30-06-2025

6901CJA10153625MJ005SB JM

PHYSICS

SECTION-I

1) In the circuit shown below, initiallly the switch is open and the capacitors are uncharged. The
ratio of current through 2Ω resistor, just after the switch is closed and a long time after the switch is

closed :

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

2) In the circuit shown in figure. The potential difference across capacitor of

(A) 5

(B)

(C) 10

(D)

3) Find the time constant (in μs) for the given RC circuits in the given order respectively+
i) ii)

iii)
R1 = 1Ω, R2 =2Ω, C1 = 4μF, C2 = 2μF

(A)

(B)

(C)

(D)

4) Calculate the amount of charge on capacitor of 4 µF. The internal resistance of battery is 1Ω :

(A) 8 µC
(B) zero
(C) 16 µC
(D) 4 µC

5) The key in circuit is closed at t = 0, then total work done by batteries till the capacitor gets fully
 0

charged is. [q = 64μC and C = 16μF]

(A) 1024 μJ
(B) 512 μJ
(C) 320 μJ
(D) 64 μJ

6) In the circuit shown in figure the reading of ideal ammeter is :-

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

7) A conducting wire has a non uniform cross section as shown in figure. A steady current flows

through it. Which of the following is true :

(A) Drift speed of electrons at A is less than at B.

(B) Drift speed of electron at A is more than at B.
(C) Magnitude of electric field at A is less than magnitude of electric field at B.
(D) Electric field is zero at every point, inside of wire.

8) In the process of measuring current through galvanometer one wishes to have 1% of circuit
current in the galvanometer coil. If coil resistance is 99 Ω, then find shunt resistance used in
parallel :-

(A) 0.1Ω
(B) 1.1Ω
(C) 1.0Ω
(D) 11.0Ω

9) A galvanometer has coil of resistance 100Ω showing a full scale deflection 50 mA. What resistance
should be added to use it as ammeter of range 10A.

(A)

(B)

(C)

(D)

10) A potentiometer wire has length of 10 m and resistance 10Ω as shown in figure. A circuit is setup

as shown. For null deflection in galvanometer length AP is :

(A) 2m
(B) 4.5 m
(C) 6.5 m
(D) 2.3 m

11) A charged capacitor of capacitance C and having charge Q is to be connected with another
uncharged capacitor of capacitance C' as shown till steady state is reached. Then the value of C' for

heat liberated through the wires to be minimum :-

(A) zero
(B) C
(C) C/2
(D) 2C

12) Diagram shows 6 identical air capacitors connected to an ideal battery (V). In this condition,
charge on capacitor C6 is Q1. Now a dielectric whose dielectric constant is 2, is put between the
plates of C4 so that it occupies the entire space between the plates. Now charge on C6 becomes Q2.
Then should be :-

(A)

(B)

(C)

(D)

13) In the circuit shown, when the switch S is closed, then (S is closed after the steady state)

(A) no charge flows through S

(B) Positive charge flows from A to B
(C) Positive charge flows from B to A
(D) Positive charge flows initially from A to B and later from B to A

14) A dielectric slab of area A passes between the capacitor plates of area 2A with a constant speed
v. The variation of current (i) through the circuit as function of time (t) can be

qualitatively represented as
(A)

(B)

(C)

(D)

15) In the figure, capacitor is completely charged and switch is closed at t = 0. The time after which

the current from the capacitor becomes 1/4th of its maximum value will be :-

(A) 2 RC ln2

(B)

(C) RC ln2

(D)

16) A block of ice of mass 20 gm is kept in a steel container of water equivalent 10gm. The
temperature of both ice & the container is –30°C. Now 30 gm water at 80°C is poured in this
container. Find the final common temperature (Sice = 0.5 cal/g°C, Swater = 1cal/g°C, Lice = 80 cal/g) :-

(A) 0° C
(B) –5.55° C
(C) 3.33° C
(D) 5.55° C

17)

In the following P–V diagram of an ideal gas, AB and CD are isothermal whereas BC and DA are
adiabatic process. The value of VB/VC is

(A) = VA /VD
(B) < VA / VD
(C) > VA / VD
(D) cannot say

18) An ideal gas is undergoing a cyclic process as shown. The ratio of pressure of gas at point C to

point A i.e.

