10-08-2025

1001CJA106216250126 JA

PART-1 : PHYSICS

SECTION-I (i)

1) Two uniform semicircular discs, each of radius R are stuck together to form a disc. Masses of the
two semicircular parts are M and 3M. The moment of inertia of the circular disc about an axis
perpendicular to its plane and passing through the centre of the formed disc. (Point 'O' in figure)

(A) 2MR2

(B)

(C)

(D)

2) A hollow sphere of radius R moves with initial linear and angular velocities as shown in the figure
on a rough horizontal surface. The angular velocity of the sphere when its linear velocity becomes

zero is :-

(A)
anticlockwise

(B)
clockwise

(C)
clockwise

(D)
anticlockwise

3) A sphere of radius r is undergoing pure rolling with constant angular velocity on a fixed curved
surface of radius R with O as centre as shown. The ratio of angles through which line PC rotates to
the angle through which line OC rotates is : (from initial position as shown in figure)
(A)

(B)

(C)

(D)

4) A wheel of radius R = 10 cm and moment of inertia ℓ = 0.05 kg-m2 is rotating about a fixed
horizontal axis O with angular velocity ω0 = 10 rad/sec. A uniform rigid rod of mass m = 3 kg and
length ℓ = 50 cm is hinged at one end A such that it can rotate about end A in a vertical plane. End
B of the rod is tied with a thread as shown in figure such that the rod is horizontal and is just in
contact with the surface of rotating wheel. Horizontal distance between axis of rotation O of cylinder
and A is equal to a = 30 cm. If the wheel stops rotating after one second after the thread has burnt,
calculate co-efficient of friction µ between the rod and the surface of the wheel. (g = 10 ms–2)

(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4

SECTION-I (ii)

1) One end of an ideal spring is fixed at point O and other end is attached to a small disc of mass m
which is given an initial velocity v0 perpendicular to its length on a smooth horizontal surface. If the

maximum elongation in spring is then ( = natural length and k = stiffness of spring)

 Velocity at maximum elongation, v =
(A)

Velocity at maximum elongation, v =

(B)

(C) 0
v =
v0 =
(D)

2) A uniform rod of mass 1kg and length L = 1m stands vertical. Rod is free to rotate about hinge at
B in x-y plane only. A force starts acting on the rod at point A. Then immediately

after force starts acting on the rod

(A)
angular acceleration of rod will be

(B)
angular acceleration of rod will be .

(C)
reaction torque of hinge on rod about B will be .
(D) reaction torque of hinge on rod about B will be 0.

3) A uniform slender rod of length L is dropped onto rigid supports at A and B. Immediately before
striking A the velocity of the rod is VC. Since support B is slightly lower than support A, the rod
strikes A before it strikes B. Assume perfectly elastic impact at both A and B. Let the angular
velocities of the rod be ω1 and ω2 and the velocities of its C.M. be v1 and v2 immediately after the rod
strikes at A and B respectively. Then (Consider only two collisions one at A and other at B)

(A) ω1 is clockwise
(B) ω2 is anticlockwise
(C) v1 is downward
(D) v2 is upward
4) On a smooth horizontal table there are two blocks and two thin light discs. The axis of one of them
is fixed, so that that disc cannot move, but can rotate about its axis. A piece of light inextensible
thread covers the pulleys, to the ends of the thread we attach identical loads of mass M each. The
threads are always taut. A constant force F is exerted on the movable disc as shown in the figure

(top view). The thread does not slip relative to the discs.

(A) The left disc does not rotate but the right disc rotates in the clockwise direction.
(B) The left disc rotates in clockwise direction and the right disc rotates in anticlockwise direction.
(C) The acceleration of both the blocks is F/2M in opposite directions.
(D) The net torque on the fixed disc is zero (or negligibly small).

5) A dumbbell is formed by connecting two identical particles each of mass m with a massless rod.
The system is placed on a horizontal smooth surface and impulse J is applied in two ways as shown
in the figure as situation-(i) and situation-(ii). Then just after the application of impulse, choose the

correct option(s).

