07-07-2025

2302CJA101021250017 JM

PART-1 : PHYSICS

SECTION-I

1) A sphere of radius R carries volume charge density proportional to the square of the distance
from the center : ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the
magnitude of the electric field is :-

(A) A/4πε0
(B) AR /40ε0
3

(C) AR /24ε0
3

(D) AR /5ε0
3

2) At any point (x, y, z) the electric potential V is volt, then electric field at x
= 1 m :-

(A)
5500 V/m
(B) 5500 V/m

(C)
V/m

(D)
V/m

3) A 1 µF capacitor is connected in the circuit shown below. The e.m.f of the cell is 2 volts and
internal resistance is 0.5 ohm. The resistors R1 and R2 have values 4 ohm and 1 ohm respectively.
The charge on the capacitor must be :-

(A) 2 µC
(B) 1µC
(C) 1.33 µC
(D) Zero

4) In the adjoining circuit diagram each resistance is of 10 Ω. The current in the arm AD will be :-
(A)

(B)

(C)

(D)

5) In the circuit shown in the figure, the switch S is initially open and the capacitor is initially
uncharged. I1, I2 and I3 represent the current in the resistance 2Ω, 4Ω and 8Ω respectively (Just after

closer of the switch)

(A) Just after the switch S is closed, I1 = 3A, I2 = 3A and I3 = 0

(B) Just after the switch S is closed, I1 = 3A, I2 = 0 and I3 = 0
(C) Just after the switch S is closed, I1 = 0.6A, I2 = 0 and I3 = 0
(D) long after the switch S is closed, I1 = I2, I3 = 0.6A

6) In the given circuit, charge Q2 on the 2µF capacitor changes as C is varied from 1µF to 3µF. Q2 as
a function of 'C' is given properly by : (figures are drawn schematically and are not to scale):

(A)
(B)

(C)

(D)

7) A milliammeter of range 10 mA and resistance 9Ω is joined in a circuit as shown. The metre gives
full-scale deflection for current I when A and B are used as its terminals, i.e., current enters at A and

leaves at B. The value of I is :-

(A) 100 mA
(B) 900 mA
(C) 1 A
(D) 1.1 A

8) A ring is made of a wire having a resistance R0 = 15 Ω . Choose the points A and B as shown in
figure at which a current carrying conductor should be connected so that the resistance R of the sub

circuit between these point is equal to Ω.

(A)

(B)

(C)

(D)

9) Total power consumption for the given circuit is :-

(A) 25 W
(B) 50 W
(C) 100 W
(D) 75 W

10) A battery of internal resistance 4Ω is connected to the network of resistance as shown. In order
that maximum power can be delivered to the network, the value of R in ohm should be :-

(A)

(B) 2

(C)

(D) 18

11) Two infinitely long straight parallel wires are 5m apart, perpendicular to the plane of paper. One
of the wire, as it passes perpendicular to the plane of paper, intersects it at A and carries current 'i'
in the downward direction. The other wire intersects the plane of paper at B and carries current 'i' in
the outward direction. O is a point in the plane of paper as shown in figure. With x and y axes as
shown, magnetic induction at O in the component form can be expressed as :-
(A)

(B)

(C)

(D)

12) Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out
of the plane of the paper as shown. The variation of magnetic field B along the line XX' is given by :

(A)

(B)

(C)

(D)

13) A long horizontal wire AB which is free to move in vertical plane and carries current I1, is in
equilibrium at a height of d meter above another parallel long wire CD, which is fixed in horizontal
plane and carries current I2, as shown in the figure. When AB is slightly pressed down, it execute

simple harmonic motion. The time period of S.H.M. is :-

(A)
T=

(B)
T=
 T=
(C)

(D) wire AB will not perform S.H.M.

14) A charged sphere of mass m and charge q starts sliding from rest on a vertical fixed circular
track of radius R from the position as shown in figure. There exists a uniform and constant
horizontal magnetic field of induction B. Find the maximum force exerted by the track on the sphere.