(A) 3
(B) 8
(C) 9
(D) 7

19) A voltmeter having a resistance of 1800 Ω is used to measure the potential difference across a
200 Ω resistor which is connected to a power supply of emf 50 V and internal resistance of 20Ω. The
percentage decrease in the potential difference across the 200Ω resistor when the voltmeter is
connected across it is :-

(A) 1%
(B) 5%
(C) 10%
(D) 25%

20) A circular portion is cut from a disc of thickness t, its resistivity is and radii of disc are a and b
(b > a). A potential difference is maintained between outer and inner cylindrical surfaces of the disc.
What is resistance of the disc ?

(A)

(B)

(C)

(D)

SECTION-II

1) A potentiometer wire of resistance per unit length (where x is distance from left
end of the potentiometer wire) is connected to four ideal batteries as shown in the figure. Resistance
of connecting wires is negligible. A cell of e.m.f. 1 volt is balanced against potential drop on
potentiometer wire. How many null point can be obtained on the wire.

2)

In the given circuit, the current flowing through the resistance 20Ω is 0.3 A, while the ammeter
reads 0.9 A. The value of R1 is ______ Ω.
3) In the experiment to determine the galvanometer resistance by half-deflection method, the plot of

vs the resistance (R) of the resistance box is shown in the figure. The figure of merit of the
galvanometer is .............. ×10–1 A/division. [The source has emf 2V]

4) As show in the figure, in steady state, the charge stored in the capacitor is……. × 10–6 C.

5) A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all
vibrational modes, the total internal energy of the system is nRT then n is

CHEMISTRY

SECTION-I

1) Given below are two statements : one is labelled as Assertion A and the other is labelled as
Reason R :
Assertion A : Acetal/Ketal is stable in basic medium.
Reason R: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic
medium.
In the light of the above statements, choose the correct answer from the options given below:

(A) A is true but R is false

(B) A is false but R is true
(C) Both A and R are true and R is the correct explanation of A
(D) Both A and R are true but R is NOT the correct explanation of A

2) A colourless oily liquid P (C2HCl3O) when treated with chlorobenzene in the presence of a
catalytic amount of sulfuric acid form a colorless, crystalline compound Q(C14H9Cl5)
which is used as insectiside. Hydrate of P is stable
Choose incorrect option

(A) Q is dichlorodiphenyltrichloroethane .
(B) P is chloral
(C) P disproportionate on treatement with NaOH.
(D) Q is Non–biodegradable pollutant

3) Identify the correct product of following reaction :

(A)

(B)

(C)

(D)

4) In the given reaction :

[X]
[X] will be :

(A) Methyl oxide

(B) Phorone
(C) 1, 3, 5-Trimethylbenzene
(D) 2-Butyne

5) Isobutyraldehyde on reaction with formaldehyde and K2CO3 gives compound ‘A’. Compound ‘A’
reacts with KCN and yields compound ‘B’, which on hydrolysis gives a stable compound ‘C’. The
compound ‘C’ is :

(A)

(B)

(C)

(D)

6) Which of the following arrangements with respect to their reactivity in nucleophilic addition
reaction is correct?

benzaldehyde < acetophenone

(A)
< p-nitrobenzaldehyde < p-tolualdehyde
acetophenone < benzaldehyde
(B)
< p-tolualdehyde < p-nitrobenzaldehyde
acetophenone < p-tolualdehyde
(C)
< benzaldehyde < p-nitrobenzaldehyde
p-nitrobenzaldehyde < benzaldehyde
(D)
< p-tolualdehyde < acetophenone

7)
The major product of the above reaction is
(A)

(B)

(C)

(D)

8)
Major product of above reaction :

(A)
(B)

(C)

(D)

9) The major product in the following reaction is

(A)

(B)

(C)

(D)

10)
is :

(A)
(B)

(C)

(D)

11) ; Product P is :

(A)

(B)

(C)
(D)

12)

Product ‘D’ will be :

(A) Aldehyde
(B) Ketone
(C) Carboxylic acid
(D) Alcohol

13) Which of the following will not give ether as product :-

(A)

(B)

(C)

(D) All of these

14) Which of the following dehydration product is incorrect ?

