27-04-2025

4001CJA101021250005 JA

PART-1 : PHYSICS

SECTION-I (i)

1) A body of elliptical cross - section is made up of material of refractive index n2. The body is
surrounded by a liquid of refractive index n1. The figure here shows a cross - section of this body. If
any ray parallel to major axis of the cross - section passes through focus of the cross - section (as
shown), then the eccentricity of the elliptical cross - section is

(A)

(B)

(C)

(D)

2) Which of the following is not a possible path for a light ray through a glass lens (or block) kept in
air ?

(A)

(B)

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 (C)

(D)

3) Standing wave produced in a metal rod of length 1m fixed at the left end is represented by the
equation

y = 10–6 sin sin 200πt

where x and y are in metre and t is in seconds.
The maximum tensile stress at the midpoint of the rod is : (Young's modulus of material of rod = 1012
N/m2)

(A)
× 106 N/m2
(B) 2π × 106 N/m2

(C)
× 106 N/m2

(D)
× 106 N/m2

4) The temperature of a mono-atomic gas in an uniform container of length ‘L' varies linearly from T
to TL as shown in the figure. If the molecular weight of the gas is M0, then the time taken by a wave

pulse in travelling from end A to end B is

(A)

(B)

(C)

(D)

SECTION-I (ii)

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 1) Consider a refracting media of refractive index kept in air and surrounded by two non
parallel surfaces such that one surface is fixed and another surface is being moved with constant
angular velocity rad/sec about point A in clockwise direction as shown in figure. Assume that
refractive media bounded by nonparallel surfaces always remains isotropic and refractive index
always remains constant. A light ray falls on fixed surface at angle of incidence 45°. Choose the

correct option(s)

Magnitude of the rate of change of angle of deviation with respect to time when is
(A)
rad/sec
(B) The rate of change of angle of deviation with respect to time when is negative
Magnitude of the rate of change of angle of emergence from moving surface with respect to
(C)
time when is rad/sec
Magnitude of the rate of change of angle of emergence from moving surface with respect to
(D)
time when is rad/sec

2) A silvered, hollow glass sphere (mirror) , has a round hole as shown in the figure. Radius of round
hole makes α (acute) angle at the centre. Into this round hole a parallel beam of ray falls,
perpendicular to the plane of the hole. Part of rays having undergone one reflection, (portion
crossing the smaller circle of radius r2) exits the sphere back through the hole. (Assume 100 %

reflection). Choose the CORRECT option.

(A)

(B)

(C)

Fraction of the power of the incoming beam, which exit through the hole after one reflection is
(D)

3) A point isotropic sonic source of sound power 1 milli watts emits sound of frequency 170 Hz in all
directions. It is known that at some moment t = 0, the displacement of air particles at a certain point
4 m away from the source is A. The amplitude of oscillation at this point is known to be 2A. The
displacement of air particles in terms of A, at a point 55 m away from the source at the moment t = 0
is d and distance of a point from the source where loudness level is 60 dB, is r. (vs = 340 m/s).

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 Consider displacement of particles to be positive if it is away from the source, then

(A)

(B)

d=
(C)

d=
(D)

4) Assume that temperature varies linearly with height near the earth’s surface. Consider the
temperature of the surface of the earth T1 and T2 at the height h above the surface. Let the velocity
of sound at earth’s surface is c.

(A)
The temperature at the distance x is

(B)
The temperature at the distance x is
The time t needed for the sound wave to reach from height x at the earth’s surface is
(C)

The time t needed for the sound wave to reach from height x at the earth’s surface is
(D)

5) The focal length of a thick glass lens in air with refractive index n, radius of curvatures r1, r2 and

vertex distance d as shown in figure is given by :

Remark : ri > 0 means that the central curvature

point Mi is on the right side of the aerial
vertex Si, ri < 0 means that the central curvature point Mi is on the left side of the aerial
vertex Si (i = 1,2). For some special applications it is required, that the focal length is independent
from the wavelength. Choose the CORRECT statement(s) :

(A) There are atmost two different wavelengths for which the same focal length can be achieved.

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 (B) If r1 – r2 = d , then focal length will be same for two wavelength. (n1 and n2
are refractive index for two wavelength being consider)
(C) For a given plano-convex lens, a specific focal length can be achieved by only one wavelength.
(D) There are atmost three different wavelengths for which the same focal length can be achieved.

