(Batches: e-SANKALP-2325 S1, T1 & P1)

IIT – JEE, 2325 Paper Code

101138
(CLASS XII)
Time: 2 Hours Maximum Marks: 126
INSTRUCTIONS
A. General
1. Write your Name, Enrolment number in the space provided on this booklet as soon as you get the paper.
2. Blank papers, clipboards, log tables, slide rules, calculators, cameras, cellular phones, pagers, and electronic
gadgets of any kind are NOT allowed in the examination hall.
3. Use a ball point pen do darken the bubbles on OMR sheet as your answer besides Name, Enrolment
number, Phase, Paper sequence, Venue, Date along with your signature on OMR sheet.

B. Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists of
three sections.
4. Section–1 (01 – 06) contains (06) Multiple Choice Questions which have Only One Correct answer. Each
question will be evaluated according to the following marking scheme.
Full Marks : +3 If only (all) the correct option(s) is (are) chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –1 In all other cases
5 Section–2 (07 – 09) contains (03) Multiple Choice Questions which have one or more than one
correct answer. Each question will be evaluated according to the following marking scheme.
Full Marks : +4 If only (all) the correct option(s) is (are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen and
both of which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –2 In all other cases.
6. Section–3 (10 – 12) contains (03) Non-Negative Numerical Value Questions, the answer to each
question is a Non-Negative Numerical Value. For each question, enter the correct numerical value
corresponding to the answer and each question carries +4 marks for correct answer. There is no
negative marking.

Enrolment No. :

Name : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Batch : . . . . . . . . . . . . . . . . . . . . . . .Session :. . . . . . . . . . . . . . . . . . . . . . Date:. . . . . . . . . . . . . . . .

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-2

PART I : PHYSICS
SECTION – 1 : (Only One Option Correct Type)

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONLY ONE option is correct.

1. If a charge q is uniformly spread on a thin non conducting square plate. The electric potential of
its centre is found to be V1 = 2 volt. If six such charged plates are joined to make a hollow cube
the potential at the centre of the cube is found to be V2 = 4 volt. Find potential at one of the
vertex of this cube. In all the case potential at infinitely distant points has been assumed zero.
(A) 2 volt (B) 4 volt
(C) 6 volt (D) 8 volt

2. The diagram shows a uniformly charge hemi-sphere of radius R. It has P1

volume charge density . If the electric field at a point 2R distance 2R
above its centre is E, then the electric field at the point which is 2R
below its centre is :      
R R      
(A) E (B) E     
6 0 12 0  
R R
(C)  E (D) E
6 0 24 0 P2

3. A thin rigid insulating ring of radius r and mass m has a very small E
gap of length (  r ) and carries a uniformly distributed charge
q. The ring is at rest in free space and a uniform electric field E in
the plane of the ring and parallel to the gap is switched on a shown
in the figure. Find the maximum angular speed of the ring
subsequent motion.
qE
(A) zero (B)
mr 2
qE 2qE
(C) 2
(D)
2mr mr 2

Space for Rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-3

4. A spherical conductor has a radius of R and charge Q. A spherical shell of thickness R and
uniform charge Q, is kept so as to be concentric to the conductor and touching the conductor.
The electric field at a distance of 1.5 R from the center of the sphere is
83Q 75Q
(A) (B)
5040R 2 5040R 2
83Q 53Q
(C) 2
(D)
2520R 2520R 2

5. A dipole consists of a light rod of length , having two point charges +q B

+q and -q of same mass m joined at ends A and B of the rod. A +q
B
uniform electric field E is present in the region and perpendicular to
the length of the rod. The rod is released from rest. The angular
velocity of the rod, when it becomes parallel to the electric field for A q q
the first time is A
qE qE
(A) (B) 2
m m
qE 1 qE
(C) (D)
2m 2 m

6. Figure shows a cross section through a very large nonconducting slab

of thickness d and uniform volume charge density . The origin of x
axis is at the slab’s center. For what value of x will the total E-field
1
within the slab be of the field just outside it?
3
d d
(A) (B)
3 6
d d
(C) (D)
9 4

Space for Rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-4

SECTION – 2 : (One or More Than One Options Correct Type)

This section contains 3 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.

