ , ⎥ and F is least at
⎣4 4 ⎦
=4–π π
x= .
4

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All India Aakash Test Series for JEE (Main)-2014 Test - 6 (Paper-I) (Answers & Hints)

79. Answer (3) 81. Answer (3)

Y
Y

2
y=x

A(1, 1)
y = –x
y = 1/x
X X
O 1/2 1
0 B(2, 0)

Required area
(3, –3)
1
⎛1 2⎞
= ∫ ⎜⎝ x − x ⎟ dx
⎠
1/2
Solving y = 2x – x2 and y = –x, we get x = 0, 3
1
Required area ⎪⎧ x 3 ⎪⎫ 7
= ⎨log x − ⎬ = loge 2 −
⎩⎪ 3 ⎪⎭
1/2 24

n
3

∫ {(2x − x }
) − ( − x ) dx
2 7
= So, p =
0
.i 24
al
82. Answer (3)
3

∫ (3 x − x
2
) dx 2
n
= 3
0 As, f(0) = f(2) = 0 and ∫ f ( x )dx = 4
ur

3 0
⎪⎧ 3 x x 3 ⎪⎫
2
= ⎨ − ⎬ 2 2

Now, ∫ xf ′( x )dx = { xf ( x )}0 − ∫ f ( x )dx

2
⎩⎪ 2 3 ⎭⎪
o

0
0 0
uj

9
= 3 3
2 = 2f (2) − = 0−
Ed

4 4
80. Answer (2)
3
= −
Y 4
83. Answer (4)

2 n
y = 4a(2a – x) 1
lim Sn = lim
n →∞ n →∞
∑r + rn
O r =1
X
2
y = 4ax
n
11
lim
n →∞
∑r r n
.
= r =1
+
n n
We have,

⎧ 4ax, if x < a 1 1
y2 = ⎨ dx 2tdt
⎩ 4a(2a − x ) if x ≥ a = ∫ =∫
t + 1)
(t = x )
0 x + x 0
t (
Therefore required area

= 2{log(t + 1)}0
a 1
16a2
= 4∫ 4ax =
3
0
= 2loge2
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Test - 6 (Paper-I) (Answers & Hints) All India Aakash Test Series for JEE (Main)-2014

84. Answer (4) 87. Answer (1)

Differentiating the given function w.r.t. x, we get Clearly f(1) = 2, f(1+) < 2, f(1–) < 2
Hence x = 1 is the point of local maxima
1
f ′( x ) = sin θ.3 tan2 x.sec 2 x + (sin θ − 1)sec 2 x Hence statement-1 is true
3
Statement-2 is also clearly true and explains
= sinθ.tan2x.sec2x + (sinθ – 1)sec2x statement-1
Now, f′(x) = 0 88. Answer (3)
1 − sin θ f(0) = 2, but clearly f(0+) < 2, f(0–) < 2. Hence x = 0
⇒ tan2 x = <0 is the point of local maxima.
sin θ
Hence statement is true.
which is not true. Therefore, f(x) has no critical
points. Statement-2: Put y = –x
85. Answer (2) f(0) = f(x) + f(–x)
Put cosx = t, so that –1 < t < 1. put x = 0, we get f(0) = 0
⇒ f(x) + f(–x) = 0 ⇒ f(x) is odd
(2 + t )2
Let φ(t ) = , −1< t < 1 2
1− t 2
⇒ ∫ (f ( x ) + f ( −x )) = 0

n
−2
Differentiating, we get

2(2t + 1)(t + 2) .i
Statement-2 is false.
al
φ′(t ) = 89. Answer (4)
(1 − t 2 )2
Statement-1: The integral can be written as
n
1
∫ e ( ( x tan x ) + ( x sec x + tan x ) ) dx = e x tan x + c
x 2
Since, –1 < t < 1, hence t = −
ur

only
2
⇒ f(x) = ex tanx ⇒ f(0) = 0
o

1 Hence statement-1 false

Also, φ′(t ) < 0 for t < −
2
uj

Statement-2: The integral can be written as

1
and φ′(t ) > 0 for t > − ⎛⎛ 1⎞ ⎛ 1 ⎞⎞ x ⎛ 1⎞
∫e
Ed

⎜ ⎜ x + ⎟ + ⎜ 1 − 2 ⎟ ⎟ dx = e ⎜ x + ⎟ + c
x
2
⎝ ⎝ x ⎠ ⎝ x ⎠ ⎠ ⎝ x ⎠
1
Hence, φ is minimum at t = − and least value =
2 1
⇒ f ( x) = x + ⇒ f(1) = 2
2 x
⎛ 1⎞
⎜2 − ⎟
⎝ 2⎠
=3.
Hence statement-2 is true
1
1− 90. Answer (3)
4
Statement-1: The integral can be written as
86. Answer (1)
π π /2 π
Statement-1 : f ′ ( x ) = 3 x 2 + 2bx + c > 0 ∫ [ sin x ] dx = ∫ 0.dx + ∫ 0.dx = 0
0 0 π /2
⇒ 4b2 – 4 × 3 × c < 0
Hence statement-1 is true
⇒ b2 < 3c
Statement-2: The integral can be written as
⇒ 0 ≤ b 2 < 3c
20 π π
Hence statement-1 is true. ∫ sin x dx = ( 20 − 10 ) ∫ sin x dx = 10 × 2 = 20
10 π 0
Statement-2 is clearly true and statement-2 explains
statement-1. Statement-2 is false.

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