6) A car is moving with a velocity towards a medium MNRP as shown, where 'v' is the speed of

sound in air. Two men A and B are moving towards each other in the medium with the velocities

and respectively. The speed of sound in medium is 7v and medium is kept at rest. The car is
emitting a sound of frequency f. Then choose the correct option(s).

(A)
The wavelength received by the man A is
(B) The wavelength received by the man A is different from the wavelength received by man B.

(C)
The ratio of frequency observed by the man A to that observed by the man B is

(D)
The frequency observed by the man B is .

SECTION-I (iii)

Common Content for Question No. 1 to 2

A transparent clear cylindrical slab of snow floats on still water body. A bird sittings on slab
looks through slab into water in search of fish. Vision through lateral surface is obscure. Radius of
slab is its height is 40 mm. Depth of water is 8m and its refractive index
1.732. Refractive index of snow is 1 < μ < 1·732.

1) Maximum area of bottom of water body the bird can view for prey is

(A) 186 m2
(B) 224 m2
(C) 308 m2
(D) 420 m2

2) If snow melts such that thickness becomes half but radius remains same, then maximum visible
area at bottom

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 (A) remains same
(B) Decreases
(C) Increases
(D) First increases then decreases

Common Content for Question No. 3 to 4

If a plane simple harmonic wave is traveling towards positive x axis its wave function is given by S
=S0 sin (ωt – kx) Here S is longitudinal displacement of any particle. If we take a plane parallel to yz
plane all particles in this plane have same phase. This plane is called wave front.
If a plane wave is traveling along direction its wave function can be written as : S = S0 sin (ωt – k
) Let's call k (a wave vector) S = S0 sin(ωt – ) Wave vector points in direction of phase
velocity. In other words, the wave vector points in the normal direction to the wavefront.

3) Consider following time wave function. S1 = S0 sin(ωt – 2x + 2y + z)

S2 = S0 sin(ωt + 2x – 2y – z) An observer is traveling towards +x axis with speed π m/s. The beat
frequency observed by the observer is :

(A)

(B) 2
(C) 1

(D)

4) Consider a wave function :

S = S0 sin(ωt – 2x + 2y + z) Here all variables are in SI units.
Select correct Statement :

(A)
Wavelength of the wave is .
(B) Particle (0, 0, 0) and (1, 1, 2) are on same wavefront.

(C)
The wave vector is at angle cos–1 from x-axis.
(D) The wave vector is at equal angle from positive x-axis and negative z-axis.

SECTION-II

1) Analyse the figure there are two different media on both sides of lens. The radius of curvature of
concave mirror is 80 cm. The focal length of equiconvex lens in the air is 20 cm. The value of "d" so

that the final image is formed at object itself is _____ (in cm.)

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 2) A ray of light is incident on a glass sphere of refractive index such that the directions of
the incident ray and emergent ray when produced meet the surface at the same one point. The value

of angle of incident is θ. Calculate the value of .

3) A composite string is made up by joining two strings of different masses per unit length m and 4m
respectively. The composite string is under the same tension. A transverse wave pulse Y = (6
mm)sin (5t + 40x), where t is in seconds and x in metres, is sent along the lighter string towards the
joint. The joint is at x = 0. The equation of the wave pulse reflected from the joint is A sin (BT – CX +

π) where A is in mm, B is per sec. and C is per meter then value of .

4) A heavy rope of length ‘ℓ’ and variable linear density is hanging from ceiling. A small jerk

is produced at bottom, time it takes to reach top is . Find b.

PART-2 : CHEMISTRY

SECTION-I (i)

1) Which of the following is correct ?

(A)

(B)

Correct order of stability is

(C)

Incorrect order of stability is

(D)

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 2) Consider following statements.

