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All India Aakash Test Series for JEE (Advanced)-2021
Test Date : 21/06/2020
ANSWERS
Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456
TEST -1A (Paper-2) - Code-C
Test -1A (Paper-2) (Code-C)_(Answers) All India Aakash Test Series for JEE (Advanced)-2021
1/13
PHYSICS CHEMISTRY MATHEMATICS
1. (A, B, C)
2. (A, C)
3. (A, C)
4. (A, C)
5. (C, D)
6. (B)
7. (A)
8. (C)
9. (C)
10. (B)
11. (A)
12. (B)
13. (D)
14. (C)
15. (D)
16. A → (Q, R, T)
B → (P, S)
C → (Q, S)
D → (Q, R)
17. A → (R)
B → (T)
C → (T)
D → (S)
18. (03)
19. (02)
20. (03)
21. (A, C, D)
22. (A, C, D)
23. (A, B, D)
24. (B)
25. (A, C)
26. (A)
27. (A)
28. (C)
29. (B)
30. (B)
31. (A)
32. (A)
33. (B)
34. (B)
35. (C)
36. A → (P, S)
B → (P, T)
C → (Q, R)
D → (P, S, T)
37. A → (P, R, T)
B → (Q, R, S)
C → (P)
D → (Q)
38. (08)
39. (04)
40. (03)
41. (B, C, D)
42. (A, B, D)
43. (A, B, C)
44. (B, C)
45. (A, B, D)
46. (A)
47. (D)
48. (C)
49. (B)
50. (D)
51. (C)
52. (D)
53. (A)
54. (C)
55. (B)
56. A → (P, S)
B → (Q, R, T)
C → (R, T)
D → (T)
57. A → (P, R, S)
B → (R, S)
C → (T)
D → (T)
58. (02)
59. (04)
60. (06)
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HINTS & SOLUTIONS PART - I (PHYSICS)
1. Answer (A, B, C)
Hint :
Let angle subtended by equal to Q, at Point P
q1 = r1Qλ
q2 = r2Qλ
1 1
2 2
=q r
q r
Solution :
V due to the elements are 1 2
2 2
andkq kq
r r
Since 1 2
1 2
=q q
r r, hence potentials are equal
⇒ Electric feed at point P has magnitude
1 21 22 2
1 2
and= =kq kq
E Er r
Hence,
1
1 11 2
22 1
2 2
1
1
= ⇒
q
r rE r
qE r
r r
2. Answer (A, C)
Hint :
Now electric field at any distance x from centre
on axis of hole be
20 0 2
– 1–2 2σ σ = ε ε +
xE
x R
2 2
02σ
=ε +
x
R x
Solution :
2 20
–
2
σ=
ε +
e xF
x R
2 20
–
2
σ=
ε +
dv e xmv
dx x R
0
2 20 3 0
–
2
σ=
ε +∫ ∫v
R
e xdxmvdv
x R
2 02 2
30
–2 2
σ = + ε R
mv ex R
[ ]0 0
–– 2
2 2σ σ
= =ε εe eR
R R
⇒ 0
σ=
εe R
vm
3. Answer (A, C)
Hint :
On closing the switches and earthing, charges on left-most and right-most surfaces will become = 0
Solution :
E = 0 inside the conductor
Since A and B connected, hence
VA = VC ( connected)
⇒ VA – VB = VC – VB
E1d = E2d
⇒ E1 = E2
⇒ 0 0
=ε εx y
A A
⇒ x = y …(i)
Also VB = VD (both earthed)
VC – VD = VC – VB
E3d = E2d
⇒ y = z …(ii)
By charge conservation,
x + y + z = Q0
⇒ 3x = Q0
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0
3= = =
Qx y z
Hence charge flown through
( ) 0 01 0
2 5
2 3= + + = + =
Q QS x y Q
Magnitude of charge flown through
0 02 0
72
3 3= + =
Q QS Q
4. Answer (A, C)
Hint :
Let speeds of A and B just before collision
By conservation of linear momentum
mVA = 2mVB
⇒ VA = 2VB …(i)
Solution :
Hence
2
2
122
1 122
= =AA
BB
mVK
K mV
By conservation of total mechanical energy
Ui + Ki = Uf + Kf
⇒ 2 2– –
010 2
+ = + +A BKQ KQ
K KR R
