FIITJEE – RBT-2 for (JEE-Advanced)

PHYSICS, CHEMISTRY & MATHEMATICS

Pattern – 3 QP CODE: 100950 PAPER - 1

Time Allotted: 3 Hours Maximum Marks: 204
BATCHES – Two Year CRP325 batches

▪ Please read the instructions carefully. You are allotted 5 minutes specifically for
this purpose.
▪ You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS
Caution: Question Paper CODE as given above MUST be correctly marked in the answer
OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results.

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Sections.
3. Section-I is Physics, Section-II is Chemistry and Section-III is Mathematics.
4. All the section can be filled in PART-A & B of OMR.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on
OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with Blue/Black Ball Point Pen for each
character of your Enrolment No. and write in ink your Name, Test Centre and other details at the
designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Two Parts.

(i) Part-A (01-04) – Contains Four (04) multiple choice questions which have ONLY ONE CORRECT answer
Each question carries +3 marks for correct answer and -1 marks for wrong answer.

(ii) PART–A (05–12) contains Eight (8) Multiple Choice Questions which have One or More Than One
Correct answer.
Full Marks: +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks: +1 For darkening a bubble corresponding to each correct option, provided NO incorrect
option is darkened.
Zero Marks: 0 If none of the bubbles is darkened.
Negative Marks: −2 In all other cases.
For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result
in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) will result in −2
marks, as a wrong option is also darkened.

(iii) Part-B (1 – 8) contains Eight (08) Numerical based questions, the answer of which maybe positive or
negative numbers or decimals TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) and each
question carries +3 marks for correct answer. There is no negative marking.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________

Enrolment Number :_______________________________________________

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 IT−2025−RBT-2-Paper-1 (PCM) JEEA

S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.

1. A ball is projected from a point in one of the two smooth parallel vertical walls against the
other in a plane perpendicular to both. After being reflected at each wall impinge again on
the second at a point in the same horizontal plane as it started from. If “e” is the coefficient of
restitution, “a” is the distance between the walls and “R” is the free range on horizontal
surface, then which of the following is correct?
(A) Re3 = a(1 + e + e2) (B) Re = a(1 + e2 + e3)
(C) Re2 = a(1 + e + e2) (D) R = a(1 + e + e2)

2. A hemisphere rests in equilibrium on a rough ground and against an equally rough wall, if the
equilibrium is limiting, then the inclination  of the base to the horizontal is
 3   + 2  
−1 8   +  
2

(A)  = sin−1   2 
(B)  = sin   2 
 8  1 +    3  1 +  
 8   + 2  
−1 3   +  
2

(C)  = cos−1   2 
(D)  = cos   2 
 3  1 +    8  1 +  

3. A cubical container of side ‘a’ and wall thickness x (x << a) is suspended in air and filled n
moles of diatomic gas (adiabatic exponent ) in a room where room temperature is T0. If at
time t = 0 gas temperature is T1 (T1 > T0), the temperature of gas ‘T’ at time ‘t’ is
6ka2 ( −1) 6ka2 ( −1)
− t − t
(A) T = T0 + T1e nRx (B) T = T0 − (T1 − T0 )e nRx

6ka2 ( +1) 6ka2 ( −1)

− t t
(C) T = T0 + (T1 + T0 )e nRx (D) T = T0 + (T1 − T0 )e nRx

Q0
4. Two identical capacitor connected as shown and having v +–
initial charge Q0 each. Separation between plates of +–
+–
capacitor is d. Suddenly the left plate of upper capacitor and
right plate of lower capacitor start moving with speed v
+–
towards left while other plate of capacitor remains fixed. The +–
current in the circuit is +–
Q0 v
Q0 v Q0 v Q0 vt
(A) (B) (C) (D) zero
2d d 2d
Space For Rough Work

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 IT−2025−RBT-2-Paper-1 (PCM) JEEA

PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).

5. A strip of wood of mass M and length  is placed on a smooth horizontal surface. An insect
of mass m starts at one end on the strip and walks to the other end in time t, moving with a
constant speed.
(A) the speed of the insect as seen from the ground is < .
t
 M 
t  M + m 
(B) the speed of the strip as seen from the ground is

 m 
(C) the speed of the strip as seen from the ground is 
t  M + m 
2
1  
(D) the total kinetic energy of the system is (m + M)  
2 t

6. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons
have maximum kinetic energy TA eV and de-Broglie wavelength A. The maximum kinetic
energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is
TB = (TA – 1.50) eV. If the de-Broglie wavelength of these photoelectrons is B = 2A then
(A) the work function of A is 2.25 eV (B) the work function of B is 4.20 eV
(C) TA = 2.00 eV (D) TB = 2.75 eV

7. A tube of length l and radius R carries a steady flow of fluid whose density is  and viscosity
is . The fluid flow velocity depends on the distance r from the axis of the tube
 r2 
as v = v0 1 −  . Choose the correct option(s) from the following.
 R 2 
1 2
(A) the volume of the fluid flowing across the section of the tube per unit time is  v0R 2
2
1
(B) the kinetic energy of the fluid within the tube's volume is  R2v02
6
4
(C) the friction force exerted on the tube the fluid is v 0
4 v0
(D) the pressure difference at the ends of the tube is
R2
Space For Rough Work

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8. In the following R − C circuit, the capacitor is in the

steady state. The initial separation of the capacitor
plates is x 0 . If at t = 0, the separation between the
plates starts changing so that a constant current
flows through R, find the velocity of the moving
plates as a function of time. The plate area is A.
−2 −2
 I /  0 A   I /  0 A  1 
2
 I / 0 A   I / 0 A  1 
(A)  
   t+  (B)    t + 
 IR −     − IR  x0 
  IR −     − IR  x0 
2
 I /  0 A   I /  0 A  1 
2 −1
 I / 0 A   I / 0 A  1 
(C)    t +  (D)  
   t + 
 IR −     − IR  x0   IR −     − IR  x0 