7. A very large slab of insulating material having face area A

and dielectric constant K=2 is placed between plates of
parallel plate capacitor and some uniform charges on
plates are given. The systems is placed in uniform
electric field. The field is perpendicular to the surface of
the material. E1=Q/A0 E2=2Q/A0 E3=3Q/A0

(A) The total charge on right surface is Q.

(B) The free charge on left surface is 3Q
(C) The bound charge on right surface is 2Q
(D) The magnitude of the applied electric field is 2Q/A0.

8. In electrostatics, the electric field is a vector quantity, while electric potential is scalar. Which of
the following statements are incorrect
(A) The electric field exists along the direction in which the potential decreases and potential
gradient is maximum.
(B) The potential does not vary along the direction perpendicular to the electric field.
(C) Work done in moving a charge on an equi-potential surface may be non-zero.
(D) Electric potential is always negative.

9. Two concentric conducting shells of radii R and 2R having charges qA B

A
and qB and potential 2v and 3v/2 respectively. (Assume potential to be
zero at infinity). Now shell B is earth by closing switch S. Let charges R
on spherical shells A & B become qA & qB respectively, then
(A) qA/qB = ½ (magnitude) 2R
(B) qA/qB = 1 (magnitude) S
(C) potential of A after earthing becomes 3/2 v
(D) potential difference between A and B after earthling becomes v/2

Space for Rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-5

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

10. A point chare q is placed at a distanc 2r from the centre o of a

conducitng charged sphere of radius r. due to the induced charges
on the sphere, find the electric potential at point p on the surface of P O q
sphere (in volt) (if kq/r = 18 volt).
2r

11. There is a non-conducting sphere of uniformly distributed charge +

density  insulated from surrounding. There is a cavity inside the sphere + +
as shown in figure. P is the centre of sphere and C is centre of cavity. B
a 2 + + A
Value of  VA  VC  is of the form , where k 0 is constant. Then value + P C
k 0 0
+
of k 0 will be? VA and VC are the electrostatic potential A and C. It is given +
+
that PC  a, AB  a , BC  a , AB is parallel to PC and BC is
2 2
perpendicular to PC.

 E y
12. Electric field given by the vector E  0 (xiˆ  yj)ˆ N/C is present in the
(0, )
 (Q, M)
x-y plane. A small ring of mass M carrying charge +Q, which can slide
freely on a smooth non conducting rod, is projected along the rod from
the point (0, ) such that it can reach the other end of the rod.
(, 0) x
Assuming there is no gravity in the region. What minimum velocity
QE0 
should be given to the ring (in m/s)? if in S.I. unit 8
M

Space for Rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-6

PART II : CHEMISTRY
SECTION – 1: (Only One Options Correct Type)
This section contains 6 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE is correct.
1. For the following reaction
O
NaBH4
O C OCH 3 
OH
 Q
Major product 
The major product (Q) is

(A) (B)
HO CH2OH O CH2OH

O O
(C) (D)
HO C OCH3 O C OH

2. Which of the following molecules will give positive iodoform test?

OH

O
(A) (B)
Ph C CH2CH3

CH3

O O H3C C OH
(C) (D)
Ph C CH2 C Ph CH3

Space for rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-7

3. Et 1. CH MgBr  excess 
3

2. H O
 P  Major 
2

Cl
CH3 OH
Et Et
(A) OH (B) CH3

CH3 CH3
CH3 OH
Et Et
(C) OH (D) CH3

CH3 OH

4. Which of the following statements is incorrect?

(A) Glycerol (1 mole) on reaction with periodic acid (2 moles) yields formaldehyde and formic
acid
(B) Glycerol on dehydration in presence of conc. H2 SO 4 yields unpleasant smelling
compound
(C) Glycerol on reaction with PCl5 gives mainly 1, 2, 3-trichloropropane
(D) Glycerol on heating with large excess of HI produces allyl iodide

Space for rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-8

5. OH
O
H


x
The major product (x) is
(A) O (B) O

(C) OH (D) OH

6. Which of the following is/are not correctly matched with major product?

(A) OH (B) OH CHO


OH
H

H 

O
OH
(C) OH O (D) CHO
OH 
H
H2 SO 4
  
OH

Space for rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-9

SECTION – 2: (Multi Correct Choice Type)

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE OR MORE may be correct.
7. Which of the following are correctly matched with major product?