⇒ 22
5+ =A B
KQK K
R
5. Answer (C, D)
Hint :
Conductors are equipotential, hence VA = VB…
Solution :
in
0 0
lim→∞
⋅ = =ε ε∫
x
q qE ds
1
∝ σ ∝ER
Hence E will not be equal
6. Answer (B)
Hint :
( )3
2 2 204
=
πε +
QxE
R x
Solution :
The net force on the dipole isrE
Prx
( ) ( )
( )
3 12 2 2 2 22 2
022 2
32 – 2
4 2
+ +
πε =+
Qqa R x R x x
R x
( ) [ ]
[ ]
12 2 2 22
32 20
– 22
4+
=πε +
Q R x R xqa
R x
[ ]
[ ]
2 2
50 2 2 2
– 22
4=
πε+
Q R xF qa
R x
7. Answer (A)
Hint :
W = ∆u
Ui = – PEcos0
Uf = – PEcosπ
Solution :
( )
( ) ( )3 1
2 2 2 22 20 0
2. 22
4
= = =
πε + πε +
qa Qx aqQxW PE
R x R x
8. Answer (C)
Hint :
2= πI
TPE
Solution :
( )
2
32 2 2
0
22
2
4
= π πε +
mT
Qxqa
R x
( )
32 2 2
042
πε += π
R x maT
qQx
9. Answer (C)
Hint :
Find arrangement of charge before and after closing switch
Solution :
Before
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Before
Considering isolated plates
Hence charge flown =5
CE
10. Answer (B)
Hint :
WDcell = qsupply× emf of cell
Solution :
WDcell = qsupply × emf of cell
12 2 12
–5 3 7
= + CE CE CE
2105
=CE
2
cell2105
=CE
WD
11. Answer (A)
Hint :
H = WDcell – ∆u
Solution :
H = WDcell – ∆u
2 2
22 1 12 1 50– –
105 2 5 2 21
= × ×
CE CE
CE
2 2 22 6 25
–105 5 21
= +CE CE CE
2
22 – 126 125105 105
+= =
CECE
12. Answer (B)
Hint :
Apply superposition principle.
Solution :
By superposition principle the given system is
equivalent to
13. Answer (D)
Hint :
( ) ( )( )
( )
0 0
,
0 0,
, – , –= ⋅∫
x y
x y
V x y V x y E dr
Solution :
( ) ( )( )
( )
0 0
,
0 0,
, – , –= ⋅∫
x y
x y
V x y V x y E dr
Let (x0, y0) be (Q, 0) and V(x0, y0) = 0
⇒ ( ) ( ),
0,0
, –= +∫x y
V x y ydx xdy
( ) ( ),
0,0
, – –= =∫x y
V x y d xy yx
Hence equation of equipotential surface is like
xy = constant which represent a hyperbola.
14. Answer (C)
Hint :
Apply Gauss’s Law.
Solution :
As position of charge q is changed, electric field
and charge distribution on inner walls will
change. Electric field outside depends upon the
rotate distance from centre of sphere for any
point that lies outside the sphere.
15. Answer (D)
Hint :
Electrostatic flied are conservative, hence WD
in closed path = 0 and also WD is path
independent
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Solution :
Electrostatic flied are conservative, hence WD in
closed path = 0 and also WD is path
independent
∴ 0⋅ =∫
E dl or2 2 2
1 1 1
, ,
, ,
0⋅ =∫
x y z
x y z
E dr
for (x1, y1, z1) = (x2, y2, z2)
( )1 2 3 0⋅ = + + ≠∫ ∫
E dl xydx yzdy xzdz
( )2 22
ˆ2 2⋅ = + + +∫ ∫
E dl y dx xy z dy yzk
( ) ( )2 2= +∫ ∫d xy d z y
2 2 2
1 1 1
, ,2 2, , 0 = + =
x y z
x y zxy z y
Hence only 2
E is conservative hence
electrostatic field
16. Answer A(Q, R, T); B(P, S); C(Q, S); D(Q, R)
Hint :
Use Field formula.
Solution :
(A)
(B)
(C)
(D)
17. Answer A(R); B(T); C(T); D(S)
Hint :
Apply KVL.