(A) 3mg + qB

(B) 3mg – qB

(C) 3mg + qB

(D) 3mg + qB

15) The magnetic field at the centre of an equilateral triangular loop of side 2L and carrying a
current i is :-

(A)

(B)

(C)

(D)

16) Three infinitely long, straight and parallel wires carrying currents are arranged as shown in

figure. The force experienced by 10 cm length of wire Q is:

(A) 1.4 × 10–4 N towards the right

(B) 1.4 × 10–4 N towards left
(C) 2.6 × 10–4 N towards the right
(D) 2.6 × 10–4 N towards the left

17) The magnetic field at the centre of a circular coil of radius r is π times that due to a long straight
wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the
circular part has radius r and straight ones are infinitely long. For same current the magnetic field

at the centre P in cases 1, 2, 3 have the ratio :-

(A)

(B)

(C)

(D)

18) Eight charge particles are placed at circumference of circle of radius r. The net electric field at

the centre of circle is :-

(A) zero

(B)

(C)

(D)

19) The two metallic circular plates of radius r are placed at a distance d apart to form a parallel
plate capacitor of capacitance C. If a dielectric slab of radius r/2 and thickness d of dielectric
constant 6 is placed between the plates of the condenser, then its capacity will be

(A) 7 C/2
(B) 3C/7
(C) 7C/3
(D) 9C/4

20) For the circuit shown in the figure, the charge on 4µF capacitor is :-
(A) 30µC
(B) 40µC
(C) 24µC
(D) 54µC

SECTION-II

1) A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field
at right angle to the field. If rd and rp are the radii of their circular paths respectively, then the ratio

will be where x is ______.

2) A positively charged particle moving with velocity v having specific charge (q/m) enters a region

of magnetic field B having width at angle 53° to the boundary of magnetic field. Find the

angle θ in the diagram (in degree).

3) An electron moves through a uniform magnetic field At a particular instant of

time, the velocity of electron is m/s. If the magnetic force acting on electron is,
where e is the charge of electron, then the value of B0 is ____ T.

4) A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure.
Both the cylinder and the cavity are infinitely long. A uniform current density J flows along the

length. If the magnitude of the magnetic field at the point P is given by then the value of N
is

5) A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring
carries a current i = 4A. A horizontal magnetic field B = 10 T is switched on at time t=0 as shown in
figure. The initial angular acceleration of the ring is N radian per second2 then N is
 PART-2 : CHEMISTRY

SECTION-I

1) For a reaction 2A + B → Product, rate law is . Concentration of reactant A at time t =

0
, where C is initial concentration of A -

(A)

(B) C0e

(C)

(D) 10C0

2)

The plot that follows Arrhenius equation is :-

[k = rate constant, T = Temperature]

(A)

(B)

(C)

(D)
3) Azo isopropane decomposes according to the equation :–

(CH3)2CHN=NCH(CH3)2(g) N2(g)+C6H14(g)
It is found to be a first order reaction. If initial pressure is PO and pressure of the mixture at time t is
(Pt) then rate constant K would be :–

(A)

(B)

(C)

(D)

4) If Ar gas is bubbled through water at 298 K, how many moles of Ar gas would dissolve in 1 litre of
water? Assume that Ar exerts a partial pressure of 0.403 bar. Given that Henry's law constant for Ar
at 298 K is 40.3 K bar

(A)
(B)
(C)
(D)

5) At 35°C, the vapour pressure of CS2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution
of CS2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the
following is :

(A) heat must be absorbed in order to produce the solution at 35°C

(B) Raoult's law is not obeyed by this system
(C) a mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL
(D) CS2 and acetone are less attracted to each other than to themselves

6) The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a
solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in
water, the concentration of XY (in mol L–1) in solution is:

(A) 6 × 10–2
(B) 4 × 10–4
(C) 16 × 10–4
(D) 4 × 10–2

7) In which of the following reaction, 2º alcohol is obtained as major product.

(A)
(B)

(C)

(D)

8) Suitable reagent for above reaction.

(A) Zn-Hg/HCl
(B) N2H4/OH / Δ
–

(C) RedP/HI
(D) All of these

9) P + Q (Major product) P and Q are related as :

(A) Chain isomer

(B) Functional group isomer
(C) Position isomer
(D) Not isomers

10) Which of the following compound will not give CO2 gas on heating ?