Velocity of centre of mass with respect to ground frame in situation-(i) and situation-(ii) are
(A)
same.
Kinetic energy of the dumbbell with respect to centre of mass frame in situation-(i) and
(B)
situation-(ii) are same.
Angular momentum of the dumbbell with respect to centre of mass in situation-(i) and situation-
(C)
(ii) are different.
Velocity of particle A in situation-(ii) is double of velocity of particle A in situation-(i) just after
(D)
the application of impulse.

6) An spherical shell of mass M, inner radius a and outer radius b is placed on a horizontal surface
with coefficient of friction µ, as shown in the figure. At some time, an impulse is applied at a height h
above the center of the shell. If h = hm then the shell rolls without slipping along the x-axis. Which of
the following statement(s) is(are) correct ?
(A) For µ ≠ 0 and a → 0, hm = b/2
(B) For μ ≠ 0 and a → b, hm = 2b/3
(C) For h = hm, the initial angular velocity is inversoely proportional to the outer radius b.
(D) For μ = 0 and h = 0, the shell always slides without rolling.

7) A horizontal disc rotates freely about a vertical axis through its centre. A ring, having the same
mass and radius as the disc, is now gently placed on the disc. After some time, they rotate with a
common angular speed.

(A) some friction exists between the disc and the ring when there is relative slipping between them.
(B) the angular momentum of the disc plus ring is conserved.

(C)
the final common angular speed is rd of the initial angular velocity of the disc.

(D)
rd of the initial kinetic energy changes to heat.

8) A wheel (to be considered as a ring) of mass m and radius R rolls without sliding on a horizontal
surface with constant velocity v. It encounters a step of height R/2 at which it ascends without

sliding

(A) the angular velocity of the ring just after it comes in contact with step is 3v/4R
the normal reaction due to the step on the wheel just after the impact is
(B)

(C) the normal reaction due to the step on the wheel increases as the wheel ascends
(D) the friction will be absent during the ascent.

SECTION-II

1) A semicircular disc of radius R and mass M is pulled by a horizontal force F so that is moves with

uniform velocity. The coefficient of friction between disc and ground is . If the angle θ = π/n
radians, then find the value of n.

2) A rod of mass m and length L( = 50/3 m) is hinged in plank of same mass m. The plank is kept at
the corner of a smooth table and rod makes θ with the vertical. The system is released from the θ =
0°. Find the velocity (in m/s) of plank when the rod makes the angle θ =180°. (g = 10 m/s2)

3) Spool of a bobbin rolls without slipping on a horizontal track. The radius of the spool is 50 cm and
that of flanges (external rim of spool) is significantly more than radius of spool. At some instant
speeds of two diametrically opposite points on the spool are and 3 m/s. Find out speed (in
m/s) of the centre of the bobbin.

4) A block enters a horizontal smooth spiral track in which the radius of the track decreases from 10
m to 5m. If the block enters the spiral at a speed of 10 m/s, what is it's speed (in m/s) at the end of

the spiral?

5) In the given figure very long plank of mass 2m is initially at rest on smooth horizontal floor while
uniform solid cylinder of mass m and radius R is placed on it with initial linear velocity v0 and no
angular velocity. Coefficient of friction between plank and cylinder is µ. Time after which slipping

between cylinder and plank comes to an end is . Find x.

6) Three identical uniform dominos are arranged as shown in the diagram. The possible values of x
such that the dominoes are in stable equilibrium is α < x < β. α, β and x are measured in mm. Then
the value of α + β (in mm) is :-

PART-2 : CHEMISTRY

SECTION-I (i)

1) Match List-I with List-II

List-I (Mixture) List-II (Purification Process)

(A) Chloroform & Aniline (I) Steam distillation

(B) Benzoic acid & Napthalene (II) Sublimation

(C) Water & Aniline (III) Distillation

(D) Napthalene & Sodium chloride (IV) Crystallisation

(A) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(B) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(C) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(D) (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

2) If on a strand of DNA the base sequence is ATTGACGCAT then the sequence transcription on RNA
would be _______.

(A) UAACUGCGUA
(B) AUUCUGCGUA
(C) UAACTGCGUA
(D) TAACTGCGTA

3) Which of the following pairs give positive Tollen’s test?

