Mock Test - 2 (Code-A) (Answers) All India Aakash Test ... · 2/1/2020 · Mock Test - 2 (Code-A)...
Transcript of Mock Test - 2 (Code-A) (Answers) All India Aakash Test ... · 2/1/2020 · Mock Test - 2 (Code-A)...
Mock Test - 2 (Code-A) (Answers) All India Aakash Test Series for JEE (Main)-2020
All India Aakash Test Series for JEE (Main)-2020
Test Date : 02/01/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
MOCK TEST - 2 - Code-A
1/14
PHYSICS CHEMISTRY MATHEMATICS
1. (1)
2. (3)
3. (2)
4. (3)
5. (3)
6. (3)
7. (3)
8. (1)
9. (3)
10. (4)
11. (4)
12. (2)
13. (3)
14. (2)
15. (1)
16. (2)
17. (2)
18. (2)
19. (3)
20. (3)
21. (17.00)
22. (41.00)
23. (19.50)
24. (10.00)
25. (18.00)
26. (3)
27. (2)
28. (4)
29. (2)
30. (2)
31. (3)
32. (3)
33. (4)
34. (4)
35. (1)
36. (4)
37. (2)
38. (4)
39. (1)
40. (3)
41. (2)
42. (2)
43. (1)
44. (2)
45. (1)
46. (01.00)
47. (18.00)
48. (30.18)
49. (06.00)
50. (11.00)
51. (3)
52. (2)
53. (3)
54. (2)
55. (3)
56. (3)
57. (1)
58. (3)
59. (2)
60. (3)
61. (2)
62. (4)
63. (4)
64. (1)
65. (2)
66. (3)
67. (1)
68. (2)
69. (4)
70. (2)
71. (05.00)
72. (12.00)
73. (00.25)
74. (00.50)
75. (-02.37)
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
2/14
1. Answer (1)
Hint : Itriangle = 2
6Mh
Sol. :
2
2
12
6 24
LM MLI
× = =
2
2 2
2
32
6 242 3
LM L MLI M
× = − × =
2
1 2 12cMLI I I∴ = + =
2. Answer (3)
Hint : vmin = ucosθ
Sol. : 2min
2
1(KE) ( cos )21 2 (20 cos ) 1442
m u= θ
= × × × θ =
12 3cos20 5
⇒ θ = =
22 sin cosRange u
gθ θ
∴ =
3 42 20 205 5
10
× × × ×=
32 12 38.4 m10×
= =
3. Answer (2)
Hint : 00
s
v vn nv v
−′ = ×−
Sol. : n1 = 300 Hz
2 0330 5 300 309 Hz330 5
c
c
v vn n
v v+ +
= × = × =− −
∴ Beats = 309 – 300 = 9 Hz.
4. Answer (3)
Hint : dE dldt
φ⋅ = −∫
Sol. : 2
2 B RE Rt
× π× π =
∆
2BRE
t⇒ =
∆
2( )E Q R t MR∴ × × × ∆ = × ω
2
2BR QR t MR
t⇒ × × ∆ = × ω
∆
2BQ
m⇒ ω =
5. Answer (3)
Hint : Linear momentum is not conserved but angular momentum is conserved.
Sol. : 2
20 3
mLmv L mL
× = + × ω
034vL
⇒ ω =
Now, 2
21 42 3 2
mL LmgL mg
× × ω = +
2202
92 33 216
vmL mgLL
⇒ × =
20 04 2v gL v gL⇒ = ⇒ =
6. Answer (3)
Hint : Theoretical
Sol. : Demodulation is carried out with rectifier and envelope detector.
7. Answer (3)
Hint : Capacitors come into parallel combination.
Sol. : Q1 = 1 × 100 = 100 µC
Q2 = 2 × 20 = 40 µC
∴ Charge on plates connected through
S1 = –100 + 40 = –60 µC
When switches are closed, the two capacitor becomes in parallel.
