10-08-2025

5001CJA101029250013 JA

PHYSICS

SECTION I

1) A conducting wire MN carrying a current I is bent into the shape as shown and placed in xy plane.
A uniform magnetic field is existing in the region. The net magnetic force experienced by

the conducting wire MN is :-

(A) 2BIR
(B) 3BIR
(C) 4BIR
(D) 6BIR

2) The flux linked with a coil at any instant ‘t’ is given by ϕ =10t2 - 50t+250.
The induced emf at t = 3 s is..

(A) 190 V
(B) –190V
(C) –10 V
(D) 10V

3) A resistor has colour code of sequence red, yellow, violet, silver. What is value of its resistance :-

(A) 24 × 107 ± 10%

(B) 24 × 107 ± 5%
(C) 42 × 107 ± 10%
(D) 42 × 107 ± 5%

4) Find the potential at point E if D is grounded, as shown in figure.

(A) 12 volt
(B) –8 volt
(C) 14 volt
(D) –20 volt

5) A straight rod of mass m and length L is suspended from the identical spring as shown in the
figure. The spring stretched by a distance of x0 due to the weight of the wire. The circuit has total
resistance RΩ. When the magnetic field perpendicular to the plane of the paper is switched on,
springs are observed to extend further by the same distance. The magnetic field strength is

(A)
; directed outward from the plane of the paper

(B)
; directed outward from the plane of the paper

(C)
; directed into the plane of the paper

(D)
; directed into the plane of the paper

6) In the circuit shown below, find the decrease in power dissipated in 2Ω resistance after switch is

closed. The battery is ideal :-

(A) 8.34 W
(B) 21.21 W
(C) 3.68 W
(D) 14.72 W

7) Figure shows 3 different capacitors, separately charged by batteries, and then connected as
shown, with initially, switch open. When switch is closed, the charge :-
(A) On all capacitors get equalised
(B) becomes zero on all capacitors
(C) On 2µF capacitor increases by 20%
(D) On 2µF capacitor decreases by 20%

8) Which of the statement is incorrect :

(A) Relative permeability of diamagnetic substance in negative.

(B) Relative permeability of paramagnetic substance have value greater than 1
Field line inside paramagnetic substance are concentrated when such material are placed in
(C)
external magnetic field of magnetic intensity.
(D) At sufficiently high temperature, a ferromagnetic substance become paramagnetic.

9) A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III
& IV. Arrange them in the decreasing order of potential energy :

(A) I > III > II > IV

(B) I > II > III > IV
(C) I > IV > II > III
(D) III > IV > I > II

10) If a coil of 40 turns and area 4.0 cm2 is suddenly removed from a magnetic field, it is observed
that a charge of 2.0 × 10–4 C flows into the coil. If the resistance of the coil is 80 Ω, the magnetic flux
density in Wb/m2 is

(A) 0.5
(B) 1.0
(C) 1.5
(D) 2.0
11) If a charged particle goes unaccelerated in a region containing electric and magnetic fields :

(A) must be parallel to

(B) must be perpendicular to Electric field

(C) must be perpendicular to

(D) E must be equal to vB.

12) A charge +q of mass m is projected in uniform magnetic field with velocity

. Pitch of path of particle is P0. Now the same charge particle is projected in uniform
magnetic field with velocity . This time pitch of path is P2 then :-

(A) P2 = 2P0
(B) P2 = 4P0
(C) P2 = P0
(D) P2 = 8P0

13) Current i = 2.5 A flows along the circle x2 + y2 = 9 cm2 (here x & y in cm) as shown. Magnetic

field at point (0, 0, 4 cm) is :-

(A)

(B)

(C)

(D)

14) A wire of resistance 18 ohm is drawn until its radius reduce th of its original radius then
resistance of the wire is

(A) 188 Ω
(B) 72 Ω
(C) 288 Ω
(D) 388 Ω

15)

Identical dielectric slabs are inserted into two identical capacitors A and B. These capacitors and a
battery are connected as shown in figure. Now the slab of capacitor B is pulled out with battery
remaining connected:

(A) During the process positive charge flows from a to b

(B) Finally charge on capacitor B will be less than that on capacitor A
(C) During the process, a work is done by the external force F, which appears as heat in the circuit
(D) During the process, internal energy of the battery decreases

16) Find Ceq for system

(A)

(B)

(C)

(D)

17) The potential of electrostatic field is given by ϕ = xy, where x & y are coordinate. Which of the
following is correct?