(A) Glucose, sucrose

(B) Glucose, fructose
(C) Hexanal, Acetophenone
(D) Fructose, sucrose

4)
Match the list

List-I List-II

(P) (1) Thermoplastic polymer

(Q) (2) Thermosetting polymers

(R) (3) Fibers

(S) (4) Elastomer

(A) P→3; Q→4; R→2; S→1

(B) P→4; Q→3; R→1; S→2
(C) P→4; Q→1; R→3; S→2
(D) P→2; Q→3; R→1; S→4

SECTION-I (ii)

1) Which of the following can be used as artificial sweetners?

(A) Aspartame
(B) Alitame
(C) Sucralose
(D) Saccharin

2) Which of the following statement(s) is(are) true ?

(A) Oxidation of glucose with bromine water gives glutamic acid

(B) The two six-membered cyclic hemiacetal forms of D-(+)-glucose are called anomers
(C) Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose
(D) Monosaccharides cannot be hydrolysed to give polyhydroxy aldehydes and ketones

3) Amino acids are synthesised from

(A) α-Halo acids by reaction with NH3

(B) Aldehydes by reaction with NH3 and cyanide ion followed by hydrolysis
(C) Pyruvic acid is treated with NH3 followed by addition of H2(Ni)
–
(D) Alcohols by reaction with NH3 and CN ion followed by hydroysis.
4) Select the correct option.

(A) Isoelectric point is the pH at which an amino acid exists primarily in its neutral form.
Isoelectric point is the average of pKa values of α-COOH amino α-NH3+ groups [ valid only for
(B)
neutral amino acid ]
(C) Glycine is optically active and chiral.
(D) For neutral amino acid the concentration of zwitter ion is maximum at its isoelectric point

5) Consider the given sequence of reaction

Select the correct statement-

(A) Na extract contains Br– and I– together

(B) B is vapour of I2
(C) The clear solution contains AgCl.
+
(D) The clear solution contain [Ag(NH3)2]

6) Choose the correct option(s) from the following

(A) Natural rubber is polyisoprene containing trans alkene units

(B) Nylon-6 has amide linkages
(C) Cellulose has only α-D-glucose units that are joined by glycosidic linkages
Teflon prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high
(D)
pressure

7) Which of the following pairs are bacteriostatic antibiotics?

(A) Penicillin, tetracycline

(B) Erythromycin chloramphenicol
(C) Ofloxacin, aminoglycosides
(D) Tetracycline, chloramphenicol

8) Correct statement(s) about is/are

(A) It gives coloured solution with neutral FeCl3 solution.

(B) It liberates H2 gas with Na metal.
(C) It gives positive 2,4-DNP test.
(D) It forms sweet smelling compound with alcohols.

SECTION-II

1) How many of below given sugars are reducing ?

(i) Glucose
(ii) Galactose
(iii) Fructose
(iv) Mannose
(v) Sucrose
(vi) Maltose
(vii) Lactose
(viii) Methyl derivative of α-D(+)-Glucose

2) A tetrapeptide has -COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl
alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible
sequences (primary structures) with -NH2 group attached to a chiral center is

3) Number of sp2 hybrid carbon atoms in one molecule of paracetamol is –

4) In Kjedahl's method, NH3 evolved from 0.5 gram organic compound is absorbed by 20 ml, 1 M
sulphuric acid solution. Remaining sulphuric acid needed 20 ml, 1 M NaOH solution for complete
neutralisation then % of N in organic compound?

5) Using the provided information in the following paper chromatogram :

the calculate Rf value of A _______ × 10–1.

6) Among the following number of condensation polymers are ?

Nylon-2-Nylon-6, Buna-N, Buna-S, Nylon-6,6, Terylene (Dacron), PVC, Polystyrene, Teflon.