Solution :
By KVL in loop-1
( )– 2 – –
– 01 3 2
+ =x x Q Q x
⇒ –6x – 4x + 2Q + 3Q – 3x = 0
⇒ –5Q = 13x …(i)
By KVL in loop-2
( )–
10 – – 02 1
=Q x x
20 – Q + x – 2x = 0
20 – Q = x …(ii)
From equation (i) and (ii)
50
20 –13
=Q
18
2013
=Q
260 13018 9
= =Q
⇒ 5 130 50
13 9 9= × =x
Now –2
=B CQx
V V
130 50– 409 9– 02 9
= =BV
409
=BV
50
–1 9
= =D Cx
V V
509
=DV
Charge on
100 130 30 103 2 – – – –
9 9 9 3µ = = = =F x Q
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18. Answer (03)
Hint :
Use COE
Solution :
By energy conservation
( )
22 2
30 0
13 – 0
2 4 48
+ = +
πε + πε
Qq Qq Rkx R
R x R
( )
2
0 0
1 112 4 32
+ =πε + πε
Qq Qqkx
R x R
( )
2
0 0
1 11– , since
2 32 4= >>
πε πε +Qq Qq
kx R xR R x
2
0 0 0
1 11 3–
2 32 4 32= =
πε πε πεQq Qq Qq
kxR R R
0
316
=πεQq
xR
Comparing R = 3
19. Answer (02)
Hint :
Now since electric field between the plates is
uniform and equal to 0
σε
, hence motion of
charged particle is SHM.
Solution :
22
= πm
Tk
( springs are in parallel)
The point of released of charged particle will be extreme position, since its velocity = 0 at this position mean position is where Fnet = 0
⇒ 0
2σ
=εq
kA
02
σ=
εq
Ak
, hence b = 2
20. Answer (03)
Hint :
2 2
,= =+
o Pkq kq
V Vr a r
Solution :
=okq
Vr
2 2
=+
Pkq
Va r
2 2
10+
= =o
P
V a rV r
⇒ 2 2
210
+=
a r
r
a2 = 90
2
3=ar
PART - II (CHEMISTRY)
21. Answer (A, C, D)
Hint :
Unit cell characteristics
(A) AgBr shows both schottky and frenkel
defects
Solution :
(B) Unit cell having crystal parameter a ≠ b ≠ c,
α=γ = 90° and β≠ 90° is monoclinic
(C) In Al2O3, O2–
ion form hexagonal unit cell
and Al3+
ion occupy 2/3rd of octahedral voids.
So each Al3+
ion is surrounded by 6 O2–
ion,
each O2–
ions is surrounded by 4 Al3+
ions
(D) In ZnS structure the Zn2+
ions occupy
alternate tetrahedral voids. The shortest
distance between two Zn2+
ions is given by
2 2a a ad
4 4 2= + =
22. Answer (A, C, D)
Hint :
The nearest and next nearest neighbours of a
lattice point in rock salt structure.
Solution :
This distance of 2nd co-ordination B–
is a
2other
statements are fact based.
23. Answer (A, B, D)
Hint :
BCC unit cell with the lattice point at the centre
of cube missing.
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Solution :
If body centre atom is removed, effective no. of
atom remain one (i.e. half of original)
So density become half
PE become half
24. Answer (B)
Hint :
Factors on which V. P of a liquid does not
depend.
Solution :
Surface area doesn’t effect V.P
25. Answer (A, C)
Hint :
Clausius clapeyron equation.
Solution :
2
1 1 2
P H 1 1log –
P 2.303R T T∆ =
3
i
760 4.606 10 1 1log –
7.6 2.303 2 T 300× = ×
3
1
1 1log100 10 –
T 300 =
1
2 1 11000 300 T
+ =
13000
T 187.5 K16
= =
= –85.5°C
26. Answer (A)
Hint :
Simple cubic unit cell.
Solution :
It is simple cubic packing
PE = 52.4%
27. Answer (A)
Hint :
Density = ( )3
A
1 Mgm/cc
N a
×
Solution :
d = 2.08 gm/cc
28. Answer (C)
Hint :
Nearest neighbours in simple cubic unit cell.
Solution :
Co-ordination
No. = 6
29. Answer (B)
Hint :
Lowering in V.P. = P°– P
Solution :
( )A
A
0.01 0.2 1i
0.01 0.2
× + α=
×= 1 + 0.08
iA = 1.08
BBi 1–
2
α=
A A
A A A
B BB B
B
P – P
P i n
i nP – P
P
°
°
°
°
=
B A A
B BA
P i n
i nP
°
°=
B
30 1.08 0.2100
1– 0.82
×=
α ×
αB = 0.2 ⇒ 20%
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30. Answer (B)
Hint :
Liquid B has more relative lowering of V.P than
liquid A.
Solution :
B has more relative lowering in V.P
31. Answer (A)
Hint :
Liquid A is denser than liquid B.