(A) Electric field at all points on positive x-axis is in positive y-direction.

(B) Electric field at all point of positive y-axis is in negative x-direction.
(C) Electric field at (2, 2) is .

(D)
Electric field at x, y makes an angle with x-axis.

18) An electric dipole with dipole moment C-m is placed in an electric field
(N/C). An external agent turns the dipole slowly until its electric dipole moment becomes

× 10–30 C-m. The work done by the external agent is equal to :–

(A) 4 × 10–28 J
(B) –4 × 10–28 J
(C) 2.8 × 10–26 J
(D) –2.8 × 10–26 J
19) A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic
vector field at a point having coordinates (x, y) in the z = 0 plane is:

(A)

(B)

(C)

(D)

20) Six identical conducting rods are joined as shown in figure. Points A and D are maintained at
temperature 200°C and 20°C respectively. The temperature of junction 'B' will be :-

(A) 120°C
(B) 140°C
(C) 100°C
(D) 80°C

SECTION II

1) A rod CD of thermal resistance 10.0 KW-1 is joined at the middle of an identical rod AB as shown
in figure. The ends A, B and D are maintained at 200°C,100°C and 125°C respectively. The heat
current in CD is P watt. The value of P is.......

2) In the figure, the flux due to magnetic field through the loop perpendicular to the plane of the coil
and directed into the paper is varying according to the relation, ϕ = t2 + 7t + 1 where f is in weber
and t is in seconds. The magnitude of the emf induced in the loop at t = 1s is (in volt).
3) A charge particle having charge q = 1C and mass 0.25 kg is projected from origin as shown in
figure in x-y plane magnetic field of 1T is present perpendicular to plane. Find speed v (in m/s) so

that particle can pass through point A(2, 0).

4) A moving-coil galvanometer has 100 turns and each turn has an area 2.0 cm2. The magnetic field
produced by the magnet is 0.01 T. The deflection in the coil is 2 radian when a current of 10 mA is
passed through it. If the torsional constant of the suspension wire is N × 10–6 N-m/rad, then find N.

5) A charged particle q is shot towards another charged particle Q which is fixed, with a speed v. It
approaches Q upto a closest distance r and then returns. If q was given a speed 2v, the closest

distance of approach would be (r/α) then value of α is?

CHEMISTRY

SECTION I

1) In the reaction xA yB, Then, x : y is

(A) 3 : 1
(B) 3 : 5
(C) 5 : 3
(D) None

2)

In the reaction :
P+Q→R+S
 the time taken for 75% reaction of P is twice the time taken for 50% reaction
of P. The concentration of Q varies with reaction time as shown in the above figure. The overall
order of the reaction is -
(A) 2
(B) 3
(C) 0
(D) 1

3) We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M,
0.01M & 0.001 M, respectively. The value of van t’ Haft factor (i) for these solutions will be in the
order.

(A) iA < iB < iC

(B) iA < iC < iB
(C) iA = iB = iC
(D) iA > iB > iC

4) Equimolar solution of two liquids A( =100torr) & B( =200 torr) has vapour pressure 180 torr.
Then which is correct for the solution :

(A) ΔHmix > 0

(B) Show negative deviation
(C) ΔHmix < 0
(D) ΔSmix < 0

5) The major product (Y) in the following reactions is :

(A)

(B)

(C)
(D)

6) Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes
the freezing point of the solution. Use the freezing point depression constant of water a 2 K kg mol–1.
The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T).
[Molecular weight of ethanol is 46 g mol–1].
Among the following, the option representing change in the freezing point is

(A)

(B)

(C)

(D)

7) The increasing order of nitration of the following compound is

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

8) In which of the following reaction only retention of configuration take place :

(A)

(B)

(C)

(D)

9)

Which one is the most reactive towards SN1 reaction?

(A) Ph–CH2–Br

(B)

(C)

(D)
10)
X and Y are respectively :

(A)

(B)

(C)

(D)

11) C6H5–C≡C–CH3 A

(A)

(B)

(C)

(D)

12) Select option with incorrect major product.

(A) C6H5CH2Cl C6H5CH2OH

(B) C6H5CH2Cl C6H5CH2OCH2C6H5

(C)
(D)

13)

For the reaction,

(A) A single stereoisomer with same configuration will be formed.