(A)

(B)

(C)
(D)

15) Which of the following is incorrect :

(A)

(B)

(C)

(D)

16)

Final major product 'W' is :

(A)

(B)
(C)

(D)

17) Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov
product?

(A) F3C – CH = CH2

(B) Cl – CH = CH2
(C) CH3O – CH = CH2
(D) H2N – CH = CH2

18) A is

(A)

(B)

(C)

(D)

19) Which of the following does not form a stable hydrate by the addition of H2O?

(A)

(B)

(C)
(D)

20) Write the product of following reaction :

(A)

(B)

(C)

(D)

SECTION-II

1) Find out the number of species which are polar as well as planar. O3, BF3, ClF3, SF4, , ,
SnCl2, , CO,

2) The number of water molecules directly bonded with central metal ion in

3) If z-axis in an internuclear axis then how many following pair of orbitals can form π-bond (2 lobes
interaction)?
4) How many of the following carbocations may undergoes hydride shifting

(I) (II)

(III) (IV)

(V) (VI)

5) Number of bromo derivatives obtained on treating ethane with excess of Br2, in diffused sunlight
is…

MATHEMATICS

SECTION-I

1) If the ellipse and hyperbola intersect orthogonally; then the value of b2

is :-

(A) 1
(B) 5
(C) 7
(D) 9

2) Let and If e and l denote the eccentricity and

the length of the latus rectum of the ellipse , then is equal to.

(A) 8
(B) 16
(C) 6
(D) 12

3) Distance between center and focus of the ellipse 4(x – 2y + 1)2 + 9(2x + y + 2)2 = 25 is :

(A)

(B)

(C)
(D)

4) The length of the major axis of the ellipse (5x – 10)2 + (5y + 15)2 = is :

(A) 10

(B)

(C)

(D) 4

5) If maximum distance of any point on the ellipse x2 + 2y2 + 2xy = 1 from its centre be r, then r is
equal to :

(A)
(B)

(C)

(D)

6) Equation of the ellipse whose axes are the axes of coordinates and which passes through the

point (2, 3) and has eccentricity is :-

(A) 3x2 + 2y2 = 35

(B) 3x2 + 2y2 = 15
(C) 2x2 + 3y2 = 35
(D) 2x2 + 3y2 = 15

7) If is an ellipse, centre at O(0, 0) and A, B, C are three points on the same ellipse, such

that, segment OA, OB, OC make angles respectively, with the positive x-axis, then the

value of is

(A)

(B)

(C) 3
(D) 4
8) Let a > b. Let E2 be another ellipse such that it touches the end points of major
axis of E1 and the foci of E2 are the end points of minor axis of E1. If E1 and E2 have same
eccentricities, then its value is :

(A)

(B)

(C)

(D)

9) The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2 – 9y2 + 32x
+ 36y – 164 = 0, and its foci is :

(A) 16x2 – 9y2 + 32x + 36y – 36 = 0

(B) 9x2 – 16y2 + 36x + 32y – 144 = 0
(C) 16x2 – 9y2 + 32x + 36y – 144 = 0
(D) 9x2 – 16y2 + 36x + 32y – 36 = 0

10) Let where r ≠ ±1. Then S represents :

(A)
A hyperbola whose eccentricity is where 0 < r < 1.

(B)
An ellipse whose eccentricity is where r > 1

(C)
A hyperbola whose eccentricity is when 0 < r < 1

(D)
An ellipse whose eccentricity is when r > 1

11) If the chord of the hyperbola x2 – y2 = 9, touches the parabola y2 = 12x, then the locus of the
middle points of these chord is :-

(A) x3 = (x – 3)y2
(B) x3 = (x + 3) y2
(C) x (x2 – y2) = 3y
(D) x3 = x – 3y2

12) The locus of the point of intersection of the lines and

(where t is a parameter) is a hyperbola whose eccentricity is
(A)
(B) 2
(C)
(D) 3

13) A hyperbola passes through the foci of the ellipse and its transverse and conjugate
axes coincide with major and minor axes of the ellipse, respectively. If the product of their
eccentricities in one, then the equation of the hyperbola is

(A)

(B)

(C) x2 – y2 = 9

(D)

14) If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which
one of the following points does not lie on this hyperbola?