9. Two identical circular coils M and N are arranged coaxially as

shown in the figure. Separation between the coils is large as M N
compared to their radii. The arrangement is viewed from left
along the common axis. The sign convention adopted is that
currents are taken to be positive when they appear to flow in
clockwise direction. Then
(A) if M carries a constant positive current and is moved towards N, a positive current is
induced in N
(B) if M carries a constant positive current and N is moved towards M, a negative current is
induced in N
(C) if a positive current in M is switched off, a positive current is momentarily induced in N
(D) if both coils carry positive currents, they will attract each other

10. An external magnetic field is decreased to zero due to which a current is induced in a
circular wire loop of radius r and resistance R placed in the field. This current will not
become zero.
(A) at the instant when external magnetic field stops changing (t = 0), the current in the loop
as a function of time for t  0 is given by i0e−2Rt/2F
(B) at the instant when B stops changing (t = 0), the current in the loop as a function of
0 IR
time t  0 is given by
2r
(C) the time in which current in loop decreases to 10−3 I0 (fromt = 0) for R = 100 and
32 ln10
and r = 5 cm is given by
1010
(D) the time in which current in loop decreases to 103 I0 (from t = 0 )for R = 100 and r = 5cm is
3 2
given by s
106

Space For Rough Work

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11. In the series L – C – R circuit, the voltage across resistance, capacitance and inductance are
30V each at frequency f = f0.
(A) It the inductor is short-circuited, the voltage across the capacitor will be 30 2 V.
30
(B) If the capacitor is short-circuited, the voltage drop across the inductor will be V.
2
(C) If the frequency is changed to 2f0, the ratio of reactance of the inductor to that of the
capacitor is 4 : 1.
(D) If the frequency is changed to 2f0, the ratio of the reactance of the inductor to that of the
capacitor is 1 : 4.

12. A wave disturbance in a medium is described by

y(x, t) = 0.02 cos (50 t + /2) cos (10 x),
where ‘x’ and ‘y’ are in metre and ‘t’ in seconds.
(A) A node occurs at x = 0.15 m
(B) An antinode occurs at x = 0.3 m
(C) The speed of the component wave is 5.0 m/s
(D) The wavelength is 0.2 m.

PART – B (Maximum Marks: 24)

(Numerical Type)
This section contains Eight (08) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

1. A machine is blowing spherical soap bubbles of different radii filled with helium gas. It is
found that if the bubbles have a radius smaller than 1 cm, then they sink to the floor in still
air. Larger bubbles float in the air. Assume that the thickness of the soap film in all bubbles is
uniform and equal. Assume that the density of soap solution is same as that of water
(=1000 kg m−3). The density of helium inside the bubbles and air are 0.18 kg m −3 and
1.23 kg m−3, respectively. If d is the thickness (in m) of the soap film of the bubbles which
are just floating, find the value of 2d.

2. Find the mass ratio MD/MH of deuterium and hydrogen (in near integer value) if their H lines
have wavelengths of 6561.01 Å and 6562 Å respectively.

P (N/m2)
3. Two moles of a monatomic ideal gas is taken
A B
through a cyclic process shown on pressure(P) - 2  10 5

temperature(T) diagram in figure. Process CA is

represented as PT = Constant. If 105 C
efficiency of given cyclic process is
3x
1− , then find the value of ‘x’. 300 K T
12ln 2 + 15

Space For Rough Work

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4. A solid sphere of radius R = 0.2 m and of negligible mass is swimming in the water tank of
base surface area A = 0.5 m2. The depth of the water in the tank is h = 1 m. Determine the
work (in joules) needed to push to push the sphere down the bottom of tank. (Assume that
no water flows out of the tank. The density of water is  = 1000 kg/m3 and g = 9.81 m/s2).

5. A small wheel, initially at rest, rolls down a ramp in

the shape of a quarter circle without slipping. The
radius of the circle is R = 1 m and  = 60°, β = 30o .
Find the height ‘x’ (in cm) reached by the wheel
after leaving the track.

6. A string is wrapped on a wheel of moment of inertia

0.20 kg m2 and radius 10 cm and goes through a 10 cm
light pulley to support a block of mass 2.0 kg as
shown in figure. Find the acceleration of the block.
(in m/s2)
2 kg

7. A battery is made by joining identical cells in m rows each row having m cells. The current
flowing in an external resistance R is i1. Now if the number of rows ‘m’ and number of cells in
m R
each row ‘n’ are interchanged, the current in external resistance is i2. If = 2 and = 4
2 r
i2
then the value of is
i1
0.25H 100

8. In the given Ac circuit, what is the power 100 50F

developed in watts?

220 V, 200 rad/s

Space For Rough Work

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S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.

1. The product A is
OH

( i) K 2CO3
⎯⎯⎯⎯⎯⎯ →A
( ii) CH3I
CH2OH
OH OCH 3

(A) (B)

CH2OCH 3 CH2OH
CH3 OCH 3

(C) (D)

CH2OH CHO

2. CH3
l3 P+Q+R
FeC
Cl 2/

Cl2 /h
v(1 e
q)
M
Which product can easily undergoes nucleophilic substitution reaction(SN1 and SN2 with
equal ease).
(A) P (B) Q
(C) R (D) M
Space For Rough Work

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3. Which is most reactive towards aromatic nucleophilic substitution reaction towards NaOH?
Cl
Cl
CH3
CH3
(A) (B)

NO 2
NO 2
Cl
Cl
NO 2
(C) (D)

O 2N NO 2
NO 2

4.
OH P
O /Na
Br 2
P4O10
Q
CH3CH2CNH2 CH
3 CO
ON
a
LiA
lH R
4

S
Choose correct statement
(A) (P) and (S) are chain isomers (B) hydrolysis of (Q) produces carboxylic acid
(C) (R) is obtained by the fastest reaction (D) (S) contains four carbon atom

PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).