(B) A racemic mixture will be produced.
(C) Meso product is formed
(D) A single stereoisomer will be formed.

14) Which of the following property is INCORRECT for Zeise's salt :

(A) The complex present has the EAN value of 86

(B) C–C bond length is greater than 134 pm that is C – C bond length in free C2H4
(C) All Pt–Cl bond are present in same plane
(D) π-donor ligand is present

15) Total number of stereo isomers of complex [Pt(gly)2] is -

(A) 4
(B) 5
(C) 6
(D) 2

16) The magnetic moment is measured in Bohr Magneton (BM).

Spin only magnetic moment of Fe in [Fe(H2O)6]3+ and [Fe(CN)6]3– complexes respectively is :

(A) 6.92 B.M. in both

(B) 4.89 B.M. and 6.92 B.M.
(C) 3.87 B.M. and 1.732 B.M.
(D) 5.92 B.M. and 1.732 B.M

17) Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the
oxidation state changes from

(A) + 3 to + 1
(B) + 6 to + 3
(C) + 2 to + 1
(D) + 6 to + 2

18) The correct order of basicity of oxides of vanadium is

(A) V2O3 > V2O4 > V2O5
(B) V2O3 > V2O5 > V2O4
(C) V2O5 > V2O4 > V2O3
(D) V2O4 > V2O3 > V2O5

19) In which of the following pairs, both the species have the same hybridization ?
(I) SF4, XeF4 (II)
(III) (IV)

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

20) Which order is wrong :-

(A) Electronegativity – P < N < O < F

(B) Ist ionisation potential – B < Be < O < N
(C) Basic property – MgO > CaO > FeO > Fe2O3
(D) Reactivity – Be < Li < K < Cs

SECTION II

1) 20 ml of 0.8 M–HA solution is neutralised by 0.2 M–NaOH solution, at 25°C. pH of resulting

solution at equivalent point is x, then value of 10x is (pKb of A– = 8, log2 = 0.3)

2) Using the following information determine the boiling point (in °C) of a mixture which contains
1560 gm benzene and 1125 gm chlorobenzene, when the external pressure is 1000 torr. Assume the
solution is ideal. [Atomic mass : Cl = 35.5]

Vapour pressure Vapour pressure of

Temperature (0°C)
of benzene(torr) chlorobenzene(torr)

80 750 120

90 1000 200

100 1350 300

110 1800 400

120 2200 540

3)
Number of equivalents of Grignard reagent (x) used in the reaction is

4) Considering that Δ0 > P, the magnetic moment (in BM) of [Ru(H2O)6]2+ would be________.

5) How many of the given molecules have pπ-dπ bond ?

SO2, NO3–, PO43–, CO2, SO3, POCl3, XeOF4, CO ?

MATHEMATICS

SECTION I

1) The sum of all the values of k for which the system of equations
4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0; has a non-trivial (non-zero) solution, is

(A) 2
(B) 4
(C) 6
(D) None

2) The value of the determinant where p ≠ q ≠ r is :-

(A) p + q + r
(B) –(p + q + r)
(C) pqr
(D) 0

3) Number of solutions of the equation is :

(A) 1
(B) 2
(C) 3
(D) 4
4) Let f(x) = and g(x) = , then :-

(A) ƒ (g(x)) is defined but g(ƒ (x)) is not defined

(B) g(ƒ (x)) is defined but ƒ (g(x)) is not defined
(C) both g(ƒ (x)) and ƒ (g(x)) are defined but they are not identical functions
(D) both g(ƒ (x)) and ƒ (g(x)) are defined and they are equal function

5) Period of function f(x) = min{sin x, |x|} + (where [.] denotes greatest integer function) is
-

(A) π/2
(B) π
(C) 2π
(D) 4π

6) If , x ≠ 0 then possible value of x is :-

(A)

(B)

(C) 1

(D)

7) If , then the value of S is-

(A) 2
(B) 3
(C) 4
(D) 12

8) Let . Then the proper set of value of x for which ƒ(x) = 0, is

(A)
(B)
(C)
(D)
9) is equal to

(A) 5
(B) 4
(C) 0
(D) D.N.E.