(A)

(B)

(C) HOOC–CH2–COOH

(D)

11) product P is

(A) CH3—CH2—CN
(B)

(C)

(D)

12)

(A) 1° Amine
(B) 2° Amine
(C) 3° Amine
(D) Acid amine

13)

Major product (C) is :

(A)

(B)

(C)
(D)

14) Among the following compounds, which will react with acetone to give a product containing

(A) C6H5NHNH2
(B) (CH3)3N
(C) C6H5NHC6H5
(D) None of these

15) Which of the following compounds give positive iodoform test?

(A)

(B)

(C)

(D)

16) m-chlorobenzaldehyde Product. Product is :

(A)

Mixture of

(B)

(C)
(D)

17) Identify A in the following chemical reaction

(A)

(B)

(C)

(D)

18) Identify the product in following reaction ?

(A)
(B)

(C)

(D)

19) SN1 reaction underoges through a carbocation intermediate as follows : -

[R = t-Bu, iso-Pr, Et, Me] (X = Cl, Br, I)

The incorrect statement is

(A) The decreasing order of rate of SN1 reaction is t-BuX > iso-PrX > EtX > MeX
(B) The decreasing order of ionisation energy is MeX > EtX > iso-PrX > t-BuX
(C) The decreasing order of energy of activation is t-BuX > iso-PrX >EtX > MeX
(D) Decreasing order of rate of SN1 is R–I > R–Br > R–Cl

20)

(A)
(B)

(C)

(D) None of these

SECTION-II

1) Examine the structural formulas of compounds given below and identify number of compounds
which show positive iodoform test on vigorous reaction with excess of alkali and I2

2) How many reaction are correct with major product?

(1)

(2) CH3CH2O⊖ + CH3–CH2–Br

(3)

(4)

(5)

(6)
3) Major (P)
In the above reaction, number moles of hydride ions consumed per mole of the reactant is 'x' and
degree of hydrogen index of product 'P' is 'y' then x + y is ?

4) Consider the following reaction, the rate expression of which is given below
A+B→C
rate = k [A]1/2 [B]1/2
The reaction is initiated by taking 1M concentration A and B each. If the rate constant (k) is 4.6 ×
10–2 s–1, then the time taken for A to become 0.1 M is_____sec. (nearest integer)

5) 0.2M aq. solution of KCl is isotonic with 0.2M K2SO4 at same temperature. What is the van't Hoff
factor of K2SO4 ?

PART-3 : MATHEMATICS

SECTION-I

1) Range of y = cos–1 ; ("where [x] denotes greatest integer less than or equal to x").

(A) [0, π]
(B) [–1, 1]

(C)

(D) {1, –1}

2) If f(x) = {x} + {x + 1} + {x +2} + ... + {x + 99} then [f( )] is ("where [x] denotes greatest
integer less than or equal to x") and "where {x} denotes fractional part of x").

(A) 5050
(B) 4950
(C) 41
(D) 14

3)

Difference between the greatest and the least values of the function

f(x) = x (ℓn x – 2) on [1, e2] is

(A) 2
(B) e
(C) e2
(D) 1

4) If ƒ(x) is an even function and g(x) is an odd function, and x2ƒ(x) – 2ƒ = g(x), then ƒ(5) is equal
to :-

(A) 5

(B)

(C) 0
(D) g(5)

5) If f(x) = then f–1(x) is equal to :-

(A)

(B)

(C)

(D) None of these

6) The graph of y = f (x) is shown below. Which could be the best

sketch of y = ?