PART - A (PHYSICS)
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
3/14
11 ( 60) 20 C
1 2⇒ = × − = − µ
+q
2
2
40 C,( ) 20 100
qQ S
= − µ
∴ ∆ = − +
= 80 µC
8. Answer (1)
Hint : Time period = 2 mqBπ
Sol. : 2 , at ,m mT tqB qBπ π
= ∴ = velocity of
particle becomes
2ˆ ˆ ˆ2 3v i j k= − −
Now, 0m eF F+ =
2( ) 0q v B qE⇒ × + =
ˆ ˆ ˆ ˆ ˆ ˆ( 2 3 ) (4 ) 12 8E i j k i j k⇒ = − − × = − +
9. Answer (3)
Hint : Use 2
0 20
, ,4
qF i B B nI Fr
= = µ =πε
Sol. : 0 2[ ] Fi
µ =
2
0 2[ ] qFr
ε =
2 2
02 2
0
F ri q−
µ ∴ = ε
2 2 2
2 2(MLT ) L
A (AT)
− ×=
×
= M2L4T–6A–4
10. Answer (4)
Hint : Theoretical
Sol. : Size of collector plate is largest.
11. Answer (4)
Hint : The leakage process is slow so use equilibrium equations.
Sol. : Tcosθ = mg
2
20
sin4
qTx
θ =πε
2
20
tan4
qx mg
⇒ θ =πε ×
2
202 4
x qmgx
⇒ =πε
3/202 mg x qπε⇒ × =
0232
mgdq dxxdt dt
πε ∴ − = ⋅ ×
0232
mga πε=
12. Answer (2)
Hint : Use deviation graph
Sol. : i i A′δ = + −
90 (initial)i i ′+ = °
( ) 90 (final)i i ′+ < °
50 90i ′+ < °
40i ′⇒ < ° but 37i ′ > °
∴ Angle of emergence can be 38°
13. Answer (3)
Hint : Use conservation of energy
Sol. : 212 3
GMm GMmmvR R
−= +
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
4/14
21 22 3
GMmmvR
= ×
4 23 3GM gRvE
⇒ = =
14. Answer (2)
Hint : Speed of efflux = 2gh
Sol. : 1 2, 2 ,h h h v gh= − = 1 22dh dhA Adt dt
− × =
and 1dhA a vdt
− = ×
2 23
A Hta g
⇒ =
15. Answer (1)
Hint : Restoring force, ∆F = kx
Sol. : Restoring force
1 1 22 2
F k x kx ∆ = × × × × =
2dmm kxdt
∴ = −
2 mTk
∴ = π
16. Answer (2)
Hint : 1 2tan ; tanC LX XR R
φ = φ =
Sol. : tan60 3LL
X X RR
° = ⇒ =
tan60 3CC
X X RR
° = ⇒ =
2 2( )C Lz R X X R∴ = + − =
rms200 2 A100
I∴ = =
17. Answer (2)
Hint : Shift = ( 1)tDd
µ −
Sol. : Shift = ( 1)tDd
µ −
∴ Number of fringes = ( 1) ( )( )tD d
d Dµ − ×
× λ
( 1) 10tµ −= =
λ
18. Answer (2)
Hint : Use shift = 11 t − × µ
Sol. : 1Shift 1 1cmt = − × = µ
21 1 20 cmu∴ = − =
20 15 60 cm(20 15)
v ×∴ = =
−
final 60 1 61cmu∴ = + =
19. Answer (3)
Hint : 20 2 2
1 2
1 1E E zn n
∆ = −
Sol. : 21 0
11 40.84
E E z ∆ = − =
20 54.4 eVE z⇒ =
24 3 0
1 1 7 (54.4)9 16 9 16
E E z→ ∆ = − = × ×
1240 9 16 470 nm7 54.4
× ×∴ λ =
×
20. Answer (3)
Hint : KEmax = hν – φ
Sol. : h kν = φ +
21. Answer (17.00)
Hint : 2
cc
vra
=
Sol. : ˆ ˆ ˆ ˆ4 , At 1 s, 4v t i j t u i j= + = = +
ˆ4a i=
2
2 2, cc
vr a a aa τ= = −
2
16 16 11716 1
tat
τ× ×
= =+
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
5/14
216 16 1 416
17 17 17ca ×∴ = − = =
16 1 17 17 m4c
ra
+∴ = =
22. Answer (41.00)