OH

(A) 
POCl
3

Py

HO

(B) 2

H SO
4



OH O
(C)
K 2Cr2O7 /H

OH

(D) conc. H SO
2
 4


8. Which of the following combination of reagents can give 3-methyl hexan-3-ol?

(A) n-C3H7MgBr + butanone followed by hydrolysis
(B) C2H5MgBr + 2-pentanone followed by hydrolysis
(C) CH3MgBr + 3-hexanone followed by hydrolysis
(D) C5H11MgBr + 3-pentanone followed by hydrolysis

HO CH3

1 H
2 4SO

 2 
P
9.
Products are
CH3 CH3

(A) (B)

OH
CH3 CH3

(C) (D)

Space for rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-10

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

10. The weight of CH4 (in grams) produced by reacting 4.6 grams of glycerol with excess of CH3MgBr
is

11. If 52 g starting material is used then calculate the mass (g) of ‘E’ produced?
H3 O PCC
Ph  CH  CH2   A 
CH2Cl2
B

1. Mg H 1. I  NaOH
2B 
2. H2 O
C  D 
2
2. H
 CHI3  E

12. How many of the following alcohols will give turbidity immediately on reaction with ZnCl2/HCl?

OH OH
H2C OH
HO
CH3
H3C OH
H3C C OH

CH3

Space for rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-11

PART III : MATHEMATICS

SECTION – 1: (Only One Options Correct Type)

This section contains 6 multiple choice type questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE is correct.

1
1 Let f ( x )  lim 2n
. Then the set of values of x for which f(x) = 0, is
n 
3 1 
 tan 2 x  5
 
(A) | 2x |  3 (B) | (2x) |  3
(C) |2x |  3 (D) | 2x |  3

1 / n(tan x )
2. The value of lim 1  [ x]  (where [.] denotes the greatest integer function) is equal to
x  / 4
(A) 0 (B) 1
(C) e (D) e1

n
k S1  n  S5  n   S3  n2 
3. Let Sk(n) = r
r 1
, then lim
n S7  n 
is equal to

1 1
(A) (B) –
6 3
2 4
(C) (D) –
3 3

 32
4. If f(x) = 0 be a quadratic equation such that f(– ) = f() = 0 and f   = – , then
 2 4
f x
lim is equal to
x  sin  sin x 

(A) 0 (B) 
(C) 2 (D) none of these

Space for rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-12

5. Let f(x) be a polynomial one-one function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy)  x, y  R –
x
x
{0}, f(1)  1, f(1) = 3. Let g(x) =  f  x   3   f  x  dx , then

4
0
(A) g(x) = 0 has exactly one root for x  (0, 1) (B) g(x) = 0 has exactly two roots for x  (0, 1)
(C) g(x)  0  x  R – {0} (D) g(x) = 0  x  R – {0}

6. A differentiable function f: R+  R satisfies f(xy) = f(x) + f(y)  x, y  R+. If f(16) = 3, then the
value of f(2) is
3 3
(A) (B)
8 4
3
(C) 3 (D)
2

SECTION – 2: (Multi Correct Choice Type)

This section contains 3 multiple choice questions. Each question has four choices (A), (B), (C) and (D)
out of which ONE OR MORE may be correct.

7. f, g, h : R  R, f(x), g(x), h(x) are all continuous, differentiable functions which satisfy the relation
f(x + y) = g(x) + h(y) for all x, y  R
(A) f(0) = f(1) (B) g(x) = h(x) + 1
(C) h(x)  g(x)  x  R – {0} if h(0)  g(0) (D) f(0) = g(0) + h(0)

8. A continuous function f  x  on R  R satisfies the relation

f  x   f  2x  y   5xy  f  3x  y   2x 2  1 x, y  R . Then, which of the following holds true?
(A) f is many one (B) f has no minima
(C) f is neither odd nor even (D) f is bounded

9. Let f : R  R, f(x) = x + ln(1 + x 2) then

(A) f is injective
(B) lim f(x)  
x 
(C) there is a point on the graph of y = f(x) where tangent is not parallel to any of the chords
(D) f is bijective

Space for rough work

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-13

SECTION – 3: (Numerical Answer Type)

This section contains THREE questions. The answer to each question is a NUMERICAL VALUE. For
each question, enter the correct numerical value (in decimal notation, truncated/rounded‐off to the
second decimal place; e.g. xxxxx.xx).