PART-3 : MATHEMATICS
 SECTION-I (i)

1) Let f(x) be a real valued cubic polynomial with real coefficients such that both f(0) and f(–1) are
odd integers. Also the leading coefficient is unity. Which of the following is/are true

(A) f(x) = 0 may have all integral roots

(B) f(x) = 0 will never have all integral roots
(C) f(x) = 0 will have all odd integral roots
(D) f(x) = 0 will have all even integral roots

2) Let set then cardinality of set A is

(A) 2022
(B) 2020
(C) 2018
(D) 2016

3) For all sets S, let |S| denotes number of elements of S and n(S) denotes number of subsets of S
including empty set and the set itself. If A, B, C are three sets satisfying
n(A) + n(B) + n(C) = n(A∪B∪C) and |A| = |B| = 100
then the value of (|A∪B∪C| – |C|)

(A) 100
(B) 1
(C) 2
(D) 200

4) If the equations ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 have two common roots then value of

equals
(where a,b,c –{0})

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

SECTION-I (ii)

1) Let f(x) = ax2 – (a2 + b)x + ab and a2 < b. Which of the following options is/are true (where a ≠ 0)

(A)
If , then f(–2) > 0
(B)
If , then f(0) < 0

(C)
If , then f(0) > 0

(D)
If , then f(2) > 0

2) Let roots of the equation 2x3 + 4x2 – 5x + 8 = 0 be α, β, γ then which of the following is/are true

(A)

(B) 1 + α, 1 + β, 1 + γ are roots of 2x3 – 2x2 – 7x + 15 = 0

(C)

(D) 1 + α, 1 + β, 1 + γ are roots of 2x3 – 6x2 – 7x – 15 = 0

3) Possible set of values of 'a' for which the equation, (x2 + x)2 + a(x2 + x) + 4 = 0 have all four
imaginary roots is/are

(A)

(B)

(C)

(D)

4) If a, b ∈ R satisfy the equation a3 + 24ab + 64 = 8b3 , then roots α, β (α < β) of equation x2 + bx +

a = 0 satisfies (provided 0 < b < 3)

(A) α ∈ (–4,–2)
(B) β ∈ (–1,2)
(C) α ∈ (–4,–2]
(D) β ∈ (–4,–2)

5) Let α, β, γ be the roots of x3 – 3x2 + 2x + 4 = 0 and y(x) =

, then

(A) y(–1) = 0.5

(B) y(–1) = –0.5
(C) y(1) = 0.25
(D) y(2) = 2

6) Consider the following sets.

*
*
*
Choose the set of correct option(s)

(A)
(B)
(C) The cardinality of the set is 10.
(D) The cardinality of the set is 10.

7) Let f(x) is a polynomial of degree n, such that (α + 1) f(α) – α = 0 ∀ α ∈ N ∪ {0}, α < n. Which of
the following is/are true

(A)
For n = 10 ; f(11) is
(B) For n = 15, f(16) is 1
(C) For n = 10 ; f(11) is 11
(D) For n = 15, f(16) is 15

8) Let a, b, c and d are four distinct real numbers satisfying the system of equations a + b = 8, ab +
c + d = 23, ad + bc = 28 and cd = 12. Which of the following is/are true

(A) (2c + 2d – 15)2 = 1

(B) ((4ab + 4cd) – 110)2 = 9
(C) 6
2
(D) (2ab – 31) + 1 = 5

SECTION-II

1)

Let a1, a2, ..... an, be real numbers such that

then find the value of

2) When x100 is divided by x2 – 3x + 2, then remainder is (2k+1 – 1)x – 2(2k – 1) where k is a numerical
quantity. Then k must be

3) w, x, y & z are non-zero numbers such that and , then sum

of all possible values of is _____

4) If real numbers x, y, & z satisfy the equations x + y + z = 5 and xy + yz + zx = 8, then largest

possible integral value of z will be

5) Let m ≠ n be two real numbers such that m2 = n + 2 and n2 = m + 2. The value of will be

6) The number of all the real and distinct solutions of equation (x2 + x – 2)3 + (2x2 – x – 1)3 = 27(x2 –
1)3 is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

Q. 1 2 3 4
A. A D A B

SECTION-I (ii)

Q. 5 6 7 8 9 10 11 12
A. A,C A,D A,B,C,D A,C,D A,C,D B,C,D A,B,D A,B,C

SECTION-II

Q. 13 14 15 16 17 18
A. 6.00 10.00 2.00 10.00 7.00 21.00

PART-2 : CHEMISTRY

SECTION-I (i)

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

SECTION-I (ii)