Hint : distancetimeavv =
Sol. : 2 0 0,1 sv t t t= − = ⇒ =
1 1
2
0 0
( )x
dx t t dt= −∫ ∫
13 2
10
1 m3 2 6t tx
⇒ = − = −
And 2 4
2
0 0
40( ) m3
x
dx t t dt= − =∫ ∫
total40 1 412 m3 6 3
S∴ = + × =
avg41 41 m/s
3 4 12V∴ = =
×
23. Answer (19.50)
Hint : ∆θ = ∆U + ∆W
Sol. : A B vQ nC T→∆ = ∆
Let at 0, ,AA T T= then 02BT T=
0 0 05 5( )2 2ABQ n R T P V∴ ∆ = × × =
0(2 )ln2BC BCQ nR T∆ = ∆ω = ×
0 0 0 02 0.7 1.4P V P V= × =
0 0 0 0 0 02.5 1.4 3.9ABCQ P V P V P V∴ ∆ = + =
⇒ 5n = 19.50
24. Answer (10.00)
Hint : dN Ndt
= α − λ
Sol. : dN Ndt
= α − λ
0 0
N tdN dtN
⇒ =α − λ∫ ∫
11Ne
α ⇒ = − λ
25. Answer (18.00)
Hint : dN Ndt
− = λ
Sol. : (238)
(206)
102383
206
NN
=
0
00.75 1.33NN
N N∴ = ⇒
0 tNe
Nλ
∴ =
10 90.72.303log (1.33)
4.5 10t⇒ = ×
×
91.8 10 yearst⇒ = ×
26. Answer (3)
Hint: A buffer system has maximum buffer capacity when pH= pKa.
Sol. : For (3), pKa = 8 – log 6.2
= 7.2
27. Answer (2)
Hint: Geminal diacids on heating lose CO2.
Sol. :
PART - B (CHEMISTRY)
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
6/14
28. Answer (4)
Hint: Ammonia causes asymmetrical cleavage.
B2H6 + NH3 → [BH2(NH3)2]+ [ –4BH ]
Sol. : LiBH4 is a reducing Agent.
29. Answer (2)
Hint: 4 3 2 2NH NO N O H O∆→ +
Sol. : NH4ClO4, NH4NO2 and (NH4)2Cr2O7
produce N2 on heating.
30. Answer (2)
Hint: Pentagonal rings increase the deviation from planarity.
Sol. : Deviation from planarity will be most severe at (C) and least severe at (B).
31. Answer (3)
Hint: Greater the energy of incident light, greater is the energy of the photoelectron.
Sol. : Blue light has greater energy than Red light.
32. Answer (3)
Hint: Aromaticity based on Huckel’s Rule.
Sol. :
33. Answer (4)
Hint: Minute living organisms dispersed in
atmosphere are examples of viable
pollutant.
Sol. : Smoke, Mist, Fumes, and Dust are non-
viable pollutants.
34. Answer (4)
Hint: Functional isomers are not metamers.
Sol. : 2° amines and 3° amines are functional
isomers.
35. Answer (1)
Hint:
Sol. :
And
36. Answer (4)
Hint: Valium is
Sol. : Valium contains cyclic amide
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
7/14
37. Answer (2)
Hint: Loss of aromaticity is the rate determining step.
Sol. : The stabilisation of the carbanionic intermediate plays a major role in rate of ArSN2.
38. Answer (4)
Hint: Cu
3 2 3573 KCH — CH — OH CH — CHO→
Sol. : In 1° and 2° alcohols, Cu causes dehydrogenation. In 3° alcohols, it causes dehydration.
39. Answer (1)
Hint: Cannizzaro reaction followed by intramolecular esterification.