5. Choose incorrect statements:

3+
(A) AlCl3 undergoes hydrolysis to form  Al (H2O )6 
(B) Amalgamated Al reacts with cold water to liberate H2
(C) The ash formed on burning magnesium in air fumes on wetting and exposing to HCl
(D) B3+ ion can exist in aqueous solution in the hydrated form
Space For Rough Work

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6. Given H0f ( C2H6 ,g) = −85kJ / mole,

H0f ( C3H8 ,g) = −104kJ / mole,
H0sub (C,s) = 718kJ / mole&B.E.of (H − H) = 436 kJ/mole.
Then, in kJ/mole, the
(A) C – C bond enthalpy is 218 (B) C – H bond enthalpy is 414
(C) C – C bond enthalpy is 345 (D) C – H bond enthalpy is 448

7. -
O
O
H3C 
O
Br
Which of the statements regarding the product formed in the above reaction is/are correct?
(A) If Br is replaced with NO2 rate of reaction increases
(B) Inversion of configuration takes place.
(C) Nucleophillic acyl substitution takes place.
(D) Configuration at chiral carbon does not change

8. Which of the following compounds are more reactive than R - C - OH towards attack of
nucleophile?
(A) O (B) O

R - C - Cl R - C - CN
(C) O (D) O

R - C - NH 2 R-C-H

9. Compound/s which contain 3c – 2e– bond.

(A) (BeH2)n (B) Al2(CH3)6
(C) B2H6 (D) Al2Cl6

10. CH3CHO + NaHSO3 → CH3CH ( OH) SO3−Na +

Reagent - X

In the above change, the reagent (X) can be

(A) CCl4 (B) HCl
(C) NaOH (D) DMF
Space For Rough Work

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11. Acetone can form an aromatic compound when treated with conc. H2SO4. Which of the
following reaction(s) do(es) not take place?
(A) oxidation (B) dehydration
(C) sulphonation (D) dehydrogenation

OH
CH3
OH
12. o
⎯⎯⎯⎯→
Cu/300 C
Product

CH 2OH
Which of the following(s) is/are minor product(s) or product(s) that are not formed at all?
HO CH3 HO CH3
O O

(A) (B)

CHO COOH
CH3 CH2
O O

(C) (D)

CHO CHO

PART – B (Maximum Marks: 24)

(Numerical Type)
This section contains Eight (08) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

1. 4 moles of a liquid A (V.P = 100 mm of Hg) was mixed with 6 moles of B(V.P = 400 mm of
x
Hg). If the mole fraction of A in the vapour above the solution is expressed as , what is the
y
value of (x + y)?

2. How many moles of O2 gas can be produced by passing eight Faraday of electricity into
sufficient water?
Space For Rough Work

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3. NH4NO3 ⎯⎯→ N2O + 2H2O

What is the n-factor of NH4NO3 in above reaction?

4. An organic compound(X) upon ozonolysis reaction, produces 2-heptanone and

formaldehyde. Reaction of (X) with HCl forms major product(Y) and minor product (Z).
Reaction of (Y) with sodium metal in presence of dry ether(Wurtz reaction) forms the only
product(P).
If x = the number of quartenary carbon atoms present in (P)
y = the number of monochloro products formed by (P) excluding stereoisomers
z = number of allylic hydrogen atoms present in (X)?
x+ y+z
What is the value of  ?
 10 

5. 11.7 g of NaCl solid is added to 176.4 g of H2O. The degree of dissociation of NaCl in the
solution is 0.8. If the relative lowering of vapour pressure of water in the solution is
expressed as x  10-2, what is the value of x?

6. In a mixed metallic oxide, the oxide ion forms the hexagonal close packing(hcp) lattice. The
M2+ ions occupy ½ of the octahedral voids and the N3+ ions occupy 1/6th of the tetrahedral
x+ y+z
voids. If the formula of the solid is written as MxNyOz, what is the value of  ?
 10 

7. A 0.5 M aqueous solution of Al2(SO4)3 is isotonic to a 0.2 M aqueous solution of glucose. If

the solute undergoes completely ionization, then the ratio of their osmotic pressure is
x
expressed as x : y. What is the value of in decimal form?
y

8. 5.35 g of a salt of a strong acid and weak base is dissolved in water to make one litre
solution. The pH of the solution is found to be 5. The crystal of the salt contains NaCl type
unit cell. What is the mass of one mole of unit cell of the salt in gram unit?
[Kb of the weak base = 10–5]
Space For Rough Work

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S
SEEC
CTTIIO
ONN –– IIiiii :: M
MAAT
THHE
EMMA
ATTIIC
CSS
PART – A (Maximum Marks: 12)
This section contains FOUR (04) questions. Each question has FOUR options. ONLY ONE of these four
options is the correct answer.

1. A varying parabola ‘S’ of latus rectum ‘4a’ touches y2 = 4ax, the axis of ‘S’ being parallel to x
– axis. The locus of vertex of ‘S’ is
(A) x + y + a = 0 (B) y2 = 16ax
(C) x + y − a = 0 (D) y2 = 8ax

2. On the open interval (–c, c), where c is a positive real number, y(x) is an infinitely
dy
differentiable solution of the differential equation = y 2 − 1 + cos x, with the initial condition
dx
y (0) = 0 . Then which one of the following is correct?
(A) y ( x ) has a local maximum at the origin
(B) y ( x ) has a local minimum at the origin.
(C) y ( x ) is strictly increasing on the open interval ( −, ) for some positive real number 
(D) y ( x ) is strictly decreasing on the open interval ( −, ) for some positive real number  .

 1 1
3. Let f : ( 0,  ) → ( 0,  ) , f (1) = 2 be a twice differentiable function such that f '   = for
 x  f (x)
all x in given domain, then f ( 3 ) =
(A) 2 (B) 10
(C) 3 3 (D) 3

4. The values that ‘a’ can take so that the function f :R → R f ( x ) = x3 + x2 + acos x
is injective
1   1
(A) a   ,   (B) a   −, − 
3   3
(C) a = 2 (D) No value of a
Space For Rough Work

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 IT−2025−RBT-2-Paper-1 (PCM) JEEA

PART – A (Maximum Marks: 32)

This section contains EIGHT (08) questions. Each question has FOUR options (A), (B), (C) and (D). ONE
OR MOER THAN ONE of these four option(s) is (are) correct answer(s).

x2 y2
5. Equation of ellipse E is + = 1 . Suppose C be a circle concentric with E intersecting the
16 4
ellipse in four points. Let A be a point on E and B be a point on C such that line joining A and
B is common tangent of ellipse E and circle C. If be maximum value of length AB and m
be its corresponding slope, then
(A) = 2 (B) m =  2
1
(C) = 6 (D) m =
2

