17-08-2025

9610ZJM802000250002 JM

PART 1 : PHYSICS

SECTION-I

1) A wire has resistance of 24 Ω and it is bent in a shape, as shown in figure. The effective resistance

between A and B is :-

(A) 24 Ω
(B) 10 Ω

(C)

(D)

2) A potentiometer wire AB is 100 cm long and has a total resistance of 10 ohm. If the galvanometer
shows zero deflection at the position C, then find the value of unknown resistance R.

(A) 4 Ω
(B) 6 Ω
(C) 8 Ω
(D) 2 Ω

3) An insulating pipe of cross-section area 'A' contains an electrolyte which has two types of ions
their charges being –e and +2e. A potential difference applied between the ends of the pipe result
in the drifting of the two types of ions, having drift speed = v (–ve ion) and v/4 (+ve ion). Both ions
have the same number per unit volume = n. The current flowing through the pipe is :

(A)
nevA
(B)
nevA

(C)
nevA

(D)
nevA

4) Two batteries, one of emf 18 volt and internal resistance 2 Ω and the other of emf 12 volt and
internal resistance 1 Ω, are connected as shown. The voltmeter V will record a reading of :–

(A) 18 volts
(B) 30 volts
(C) 14 volts
(D) 15 volts

5) The circuit below is made up using identical light bulbs. The light bulbs of maximum brightness of

the following will be :-

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

6) A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can
be adjusted to any value greater than or equal to zero. A graph is plotted between the current (i)
passing through the resistance and potential difference (V) across it. Select the correct alternative.

(A) internal resistance of battery is 5Ω

(B) emf of the battery is 20V
(C) maximum current which can be taken from the battery is 4A
(D) V- i graph can never be a straight line as shown in figure.
7) Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are
brought close together to form a parallel plate capacitor with capacitance C, the potential difference
between them is :-

(A)

(B)

(C)

(D)

8) Find out equivalent capacitance between A and B.

(A)

(B)

(C)

(D)

9) Find the charge on the capacitor C in the following circuit (in steady state):-

(A) 12 µC
(B) 14 µC
(C) 20 µC
(D) 18 µC

10) Four capacitors of capacitance 10µF and a battery of 200V are arranged as shown. How much

charge will flow through AB after the switch S is closed ?

(A) 4.5 × 10–3 C
(B) 6 × 10–3 C
(C) 1.5 × 10–3 C
(D) 3 × 10–3 C

11) In the figure below, what is the potential difference between the point A and B and between B
and C respectively in steady state :-

(A) VAB = 100 volt, VBC = 100 volt

(B) VAB = 75 volt, VBC = 25 volt
(C) VAB = 25 volt, VBC = 75 volt
(D) VAB = 50 volt, VBC = 50 volt

12) In the following circuit, the resultant capacitance between A and B is 1 µF. Then value of C is :-

(A)
µF

(B)
µF

(C)
µF

(D)
µF

13) A toy plane is flying horizontally at 10 m/s at height 80 m above the ground towards a man on
the ground as shown. When plane is 60m away horizontally from man, a ball is dropped by the plane.
Find speed of man (assumed uniform) so that he is able to catch the ball just before it strikes the

ground (g = 10 / s2) :-
(A) 10 m/s
(B) 15 m/s
(C) 5 m/s
(D) 7.5 m/s

14)

Which of the following represents uniformly accelerated motion :-

(A)

(B)

(C)

(D)

15) Two stones are thrown vertically upwards simultaneously from the same point on the ground
with initial speed u1 = 30 m/s and u2 = 50 m/s. Which of the following curve represent variation of
relative velocity of second stone with respect to first with time (t). Assume that stones do not
rebound after hitting

(A)

(B)

(C)
(D)

16) Two balls are projected from an inclined plane from point Q & P respectively as shown. Select

the incorrect statement:- (During the time both remain in air)

(A) Path of A w.r.t. B is straight line

(B) Velocity of A w.r.t. B is constant
(C) Path of A w.r.t. B is parabolic
(D) Acceleration of A w.r.t. B is zero

17)

The final position of the particle if the particle starts from x = –15 m is :

(A) 10 m
(B) 5 m
(C) – 5m
(D) –15 m

18) A particle is projected vertically upward with initial velocity 25 ms-1. During third second of its
motion, which of the following statement is correct? (g = 10 m/s2)
(A) Displacement of the particle is 30 m
(B) Distance covered by the particle is 30 m
(C) Distance covered by the particle is 2.5 m
(D) Distance covered by the particle is 5 m

19) The velocity–time graph of a linear motion is shown in figure. The displacement from the origin

till 8 sec., is (initially particle was at origin) :-

(A) 5 m
(B) 16 m
(C) 8 m
(D) 6 m

20) If a body travels half of its total path in the last second of its fall from rest under gravity. Find
the height of its fall (approximately) (take g = 9.8 m/s2).