Sol. :
40. Answer (3)
Hint: The reaction is electrophilic aromatic substitution.
Sol. : Greater the electrophilic nature, greater is the rate of reaction.
41. Answer (2)
Hint:
( )2H O3 2
3 –3 3 6FeFe (OH) HCH COO CH COO + ++ + → +
Sol.:( )
( ) ( )3 2 2
2 3 3
3 6Fe (OH) H O
Fe O
CH C
H CH COO CH COOH
OO
H
+
+
+ →
+ +
42. Answer (2)
Hint : is a copolymer.
Sol. : It is urea formaldehyde resin.
43. Answer (1)
Hint: Pepsin and trypsin convert proteins into amino acids.
Sol. : Invertase acts on sucrose and maltase acts on Maltose.
44. Answer (2)
Hint: (N atoms in Y) × 3 = 3 × 2.
⇒ N atoms in Y = 2
Sol. : N3H has O.S of N as – 13
N2H4 has O.S of N as –2
KMnO4 cannot reduce N .
45. Answer (1)
Hint: Configuration of chiral carbon does not change as no bond is broken.
Sol. : L-Lactic acid has S configuration. So both the chiral centre in dilactide should have S configuration.
46. Answer (01.00)
Hint: Rate law ; [ ]2d N Odt
= [ ]2 2
3
NO NHK
H O+
Sol.: Step (ii) is RDS.
∴ Rate = 3 2K [NO NH ]−
and Keq = [ ] [ ]
–2 31
2 2 2 2
NO NH H OKK NO NH H O
+ =
[ ] –– 21 2
22 3
NO NHK H ONO NH
K H O+
∴ =
1 3 2 22
2 3
K K [NO NH ]Rate [H O]K [H O ]+
∴ =
2 2
3
[NO NH ]K[H O ]+
=
In a buffer solution [H3O+] = constant
So, order = 1
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
8/14
47. Answer (18.00)
Hint: ‘a’ is insignificant
P (Vm – b) = RT
Sol. : Vm = 10b
⇒ b = mV10
So, m9PV RT
10 =
mPV 10RT 9
⇒ =
R
i
V 10ZV 9
∴ = =
Ri
9VV10
⇒ =
⇒ Vi = 9b
48. Answer (30.18)
Hint: Csystem = Ccalorimeter+ Cwater
Sol. : Total heat liberated (Q)
= 0.6 3050 15 kJ122
× =
Q 15000Total heat capacity 6000T 2.5
∴ = = =∆
∴ 710 × 4.2 + C = 6000
C = 3018 JK–1
49. Answer (06.00)
Hint: Diacids can be geminal (1, 1), vicinal (1, 2) or isolated (1, 3)
Sol. :
50. Answer (11.00)
Hint: B is exocyclic alkene.
Sol. :
51. Answer (3)
Hint : Coefficient of x9 in 10 10( 1) (1 )dx xdx
+ ⋅ +
Sol. : 10 2 100 1 2 10(1 ) ...x C C x C x C x+ = + + + +
Differentiating w.r.t. x 9 2
1 2 39
10
10(1 ) 1 2 3 ...