6. Let S = w1,w 2 ,.... be a sample space associated to a random experiment. Let

P ( w n−1 )
P ( wn ) = ,n  2 . Let A = 2k + 3 :k,  N and B = wn :n  A . Then
2
1 3
(A) P ( w n ) = (B) P (B ) =
2n−1 32
1 3
(C) P ( w n ) = n (D) P (B ) =
2 64

7. If 2z + 3 = 3z + 4 , then
7
(A) z min = 2 (B) z max =
5
(C) z max = 2 (D) z min =1

8. Let f : X → Y be a function. If P  X and Q  Y, define f (p ) = f ( x ) : x  P and

f −1 ( Q ) = x :f ( x )  Q . Then the true statement(s) is/are:
(A) f ( f −1 ( Q ) ) = Q (B) f −1 ( Q)  f −1 (R) = f −1 (Q  R)
(C) f −1 ( f (P ) ) = P (D) f −1 ( Q)  f −1 (R) = f −1 (Q  R)

Consider function f : A → B and g :B → C ( A,B,C  R ) such that ( gof ) exists then

−1
9.
(A) f is ONTO and g is ONE – ONE (B) f is ONE – ONE
(C) g is ONTO (D) f and g are both ONTO
Space For Rough Work

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10. The function f ( x ) is differentiable, continuous and f ( x )  0 for all x on the interval  4, 8 .

( f ' ( x ))
2
8
1 1
Also f ( 4 ) = , f ( 8 ) = and  f ( x) dx = 1. Find f ( 6 ) .
( )
4
4 2 4

1 1
(A) f ( 6 ) = (B) f ( 6 ) =
9 3
2
(C) f ( 5 ) = (D) f ( 5 ) = 3
7

z1 − z2
11. A relation R on the set of complex numbers is defined by z1 R z 2 if and only if is
z1 + z2
real. Then
(A) R is reflexive relation (B) R is symmetric relation
(C) R is transitive relation (D) R is equivalence relation

12. Consider 4 persons A, B, C, D

(A) If A, B, C, D each have four houses then number of ways in which they can enter these
houses so that none of them enter their own house is 9
(B) If A has two houses whereas B, C, D have one house each then number of ways in
which they can enter these houses such that no one enters their own house is 44
(C) If A, B, C, D each have four houses then number of ways in which they can enter these
houses so that none of them enter their own house is 16
(D) If A has two houses whereas B, C, D have one house each then number of ways in
which they can enter these houses such that no one enters their own house is 42

PART – B (Maximum Marks: 24)

(Numerical Type)
This section contains Eight (08) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals to TWO decimal places (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30).

1. If number of arrangements of letters of the word MISSISSIPPI in which the first S precedes
the first I equals , then equals
10

2. For all real numbers x, f ( x ) is an increasing function that is differentiable and satisfies the
following conditions
2
5
(i) f (1) = 1,  f ( x ) dx =
1
4
(ii) Let g ( x ) be inverse of f ( x ) , then for all x  1,g ( 2x ) = 2f ( x ) .
8
p
If  x f ' ( x ) dx = q ; HCF (p,q) = 1, then p + q =
1

Space For Rough Work

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 99 
  10 + n 
3. The value of  n99=1  is equal to (here [.] represents greatest integer function) is
 10 − n 
 
n =1

equal to

4. If the length of the shortest distance between any two opposite edges of the tetrahedron
formed by the planes whose equations are y + z = 0, z + x = 0, x + y = 0 and x + y + z = 3
is  , then the value of 2 is equal to

5. f is differentiable function such that f ( f ( x ) ) = x where x  0,1 . Also, f ( 0) = 1 . Then the

1
value of 2024  ( x − f ( x ) )
2023
dx is
0

6. Let positive vector of points A, B and C of triangle ABC respectively be ˆi + ˆj + 2k,ˆ ˆi + 2jˆ + kˆ
and 2iˆ + ˆj + kˆ . Let I1,I2 and I be the lengths of perpendiculars drawn from the orthocenter
3

‘O’ on the sides AB, BC and CA, then (l1 + l2 + l3 ) equals.

1 f ( x)
7. Let f ( x ) = x sin x − (1 − cos x ) . The smallest positive integer k such that lim k  0 is
2 x →0 x

8. Consider the real valued function h : 0,1,2,......,100 → R such that h ( 0) = 5, h (100) = 20

and satisfying h (i ) =
1
2
(h (i + 1) + h (i − 1) ) for every i = 1,2,.......,99 . Then the value of h(1) is
Space For Rough Work

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Q
Q.. P
PCCooddee:: 110000995500
Answers
S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – A
1. C 2. B 3. D 4. B
5. AC 6. ABC 7. BCD 8. B
9. BCD 10. AC 11. BC 12. ABCD
PART – B
1. 7 2. 2 3. 7 4. 274.00
5. 4.58 6. 0.89 7. 1.50 8. 629.20

S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – A
1. B 2. D 3. D 4. B
5. AD 6. BC 7. ACD 8. ABD
9. ABC 10. BC 11. ACD 12. ABD
PART – B
1. 8 2. 2 3. 4 4. 1.4
5. 3.6 6. 1.1 7. 12.5 8. 214

S
SEEC
CTTIIO
ONN –– IIIIII :: M
MAATTH
HEEM
MAATTIIC
CSS
PART – A
1. D 2. D 3. B 4. C
5. AD 6. CD 7. BD 8. BD
9. BC 10. BC 11. ABCD 12. AD
PART – B
1. 924 2. 143 3. 2 4. 2.82
5. 1 6. 1.22 7. 4.00 8. 5.15

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Answers & Solutions

S
SEEC
CTTIIO
ONN –– II :: P
PHHY
YSSIIC
CSS
PART – A
1. C
a q a 2uy
Sol. + + 2 = …(1)
ux eux e ux g
Horizontal range:
2.uxuy
R= …(2)
g

2. B
Sol. N1 = N2 …(1)
W = N1 + N2 …(2) N1
By (1) and (2)
W
N1 = …(2) N1
1 + 2 O
Balancing torque about ‘O’
3R
N1 × R + N2 × R = W  sin 
8 N2