(P) Resonance is possible in

(Q) Delocalization of lone pair is possible in

(R) In carbocation , σ – (vacant p-orbital) type conjugation is involved

(S) In compound hyperconjugation is not possible

(T) Resonance hybrid is a real structure
(U) There is equilibrium in between resonating structures
(V) Resonance is a permanent effect.
If number '1' is assigned to incorrect statement and number '0' is assigned to correct statement then
for statements P, Q, R, S, T, U and V; correct option regarding set of numbers will be

(A) 0, 0, 1, 1, 1, 0, 1
(B) 0, 0, 1, 1, 0, 1, 0
(C) 1, 0, 1, 1, 1, 0, 1
(D) 1, 1, 0, 0, 0, 1, 0

3) The correct statement about the following molecule is:

(A) Molecule is chiral and does not possess chiral centre

(B) Molecule is chiral and possess chiral centre
(C) Molecule is achiral due to plane of symmetry
(D) Molecule is achiral due to axis of symmetry

4) Consider the following compounds

(1) (2)
Which option satisfy for the above two?

(A) Both can show geometrical as well as optical isomerism

(B) Both can show optical isomerism

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 (C) Both are chiral but only 2nd can show geometrical isomerism
(D) Both can show geometrical isomerism

SECTION-I (ii)

1) Which of the following is/are less reactive than toluene towards EAS reaction.

(A)

(B)

(C)

(D)

2) Which of the following statement(s) is/are correct regarding Azulene?

(A) Azulene is an isomer of naphthalene and is blue in colour

(B) Azulene has very high dipole moment while naphthalene has zero dipole moment
(C) Seven membered ring of azulene is electrophilic and five membered ring is nucleophilic
(D) Seven membered ring of azulene is nucleophilic and five membered ring is electrophilic

3) Identify reactions which are not feasible.

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 (A)

(B)

(C)

(D)

4) Select the incorrect one

(A) Conjugated acid of pyrrole is antiaromatic

(B)
can’t tautomerise.
– –
(C) CH3–S is less stable than CH3–O .

(D)

Optically pure racemises during tautomerisation

5) Which of the following is/are correct statements ?

(A)

are Functional isomers

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 (B)

are Metamers

(C)

are Positional isomers

(D)
are Chain isomers.

6)
Select correct statement for above pairs.

(A) Compound A & B are geometrical isomers

(B) Compound C & D are enantiomeric compounds
(C) Compound A & B are not diastereomers
(D) Compound C & D are not geometrical isomers

SECTION-I (iii)

Common Content for Question No. 1 to 2

The compound shows below in the diagram, on treatment with acid catalyst isomerizes and
equilibrium is established among the stereoisomers.

, where X ≠ H or CH3
The equilibrium mixture contains all the three stereoisomers of this compound. Specific rotation of
pure dextro isomer is +62° while the specific rotation of the equilibrium mixture is +22°. Also the
equilibrium mixture contains 20% of the meso isomers. Answer the following questions based on the
above information.

1) Composition of equilibrium mixture is :

(A) 20% meso, 58% dextro and 22% laevo

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 (B) 20% meso, 60% dextro and 20% laevo
(C) 20% meso, 54% dextro and 26% laevo
(D) 20% meso, 30% dextro and 50% laevo

2) The percentage of racemic mixture present in the equilibrium mixture is :

(A) 35%
(B) 44%
(C) 65%
(D) 80%

Common Content for Question No. 3 to 4

If bulky groups are attached at ortho position to a group present on aromatic ring then resonance is
greatly reduced. The phenomenon is described as steric inhibition to resonance (SIR). SIR affects
many physical & chemical properties of compounds like bond length, acidic strength, basic strength,
reactivity etc.

3) The acid strength order is

(a) (b) (c) (d)

(A) a>b>c>d
(B) a>c>b>d
(C) a>b>d>c
(D) b>a>d>c

4) The benzene rings with the most and the least π–electron density respectively :

(1) (2) (3) (4)

(A) 3, 1
(B) 3, 2
(C) 1, 2
(D) 1, 4

SECTION-II

1) Consider the following sequence of reactions to produce major product (A)

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 Molar mass of product (A) is ______ g mol–1.
(Given molar mass in g mol–1 of C : 12, H : 1, O : 16, Br : 80, N : 14, P : 31)

2) Number of compounds from the following which cannot undergo Friedel-Crafts reactions is :____
toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, m-nitroaniline, m-dinitrobenzene

3) The number of different chain isomers for C7H16 is __________.

4)
(a) Total number of chiral centres in the above molecule = x
(b) Total number of stereo isomers of the above molecule = y
Find the value of x + y = ?