Q. 23 24 25 26 27 28 29 30
A. A,B,C,D B,C,D A,B,C A,B,D A,B,D B,D B,D A,B,C

SECTION-II

Q. 31 32 33 34 35 36
A. 6.00 4.00 7.00 56.00 4.00 3.00

PART-3 : MATHEMATICS

SECTION-I (i)

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

SECTION-I (ii)

Q. 41 42 43 44 45 46 47 48
A. A,B,C A,B B,C B,C A,C,D A,D A,B A,C
 SECTION-II

Q. 49 50 51 52 53 54
A. 5050.00 99.00 6.00 2.00 0.25 4.00
 SOLUTIONS

PART-1 : PHYSICS

1)

(i) The MOI of a uniform semicircular disc about axis AOB shown in figure is and of

another disc =
So MOI of the composite disc about
AOB is

0
I =

2)
fr = µmg
a = µg
time at which V become zero

0 = V – µgt ⇒

by τ = Iα ⇒

3) (R + r)ω = rω'
(R + r)ωdt = rω'dt

4)

α = 10 rad/sec
τ = µK · N · R = Iα
⇒ µK · 25 = 5
⇒ µK = 0.2

5)

Correct Answer is (A,C)

6)

Correct Answer is (A,D)

7)

Correct Answer is (A,B,C,D)

8)

F = 2T ⇒ T =

acceleration of blocks,
both blocks have same acceleration but direction is opposite.
Net torque on fixed disc is zero. Torque due to tensions in the string cancel.
Acceleration of left pulley

As the rope does not slip over pulley

a0 – αR = a2

⇒ ⇒α=0
Hence left pulley does not rotate
Right pulley rotates in clockwise direction.

9) In both situations

In situations (ii)

10)
J0 = mv .......(1)
J0hm = Icω ........(2)
v = ωR ........(3)

⇒ hm =

(A) If a = 0 IC = &R=b ∴ hm =

(B) If a = b IC = mb2 & R = b ∴ hm =

(C) v = ⇒ω=
(D) Force is acting on COM ∴ No rotation.

11)

By conservation of angular momentum,

I1ωi = (I1 + I2)ωf
ωf = ω1/3

Ratio of the heat produced to initial kinetic energy =

12)
Angular momentum about IAOR remains conserved. [no net external torque]
After hitting the step, point O behaves as the instantaneous axis of rotation. Let w; be the
initial angular velocity about centre and ωi be the angular velocity about point O immediately
after collision.
Then, Li = Lf
Icωi + mvir⊥ = IoW

Now, just after impact

mg

[∵ vc = Rω]
Using (1) in (2).

As the wheel ascends, q (which was initially 60o) decreases and Vc decreases [from
conservation of mechanical energy]

Thus, N increases.

13)

f ⇒ F = µmg
Balance torque of F and Mg about 'O'.
F (R – Rsinθ) = Mg D sinθ

1 – sinθ = sinθ

sin θ =

θ = 30° =

14)

There is no horizontal force, momentum is conserved

mv1 – mv2 = 0
for hinged point

Energy conservation,

Solving, .

15)

(2 r w)2 = (r1w)2 + (r1w)2

16)

Energy is conserved vi = vf
17)
For cylinder : μmg = macm
acm = μg ←

For plank μmg = 2ma

At time t

When slipping ends : vcm – Rw = v

18) COM of the two dominoes is

x0 = 15.5 mm
For system to remain in equilibrium the COM of upper two dominoes must be over the
thickness of third domino.

x < x0 < x + 10
or x < 15.5 mm
 and 15.5 < x + 10 ⇒ x > 5.5 mm

PART-2 : CHEMISTRY

19)

(A) is Distillation
(B) is Crystallisation
(C) is Steam distillation
(D) is Sublimation

20) Thynine base of DNA is replaced by uracil in RNA

21) Glucose and fructose both give Tollen's test positive.

Glucose having –CHO group.
Fructose is α-Hydroxy ketone.

22) The correct answer is option (B)

23) Aspartame, alitame, sucralose, saccharin all are artificial sweetener.

24) (1) False :

(2) True : Six member hemiacetal on anomeric carbon gives α-D glucose & β-D glucose.

(3) True :

(4) True : Monosaccharide cannot be hydrolysed to give polyhydroxy aldehydes and ketones
25) (A)

(C)

26) (A) It is defination of isoelectric point.