8   + 2 
Sin  =   N2
3  1 + 2  W

3. D
dQ K(6a2 ) (T − T0 )  R  dT
Sol. = = n 
dt x   − 1  dt
6Ka2 (  − 1)
T t
dT
 T − T0 = nRx o dt
T1
6Ka2 ( −1)t
T = T0 + (T1 − T0 )e nRx

4. B
q1 q2
Sol. = ; q1 + q2 = 2Q0
C1 C2
A  A
C1 = 0 ; C2 = 0
d0 + vt d0 − vt
q1 d0 − vt
=
q2 d0 + vt
 d − vt 
q2  0  + q2 = 2Q0
 d0 + vt 
 2d0 
q2   = 2Q0
 d0 + vt 
2Q0
q2 = ( d0 + vt )
2d0
dq Qv
I= 2 = 0
dt d0

5. AC

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Sol. Velocity of insect w.r.t. strip =

t
Let strip moves with speed 
Initial momentum was 0
 
 0 = m  + v  + Mv
t 
m /t m /t
v=− i.e. toward left
M+m M+m
m /t
 Velocity of insect w.r.t. ground = − < .
t M+m t

6. ABC
Sol. KE = h – 

7. BCD
Sol. Let dV be the volume flowing per second through the cylindrical shell of thickness dr then,
 r2   r3 
dV = (2rdr)v0 1 − 2  = 2v 0  r − 2  dr and the total volume,
 R   R 
 
R
 r3  R2 R2 v0

0

V = 2v0  r − 2  dr = 2v0
R  4
=
2

Let, dK be the kinetic energy, within the above cylindrical shell. Then
1 1
dK = (dm)v 2 = (2r )v 2dr
2 2
1  r2 
= (2 )rv 20 1 − 2  dr
2  R 

 2r 3 r5 
= lv20  r − 2 + 4  dr
 R R 
Hence, kinetic energy of the fluid,
R
 2r3 r5  R2v02

0

K = lv20  r − 2 + 4
R R
 dr =
 6
Hence frictional force is the shearing force on the tube, exerted by the fluid, which
dv
equals −S , where S is the surface area of tube.
dr

r2  dv r
Given, v = v0 1 − 2 

So = −2v 0 2
 R  dr R
dv 2v
and at r = R, =− 0
dr R
Then, viscous force is given by,
 dv 
dF = −(2Rl)  
 dr r =R
 2v 
= −2Rl  − 0  = 4lv0
 R 
dv
Taking a cylindrical shell of thickness dr and radius r Viscous force, F = −(2rl) ,
dr

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dv
Let,  P be the pressure difference, the net force on the element = Pr2 + 2lr
dr
But, since the flow is steady, Fnet = 0
dv 2lr  2v r 
−2lr  0 2  4lv
or P = dr =  R 
= 0
2 2 2
r r R

8. B
Sol. Let q be the instantaneous charge on the capacitor when a steady current I flows through
the circuit. Appling KVL on the circuit, we have
q qx  0 A 
= + IR or  = + IR  C=  .....(i)
C 0 A  x 
Differentiating eqn. (i) with respect to time, we get
q  dx  Ix
0=   + +0 ( q = It)
0 A  dt  0 A
Ix  dx 
or q = − where = velocity  .....(ii)
(dx / dt)  dt 
From equationn (ii) substituting the value of q in equation (i). we have
x2
 = −I + IR
0 A(dx /dt)
dx  −I/ 0 A  2
or =v= x ....(iii)
dt   − IR 
dx  I/ 0 A 
or − =
2  dt ....(iv)
x   − IR 
Integrating the above expression w.r.t. time, we get
1 1  I/ 0 A 
− = t ....(v)
x x0   − IR 
From eqn (iii) and (v), we get
−2
 I/ 0 A   I/ 0 A  1 
v=   t + 
 IR −     − IR  x0 

9. BCD
Sol. Apply Faraday law for direction of induced current.

10. AC
Sol. Flux linked with loop due to its own magnetic field,
0 I 2  rI
= (r ) = 0
2r 2
d  r dI
emf induced = ==− 0
dt 2 dt
 0 r dI
I= =−
R 2R dt
1 t
dI 2R

I0
I
=− 
0 r
0
.dt;I = I0 e −2Rt/0 r

2Rt
−
0 r
−3
Now, 10 I0 = I0e = i0e−2Rt/0r
32 ln10
which gives t = s.
1010

11. BC
vL = vC = vR;

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 xL = xC = R
when inductor is short circuited
Z= R2 2
xC = 2R
30 30
I=
Z 2R
30 30
VL = ixL = ×R=
2R 2
 (A) is incorrect and with similar calculations (B) will be correct.
Here f0 is the resonance frequency as vL = vC
1
0 = 2f0 =
LC
xL L
= = 2LC
xC 1 C
Given f = 2f0
  = 20
xL
 =4
xC
 (C) is also correct.

12. ABCD
Sol. y = 2a cos (t + ) cos (Kx) comparison with given equation gives
2 1
K= = 10   = m
 5
 = 2f = 50  f = 25 Hz
 v = 5 ms–1
at x = 0.15 m
cos (10  × 0.15) = cos (1.5)
= –1 for all t
at x = 0.3
cos (10  × 0.3) = cos 3 = – 0 for all t

PART – B
1. 7
Sol. For equilibrium of bubble.
4 4
4R2 d solg + R3 He g = R3air g
3 3
d → Thickness of soap solution.
 d = 3.5 m.