10 ...(i)
x C C x C x
C x
+ = ⋅ + ⋅ + ⋅ +
+ ⋅
10 10 9 80 1 2 10( 1) ...x C x C x C x C+ = + + + +
…(ii)
Comparing the coefficient of x9 in (i) × (ii)
19 2 2 2 29 1 2 3 1010 1 2 3 ... 10C C C C C⋅ = ⋅ + ⋅ + ⋅ + + ⋅
52. Answer (2)
Hint : Favourable cases are (2, 2, 1), (1, 2, 3), (2, 2, 2), (1, 3, 3) and (2, 2, 3), (1, 1, 3)
Sol. : Total favourable cases are
(1, 2, 2), (1, 2, 3), (2, 2, 2), (1, 3, 3) and (2, 2, 3), (1, 1, 3)
PART - C (MATHEMATICS)
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
9/14
Required probability
1 3 3 1 3 2 3 3 33 66 6 6 6 6 6 6 6 6
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
1 2 2 3 3 2 1 33 36 6 6 6 6 6 36 4
+ ⋅ ⋅ + ⋅ ⋅ + =
53. Answer (3)
Hint : 62 0, 14 1
x mm
= − ⇒ = −+
Sol. : x + 4y = 8m and y = mx + 1
x + 4mx + 4 = 8m
8 4 624 1 4 1mxm m
−= = −
+ +
624 1m
− +
is an integer then 4m + 1
= ±1, ±2, ±3, ±6 but in is also integer then m = 0, –1
54. Answer (2)
Hint : tan75° = 2 3+
Sol. : Let height of the flag pole be h
50 50tan603
xx
° = ⇒ =
Also, 50tan75 hx+
° =
502 3 503
h+⇒ + =
100 50 503
h⇒ + = +
1003
h =
55. Answer (3)
Hint : |z1 + z2| ≤ |z1| + |z2| and equality holds when z1, z2 and origin are collinear and z1, z2 lie on the same side of origin.
Sol. : |z1 – z2| = |z1 + z2| ⇒ OP ⊥ OQ because z1 lies on the perpendicular bisector of the line segment joining z2 and –z2.
|z1 + iz3| = |z1| + |iz3| ⇒ z1 and iz3 are collinear with origin.
⇒ OP ⊥ OR
So Q and R will be collinear with O.
56. Answer (3)
Hint : Use sine rule and cosine rule.
Sol. :
coscot sin
cos sin cos sincot cotsin sin
AA A
B C C BB CB C
=⋅ ++
2cos sin sin sin sin cossin( ) sin sin
A B C B C AB C A A
⋅ ⋅= = ⋅
+ ⋅
2 2 2 2 2 2
2 2 22 2
bc b c a b c abca a
+ − + −= ⋅ = =
57. Answer (1)
Hint : S1 must be independent of α.
Sol. : Let point P is (h, k)
Length of tangent = 1S
2 2 2( 2 ) (3 ) ( 4)h k h h k h k= + + + α + + α − −
3h + k = 0
and h – k = 4
h = 1
k = –3
Point P(1, –3)
2 21 ( 3) 10OP = + − =
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
10/14
58. Answer (3)
Hint : Chord with given mid-point T = S1
Sol. : Equation of chord T = S1
1 11 14 3 4 3x y
⇒ + − = + −
3 4 7x y⇒ + =
On solving this line with the equation of ellipse
22 1 3 7 14 3 4x x − + =
−
2 212 (9 42 49) 48x x x⇒ + − + =
221 42 1 0x x⇒ − + =
Let the point of intersections be (x1, y1) and (x2, y2)
2 21 2
1 80( ) (2) 421 21
x x − = − =
Also 1 2 1 23| | | |4
y y x x− = −
So 21 2
45( )21
y y− =
Length of chord
2 21 2 1 2
125( ) ( )21
x x y y= − + − =
59. Answer (2)
Hint : 2
2( )ixx
Nσ = −∑
Sol. : Incorrect 7x =
⇒ Incorrect 7 6 42ix = × =∑
So correct 42 9 6 39ix = − + =∑
⇒ correct 39 136 2
x = =
Incorrect σ = 1
22Incorrect
(7) 16
ix− =∑
Incorrect 2 300ix =∑
Correct 2 2 2300 9 6 255ix = − + =∑
So correct
22correct
(correct ( ))6
ixxσ = −∑
2255 13 0.5
6 2 = − =
60. Answer (3)
Hint : lim
lim
1lim ( )n
n
bnb
n r a an
rf f x dxn n
→∞
→∞
→∞ =
= ∑ ∫
Sol. : 3 31
n
nr
rIn r=
=+
∑
31
1lim lim1
n
nn n r
rnI
n rn
→∞ →∞ =
= +
∑
1
3/23
0
let1
x dx x tx
= =+∫
1
1/22
0
2 23 31
dt x dx dtt
= =+
∫
1
20
2 ln 13
t t = + +
( )2 ln 1 23
= +
61. Answer (2)
Hint : Check L.H.D. and R.H.D. at x = 1 and x = 3.