(B)
(C) Incorrect
(D) Neutral amino has number of basic groups acid equal to number of acidic group. So,
concentration of zwittor ion is maximum at isoelectric point

27)

Clear solution is [Ag(NH3)2]+ B is the vapour of I2.

28) (B) Nylon-6 has amide linkages

(D) Teflon prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high
pressure

29) Bacteriostatic
Erythromycin
Tetracycline
Chloramphenicol
 30)

31)

The correct answer is (6)

32)

The correct answer is (4)

33)
Stared carbon are sp2 hybrid = 7.

34) Let mass of N = x gram.

+ 20 × 1 = 20 × 1 × 2
x = 20 × 14 × 10–3

% of N =
= 56%

35) Rf =
on chromatogram distance travelled by compound is → 2 cm
Distance travelled by solvent = 5 cm

So Rf =

36) Condensation polymers : Nylon-6,6, Terylene (Dacron), Nylon-2-Nylon-6

PART-3 : MATHEMATICS

37)
 f(x) = a x3 + bx2 + cx + d
a = 1 (given)
f(0) = d (odd)
f(–1) = –1 + b – c + d(odd) ⇒ –1 + b – c + d = odd

⇒
⇒ b – c = odd ...(i)
If all the roots (say α,β,γ) of f(x) are integers
⇒ αβγ = –d (odd integer) ⇒ α,β,γ all should be odd
now
–b = α + β + γ ⇒ –b should be odd too
αβ + βγ + γα = c ⇒ c should be odd too
⇒ b – c = even ...(ii)
From (i) & (ii) we have a contradiction.

38) complementary counting

Let a,b ∈ {1,2,......2022} such that

⇒ a2 b2 – a2b + 2a2 – 2b2 + 2b – 4 = a2b2 – ab2 + 2b2 – 2a2 + 2a – 4

⇒ ab2 – a2b + 4a2 – 4b2 + 2b – 2a = 0
⇒ (b – a) (ab – 4a – 4b – 2) = 0
⇒ (a – 4) (b – 4) = 14

a–4 b–4

1 14
2 7
–1 –14
–2 –7

(6,11)
(3,0)×
+ (2,–3)×
Answer = 2022 – 4 + 2
= 2020

39) ∵ n(A) + n(B) + n(C) = n(A∪B∪C)

⇒ 2|A| + 2|B| + 2|C| = 2|A∪B∪C|
⇒ 2100 + 2100 + 2|C| = 2|A∪B∪C|
⇒ 2101 + 2|C| = 2|A∪B∪C|
Clearly |C| = 101 and |A∪B∪C| = 102
∴ |A∪B∪C| – |C| = 1

40) x3 + 3x2 + 3x + 2 = (x + 2) (x2 + x + 1)

Hence, ax2 + bx + c = 0 and x2 + x + 1 = 0 have common roots
∴

41)

Here f(a) = 0

∴ other root is .
Also a2 < b ⇒ b > 0
Case-I Case-II
a>0 a<0

∴ Option A ⇒ ⇒ case- I

Option B ⇒ ⇒ case- II

Option C ⇒ ⇒ case- I

Option D ⇒ ⇒ case- II

42) 2x3 + 4x2 - 5x + 8 = 2(x - α) (x - β) (x - γ)

Put x = 2, we get
30 = 2(2 - α) (2 - β) (2 - γ)
Put x = 1, we get
9 = 2(1 - α) (1 - β) (1 - γ)
Put x = –1
15 = 2(-1 - α) (-1 - β) (-1 - γ)
Now multiply all,

(Option A is correct)
Let 1 + α = x ⇒ α = x - 1
2α3 + 4α2 - 5α + 8 = 0
Put α = x – 1
⇒ 2(x - 1)3 + 4(x - 1)2 - 5(x - 1) + 8 = 0
⇒2(x3 - 3x2 + 3x - 1) + 4(x2 - 2x + 1) - 5x + 5 + 8 = 0
⇒ 2x3 - 6x2 + 6x – 2 + 4x2 - 8x + 4 - 5x + 13 = 0
⇒ 2x3 - 2x2 - 7x + 15 = 0
43)
Put x2 + x = t
t2 + at + 4 = 0 doesn't

have real root in

Case-I
⇒

Case-II
D < 0 ⇒ a ∈ (–4,4)

Thus final Ans.