(A)

(B)

(C)

(D)

15) Let P be a point on the hyperbola , in the first quadrant such that the area of
triangle formed by P and the two foci of H is . Then, the square of the distance of P from the
origin is :

(A) 18
(B) 26
(C) 22
(D) 20

16) If PSP' is a focal chord of an ellipse having eccentricity . S and S' are its focii and its auxiliary

circle is (x – 1)2 + (y – 2)2 = 25. If SP = , then S'P' is -

(A) 6

(B)
(C)

(D) 4

17) Let P1 be a parabola with vertex (1, 8) and focus (3, –1) and P2 is the mirror image of P1 with
respect to the line 3x – y + 10 = 0. If the focus of P2 is (α, β) and directrix of P2 is 7x + 6y = γ, then α
+ β + γ is equal to

(A) 125
(B) 137
(C) 52
(D) 119

18) If the system of equations

x+y+z = 5
x+2y+3z = 9
x+3y+αz = β
has infinitely many solutions, then β–α equals:

(A) 5
(B) 18
(C) 21
(D) 8

19) Let for , |A| = 2.

If |2adj (2adj (2A))| = 32n, then 3n + α is equal to

(A) 10
(B) 9
(C) 12
(D) 11

20) Let A be a 3 × 3 matrix with det(A) = 4. Let denote the ith row of A. If a matrix B is obtained
by performing the operation R2 → 2R2 + 5R3 on 2A, then det(B) is equal to :

(A) 16
(B) 80
(C) 128
(D) 64

SECTION-II
1) If A = , B = adj (A) and C = 3A then is :

2)

Let the curve C be the mirror image of the parabola y2 = 4x with respect to the line x + y + 4 = 0. If
A and B are the points of intersection of C with the line y = –5, then the distance between A and B is

3) If A and B are the foci of an ellipse

and P is any point on it, then PA + PB is

4) The number of ordered triplets (x, y, z) satisfying the given equation, (where x, y, z are positive

integers) is :

5) Let = 1 (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is

the maximum value of the function, ϕ(t) = + t – t2, then a2 + b2 is equal to :

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

SECTION-II

Q. 21 22 23 24 25
A. 2 30 5 10 11

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

SECTION-II

Q. 46 47 48 49 50
A. 6 4 2 3 9

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

SECTION-II

Q. 71 72 73 74 75
A. 8 4 4 3 126
 SOLUTIONS

PHYSICS

1)

3:1

2)

3)

r = CR
r1 = (C1 + C2) (R1 + R2) = 18μs

r2 =

r3 = (C1+C2) = (6) =

4) On simplifying circuit we get

No current in upper wire.

∴ VAB = × 4 = 4 v.
∴ θ = (Ceq)v
⇒ 2 × 4 = 8 µC
5)

B
64 μJ

6) The equivalent resistance between A and B = 4Ω.

Current I =
The current divides into different branches as shown in the figure. It is clear that the ammeter

will read . So the correct choice is (1)

7) by i = nAeVd
⇒ AB > AA ⇒ Vd|B < Vd|A

8)

I = Ia

100 =

9)
(0.05) (100) = (10–0.5)R

10) (R = 10 Ω , l = 10 m)

i=

eq.
__________ =

VAP = i × RAP

=
50 (AP) = 115

AP =
correction in dia 8v → 50 V

11) At steady state C and C' are in parallel

Uf = , Ui = 2

ΔH = Uf – Ui = ;

ΔH will be minimum when ⇒ C' = 0

12)

Q initially =
Q finally =

Qi =

Qf =
Qi : Qf = 3 : 4
13) For two conductors in series, their potential differences are proportional to their
resistances. For two capacitors in series, their potential differences are inversely proportional
to their capacitances. Hence, A and B are at the same potential and no charge will flow
between them.