PART-3 : MATHEMATICS

SECTION-I (i)

1) Let R1 and R2 be two relations defined as follows :

2 2 2
R1 = {(a,b) ∈ R : a + b ∈ Q } and
R2 = {(a,b) ∈ R2 : a2 + b2 ∉ Q},
where Q is the set of all rational numbers. Then :

(A) R2 is transitive but R1 is not transitive

(B) R1 is transitive but R2 is not transitive
(C) R1 and R2 are both transitive
(D) Neither R1 nor R2 is transitive

2) For the function , if n(d) denotes the number of integers which are not in its
domain and n(r) denotes the number of integers which are not in its range, then n(d) + n(r) is equal
to -

(A) 2
(B) 3
(C) 4
(D) Infinite

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 3) Number of integers in the range of f(x) = [sin x] + [cos x] + [sin x + cos x]
(where [.] denotes greatest integer function)

(A) 5
(B) 7
(C) 8
(D) 9

4) Let Tr be the rth term of an A.P. whose first term is and common difference is 1, then

(A)

(B)

(C)

(D)

SECTION-I (ii)

1) Let N be the set of natural numbers and a relation R on N be defined by

R = {(x,y) ∈ N × N : x3 – 3x2y – xy2 + 3y3 = 0}. Then the relation R is :

(A) symmetric but neither reflexive nor transitive

(B) reflexive
(C) not transitive
(D) an equivalence relation

2) If f(x) = sin ℓn , then

(A) domain of f(x) is (– 2, 1)

(B) domain of f(x) is [–1, 1]
(C) range of f(x) is [–1, 1]
(D) range of f(x) is [–1, 1)

3) Consider a function (where [k] denote greatest integer function),

then -

(A) number of integer in the domain is 4

(B) number of integer in the domain is 5
(C) number of elements in the range of function is 2

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 (D) number of elements in the range of function is 4

4) For the equation |x – 1| + |x – 3| – 2|x – 2| = λx, which of the following is(are) correct ?

(A) For λ = 1, the equation has 2 distinct real solution.

(B) For λ ∈ (0,1), the equation has 3 distinct real solutions.
(C) For λ ∈ (1,2), then equation has 3 distinct real soutions.
(D) For λ = 0, the equation has infinite solutions.

5) Let a1,a2,a3, ..., a100 be in Arithmetic Progression and h1,h2, ..., h100 be in Harmonic Progression. If a1
= h1= 2 and a100 = h100 = 23, then which of the following is/are correct

(A) a5 + a96 + a10 + a91 = 50

(B) h5 + h10 + h91 + h96 = 50
(C) a1h100 + a4h97 = 92
(D) a2h99 + a5h96 + a8h93 = 138

6) Let ordered pair (x, y) satisfy the equation

then which of the following is/are correct?

(A) Only 1 ordered pair of (x, y) is possible

(B) More than 1 ordered pair of (x, y) are possible
(C) In one of the ordered pair |x|+|y| = 1
(D) In one of the ordered pair x – y = 0

SECTION-I (iii)

Common Content for Question No. 1 to 2

Let p, q, r be positive integers such that is an integer. Also p, q, r (taken in that order) are in
geometric progression and the arithmetic mean of p – 2, q + 5, r + 4 is (2q – p – 1).

1)

The value of p is equal to

(A) 3
(B) 4
(C) 5
(D) 6

2)

The number of ordered triplet (p, q, r) is equal to

(A) 1

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 (B) 2
(C) 3
(D) 4

Common Content for Question No. 3 to 4

Consider the function ƒ(x) = 7x – x2 – 10.

3) Number of integers in domain of g(x) = logef([x]), (where [ . ] represents greatest

integer function)

(A) 2
(B) 3
(C) 1
(D) 10

4) The range of the function (where sgn represents signum function)

(A) {0, 1}
(B) {1}
(C) {–1, 1}
(D) {–1, 0, 1}

SECTION-II

1) If [x] and {x} denote the integral part and fractional part of a real number x, then the number of
negative real numbers x for which is :

2) If f(x) = [9x – 3x + 1] ∀ x∈(– ∞, 1) ([·] denotes the greatest integer function) then number of
elements in range of f(x) are

3) Let f(x) = (x + 1)(x + 2)(x + 3)(x + 4) + 5 where x ∈ [–6, 6]. If the range of the function is [a, b]
where a, b ∈ N then find the value of (a + b).