Sol. : f(x) = (x – 3)(x – 4)|(x – 1)(x –3)| + |x – 3|sinπx
Let f(x) = g(x) + h(x)
Where g(x) = (x – 3)(x – 4)|(x – 1)(x – 3)|
g(x) is not differentiable at x = 1 only.
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
11/14
( x = 3 is the repeated root of g(x), so it will be
differentiable at x = 3)
h(x) = |x – 3|sinπx is always differentiable
( x = 3 is repeated root of h(x) = 0)
So f(x) is not differentiable at x = 1 only.
62. Answer (4)
Hint : 2
(2) lim ( )x
f f x→
>
Sol. : For x > 2
f(x) is decreasing
So it cannot have a local minima at x = 2 for any value of k.
63. Answer (4)
Hint : A is skew symmetric matrix.
Sol. : 0ij jk kia a a+ + =
Put 3 0iii j k a= = =
0 1, 2, 3iia i⇒ = ∀ =
Put 0ij ji iii k a a a= + + =
ij jia a⇒ = −
So matrix A will be skew symmetric matrix. Then
sum of all elements 1 , 3
0iji j
a≤ ≤
=∑
Also Tr(A) = 0 and det(A) = 0.
64. Answer (1)
Hint : ( ) ( )a b d c b d× × = × ×
Sol. : a b c b× = ×
( ) ( )a b d c b d⇒ × × = × ×
( ) ( ) ( ) ( )a d b b d a c d b b d c⇒ ⋅ − ⋅ = ⋅ − ⋅
0a d⋅ =
c da c bb d
⋅⇒ = − ⋅
65. Answer (2)
Hint : ∆ABC is right angled isosceles triangle.
Sol. : AB = 2, BC = 2 and AC = 2
So, ∆ABC is right angled isosceles triangle.
12
ABR = = and
1 2 2 122 12 2 2
2
rS
⋅ ⋅∆= = =
++
66. Answer (3)
Hint : ( 1) let ( 2)1 ( 2)
xx
xe x dx x e t
x e
−−
−+
+ =+ +∫
Sol. : 1 ( 1)( 2) 1 ( 2)
x
x xx x edx dx
e x x e
−
−+ +
=+ + + +∫ ∫
Let (x + 2)e–x = t
(–(x + 2)e–x + e–x)dx = dt
(x + 1)e–x dx = –dt
1dt
t= −
+∫
= –ln(1 + t) + c
= –ln(1 + (x + 2)e–x) + c
= x – ln(ex + x + 2) + c
67. Answer (1)
Hint : Use 0 0
( ) ( )f x dx f x dxπ π
= π −∫ ∫
Sol. : Let 2
/20
sinx
xI dxe e
π
π=+∫ …(i)
2 2
/2 /2 /20 0
sin sin( )
x
x xx x eI dx dx
e e e e e
π π
π π− π π⋅
= =+ +∫ ∫ …(ii)
(i) + (ii)
2 /2
/2 /20
sin ( )2( )
x
xx e eI dx
e e e
π π
π π+
=+∫
2/2
0
12 sinI x dxe
π
π⇒ = ∫
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
12/14
/20
1 1 cos222
xI dxe
π
π− ⇒ =
∫
/222
Ieππ
⇒ =
/214
I e−π⇒ = π
68. Answer (2)
Hint : Draw graph
Sol. : , (2,8)x y ∈
Then [ ]3 3 1x y = =
So, curves are y2 = 2x and x2 = 2y
Area = 4 8 82
2 4 2
(12 ) 22x dx x dx x dx+ − −∫ ∫ ∫
8 843 2
3/2
2 24
2 2126 2 3x xx x
= + − − ⋅
56 2 2(48 24) (16 2 2 2)6 3
= + − − −
443
=
69. Answer (4)
Hint : Variable separable form
Sol. : 2 2( )x dy yd x ydx+ =
2
22 2
( ) 1ln( )d x y dx x y cxx y x
⇒ = ⇒ = − +
When x = 1, y = 1 ⇒ c = 1
When 1 12; ln(4 ) 12 2
x y= = − =
4ey⇒ =
70. Answer (2)
Hint : Truth table
Sol. :
71. Answer (05.00)
Hint : Find roots of both equations.