1
10. Let f(x) = | x |c sin  x | tan x |d , x  0, f(0) = 0, be differentiable at x = 0. The least possible value
x
of (c + d) is ____________ (where (.) denotes the least integer function)

  1, x  0

11. Let f(x) =  0, x  0
 1 x0

and g(x) = sinx + cosx
Then points of discontinuity of f(g)(x) in (0, 2) is _________.

 2 2 
12. If k = lim sec 1    exist, then the minimum value of [||] is ____________ (where [.]
x 1  ln x x  1 
denotes the greatest integer function)

Space for rough work

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-14

(Batches: e-SANKALP2325 S1, T1 & P1)

IIT – JEE, 2325 Paper Code

(CLASS XII) 101138
ANSWERS
PHYSICS (PART-I)

1. B 2. B 3. A 4. B
5. B 6 B 7. A, B, C, D 8. C, D
9. A, B, D 10. 3.00 11. 6.00 12. 2.00

CHEMISTRY (PART-II)

1. C 2. C 3. A 4. D
5. C 6. D 7. A, B, C 8. A, B. C
9. C, D 10. 2.40 11. 56.50 12. 4.00

MATHEMATICS (PART-III)
1. A 2. B 3. D 4. C
5. D 6. B 7. A, C, D 8. A, B
9. A, C, D 10. 2.00 11. 2.00 12. 1.00

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-15

HINTS AND SOLUTION

PHYSICS (PART-I)
5. W  u
m 2 2
  q E
4
qE
= 2
m

6. The E-field varies linearly with x, reaching its maximum value at the surface (x = d/2) and remains
uniform outside.

a a
11. 
3 o 2

E0 x 2 E0 y 2
12. For the electric field, V    K
2 2
QE0 2 1 QE0 2
from energy conservation   k  Mv 2    2 + K
 2 2  8
v = 2 m/s

CHEMISTRY (PART-II)

1. NaBH4 reduces ketones but does not reduce esters.

O CH3
O
Et
3CH MgBr

 Et
3. Cl
Cl
NGP
CH3 CH3
Et Et

O 3 CH MgBr O



CH3

CH3
Et
OH
2

H O
 P 
CH3

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-16

CH2OH
4. (A) CHOH  2HIO 4  2HCHO  HCOOH  2HI  H2 O

CH2OH
CH2OH CH2 CH2
CHOH 2 Conc. H SO
4 Tautomerises

or
 C   CH
P2 O5  Heat 
(B) CH2OH Dehydrating agent  CHOH
H2O CHO
Unstable Acrolein
 unpleasant smell
CH2OH CH2

CHOH  3HI 

 CH  I2
(D)
CH2OH CH2I
Allyl iodide
Allyl iodide formed further reacts with HI.
CH2 CH3 CH3
CH3
CH  HI   CHI 
 I2
 CH Hi
  CHI
CH2I CH2I CH3
CH2
Unstable Propene Isopropyl iodide
8. O
OH
H3 O 
MgBr + 

O
OH
H O
MgBr + 
3


OH

H3 O
CH3MgBr +  

O
H3C OH H3C OH 2 CH3 CH3 CH3

H H2 O 
9.    H
  

H H
H2C OH
10. CH MgBr
HC OH excess  3CH4
3

H2C OH
48
Weight of CH4 produced   4.6
92
= 2.4 grams

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-17

MATHEMATICS (PART-III)
1. A
2
3 
f(x) = 0 if and only if  tan1 2 x  > 1
  
  3 3
 tan-12x > or tan-12x < –  x > or x  
3 3 2 2
 |2x| > 3.