(A) 57 metre
(B) 65 metre
(C) 50 metre
(D) 70 metre

SECTION-II

1) The ball is projected at an angle 53° from horizontal with speed 50 m/s then at the time t = t0 sec

velocity of ball will be perpendicular to its initial velocity. Then find .

2) Find the current flowing (in Amp) through the segment AB of the circuit shown in figure.

3) Each element in the finite chain of resistors shown in the figure is 1Ω. A current of 1 A flows
through the final element. The potential difference V across the input terminals of the chain is

(30+n) volt. Find n.

4) A canon ball is thrown with an initial velocity of 40m/s at an angle of 53º to the horizontal. After
some time, the ball is seen to travel at an angle of 37º below horizontal. The time elapsed(in second)
till then will be (g=10m/s2)

5) The circuit shown in the figure consists of a charged capacitor of capacity and a charge of
. At time t = 0, when the key is closed, the value of current flowing through the resistor is
'x' μA.The value of 'x' to the nearest integer is ________.

PART 2 : CHEMISTRY

SECTION-I

1) An ideal solution is formed by mixing two volatile liquids A and B. XA and XB are the mole fractions
of A and B respectively in the solution and YA and YB are the mole fractions of A and B respectively in

the vapour phase. A plot of along y-axis against along x-axis gives a straight line. What is the
slope of the straight line?

(A)

(B)

(C)
(D)

2) An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-
volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same
temperature?

(A) 64 torr
(B) 80 torr
(C) 100 torr
(D) 400 torr

3) Which one of the following statements is false ?

The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is BaCl2 >
(A)
KCl > CH3COOH > sucrose
(B) Isotonic solutions are those solutions which have the same osmotic pressure
 Raoult's law states that the vapour pressure of a component over a solution is proportional to its
(C)
mole fraction in liquid state
Two sucrose solutions of same molality prepared in different solvents will have the same
(D)
freezing point depression

4) The boiling point of an aqueous solution of a non volatile solute is 100.15 °C. What is the freezing
point of an aqueous solution obtained by diluting the above solution with an equal volume of water ?
The values of Kb and Kf for water are 0.5 and 1.8 K molality–1 :

(A) –0.54 °C
(B) –0.512 °C
(C) –0.27 °C
(D) –1.86 °C

5) What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non-volatile
solute (molar mass 256 g mol–1) and the decrease in freezing point is 0.40 K ?

(A) 5.12 K kg mol–1

(B) 4.43 K kg mol–1
(C) 1.86 K kg mol–1
(D) 3.72 K kg mol–1

6) The vapour pressure of ethanol and methanol are 42.0 mm and 88.5 mm Hg respectively. An ideal
solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The
mole fraction of methanol in the vapour is :

(A) 0.467
(B) 0.502
(C) 0.513
(D) 0.556

7) What would be the maximum number of emission lines for atomic hydrogen that you would expect
to see with naked eye if the only electronic energy levels involved are those as shown in figure :-

(A) 15
(B) 10
(C) 6
(D) 4

8) What is the ratio of the De-Broglie wave lengths for electrons accelerated through 200 volts
and 50 volts :-
(A) 1 : 2
(B) 2 : 1
(C) 3 : 10
(D) 10 : 3

9) (i) 26Fe54 , 26Fe56, 26Fe57, 26Fe58 (a) Isotopes

(ii) 1H3, 2He3 (b) Isotones
76 77
(iii) 32Ge , 33As (c) Isodiaphers
235 231
(iv) 92U , 90Th (d) Isobars
1 2 3
(v) 1H , 1D , 1T
Match the above correct terms:-

(A) [(i), - a], [(ii) - d], [(iii) - b], [(iv) - c], [(v) - a]
(B) [(i) - a] [(ii) - d], [(iii) - d] [(iv) - c] [v - a]
(C) [v -a] [(iv) - c]. [(iii) - d] [(ii) - b] [(i) - a]
(D) None of them

10) For H atom, the energy required for the removal of electron from various sub-shells is given as

under:- The order of the energies would be :-

(A) E1 > E2 > E3

(B) E3 > E2 > E1
(C) E1 = E2 = E3
(D) None of these

11) Choose the correct statements of the following

(a) Copper can displace iron from FeSO4 solution
(b) Iron can displace copper from CuSO4 solution
(c) Silver can displace copper from CuSO4 solution
(d) Iron can displace silver from AgNO3 solution

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

12)

The e.m.f. of following cell at 300K, when

[Cl–] = [Br2] = [Br–] = 0.01M and Cl2(g) at 1 atm.