14)

15) Here Req = R' =

Now,

or

16)

20[95 + ] + 10 × 1 × (T + 30) = 2400 – 30T

20 × 95 + 10T + 10T + 300 = 2400 – 30T
50T = 200
T = 4°C

17) AB → isothermal
PA VA = PB VB ...(i)
BC → Adiabatic
PB VBγ = PC VCγ ...(ii)
CD → Isothermal
PCVC = PDVD ...(iii)
DA → Adiabatic
PDVDγ = PA VAγ ...(iv)
From (i), (ii), (iii) and (iv)

0
18) Slope of AC ⇒ PC = 3P

and
19)

Current I =
Potential difference across 200 resistor is

V=
When the voltmeter is connected across the 200 Ω resistor, the effective resistance becomes

The current in the circuit becomes

I' =
The potential difference becomes

V' =
Decrease in p.d. = V – V'

= – 45 =
∴ Percentage decrease

20)

dR = (All in series)

21) Null point will be achieved when ΔV = 1V

∴ Total of 2 null points will be there.
 22)
Given, i1 = 0.3 A, i1 + i2 + i3 = 0.9 A
So, VAB = i1 × 20Ω = 20 × 0.3 V = 6 V

i2 =

i1 + i2 + i3 =

i3 = 0.2 A
So, i3 × R1 = 6 V
(0.2)R1 = 6

23) i = Kθ

Slope =

24)

= 10 µC

25) 11

CHEMISTRY

26) For Assertion :Acetal and ketals are basically ethers hence they must be stable in basic
medium but should break down in acidic medium.
Hence assertion is correct.
For reason: Alkoxide ion (RO⊝) is not considered a good leaving group hence reason must be
false.
27) P (CCl3CHO) chloral + Q( )

Q is Non–biodegradable pollutant
(dichlorodiphenyltrichloroethane .)
CCl3CHO + NaOH → CHCl3 +HCOONa

28)

29) It is example of acid catalytic aldol condensation.

30)

31) The rate of nucleophilic addition decreased due to steric crowding around carbonyl carbon
& increased by electron withdrawing group if the steric crowding is same hence the reactivity
towards nucleophilic addition will be
32)

33)
(product after 1)

34)
(i) Grignard prefer to give nucleophilic addition on polar π-bond and form anion intermediate.

(ii) In next step anion give intramolecular nucleophilic substitution reaction & form 5
membered ring.
35)
Intramolecular aldol

36)

37) The complete reaction sequence is

38)

39)

40) Aldehyde (not acid) will be obtained on reductive ozonolysis by O3/Zn/H2O.

41)
42)

43)

44)

45)
46)

47)

48)

s + px - No bond
dxy + dxy = δ bond
dyz + dyz = π bond
py + py - π bond
px + pz - no bond
s + s - σ bond

49) (I), (IV) & (V)

50) CH3 – CH3 + Br2 (Excess)

Monobromo
 Dibromo

Tribromo

Tretrabromo

Pentabromo

Hexabromo

MATHEMATICS

51) For hyperbola a = , b =

and e = = =

⇒ ae = =2
⇒ focus of ellipse is (2, 0)

⇒ a = =2
9 – b = 4 ⇒ b2 = 5
2
⇒

52)

=8
53) Given equation is
4 (x – 2y + 1)2 + 9 (2x + y + 2)2 = 25
Substituting

and

we have

eccentricity
distance between center and focus = ae

54) (5x–10)2 + (5y + 15)2 =

⇒(x–2)2 + (y + 3)2 =

⇒ is an ellipse, whose focus is (2,-3) directrix 3x– 4y + 7 =

0 and eccentricity is

Length of perpendicular from focus is directrix is :

so length of major axis is :

55) Here centre of the ellipse is (0,0)

Let P(rcosθ,rsinθ) be any point on the given ellipse then r2cos2θ + 2r2sin2θ + 2r2sinθ cosθ = 1.
56) Let equation of ellipse is

it passes through (2, 3) ⇒ …(i)

and ⇒ …(ii)

Solving (i) and (ii) gives ,

⇒ ellipse is 2x2 + 3y2 = 35

57)