10) Consider the function

, then

(A)

(B)
(C) g(x) is continuous for all x except at x = 0
(D) g(x) is differentiable for all x except at x = 0

11) The value of k, if

is both continuous & differentiable is -

(A) 0

(B)

(C)

(D) does not exist

12) For x ∈ [–2,4], function ƒ(x) = [log2(x2 – 2x + 5)] discontinuous at (where [.] represents greatest
integer function)

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

13) Let ƒ(x) = , then (where [.] represents greatest integer function)

(A) ƒ(x) is an even function

(B) ƒ(x) is an odd function
(C) ƒ(x) is aperiodic
(D) None of these
14)

If a derivable function y = ƒ(x) satisfies the condition ƒ(x) + ƒ(y) = ƒ(x)ƒ(y) + ƒ(xy) ∀ x,y ∈ R where ƒ(1)
= 0 and ƒ'(1) = –2, then the number of solution(s) of the equation ƒ(x) = 2

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

15)

Number of points in (0, ∞) where function f(x) = sinπx |(ln x) (x + 1) (ex – e2)| is non differentiable, is

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

16) An are PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If
, and then α, β2 are the roots of the equation

(A) x2 – x – 2 = 0
(B) 3x2 + 2x – 1 = 0
(C) x2 + x – 2 = 0
(D) 3x2 – 2x – 1 = 0

17) Let and . If a vector satisfies and

then is equal to

(A) 413
(B) 423
(C) 323
(D) 313

18) The lines are coplanar if

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

19) Let and If c is a vector such that and the

angle between and is 30°, then
(A)

(B)

(C) 2
(D) 3

20) If the non zero vectors are perpendicular to each other, then the solution of the equation.
is :
where x is a scalar

(A)

(B)

(C)
(D)

SECTION II

1) Let are three vectors of which every pair is non collinear, and the vectors and
are collinear with are respectively. If = 1, then find .

2)

Three lines are given by

ℝ
ℝ and
ℝ
Let the lines cut the plane x + y + z = 1 at the points A, B and C respectively. If the area of the
triangle ABC is Δ then the value of (6Δ)2 equals ___

3) A plane which always remains at a constant distance 4 units from origin & it's normal is equally
inclined with the coordinate axes, then sum of length of intercepts cuts by the plane on the
coordinate axes is equal to units, then ℓ :-

4) If ƒ(x) = , the value of ƒ(0) so that ƒ(x) is continuous at x = 0

5) Let , then value of determinant

is :
 ANSWER KEYS

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. D C A B A D D A C B B C A C A B B C A B

SECTION II

Q. 21 22 23 24 25
A. 2.00 9.00 8.00 1.00 4.00

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. C D A A C D B A D A A C D A D D B A C C

SECTION II

Q. 46 47 48 49 50
A. 96 100 5.00 0.00 5.00

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 D D D C D A A A C D A A A D A A C B A

SECTION II

Q. 71 72 73 74 75
A. 6 0.75 12 2.00 0.00
 SOLUTIONS

PHYSICS

1) Since is constant in the region

The net magnetic force experienced by the conducting wire MN is

Where,

2)

E(t) = –

E(t) =
E(t) = – (20t – 50)
Here, Substitute t = 3s
E (C) = – (20 × 3 – 50)
E(C) = – (60–50)
E(C) = –10V

3) Red – 2 (1st digit)

Yellow – 4 (2nd digit)
Voilet – 7 (multiplier) ⇒ 107
Silver – 10% tolerance

4)
⇒ VD = 0
∴ VE = –4 × 2 = –8 volt

5)
When switch is open then,
 ...(1)
When switch is closed, spring elongate by swice of initial force on rod due to magnetic field
must be along downward direction.

so, direction of magnetic field should be along outward from plane of the paper (By right hand
thumb rule)

FB = × i × B =
⇒ FB + mg = 4kx0
⇒ FB + mg = 2(2kx0)
⇒ FB + mg = 2mg [From eqn–1)

0
6) i =
P0 = 50 W

iT =

i2 = 7 × = 4.2 A
P = 4.22 × 2 = 35.28 W. ΔP = 14.72 W

7)

after redistribution,
Q = 0.8
charge on 2F capacitor reduces by 20 %

8) Relative permeability,
µr = 1 + x
x is susceptibility

9)

Concept : Magnetic potential energy

∴ As θ increases υ↑.
I > IV > II > III

10)

B = 1.0

11)

12) Pitch = (Component of along )

⇒ P2 = P0

13) Magnetic field on the axis of a circular loop

14)

Correct Answer is
(C) 288 Ω

15)

Ceq decrease
So charge decrease.