Solution :
Liquid A comes 1st as it has high density
32. Answer (A)
Hint :
Two dimensional solid having a lattice point at
the centre of a square unit cell.
Solution :
2a 4r=
4
a r2
=
2 216a r
2=
= 8r2
2 2
2 2
2 r 2 rPE
4a 8r
π π π= = =
% PE = 78.5%
33. Answer (B)
Hint :
NaCl lattice with certain lattice points missing
which will change its formula.
Solution :
Removal of all ions present at one C3-axis
1
Na 124
+ = ×
34. Answer (B)
Hint :
Micro-organism do not grow in a pickle due to dehydration.
Solution :
Due to dehydration of microorganism exosmosis.
35. Answer (C)
Hint :
Osmotic pressure of a solution at a certain
temperature ∝ ic.
Solution :
1 2
1 1 2 2i C i C
π π=
2
2.5 4.52 0.1 i 0.2
=× ×
24.5
i 1.82.5
= =
–3 3CH COOH CH COO H++
i = 1 – α + 2α
i = 1 + α
1.8 = 1 + α
α= 0.8
[H+] = Cα
= 0.2 × 0.8
= 0.16
[H+] = 16 × 10
–2
pH = – log16 + 2
= –4 × 0.3 + 2
= –1.2 + 2
= 0.8
36. Answer A(P, S); B(P, T); C(Q, R); D(P, S, T)
Hint :
In ionic compounds, generally anions form
space lattice and cation occupy voids except
CaF2
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Solution :
(A) P, S (B) P, T (C) Q, R (D) P, S, T
37. Answer A(P, R, T); B(Q, R, S); C(P); D(Q)
Hint :
Concentration terms not involving volume are independent of temperature.
Solution :
(A) P, R, T (B) Q, R, S (C) P (D) Q
38. Answer (08)
Hint :
Boron exists as B12 icosahedron molecular form.
Solution :
B12 has 12 corners and 20 triangular faces
∴ X = 20 and Y = 12
X – Y = 08
39. Answer (04)
Hint :
vant Hoff factor takes into account any dissociation or association of solute particles.
Solution :
60400
3 2–2 3
0 020
A B 2A 3B+ +
( )200 54
2 3 2 3 420 0
4 A B A B
=
40 60 5
a20 20
+ +=
+
= 2.625
a + 1.375 = 2.625 + 1.375 = 4
40. Answer (03)
Hint :
∆Tf = ikfm
Solution :
CoCl3.yNH3
∆Tf = ikfm
0.0558 = i × 1.86 × 0.01
i = 3
PART - III (MATHEMATICS) 41. Answer (B, C, D)
Hint :
Draw the graphs of different functions.
Solution :
(A) Take –1 –11 1sin tan , – ,
2 2x xπ π = = θ θ ∈
⇒ 1
sin tan 0x
θ = = θ ≠
⇒ cos 1 – ,2 2 2π π π θ = ⇒ θ = ∈
∴ no solution
(B) Given ( )–1 –11 1 1sin cos ; 0, 1
x x x= ∈
–1 –11 1sin sin
2x xπ
+ =
⇒ –1 –11 12sin sin
2 4x xπ π
= ⇒ =
⇒ 2x = only solution
(C) –1 –11 1 1sin cosec –1or 1
x x x= ⇒ =
x = –1
or 1 ⇒ 2 solutions
(D) Take –1 –11 1sin cot ; 0,
2x xπ = = θ θ ∈
⇒ [ )1sin cot 0; 1,x
xθ = θ = > ∈ ∞
⇒ cos2θ + cos θ –1 = 0
⇒ 5 – 1 –1– 5
cos cos –12 2
θ = θ ≠ <
⇒ 2
5 – 1 5 – 1sin 1–
2 2
θ = =
⇒ 2
5 – 1x = is only solution
42. Answer (A, B, D)
Hint :
Domain of two functions in addition form is
intersection of domains of given two functions.