44) a3 – 8b3 + 64 = –24ab

a3 + (–2b)3 + (4)3 = 3(a) (–2b) (4)
⇒ 4 –2b + a = 0
⇒ x = –2 is a root
α + β = –b
0<b<3
0 > –b > –3
0 > α + β > –3
0 > –2 + β > –3
2 > β > –1

45) On solving,
Also x3 – 3x2 + 2x + 4 = (x – α) (x – β) (x – γ)
At x = 2,

46) * Option (a) :

Here, and
Therefore,
Hence option (a) is correct.
* Option (b) :
Here, , and

Therefore,
Hence option (b) is incorrect.
* Option (c) : The cardinality of the set is 10.
Here, = {1,2,3,4,5,6,7,8,9}. So the cardinality of the set is 9 not 10.
Hence option (c) is incorrect.
* Option (d) : The cardinality of the set is 10.
Heere, = {1,2,3,4,5,6,7,8,9}, and = {10}
Now, . So the cardinality of the
set is 10.
Hence option (d) is correct.

47) g(x) = (x + 1) f(x) – x

= K x(x – 1)...... (x – n)
For n = 10
g(x) = Kx (x – 1) ..... (x – 10)
(x + 1) f(x) – x = Kx(x – 1) .... (x – 10)
Put x = –1
0 + 1 = K(–1)(–2) .... (–11)

48) (x2 + ax + c) (x2 + bx + d)

= x4 + (a + b)x3 + (c + d + ab)x2 + (ad + bc)x + cd
= x4 + 8x3 + 23x2 + 28x + 12
= (x + 1)(x + 2)2(x + 3)

49)

a1 = 1, a2 = 2, .... an = n

50) Given that, when x100 is divided by x2 – 3x + 2, the remainder is (2k+1 – 1)x – 2 (2k –1)
so x100 = q(x) (x2 – 3x + 2) + ax + b
(where q(x) isquotient and ax + b is remainder also x100 is dividing by quadratic so remainder
is in the form of ax + b).
By putting x = 1,
1100 = a + b ⇒ a + b = 1 ....(i)
By puttting x = 2,
2100 = 2a + b ⇒ 2a + b = 2100 ....(ii)
from (i) and (ii), we get
⇒ a = 2100 – 1, b = 2 – 2100
⇒ remainder = (2100 – 1)x + (2 – 2100)
= (299+1 – 1)x – 2 (299 – 1)
⇒ k = 99

51) Let a = , b = , c = and d =

a+b+c+d=6
and ab + bc + cd + ad = 8
⇒ (a + c) (b + d) = 8
⇒ (a + c) (6 – a – c) = 8
⇒ (a + c)2 – 6(a + c) + 8 = 0
⇒ a + c = 2 or 4

52) x = 5 – (y + z)
yz + x(y + z) = 8
⇒ y2 + y(z – 5) + z2 – 5z + 8 = 0
for real y, Δ ≥ 0.
⇒ (z – 5)2 – 4(z2 – 5z + 8) ≥ 0

⇒ ⇒

53) m2 = n + 2
n2 = m + 2
m2 – n2 = n – m
∵ m ≠ n ⇒ m + n = –1
Also m2 + n2 = m + n + 4
⇒ (m + n)2 – 2mn = m + n + 4
⇒ 1 – 2mn = 3 ⇒ mn = –1

= 0.25

54) (x – 1)3(x + 2)3 + (2x + 1)3(x – 1)3

= 27(x – 1)3(x + 1)3
⇒ (x + 2)3 + (2x + 1)3 = 27(x + 1)3 [x = 1]
⇒ 18x3 + 63x2 + 63x + 18 = 0
⇒ 2x3 + 7x2 + 7x + 2 = 0
⇒ 2(x3 + 1) + 7x(x + 1) = 0
⇒ (x + 1)(2x2 + 5x + 2) = 0
⇒ (x + 1)(2x + 1)(x + 2) = 0
⇒ x = –1, x = –1/2, x = –2
x = –1, –1/2, –2, 1