16) Question Explanation :

two Spheres with charges + q and –q and radius a, b respectively is given the distance
between the two is d we have to find the equivalent capacitance.

Concept used :
it is based on the concept of capacitance is equals to magnitude of charge upon potential
difference between the two conductors
Formula used :

Calculation :

VB =
V+ = PD of +vely charged sphere
V– = PD of –vely charged sphere

Hence, the correct answer is option (2).

17)

18) Work done = Uf – Ui =

= –(–16 × 10–27) – (–12 × 10–27) = 2.8 × 10–26 J

19) Magnetic field at P is perpendicular to OP in the direction shown in figure.

So we have

Where
20)
Req = 3R

i=
TB = 200 – 60 = 140°C

21)
ºC

22)

Correct Answer is 9.00

23)
2 = 2R sin θ

24) = BiNA
= Cθ
By comparing
Cθ = BiNA
 C = 1 × 10–6 N-m/rad

25)
From the given data, using energy conservation

When particle is shot with a speed 2V, Let distance of closest apporach = x

⇒x=

CHEMISTRY

26)
log3 + log rA = log rB + log5
log (3 × rA) = log (5 × rB)

3rA = 5rB ;
x:y=5:3

27) r ∝ [P]1 for first order

0
r ∝ [Q] for zero order At = A0 – kt
0
r = k[P]1[Q]

28)

Values of i (for different conc. of a Salt)

Salt
0.1 M 0.01 M 0.001 M
NaCl 1.87 1.94 1.94
i approach 2 as the solution become very dilute.

29) Fact

30)
 31)

∴ (D) is appropriate.

32) activating nature: –OCH3 > –CH3 > – Cl

+R +H –I
+1

33)

Explain Question :
Asking the stereochemical concept [retention of configuration] in given reactions of alcohol.

Concept :
This question is based on SN2 mechanism.
Stereochemistry of SN2 reaction → Inversion of configuration (Walden inversion)

Solution :
If nucleophile attack at chiral atom in SN2 reaction then inversion of configuration is observed
otherwise retention of configuration is seen in the product.

(1)

(2) Inversion in product

(3) Inversion in product

(4) Inversion in product

Final Answer : (1)

34)

Correct answer is

(D)

35) Question Explanation:

Question is asking the major elimination products obtained by two different bases.

Concept:
In E2 elimination β-H removed by strong base.

Solution:
Highly crowded base (like t-BuOK or Et3COK) prefers removal of less hindered β-H while
normal or small size bases prefer removal of most hindered β-H in order to get the most stable
alkene.
Final Answer:
The correct option is (1)

36)

37) Explain Question :

Question is asking the reaction in which given product is not major product.
Concept :
This question is based on SN2 reaction of alkyl halides.
Solution :
(a) Alkyl halides in presence of moist Ag2O or AgOH give alcohol.

(b) Alkyl halides in presence of dry Ag2O give ether.

(c) KNO2 is an ambidented nucleophile, it gives alkanenitrite as the major product.

(d) AgCN is an ambidented nucleophile and it has more covalent nature so lone pair of
nitrogen acts as nucleophile and alkyl isocyanide is formed.

In option (3) given major product is wrong.

Final Answer : (3)

38)
* SN2 product will be formed
* Polar aprotic solvent is used, it favours the SN2 Mechanism
* In SN2 Nucleophile (here HS–) attack from back side.
* Inversion of product will takes place

* New product
* A single stereo isomer will be formed
* Since product is altogether a different compound we cannot say about it dextro or
levorotatory nature.
Only Option D is correct.

39)

The complex present has the EAN value of 84

C–C bond length is greater than 134 pm that is C – C bond length in free C2H4
All Pt–Cl bond are present in same plane
π-donor ligand is present

40) [M(AB)2] shows two G.I. and both are optically active so total S.I are 4.

41) [Fe(H2O)6]3+
0
Fe3+ ⇒ [Ar] 3d5 4s

No pairing
∴ Unpaired e– = 5

[Fe(CN)6]–3
Fe3+ ⇒ [Ar] 3d5 4s°

Pairing occur due to strong field ligand CN–

∴ Unpaired e– ⇒ 1

42)

43) With increase in % of oxygen acidic nature of oxide of an element increase and basic
nature decreases

44)

45)

Which order is wrong.