4) The 4th term of GP is 500 and its common ratio is , denote the sum of the first n terms of

this GP. If S6 > S5 + 1 and S7 < S6 + then the number of possible values of m is ______

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 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

Q. 1 2 3 4
A. A C C A

SECTION-I (ii)

Q. 5 6 7 8 9 10
A. B B,D A,C A,C A,B,C A,C,D

SECTION-I (iii)

Q. 11 12 13 14
A. C A B A

SECTION-II

Q. 15 16 17 18
A. 30 2 100 8

PART-2 : CHEMISTRY

SECTION-I (i)

Q. 19 20 21 22
A. D D A B

SECTION-I (ii)

Q. 23 24 25 26 27 28
A. A,C,D A,B,C A,B A,B,C,D A,B,D A,B,C,D

SECTION-I (iii)

Q. 29 30 31 32
A. A B C A

SECTION-II

Q. 33 34 35 36
A. 171 4 9 12

PART-3 : MATHEMATICS

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 SECTION-I (i)

Q. 37 38 39 40
A. D C A A

SECTION-I (ii)

Q. 41 42 43 44 45 46
A. B,C A,C A,C A,B,D A,C,D A,C,D

SECTION-I (iii)

Q. 47 48 49 50
A. C B A A

SECTION-II

Q. 51 52 53 54
A. 0 7 5049 12

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 SOLUTIONS

PART-1 : PHYSICS

3)

4)

6)

90 – α + θ + θ + θ = 90

3θ = α ⇒
sin 30 = 3 sinθ – 4 sin3 θ

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 Fraction of power =

7)

as λ = 2 meter, A' will be –ve.

9) (A) The refractive index n is a function of the wavelength λ, i.e. n = n (2). According to the
given formula for the focal length f (see above) which for a given f yields to an equation
quadratic in n there are at most two different wavelengths (indices of refraction) for the same
focal length. (B) If the focal length is the same for two different wavelengths, then the
equation f(λ1) = f(λ2) or f (n1) = f(n2)
holds. Using the given equation for the focal length it follows from equation (1) :

Algebraic calculations lead to :

10)

Frequency of surface MN(f') =

11)

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 Question (7,8)

(mg – N) R =
mg – N = 4 mRα –1
For Bead B, N = mαR
⇒ mg – mRα = 4mRα

⇒ g = 5Rα ⇒
∴ Acceleration of bead A =

Acceleration of bead B

13) Component of velocity along the wave vector = =

Time between time consecutive beat =

=
Frequency = 2 Ans.

14) (A)

∴λ=

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 (D) for angle between wave vector and +ve x-axis
= 2 ⇒ (3) (1) cosθ = 2

⇒ cosθ =
for angle between wave vector and –ve y-axis
= 2 ⇒ (3) (1) cosθ = 2

⇒ cosθ =

16)

By Geometry,
and

17)

V2 < V1
⇒ 2nd is denser ⇒ phase change of π
wave reflected from denser medium

⇒ eqn ⇒ – (2mm) sin (5t – 40x)

18) Tension at x = weight of rope below that point

T=

v=

v=

= =

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 PART-2 : CHEMISTRY

21) No chiral centre is present. POS is absent and so the molecule is chiral

22) In both the cases rings are perpendicular to each other thus, due to absence of any
symmetry element both are optically active.

24) Seven membered ring of azulene is electrophilic and five membered ring is nucleophilic

25)

Aniline (highly active and basic -NH2 group) and nitrobenzene (highly deactivating -NO2 group)
cannot show Friedel craft reactions.

30)

Let total moles of all stereoisomers = 100

There are total 3 stereoisomers.

One is dextrorotatory and other is
laevorotatory and one stereoisomer
is optically inactive.
Let (d)-stereoisomer = α moles
(l )-isomer = (80 – α) mole
meso-stereoisomer = 20 mole

Observed rotation = 22°

=
α = 57.44 58
∴ % of (d)-stereoisomer = 58
% of (l )-stereoisomer = 22
% of meso-stereoisomer = 20
⇒ % of Racemic mixture
= % of d + % of (l )
(Equimolar 'd' and 'l ' form)
= 22 × 2
= 44%

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 33)
Molar mass of product (C7H7Br) (A) is 171 g mol–1

34) Compounds which can not undergo Friedel Crafts reaction are

35)