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Solution :
( ) –12sin –2
f x xπ
=
–1 –1 1: 2sin – 0 sin 1
2 4 2fD x x xπ π
≥ ⇒ ≥ ⇒ ≤ ≤
( ) –12 tan –2
g x xπ
=
–1: tan 14gD x xπ
≥ ⇒ ≤ ≤ ∞
( ) –12sec –2
h x xπ
=
–1: sec4hD xπ
≥ ( ] )– , – 1 2,x ⇒ ∈ ∞ ∪ ∞
A. {1}f gD + =
B. )2,g hD += ∞
C. n gD + = φ
D. f g hD + + = φ
43. Answer (A, B, C)
Hint :
Use replacement property
Solution :
Putting x = 25 and y = 2
(f(50))2 = 25.(f(2)
2
= 25.6
2
∴ f(50) = 30 or –30
and by putting 1
xy
= we get
( ) ( )2 211f f y
y=
also f(2) = 6 so by putting y = 2
we have ( ) ( )2 21 11 2 36 18
2 2f f= = × =
( ) 31 18 2f = =
∴ ( ) 3 2f x x=
44. Answer (B, C)
Hint :
2
–1 –1 –13 – 3 1cos cos – cos
2 2 2x x
x + =
.
When12
x<
Solution :
Put x = cosθ
( ) ( )–1 –1cos cos cos cos –3
f x π = θ + θ
, 0
3 3
2 – ,3 3
π π < θ <= π π θ < θ ≤ π
45. Answer (A, B, D)
Hint :
Odd power of odd function is odd and even
power of odd function is even.
Solution :
( ) log cot –4 2
n
f x π π =
46. Answer (A)
Hint :
Use replacement property two time to get
function
47. Answer (D)
Hint :
( ) 1 1 – 1–
2 1–x
f x xx x
= +
48. Answer (C)
Hint :
Put 12
x = in used functions
Solution for Q. No. 46, 47 & 48
Given that ( ) – 1xf x f x
x + =
…(i)
For all x≠ 0, 1
Replacing x with ( )– 1x
x both sides, we get that
( )
– 1– 1
– 1 – 1– 1
+ =
xx xxf f
xx xx
That is – 1 1 – 1
1–x x
f fx x x
+ =
…(ii)
Again replacing x with ( )– 1x
xin this, we get
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( )1 11– 1–
f f xx x
+ =
…(iii)
Then by taking equation (i) + equation (iii) – equation (ii), we get that
( ) 1 – 12 –
1–x
f x xx x
= +
or ( ) 1 1 – 1–
2 1–x
f x xx x
= +
Substituting the values x = –1 and 12
in above
equation we get the solution for (ii) and (iii).
49. Answer (B)
Hint :
a3 + b
3 = (a + b)
3 – 3ab(a + b)
Solution :
( ) ( )3–1 –1 –1 –1 –1 –1tan cot – 3 tan cot tan cot= + +A x x x x x x
3
–1 –13– tan – tan
3 2 2x x
π π π =
23
–13tan –
3 2 4x
π π π = +
3 3
3 8A
π π≤ <
50. Answer (D)
Hint :
a2 + b
2 = (a + b)
2 – 2ab
Solution :
( )2–1 –1 –1 –1sin cos – 2sin – sin2
B t t t tπ = +
22
–12 sin –8 4
tπ π = +
∴ 2 2 2
max 28 16 4
Bπ π π
= + =
51. Answer (C)
Hint :
cot–1
(cotx) = x if x∈ (0, π)
Solution :
Here 4
λ π=
µ
∴ –1 –1– 3cot cot cot cot –
4 4 λ µπ π π = = π
52. Answer (D)
Hint :
log log logb c ca b a⋅ =
Solution :
Given ( )( )
2 2 32
log log log 11 11–3
f xf x
fx
= ⋅ +
⇒ ( )( )3
2 2log log log 11
f xf x
fx
= ⋅ +
⇒ ( )( )
( ) ( )1 11
1f x
f x f x f f f xx xf
x
= + ⇒ ⋅ = +
⇒ f(x) = 1 ± x4
∴ f(5) =1 ± 54 = 126
∴ f(x) = 1 + x3
∴ f(10) = 1001
53. Answer (A)
Hint :
–1 –1
2sin tan
1–
xx
x=
Solution :
( )
–1 –1– – 1 – – 1sin tan
1 – 11
r r r r
r rr r
= ++
–1 –1tan – tan – 1r r=
∴
( )
( )–1 –1 –1
1 1
– – 1sin tan – tan – 1
1
n n
r r
r rr r
r r= =
= +
∑ ∑