(C) Basic property – MgO > CaO > FeO > Fe2O3
↓
wrong order
Because correct order is,
⇒ CaO > MgO > FeO > Fe2O3
Basic nature of oxides :- when in peridic table the distance between the element and oxygen
increases,
Basis character increases.
(CaO > MgO : Basic property).

46) neq HA = neq NaOH

or
⇒ V = 80 ml
16mmol 16mmol 0
on complete reaction, millimole of A– = 16

∴ [A–] =

pH = 7 + (pKa + logC) = 7 + [(14 – 8) + log 0.16] = 9.6

47) moles of benzene in solution = = 20 mol

moles of chlorobenzene in solution = = 10 mol

⇒ XBenzene = mole fraction of benzene in liquid solution

= =
⇒ Xchlorobenzene = mole fraction of chlorobenzene in liquid solution

= 1 – XBenzene =
for boiling, the vapour pressure exerted by the solution should be equal to that of the external
pressure. The data given in question gives a vapour pressure of solution equal to 1000 torr at
100°C.
Ptotal =

= 1000 torr
(We have to check total vapour pressure in each case & find the case for which it equals 1000
torr)

48) 3 moles of PhMgBr act as a base & 2 moles of PhMgBr as nucleophile

49) [Ru(H2O)6]2+
so, pairing of e– will take place
Ru = [Kr]5s2 4d6
Ru2+ = [Kr] 4d6
2+
Ru = [Kr]
Unpaired e– = Zero.
So, magnetic moment =
=
 50) , ,

, ,

These molecules have pπ-dπ bond.

MATHEMATICS

51)

4x + ky + 2z = 0
kx + 4y + z = 0
2x + 2y + z = 0
for non trival solution

4(4 – 2) – k (k – 2) + 2 (2k – 8) = 0
8 – k2 + 2 k + 4 k – 16 = 0
k2 – 6k + 8 = 0
k2 – 2k – 4k + 8 = 0
k(k – 2) –4 (k – 2) = 0
(k – 4) (k – 2) = 0
k = 2, 4
sum of values of k = 2 + 4 = 6

52)
(interchanged row and columas)
= –A ⇒ A = 0

53)

four solutions are possible.

54)

Both are defined & identical

55)

56) ⇒

⇒
Now cotθ = sinα

⇒ 1 + 6x2 = 6(1 – x2)

57) nth term of series

is Tn = tan–1

= tan–1

⇒ Sn = tan–1 – tan–1

⇒ S∞ = – tan–1

⇒ S = tan

58)

59)

60)
and
61) 3 – 2k – 2 = 0

for continuity for differentibility

ƒ'(1–) = ƒ'(1+)
1 = –2k

∴ No value of k.

62)

x2 – 2x + 5 = (x – 1)2 + 4 ∈ [4, 13]

for x ∈ [–2, 4]
y = [log2(x2 – 2x + 5)] ∈ {2, 3} is continuous for log2(x2 – 2x + 5) = 2
but discontinuous for log2(x2 – 2x + 5) = 3
⇒ 2 values of x

63)

⇒ will be an even function (symmetric in opposite quadrants)

64)

ƒ(x) + ƒ(y) = ƒ(x)ƒ(y) + ƒ(xy)

partial differentiation wrt x
ƒ'(x) = ƒ'(x)ƒ(y) + ƒ'(xy).(y)
put x = 1
–2 = –2ƒ(y) + ƒ'(y)y
get ƒ(y) = 1 – y2 ⇒ ƒ(x) = 1 – x2
Now, number of solution(s) of ƒ(x) = 2
1 – x2 = 2 ⇒ x2 = –1 ⇒ x ∈ ϕ

65) ln x, x + 1, ex – e2 becomes zero at x = 1, –1 and 2 respectively, for which sin px is also zero
Hence f(x) is derivable at these points.
∴ number of points of non derivability = 0

66)

67)

Also

68) lines are coplanar

S.D. = 0

⇒
⇒
⇒
⇒
⇒
⇒

69)

70)
take cross product with

Since are non coplanar

and y = 0

71)

Given

Thus,

72)

A(1, 0, 0),

Hence,

So,
⇒

73) x + y + z = a
distance from (0, 0, 0) is 4.

∴ ⇒a=

74)

= 0 + 2 =2

75) Determinant value of every odd order skew symmetric matrix is zero.