36) x = 4
y=8

PART-3 : MATHEMATICS

37) Let a2 + b2 ∈ Q and b2 + c2 ∈ Q

eg. and
2 2
a + b = 14 ∈ Q

Let
b2 + c2 = 20 ∈ Q

But

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 for R2 Let a2 = 1, b2 = and c2 = 2
a2 + b2 ∉ Q and b2 + c2 ∉ Q
But a2 + c2 ∈ Q

38)
Domain
ex – 1 ≠ 0
ex ≠ 1 ⇒ x ≠ 0
n(d) = 1

⇒ y⋅ex – y = ex + 1

⇒
Range

⇒ y∈(–∞, –1) ∪ (1, ∞)

y ≠ –1, 0, 1
n(r) = 3
h(d) + n(r) = 4

39) f(x) = [sin x] + [cos x] +

Period = 2π, x ∈ [0, 2n]

Case-1 : , f(x) = 1

Case-2 : , f(x) = –1

Case-3 : , f(x) = –2

Case-4 : , f(x) = –4

Case-5 : , f(x) = –2

Case-6 : , f(x) = –1
at x = 0, f(x) = 2

at x = , f(x) = 2

at x = , f(x) = –1
at x = π, f(x) = –2

at x = , f(x) = –2

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 at x = , f(x) = –1
at x = 2π, f(x) = 2
Range of f(x) = {–1, 1, –2, –4, 2}

41) x3 – 3x2y – xy2 + 3y3 = 0

⇒ x(x2 – y2) – 3y (x2 – y2) = 0
⇒ (x – 3y) (x – y) (x + y) = 0
Now, x = y ∀ (x, y) ∈ N × N so reflexive
See, (3,1) satisfies but (1,3) does not. Also (3,1) & (1, –1) satisfies but (3, –1) does not satisfy
∴ not symmetric & transitive

42) f(x) = sin log

4 – x2 > 0 or x ∈ (–2, 2) and

D : (–2, 1)
R : [–1, 1]

43) f(x) is defined only when

[x] – 1 ≥ 0 & 4 – [x] ≥ 0
i.e.

& its range is

44) f(x) = |x – 1| + |x – 3| – 3|x – 2|

45)

a5 + a96 = a10 + a91 = a1 + a100

a5 + a96 = a10 + a91 = 2 + 23 = 25
Property : If between two numbers n arithmetic means and n harmonic means are inserted
then product of kth A.M. from beginning and kth harmonic mean from end is always constant
and equals to product of two numbers. i.e. AK.Hn – K + 1 = constant a1h100 = a2h99 = ... = 2 × 23 =

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 46
Thus C, D are correct.

46) Since are both positive we can apply

AM–GM to obtain

However,

Hence we must have this is true for only

x=y= only.

47)

48)

p, q, r → G.P.
A, AR, AR2 → G.P. (R ∈ I)

Given, = 2AR – A – 1
A (1 + R + R ) + 7 = 6AR – 3A – 3 ⇒ A (R2 – 5R + 4) = – 10
2

⇒ A (R – 4) (R – 1) = – 10 ⇒ A = ∈ I+
Since A is a positive integer
∴ R = 2 or R = 3
If R = 2 and R = 3, then A = 5
∴ Value of p = 5
Numbers are 5, 10, 20 or 5, 15, 45
∴ Total number of ordered triplets of (p, q, r) are 2. Ans.

49)

g(x) = loge(7x – x2 – 10)

7x – x2 – 10 > 0 ⇒ x2 – 7x + 10 < 0
x ∈ (2, 5)

50) ƒ(x) = –x2 + 7x – 10

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 g(x) = loge ƒ(x)

51) 2{x} – {x} = x + {x}

2x – 3{x} = x + {x}
x = 4{x}
x = [0, 4)
so x can't be –ve

52) 9x – 3x + 1 = ∀ x∈(–∞, 1)
∴ Range of f(x) = {0, 1, 2, 3, 4, 5, 6}

53)

let

54) T4 = 500 where a = first term,

r = common ratio =
ar3 = 500

Sn – Sn–1 = arn–1

and S7 –S6 <

m3 > 103

m > 10 ……(2)
2
m < 500 …….(1)
From (1) and (2)
m = 11, 12, 13…………..,22
So number of possible values of m is 12

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