E : x2 + 4y2 = 4

OA : y =
OB : y =

OC : y =

Solving OA & E : 3y2 + 4y2 = 4 ⇒ y =

x=

Solving OB & E : 13x2 = 4 ⇒ x = ,y=

Solving OC & E : 7y2 = 4 ⇒ y =

x=

(OA)2 = (OB)2 = (OC)2 =

⇒
58)

⇒ c2 =
Also b = ce

⇒c=

⇒e=
⇒ e2 + e – 1 = 0

59) Given hyperbola is

16(x + 1)2 – 9(y – 2)2 = 164 + 16 – 36 = 144

Eccentricity,
⇒ foci are (4, 2) and (–6, 2)

Let the centroid be (h, k) & A(α, β) be point on hyperbola

So
⇒ α = 3h + 2, β = 3k – 4
(α, β) lies on hyperbola so
16(3h + 2 + 1)2 – 9(3k – 4 – 2)2 = 144
⇒ 144(h + 1)2 – 81(k – 2)2 = 144
⇒ 16(h2 + 2h + 1) – 9(k2 – 4k + 4) = 16
⇒ 16x2 – 9y2 + 32x + 36y – 36 = 0

60)

for r > 1 ,

61) Chord of the hypererbola x2 – y2 = a2 with middle point (h, k), is hx – ky = h2 – k2

⇒ y= .......(1)

As (1) touches y2 = 4ax, condition of tangency gives

c= ⇒ =
⇒ x (y – x ) = 3y ⇒ x3 = y2(x – 3)
2 2 2

62)

⇒ t= ....(1)

⇒ t= ....(2)
From (1) and (2), the locus of the point of intersection of the lines is,

3x2 – y2 = 48

=1
The above equation is a hyperbola of the form

Eccentricity, e =

e= =2

63)

For ellipse

for hyperbola
Let hyperbola be

∵ it passes through (3,0) ⇒

⇒ a2 = 9
⇒ b2 = a2(e2 – 1)

∴ Hyperbola is

... option (B).

64) ae = 3, , , b2 = 5
65)

a2 = 9, b2 = 4

Area of

Distance of P from origin =

=

66)
radius of auxillary circle = a = 5

a = 5,
harmonic mean of PS & P'S is Semilatus Rectum

P'S + P'S' = 2a = 10 ⇒ P'S' = 6

67) Image of vertex of P1 in 3x – y + 10 = 0 will be vertex of P2

∴ vertex of P2 = (–2, 9)
Similarly focus of P2 is (–9, 3) ∴ α = –9 and β = 3

Slope of directrix

Equation of directrix y – 15
7x + 6y = 125 ∴ γ = 125.
68)
for infinite solutions D = 0 ⇒ α = 5

⇒
⇒
on β = 13 we get Dy = Dz = 0
α = 5, β = 13

69)

|A| = 2,

1(6 – 1) – 2 (2a – 1) + 3(a – 3) = 2

5–4a+2+3a–9=2

–a–4=0

a = –4

8|Adj(2Adj(2A))|

8|Adj(2×22 Adj (A))|

8|Adj(23AdjA)|

8|26Adj(AdjA)|

23(26)3|Adj(Adj)|

23 . 218 |A|4

221 . 24 = 225 = (25)5 = (32)5

n=5

a = –4
70) |A| = 4
⇒ |2A| = 23 × 4 = 32
B is obtained by R2 → 2R2 + 5R3
⇒ |B| = 2 × 32 = 64
option (4)

71)

72)

Let a point (t2, 2t) on the parabola y2 = 4x

∵ Image of this point w.r.t. x + y + 4 = 0.
is (–2t–4, –t2–4)
Hence parabola C becomes (x + 4)2+4y+16=0
∵ It intersects line y = – 5
Hence (x + 4)2 + 4(–5) + 16 = 0
⇒ x = ± 2.
Hence points A and B are (–2, –5) and (–6, –5)
⇒ AB = 4.

73) PA + PB = 2a = length of major axis

74) Multiply R1 by x; R2 by y and R3 by z and divide the determinant by xyz

75) (a > b) ; ⇒ b2 = 5a ...(i)

Now,

⇒ ... (ii)
2
⇒ a = 81 (from (i) & (ii))
So, a2 + b2 = 81 + 45 = 126