2. 2
Sol. Use formula in terms of reduced mass
1 R z 2  1 1
=  2 − 2
  m   n1 n2 
1 + M 
 
m
1+
D MD
 =
H m
1+
MH
MD
 2.0
MH

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3. 7
Sol. For process AB TA = 300 K , TB = 600 K
W = nRT = nR(TB − TA ) = 300 nR = 600R.
5
Q = n Cp T = 2  R (300) = 1500R.
2
vf p
For process BC W = nRT ln = nRT ln i = nRTln2 = 1200Rln2
vc pf
Q = W = 1200Rln2
300 K 2nRT
For process CA W =  P dV =  dT.
600 T K
= −2nR(300) = −1200R.
Q = nC V  T + W
3
= 2  R( −300) − 1200R.
2
= −900R − 1200R = −2100R
600R + 1200Rln 2 − 1200R
=
1500R + 1200Rln 2
21
= 1−  x = 7.
12ln 2 + 15

4. 274.00
 4 
 R3 
4  x 4 3
Sol. W = R g h − R +  = R g h − R +
3 3

3  2 3  2A 
 

5. 4.58
Sol. h = (R – r) [cos 30° – cos 60°]
1 1
mgh = mv 2 + mr 2 2
2 2
2 2 o
v sin 30
x=
2g
6. 0.89
Sol. 2g – T = 2a …(i)
TR = I …(ii)
a = R …(iii)
Ia
From (ii) and (iii) T = 2
R
 I 
 2g = a  2 + 2 
 R 
2g 2  9.8 9.8
 a= = =  0.89 m/s2
 I  0.2 11
 2 + 2  2 + 0.01
 R 

7. 1.50
E E
Sol. i1 = & i2 =
r R r R
+ +
m n n m
r R m R
+ 1+ 
i2 m n n r = 3 = 1.5
= =
i1 r R m R 2
+ +
n m n r

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8. 629.20
Sol. XL = 50 , XC = 100 
Z1 = 100 – 50 I ; Z2 = 100 + 100 i (i = −1)
Z1Z2 100
Z= = (13 + i)
Z1 + Z2 17
100 13
Z= 170 ; cos  =
17 170
2
V
P= . cos  = 629.20
Z

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S
SEEC
CTTIIO
ONN –– IIII :: C
CHHE
EMMIIS
STTR
RYY
PART – A
1. B
Sol. OH O OCH 3

K 2CO3
⎯⎯⎯⎯⎯ → + CH3I ⎯⎯
→

CH2OH CH2OH CH2OH

2. D
Sol. P, Q and R are formed by EAS reaction M is formed by free radical chlorination.

3. D
Sol. Electron withdrawing groups favour nucleophilic substitution when they are ortho and para to
leaving group.

4. B
Sol. P = CH3CH2NH2, Q = CH3CH2CN, R = Not formed, S = CH3CH2CH2NH2.

5. AD
Sol. AlCl3 + 3H2O ⎯⎯⎯⎯
Hydrolysis
→ Al ( OH)3 + 3HCl
B3+ cannot exist in aqueous media

6. BC
Sol. 2C+ 3H2 → C2 H6
( s) ( g) ( g)

H = 2  H
0
f
0
sub + 3  B.E (H − H)  − B.E. ( C − C ) + 6  B.E ( C − H ) 
 −85 = ( 2  718 ) + ( 3  436 )  − ( x + 6y )
 x + 6y = 2829 …..1
Similarly for C3H8 ( g)
2x + 8y = 4002 …..2
Solving (1) & (2), x = 345
y = 414

7. ACD
Sol. - O
O
H3C O O O-

O CH3
Br
Br -
O

H3C +
O
O
Br

8. ABD
O

Sol. R - C - NH 2 has least +ve charge density among all of the given compound.

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9. ABC
Sol. Bridge bonding
Al2Cl6  3c – 4e– Bond

10. BC
Sol. Original carbonyl compound can be generated from product by weak acid or base treatment.

11. ACD
CH3

Sol. 3CH3 COCH3 ⎯⎯⎯⎯⎯

Conc.H2SO4
→
Heat

H3C CH3

12. ABD
Sol. Dehydrogenation reaction takes place.

PART – B
1. 8
 4   6 
Sol. PT = p0A x A + pB0 xB = 100    + 400    = 280 mm of Hg
6+ 4 6+ 4
pA = yAPT
or, p0A x A = y APT or, 100  0.4 = yA  280
40 1
or, y A = =
280 7
x 1
 =
y 7
so, (x + y) = 8

2. 2
Sol. 2O2− ⎯⎯→ O2 + 4e−
One mole of O2 produced by passing 4F electricity.
 Two moles of O2 will be produced by passing 8F electricity.

3. 4
Sol. The oxidation number of N changes from -3 to +1 and +5 to +1.

4. 1.4
Sol. x = 2, y = 7, z = 5
CH2

X is CH 3 - C - CH 2 - CH 2 - CH 2 - CH 2 - CH 3

5. 3.6
po − p  0.2 
Sol. = iXsolute = (1 +  )  
p o
 10 
= (1 + 0.8) (0.02) = 0.036
x  10–2 = 0.036 = 3.6  10–2
 x = 3.6

6. 1.1
Sol. Formula of solid M3N2O6
x+y+z 3+2+6
 = = 1.1
10 10

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7. 12.5
Sol.  of Al2(SO4)3 =  of glucose
or, iC1RT = C2RT
or, iC1 = C2
or, 5  0.5 = 0.2
5  0.5
Ratio = = 12.5
0.2

8. 214
1  kw 1
Sol. pH = p − pkb − logC = 14 − 5 − logC = 5
2   2
or logC = -1 or C = 10-1 = 0.1 M
Mass of 0.1 mole of salt = 5.35 g
Mass of one mole of salt = 53.5 g
NaCl unit cell contains 4 units of NaCl
The unit cell of the salt contains 4 units of the salt
Mass of one mole of unit cell = 53.5  4 = 214

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S
SEEC
CTTIIO
ONN –– IIIIII :: M
MAATTH
HEEM
MAATTIIC
CSS
PART – A
1. D
Sol. Let the vertex he (h, k).
So, ‘S’ can be assumed as ( y − k ) = −4a ( x − h )
2

Solving with y2 = 4ax

y2 − 2ky + k2 = −y2 + 4ah
 2y2 − 2ky + k2 − 4ah = 0
Disc. = 0 ( the parabolas touch each other)
(
4k 2 − 8 k 2 − 4ah = 0 )
−k + 8ah = 0
2

 y2 = 8ax

2. D
Sol. y ( 0) = y ' (0) = y " (0) = 0
y "' ( 0 ) = −1
 y "' ( x ) is negative in the neighbourhood of x = 0
 y " ( x ) is a decreasing function in the neighbourhood of x = 0.
 For x  0 For x  0
y " ( x )  y " (0 ) y " ( x )  y " (0 )
 y " ( x)  0 y " ( x)  0
Clearly, the concavity of curve changes from positive to negative across x = 0
So x = 0 is point of inflection
So the answer should therefore be (D).
Also, by n – th derivative test if f n ( c )  0 (n is odd) then x = c is neither point of maxima nor
minima.