Sol. : 2 ( 3) 2 2 0x k x k− + + + =
2( 3) ( 6 9) 8 8
2k k k kx + ± + + − −
⇒ =
( 3) ( 1)2
k kx + ± −⇒ =
1, 2x k⇒ = +
2 2 2( 4) 3 3 0x k x k− + + + =
2 2 2 2( 4) ( 4) 4(3 3)
2k k kx + ± + − +
⇒ =
2 2 2( 4) ( 2)
2k kx + ± −
⇒ =
2 1, 3x k⇒ = +
21,{2, 3, 1}k kA B +∪ = +
A B∪ contains only 3 elements then
k + 1 = k2 + 1 ⇒ k = 0, 1
k + 1 = 2 ⇒ k = 1
k2 + 1 = 2 ⇒ k = 1, –1
Mock Test - 2 (Code-A) (Hints & Solutions) All India Aakash Test Series for JEE (Main)-2020
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
13/14
k + 1 = 3 ⇒ k = 2
k2 + 1 = 3 ⇒ 2, 2k = −
But at k = 1, A ∪ B contains exactly 2 elements
So k = 0, 2, –1, 2, 2−
72. Answer (12.00)
Hint : loga ba b=
Sol. : ( ) 232 2
log 3log 5log 3 log 52 2 5a = = =
( ) 33 5 3
log 5log 5 log 7 log 73 3 7b = = =
73. Answer (00.25)
Hint : If normal to the parabola at P(t1) cuts it
again at Q(t2) then 2 11
2t tt
= − −
Sol. : 21 1( , 2 )L at at where t1 = 1
If normal cuts the parabola at 22 2( , 2 )M at at
Then 2 11
2 3t tt
= − − = −
Tangents at L and M intersect at N then
1 2 1 2( , ( ))N at t a t t+
So the vertices of ∆LMN are 1 1 9 3, , ,4 2 4 2
L M −
and 3 1,4 2
N − −
Area of ∆LMN =
1 1 14 2
1 9 3 1 22 4 2
3 1 14 2
− =
− −
74. Answer (00.50)
Hint : Let 1,P tt
then
: 2xT ytt
+ = and 221: yN xt t
t t− = −
Sol. : Let point P as 1,tt
Equation of tangent : T = 0
2x ytt
+ =
2(2 ,0) and 0,A t Bt
Equation of normal
221yxt t
t t− = −
4
31,0tC
t −
and 4 10, tDt
−−
Area of 4
31 1 122
tPAC ttt
− ∆ = ⋅ − ⋅
4
41 12
tt
+=
Area of 41 2 1
2tPBD t
t t −
∆ = + ⋅
4 12
t +=
1 1area of area ofPAC PBD
+∆ ∆
4
4 42 2 2
1 1t
t t= + =
+ +
75. Answer (-02.37)
Hint : 3( ) 1f x x= −
All India Aakash Test Series for JEE (Main)-2020 Mock Test - 2 (Code-A) (Hints & Solutions)
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
14/14
Sol. : If 1 1( ) ( )f x f f x fx x
+ = ⋅
1 1( ) ( ) 1 1f x f f x fx x
⋅ − − + =
1( ( ) 1) 1 1f x fx
− − =
Let ( ) 1 ( )f x g x− =
Then 1( ) 1g x gx
⋅ =
( ) ng x x⇒ = ±
So ( ) 1 nf x x= ±
( 2) 9f − =
Then 3( ) 1f x x= −
3 192 8
f − =