(A)

(B)

(C)
(D)

7) The solution set of the equation is

(A)

(B)

(C)

(D) None of these

8) If curve and intersect orthogonally then the value of a is -

(A)

(B)

(C) 2
(D) 3

9)

Let g(x) = f(log x) + f(2 – log x)and f "(x) < 0 ∀ x ∈ (0, 3).
Then find the interval in which g(x) increases : –

(A) (0, 1)
(B) (1, 2)
(C) (2, 3)
(D) (0, e)

10) The height of a right circular cone of maximum volume inscribed in a sphere of diameter a is :–

(A)

(B)

(C)

(D)

11) The equation of normal to the curve , where it cuts x-axis is -

(A)
(B)
(C)
(D)

12) Let ƒ(x) be a quadratic function such that ƒ(0) = ƒ(1) = 0 & ƒ(2) = 1, then is
equal to

(A)

(B) π
(C) 2π
(D) 4π

13)

If f(x) is a polynomial satisfying

f(x)f(1/x) = f(x) + f(1/x) and f(2) > 1 then f(x) is :-

(A) 2
(B) 1
(C) -1
(D) 0

14) is :- [Where [·] → G.I.F.]

(A) 1
(B) 0
(C) doesn't exist
(D) –1

15) If , , then dy/dx at t = 2 is

(A)

(B)

(C)

(D)

16)
If f is continuous at x = 2 then λ is (where [·] denotes greatest integer)

(A) –1
(B) 0
(C) 1
(D) 2

17)

(A)
(B) 0
(C) 1

(D)

18) In [0, 1] Lagranges Mean Value theorem is NOT applicable to

(A)

(B)

(C) f(x) = x|x|

(D) f(x) = |x|

19) The function f(x) = [x]2 – [x2] (where [y] is the greatest integer less than or equal to y), is
discontinuous at :

(A) all integers

(B) all integers except 0 & 1
(C) all integers except 0
(D) all integers except 1

20) Let f(x) = |2x2 + 5|x| – 3|, x ∈ R. If m and n denote the number of points where f is not continuous
and not differentiable respectively, then m + n is equal to :

(A) 5
(B) 2
(C) 0
(D) 3
 SECTION-II

1) The normal to the curve at P(1,1) meets the curve again at Q(a,b). then value of
a+b=

2) Let f(x) be a cubic polynomial with f(1) = –10,

f(–1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = –1. Then f(3) is equal to
_____.

3) Let f(x) = x2 – 4x – 3, x > 2 and g(x) be the inverse of f(x). Then the value of where f(x) = 2,
is (here, g' represents the first derivative of g)

4) The value of f(0) so that the function is continuous everywhere is k, then

value of 8k is

5) Let [x] be the greatest integer ≤ x. Then the number of points in the interval (–2), where the
function f(x) = | [x] | + is discontinuous, is________.
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. B B C A B D C A C B B A A A A A A B D C

SECTION-II

Q. 21 22 23 24 25
A. 2 90 5 5 40

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. A B A D C A C D D D C A C A B B C D C A

SECTION-II

Q. 46 47 48 49 50
A. 7 5 8 50 2

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. C C B C B B A B D A C C A A C A D A D D

SECTION-II

Q. 71 72 73 74 75
A. 2 22 6 1 2
 SOLUTIONS

PART-1 : PHYSICS

1)

2)

3) VC = = iR2 =

4)

This is balanced wheat stone bridge.

OR

Resistance of A to E & A to C are equal

So, i1 = i – i1

⇒ ,
This is current through AD.

5) t1 = 0
6)

Ceq =
Q = Ceq.E

Q2=

Q2/when C = 1µF = 2E

Q2/when C = 3µF = 2E

7) ig = 10 mA = 0.01A
VA = VB = (I – ig)0.1 = ig × 9.9
or I × 0.1 = 10ig
8)

9)

10) For maximum power

rint = Rext

Rext = 2R = rint = 4
⇒ R = 2Ω
11)
M.F. in –ve x direction

12)
Towards left of both wires direction of B is downward and at mid point between two wires,
magnetic field is zero.