3. B
 1
Sol. f '   f (x) = 1
x
1  1
Replace x by , we get f ' ( x ) f   = 1
x x
 1  1  1 
Differentiating f " ( x ) f   + f ' ( x ) f '    − 2  = 0
x  x  x 
f " (x) f '(x) 1
− . =0
f ' ( x ) f ( x ) x2
f " (x) f (x) 1
 f ( x ) dx =  dx
( f ' ( x ))
2
x2

1  1  1
− f (x) −   − f ' ( x ) dx = − + c
f ' (x)  f ' (x) 
 x
f (x) 1
− +x=− +c
f '(x) x
f (1) 2
Put x = 1  − + 1 = −1 + c  c = 2 − =2−2=0
f ' (1) 1
2

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f (x) 1
 =x+
f '(x) x
f '(x)

f (x)
=
x
x +1
2
 n ( f ( x )) =
1
2
n 1 + x2 + c ' ( )
1
n 2= n2 + c '  c ' = 0
2
 f ( x ) = 1 + x 2  f ( 3 ) = 10

4. C
Sol. f ' ( x ) = 3x2 + 2x − asin x; f " ( x ) = 6x + 2 − acos x
Clearly f ' ( 0) = 0
So, for f to be injective f ' ( x ) must attain minima at x = 0
 f " (0) = 0  6 (0) + 2 − a = 0  a = 2

5. AD
Sol.

A ( acos , bsin )

O
D

AD is normal to ellipse at A  Equation of AD is asec  − bycosec  = a2 − b2

a2 − b2 a2 − b2 a2 − b2
 AB = OD = = 
a2 sec 2  + b2 cos ec 2  (a + b)
2
+ ( a tan  − b cot  )
2 a+b
 AB  a − b  AB  4 − 2  AB  2  = 2
b 1 1
= 2 happens when a tan  − bcot  = 0  tan2  = =  tan  =
a 2 2
cot  1
Slope of tangent at A = − =−
2 2

6. CD

Sol. Since P is a probability distribution, we have P ( w ) = 1
n=1
n

1 1 1
P ( wn ) = ( w n−1 ) = 2 P ( w n−2 ) = ... = n−1 P ( w1 )
2 2 2
  
1 1

n=1
P ( w n ) = 1  n=1 2
n−1
P ( w 1 ) = 1  P ( w 1  n−1 = 1  P ( w1 )( 2 ) = 1
)
n=1 2

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1
So, P ( w1 ) =
2
Clearly, the set A contains all natural numbers greater than 4 (excluding 6)
So, P (B ) = 1 − P ( w1 ) + P ( w 2 ) + P ( w 3 ) + P ( w 4 ) + P ( w 6 )
1 1 1 1 1 3
= 1−  + + + + =
 2 4 8 16 64  64

7. BD
Sol.
2 2 2
(
2z + 3 = 3z + 4  4 z + 9 + 6 z + z = 9 z + 16 + 12 z + z ) 2
( )
( )
 5 z + 7 = −6 z + z = −12Re ( z )  12 z
2

 5 z − 12 z + 7  0  ( 5 z − 7 )( z − 1)  0  1  z 
2 7
5

8. BD
Sol.
(B) If x  f −1 ( Q)  f −1 (R)  x  f −1 (Q) OR x  f −1 (R)  f ( x )  Q OR f ( x )  R
 f ( x )  Q  R  x  f −1 (Q  R )
If x  f −1 ( Q  R)  f ( x )  Q or f ( x )  R  x  f −1 (Q) or x  f −1 (R )  x  ( f −1 ( Q )  f −1 (R ) )
 f −1 ( Q)  f −1 (R) = f −1 (Q  R)

(D) If x  f −1 ( Q)  f −1 (R ) then x  f −1 ( Q) and x  f −1 (R)  f ( x )  Q and f ( x )  R .

 f ( x )  Q  R  x  f −1 (Q  R )
If x  f −1 ( Q  R) then f ( x )  Q  R  f ( x )  Q and f ( x )  R  x  f −1 (Q) and x  f −1 (R )
( )
 x  f −1Q  f −1 (R )
 f −1 ( Q)  f −1 (R) = f −1 (Q  R)

(A) Consider f :1,2,3 → 1,2,3 If y  f ( f −1 ( Q ) ) If y  Q

where f (1) = f ( 2) = f (3 ) = 1 the f −1 (  )  f −1 ( Q) then f −1 (  ) is not possible
not possible for y = 2,3 since f is not surjective

(C) Same counter example as in A can be used

9. BC
If ( gof ) exists then gof must be one – one and onto
−1
Sol.
gof is one – one  f is one – one
gof is onto  g is onto
Consequently, since ( gof ) exists, then gof is one – one and onto, hence f is one – one and
−1

g is onto

Claim 1: gof if one – one  f is one – one

Proof Let a 1 and a 2 be two elements of the set A.
f ( a1 ) = f (a2 )  gof (a1 ) = gof (a2 ) since gof is one – one we get that a1 = a2
So, we have proved that f ( a1 ) = f ( a2 ) implies f is one – one

Claim 2 : gof is ONTO  g is ONTO

Proof Since gof is onto, it follows that for any c  C, there exists a  A such that gof ( a) = c
Consequently, for any c  C there exists b = f ( a )  B such that g (b) = gof ( a) = c

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Hence g is ONTO

10. BC
f ' (x) 1
Sol. Let g ( x ) =   g ( x ) dx = − +c
( f ( x ))
2
f (x)
8
1 1
  g ( x ) dx = − + = −2 + 4 = 2
4
f (8) f ( 4)
8

 (g ( x ))
2
Also, dx = 1
4
2
 1
8
   g(x) −  = 0
4
2
1
 g(x) =
2
f ' (x) 1
 =
( f ( x )) 2
2