2. B
The given limit can be re-written as
[ x]
lim
1 /[ x ] [ x ] / n(tan x )
lim
x  / 4
(1  [x])  e x   / 4 n(tan x ) 1

[ x] 
as = 0 in the deleted neighbourhood of .
n(tan x ) 4

3. D
Sk  n  1
Divide numerator and denominator by n8 and use lim k 1
 .
n n k 1

4. C
From given f(x) = x2 – 2
x 2  2    h 2   2
lim  lim
x  sin  sin x  x  sin sin    h 

2h  h2
= lim
h 0  sin  sinh 

h  2
= lim  2 .
h 0  sin  sinh  sinh

sinh h

5. D
Put x = y = 1  f(1) = 2
1  1  1
Put y =  f(x) + f   = f(x) f 
x x x
 f(x) = x3 + 1
 g(x) = 0  x  R – {0}.

6. B
f(xy) = f(x) + f(y)  x, y  R+
  h 
f  x 1    f  x 
f  x  h  f  x   x 
f(x) = lim  lim 
h0 h h 0 h
 h
f  x   f 1    f  x 
 f(x) = lim  x
h0 h

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e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-18

 h
f  1    f 1
 f(x) = lim  x ;  f(1) = 0
h0 h
x 
x
1
 f(x) = f  1
x
 f(x) = f(1) ln(x) + lnc
if x = 1, then c = 1
 f(x) = f(1) ln(x)
 f(16) = f(1) ln(16)
f 16  3
Now f(2) = f(1) ln2 = ln 2  .
ln 24 4

7. A, C, D
f(x + y) = g(x) + h(y) …(1)
put x = 0
f(y) = g(0) + h(y)  h(y) = f(y) – g(0) …(2)
put y = 0
f(x) = g(x) + h(0)  g(x) = f(x) – h(0) …(3)
 put (2) and (3) in (1)
f(x + y) = f(y) + f(x) – h(0) – g(0)
Define C1(x) = f(x) – h(0) –g(0)  x  R.
 C1(x + y) = C1(x) + C1(y)
The solution of this functional. Equation is obtained by differentiation through first principles as
C1(x) = Cx (where ‘c’ is a constant)
 f(x) = C1(x) + h(0) + g(0) = cx + h(0) + g(0)
g(x) = f(x) – h(0) = cx + g(0)
h(x) = f(x) – g(0) = cx + h(0)
f(0) = f(1) = c

8. A, B
x
Let 2x  y  3x  y  y 
2
2
 5x  5x  5x  2
 f x  f    f   2x  1
 2  2  2 

x2
f  x  1
2

9. A, C, D
2x  x  12
f(x)  x  ln 1  x 2   f '(x)  1   0 xR
1  x2 1  x2
So, f(x) is injective and at x = –1, tangent is not parallel to any of the chords
 1  x2   2x   2 
Again lim f(x)  lim ln  e x  1  x 2    lim ln   x  = lim ln   x   lim ln   x   
x  x  x   e  x   e  x   e 
So, f is surjective also.

10. 2

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 e-Sankalp2325 S1, T1 & P1-XII-PCM-(101138)-19

 c  1 d 
 | h | sin    h | tanh | 0   c 1  1 
Rf(0) = lim   h    lim   h  sin    | tanh |d 
h0  h  h 0  h 
= 0 or 1 if c > 1 and d  0
 c  1  d 
 | h | sin    h | tanh | 0    1 
Lf(0) = lim   h    lim  | h |c 1 sin    | tanh |d 
h0  h  h 0  h 
= 0 or 1 if c – 1 > 0 and d  0
 Rf(0) = Lf(0)  c – 1 > 0 and d  0
c+d>1
 (c + d)min = 2.

11. 2
 3 7  
 , 
4 4
3 7
f(g(x)) = 1 when 0 < x < or  x  2
4 4
3 7
= 0 when x = ,
4 4
3 7
= -1 when x
4 4
3  7
clearly f(g(x)) is not continuous at x = ,
4 4

12. 1
 2 2 
lim sec 1   
x 1  ln x x  1 
let x – 1 = t  x = t + 1  as x  1, t  0
 2 2  2
1    t 
lim sec 1     lim sec    1 
t 0  ln  t  1 t  t 0  t  ln  t  1  
 2 
= lim sec 1  
t 0  2 
2
  1 |  | 2
2
Hence minimum value of [||] = 1.

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