13)
during eqm :
F mag = Fgrav ...(1)

...(1)
when press
FRestoring = Fmag – Fgrav

= =

=
Since x << d
Accelaratia =

Let ω2 = , then

14)
Magnetic force on sphere Fm = qvB
(directed radially outward)

∵ N – mg sin θ – qvB =

⇒N= + mg sin θ + qvB

Hence, at θ = π/2 we get Nmax

= + mg + qB = 3mg + qB

15) = tan30°

r=

B= = =

16)
Fnet = F1 – F2

17) Case 1 B1 =

Case 2 B2 =

Case 3 B3 =

18) Field at centre due to charge at point 1,5 & 3,7 will cancel
each other field due to charge at point (2), (6), (8) & (4) will be given as

19) Area of the given metallic plate A = πr2

Area of the dielectric plate

Uncovered area of the metallic plates

The given situation is equivalent to a parallel combination of two capacitor. One capacitor (C’)

is filled with a dielectric medium (K = 6) having area while the other capacitor (C’’) is air
filled having area

Hence

20) q = CeqV

= 24µC

21)

x=2
 22)

23)

24)

(field due to complete cylinder)

25)

By Taking right as +x and up as

+y direction we have

PART-2 : CHEMISTRY
26) K =

K=K

∴C=

27) log K = logA –

28)
Total pressure at time t = P0 – x + x = Pt
P0 + x = Pt
x = Pt – P0
at time t pressure of azo isopropane
P0 – x = P0 – (Pt –P0) = 2P0 - Pt

K=

29) As per Henry's law

[ 1 litre of H2O= 55.56 mol]

mol

30) The vapour pressure of mixture (PT = 600 mm Hg) is greater than the individual vapour
pressure of its constituents (Vapour pressure of CS2 = 512 mm Hg, acetone = 344 mm Hg).
Hence, the solution formed shows positive deviation from Raoult's law.
(1)
(2) Raoult's law is not obeyed
(3) Δsol. Volume > 0
(4) CS2 and Acetone are less attracted to each other than to themselves.

31) For XY;

XY X⊕ + Y⊝; i = 2;
For BaCl2;
BaCl2 Ba2⊕ + 2Cl⊝; i = 3;
According to osmotic pressure equation; π = CRT × i; where
π = osmotic pressure of the solution
C = molarity of the solution
R = 0.0821 l-atm/mol-k
T = Temperature of the solution
i = Van't Hoff factor

Given : =
⇒ (CRT × i)XY =
⇒ CXY × 2 = 4 × 0.01 × 3
⇒ CXY = 0.06 M

32) Explain Question :

Asking about the reaction in which 2° alcohol is the major product among the following.
Concept :
1. Electrophilic addition of alkene
2. Nucleophilic substitution followed by nucleophilic addition in reaction of acid halide with
Grignard reagent.
Solution :

Conclusion → 2° Alcohol is obtained as a major product in hydroboration oxidation reaction of

alkene in option 3.
Final Answer : (3)

33)
34)
P & Q are not the isomers of each other.

35) gives cyclic anhydride on heating.

36)

37)

38) Soln/Explanation:

Final Answer:

The correct option is (3).

39)

Question Explanation:
Question is asking about reaction of acetone with given compound to generate

Concept:

This question is based on reaction of carbonyl compound with H2N–Z by addition-elimination

mechanism.

Solution:

Ammonia derivation of type H2N–Z, adds on carbonyl compound followed by elimination of

water to generate

(1)
(2) No reaction (3° Amine)
(3) Only addition (2° Amine)

Final Answer:

The correct option is (1).

40) Carbonyl compound which has group will give positive haloform test.

41)

Ans is (B)
42)

43)

44)

Rate of SN1 reaction is a

t Bu X > iso - PrX > EtX > MeX
So order of activation energy
t – BuX < iso PrX < EtX < MeX
order of ionisation energy
t BuX < iso PrX < EtX < MeX

45)
NBS Brominates Allyl position only.

46)

(+)ve idoform test is given by compounds having grp or

47)

1,2,4,5,6

48)

x = 6 ; product 'P' is

49)

4.6 × 10–2 =
t = 50 sec.

50)
 0.2 × 2 = 0.2 × x
x=2

PART-3 : MATHEMATICS

51)
∴ [x] ≠ 0, [x] ≠ 1
x ∉ [0, 2)
∴ D = [2, ∞)

for all x ∈ [2, ∞]

Range =

Range =

52) f(x) = 100 {x}

= 41.4

53) y = x (ℓn x – 2)

y' = x + (ℓn x – 2) = ℓn x – 1

= ℓn x – 1 = 0 ⇒ x = e
now f(1) = – 2
f(e) = –e (least)
f(e2) = 0 (greatest)

∴ difference = 0 –(–e) = e Ans.