1 x
− = +d
f (x) 2
1
 d = −6 using f ( 4 ) =
4
2
 f (x) =
12 − x
1
 f (6) =
3

11. ABCD
z1 − z2
Sol. z1 R z1  = 0 = a real number  R is reflexive
z1 + z2

z1 − z2 z − z1  z − z2 
If z1 R z 2 holds i.e. is real then clearly 2 = − 1  is also real
z1 + z2 z 2 + z1  z1 + z 2 
 R is symmetric

z1 − z2 z − z2 z1 − z2
If is real i.e. 1 =
z1 + z2 z1 + z2 z1 + z2

z1 z1 + z1 z2 − z1 z2 − z2 z2 = z1 z1 − z1 z2 + z1z2 − z2 z2

z1 z = z1 z2
z1 z2
 =
z1 z2
z 2 − z3 z z
So, if is real, in same manner we have 2 = 3
z 2 + z3 z2 z3
z1 z 2 z3
 = =  R is transitive.
z1 z 2 z3

12. AD
Sol. Derangement of 4 persons in 4 houses = D4 = 9 ways. Hence option A.

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If A has two houses. Then let there be a hypothetical person E. So that A, B, C, D, E have
one house each now.
Among the total derangements (D5 = 44 ) . A is equally likely to go to B, C, D, E houses. So

1
A does not go to E’s house in D5 ways.
4
1 1
 Number of ways = D5 − D5 = 44 − ( 44 ) = 33 ways
4 4
But it is OK, for E to go to A’C house, it means one of A’s houses is occupied.
So, number of derangements in this case = D4 = 9
 Required number of ways = 33 + 9 = 42 ways

PART – B
1. 924
Sol. Consider the first S, I, P
* S * I * P *
The second P can only be placed in the last gap and hence we have * S * I * P *
The three I’s can be placed only in the last three gaps.
Choosing the places for I’s is put 3 identical object into 3 boxes i.e.
3 +3 −1
C3−1 = 5C2 = 10 ways ( x1 + x2 + x3 = 3 )
We have so far placed 7 letters. We have to put 3S’s in the gaps created by these. However,
for placing the 3 – S’s, we cannot choose the gap before the first S and hence we have 7 –
gaps to choose from
This can be done in 7+3−1C2−1 = 9C6 = 84 ways .
Only, the letter M remains to be placed. M can go in any of these 11 – gaps.
 Required number of ways = 10  84  11 = 9240

2. 143
Sol. f ( 2) = g ( 2),f ( 4) = g ( 4),f (8 ) = g (8 )
8 8 8

 x f ' ( x ) = x f ( x ) 1 −  f ( x ) dx = 8f (8 ) − f (1) = 63 −  f ( x ) dx
8

1 1 1
2 2
g ( 2x ) 4
g(t)
 f ( x ) dx = 
1 1
2
dx = 
2
4
dt ( t = 2x )
2 4 4
5 5 1
 f ( x ) dx =
1
4
 =  g ( t ) dt   g ( x ) dx = 5
4 42 2
4 4 4

 f ( x ) dx +  g ( x ) dx = 16 − 4 = 12   f ( x ) dx = 7
2 2 2
4
g ( 2x ) 4
g(t) 8

 dx = 7   dt = 7   g ( x ) dx = 28
2
2 2
4 4
8 8
  f ( x ) dx +  g ( x ) dx = 64 − 16 = 48
4 4
8
  f ( x ) dx = 48 − 28 = 20
4
8
2 4 8

  x f ' ( x ) dx = 63 −   f ( x ) dx +  f ( x ) dx +  f ( x ) dx 
1 1 2 4 
 5  139 p
= 63 −  + 7 + 20 = =
 4  4 q

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 p + q = 143

3. 2
Sol. A = 10 + 1 + 10 + 2 + ...... + 10 + 99

B = 10 − 1 + 10 − 2 + ...... + 10 − 99
A A −B
= 1+
B B

( 10 + )
99
A −B =  i − 10 − i
I=1

Consider, for each i

( 10 + )
2
i − 10 − i = 10 + i + 10 − i − 2 10 2 − i

= 20 − 2 100 − i

 10 + i − 10 − i = 20 − 2 100 − i = 2 10 − 100 − i

( )
99 99 99
A −B =  10 + i − 10 − i = 2  10 − 100 − i = 2  10 − i =B 2
i=1 i=1 i=1

A −B
= 2
B

4. 2.82
Sol. The equation of two edges can be taken to be
x−0 y−0 z−0 x −0 y −0 z− 3
= = and = =
1 1 −1 1 −1 0
The shortest distance between them is 2 .

5. 1
Sol. x = f ( t )  dx = f ' ( t ) dt
0
I =  (f (t) − t) f ' ( t ) dt
2023

1
1
= − ( t − f ( t ) ) f ' ( t ) dt
2023

0
1
= − ( t − f ( t )) ( −1 + f ' ( t ) + 1) dt
2023

0
1
= + ( t − f ( t )) (1 − f ' ( t ) ) dt − I
2023

( f − f ( t ))
2024 1
1
2I =  ( t − f ( t ) ) (1 − f ' ( t ) )
2023 2
dt = =
0
2024 2024
0

6. 1.22
Sol. Clearly, triangle formed by the given points ˆi + ˆj + 2k,
ˆ ˆi + 2jˆ + kˆ and 2iˆ + ˆj + kˆ is equilateral
as AB = BC = AC = 2 .
 Distance of orthcentre ‘O’ from the sides is equal to inradius of the triangle.

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3
( 2)
2

 1
 l1 = l2 = l3 = inradius = r = = 4 =
s 4
2
( 2) 6

3 6
 (l1 + l2 + l3 ) = =
6 2

7. 4.00
1
f ( x) x sin x − 1 + cos x
Sol. lim k = lim 2
x →0 x x →0 xk
1 1
sin x + x cos x − sin x
= lim 2 2
x →0 k xk −1
1 1 1
cos x + cos x −  sin x − cos x
= lim 2 2 2
x →0 k (k − 1) xk −2
1
− sin x
= lim 2
x →0 k ( k − 1) x −3

1
f (x) −
2 −1
If k − 3 = 1 i.e. k = 4 , lim = =
x →0 xk 4  3 24

8. 5.15
Sol. Clearly h (i) is an A.P.
h (100) = h ( 0) + 100d  100d = 20 − 5 = 15  d = 0.15
 h (1) = 5.15

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