–1tan n=
54. Answer (C)
Hint :
Cubic function may have their maximum and
minimum values
Solution :
Let ( ) ( )3 – 3 6
, 0 12
x xg x g x x
+ ′= = ⇒ = ±
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and g(1) = 2 and g(–1) = 4
∴ y = 2, y = 4 touch y = g(x) at x = 1 and
x = –1 respectively
and g(x) = 2 ⇒ x3 – 3x + 2 = 0
⇒ (x – 1)2 (x + 2) = 0
and g(x) = 4 ⇒ x3 – 3x – 2 = 0
⇒ (x + 1)2 (x – 2) = 0
and g(x) = 2 ⇒ x = 1, 1, –2
and g(x) = 4 ⇒ x = –1, –1, 2
Now due to (x – α) a void, will be
created in y = g(x), for the given
Condition to be true that void can be created
( ) ( )– , – 2 2,x∀ ∈ ∞ ∪ ∞
∴ α∈ [–2, 2]
55. Answer (B)
Hint :
Let 2
– 1
– 1
xy
p x=
+ then find range of f(x)
Solution :
( )22
– 1– 1 – 1 0
– 1
xy x y x y p
p x= ⇒ + + =
+
A3x ∈R, ∴ D≥ 0 ⇒ 1 + 4y(y(p + 1)+1)≥0
⇒ 4y2(p + 1) + 4y + 1 > 0
∴ 1
–1, –3
y ∉
So, ( )2 14 1 4 1 0 –1, –
3y p y y + + + < ∀ ∈
⇒ (2y + 1)2 + 4y2p< 0
⇒ 22 1 1
– –1, –2 3
+ < ∀ ∈
yp y
y
Hence, 2
min
2 1–
2y
py
+ <
Now, maximum value of 22 1
2y
y+
occurs at
y = –1 and is equal to 1
–4
∴ 1
–4
p <
56. Answer A(P, S); B(Q, R, T); C(R, T); D(T)
Hint :
Draw graph of the functions
Solution :
(A) ( ) [ ]–13 sin ; –1, 1 ; odd function= + =ff x x x D
Also, f(x) is an increasing so, f(x) is one-one
function
(B) ( ) 1– | |sgn ;
1 | | fx
f x D Rx
= = +
Rf = {–1, 0, 1} : even function
(C) ( ) 8 – 2 – 2f x x x=
For domain of f(x), 8 – 2x – x2≥ 0
⇒ x2 + 2x – 8 ≤ 0 ⇒ (x + 4) (x – 2) ≤ 0
⇒ x ∈ [–4, 2]
∴ Rf = [0, 3]
(D) ( )[ ]–
| | – | |{ }
2– 2 2 – 2 0 0
2
xx x x
xf x x= = = ∀ ≤
57. Answer A(P, R, S); B(R, S); C(T); D(T)
Hint :
Type of functions property will be used.
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Solution :
(A) ( )–
cos or2
x xb bf x ax
+= where b> 0
∴ f(x) is even, into and many one function
(B) f(x) is many one, into function
(C) f(x) = x3 + 1, which is bijective function
(D) f(x) = 2x + sinx is a bijective function
58. Answer (02)
Hint :
Here
1993
( ) –2
f x x =
Solution :
( )194 2 –2 91
4cos – 2cos – 4 2sin –2
f x x x x x x = +
( )192 91
1 cos2 – 2cos2 – cos4 –2
x x x x = +
192 2 2 2 91
1 cos 2 2cos – 2cos – cos 22
x x x x x = + + +
1993
–2
x =
∴ f(f(x)) = x
Then f(f(2)) = 2
59. Answer (04)
Hint :
–1 –1 –1tan tan tan1–x y
x yxy
+ + =
if xy < 1
Solution :
–1 –13sin2 tantan tan
5 3cos2 4α α + + α
–1 –12
6 tan tantan tan
48 2tan
α α = + + α
2
2
2 tan1
16 4 tan
α<
+ α
= tan–1(tanα)
= α
60. Answer (06)
Hint :
( ) ( ) ( ) ( )2 – 4
2 – 4 2 – 4x x
x x y y
f yf xf x k= ⇒ =
Solution :
Given, f(x + y) = 2xf(y) + 4
yf(x) …(i)
Also, f(y + x) = 2yf(x) + 4
xf(y) …(ii)
⇒ ( ) ( )
4 – 2 4 – 2x x y y
f yf x= = λ
⇒ f(x) = λ(4x – 2x)
f(1) = 2 = k .2 ⇒k = 1
∴ f(x) = 4x – 2x
f′(n) = 4xlog4 – 2xlog2
f′(2) = 16log4 – 4log2 = 28 log2
∴ k = 28
So, no. of factors of k are 6