54) x2ƒ(x) – 2ƒ = g(x) ......(1)

⇒ – 2ƒ(x) = g

⇒ 2ƒ – 4x2ƒ(x) = 2x2 g ...... (2)

(1) + (2) ⇒ –3x2ƒ(x) = g(x) + 2x2g

⇒ ƒ(x) = – .....(3)

⇒ ƒ(–x) = – = [∵ g(x) is odd] by (3)

= –ƒ(x)
⇒ ƒ(x) is odd function
But ƒ(x) is given as an even function, so ƒ(x) = 0 ⇒ ƒ(5) = 0

55) Let f–1(x) = y

f(y) = x

56) y = f (x) has a maximum value, which is greater than zero at x = 0, so will have a

minimum value which is greater than zero, at x = 1. Thus has a maximum value at x =
1, which is less than zero, then one units shifts the curve up.]

57)

Here, x > 0 and

⇒
⇒
Put
∴
∴
Then

58)

Two curves intersect orthogonally if

..... (i)
Now eliminating y from the given equations we have
 ..... (ii)
Eliminating x2 from (i) and (ii) we get

59)

g'(x) =
∵ x>0
∴ f '(log x) – f '(2 – log x) > 0
⇒ f '(log x) > f '(2 – log x)
⇒ log x < (2 – log x)
⇒ 2 log x < 2
⇒ log x < 1
⇒0<x<e

60)
OD = x

v(x) = π

v(x) is max at x =

∴ height =

61) x + y = xy
put y = 0
x+0=1
∴ x=1
point is (1, 0)
Now
x + y – xy = 0
∴ mN = 1
Eq of normal is
y – 0 = 1(x – 1)
y=x–1

62) From given conditions

=
= 2π

63) f(x) = 1± xn, f(2) > 1 ∴ f(n) = 1+xn

64) [cos x] = 0
0
so |x|[cosx] = |x| = 1

65) We have

Therefore

so that

66)

LHL =

RHL =
For continuity ; LHL = RHL = f(2)
So λ = –1

67)

For x ≠ –2,
So,

68) The function defined in (A) is not differentiable at x = 1/2.

69) f(x) = [x]2 – [x2]

We know that greatest integer is discontinuous at integers but here we have to check at
integers.
First so we will check at x = 0 and x = 1 at x = 0

= (–1)2 – 0 = 1

=0–0=0
LHL RHL discontinuous at x = 0
Now at x = 1

= 02 – [1–] = 0

=1–1=0
f(1) = 0
LHL = RHL = f(1) so continuous at x=1 at all other integers it will discontinuous.

70) f(x) = |2x2 + 5|x| – 3|

Graph of y = |2x2 + 5x – 3|
Graph of f(x)

Number of points of discontinuity = 0 = m

Number of points of non-differentiability = 3 = n

71) [JEE Mains-15]

Normal at (1,1) will be x +y = 2

Now,

72) F'(x) = a(x – 1)(x + 3)

F"(x) = 6a(x + 1)
F'(x) = 3a(x + 1)2 + b
F'(1) = 0 ⇒b = –12a
F(x) = a(x + 1)3 – 12ax + c
= (x + 1)3 – 12x – 6
F(3) = 64 – 36 – 6 = 22

73)
Differentiating, we get,

when f(x) = 2
74) If f(x) is continuous at x = 0

75) Need to check at doubtful points

discont at x ∈ I only
at x = – 1 ⇒ f (– 1+) = 1 + 0 = 1
⇒ f (– 1–) = 2 + 1 = 3
at x = 0 ⇒ f (0+) = 0 + 0 = 0
⇒ f (0–) = 1 + 1 = 2
at x = 1 ⇒ f (1+) = 1 + 0 = 1
⇒ f (1–) = 0 + 1 = 1
discont. at two points