Cl2(g) + 2Br–(aq.) —→ 2Cl–(aq) + Br2(aq.) E° = 0.29V

(A) +0.29V
(B) 0.23V
(C) –0.35V
(D) 0.35V

13)

Calculate the equilibrium constant for

Hg22+ —→ Hg + Hg2+
Given :

(log2 = 0.3, log3 = 0.4) (100.7 5)

(A) 5 × 10–3
(B) 6 × 10–3
(C) 7 × 10–3
(D) None of these

14) A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of
2.768 × 10–3 ohm–1 cm–1. If its resistance is 82.4 ohm at 25°C, the cell constant is :-

(A) 0.382 cm–1

(B) 0.2281 cm–1
(C) 0.581 cm–1
(D) 0.928 cm–1

15) A cell is constituted as follows

Pt|H2 (1 atm) |HA1| |HA2| H2 (1 atm) | Pt
The pH of two acids solutions HA1 and HA2 are 5 and 3 respectively. The emf of the cell is -

(A) 0.059 V
(B) 0.0295 V
(C) 0.118 V
(D) – 0.118 V

16) Which of the following statements is incorrect ?

(A) The second ionization energy of sulphur is greater than that of chlorine
(B) The third ionization energy of phosphorus is greater than that of magnesium
(C) The first ionization energy of aluminium is less than as that of gallium
(D) The second ionization energy for boron is greater than that of carbon

17) Select correct statement :-

(A) La and Ac belong to f-block
(B) An element having atomic No.31 belongs to 3rd period.
(C) General outermost shell e– configuration of d-block element is ns1-2(n-1)d1-10
(D) All actinoids are man-made elemetns.

18) Of the following, which one is correct statement ?

(A) Ionic radius of a metal is same as its atomic radius

(B) The ionic radius of a metal is greater than its atomic radius
(C) The atomic radius of a non-metal is more than its ionic radius
(D) The ionic radius of a metal is less than its atomic radius.

19) Which is mismatched regarding the position of the element as given below ?

(A) (Z = 89) – f block, 6th period

(B) (Z = 100) – f block, 7th period
(C) (Z = 115) – d block, 7th period
(D) Both (1) and (3)

20) The correct order of acidic strength is :

(A) Cl2O7 > SO3 > P4O10

(B) CO2 > N2O5 > SO3
(C) Na2O > MgO > Al2O3
(D) K2O > CaO > MgO

SECTION-II

1) Choose the number of correct statements(s) from the following:

(a) Ist ionisation potential of ‘B’ is higher than that of ‘Be’
(b) Electron affinity of ‘O’ is higher than that of ‘Be’
(c) [Ar] 4s23d3 is the electronic configuration of Mn2+
(d) IInd ionisation potential of Na > Ist ionisation potential of Na.
(e) Ist ionisation potential of N > IInd ionisation potential of N.
(f) Electronegativity of Cl > Eelectronegativity of F.
(g) C → C2+ change is called IInd ionisation potential of carbon.
(h) Energy is required to convert He → He+
(i) Conversion of O → O2– is exothermic.

2) Number of pairs for which size of 1st element is bigger than 2nd.
[Li+, Mg2+] [Ne, Ar] [Be, B] [O2–, F–]
[Cl–, Cl] [F–, H–] [Na+, Mg+2]

3) The magnitude of potential energy of electron in nth excited state of He+ ion is times the kinetic
energy of electron in the first excited state of Li2+ ion. Find "n"

4) What is the sum of van’t Hoff factor for Ca(NO3)2 and K4[Fe(CN)6] if they are 60% and 70%
dissociated respectively?

5) Calculate Ecell for following

[Given ]
Fill your answer by multiplying it with 100.

PART 3 : MATHEMATICS

SECTION-I

1) If the product of all solutions of equation can be expressed in the lowest

form as then find the value of (m + n)

(A) 1204
(B) 1
(C) 24
(D) 4041

2) If are roots of the equation ,

then, the value of is

(A)

(B)

(C)
(D)

3) Solve the inequality

(A) x ∈ (0, ∞)
(B) x ∈ (1, ∞)
(C) x ∈ (0, 1) ∪ x ∈ (2, ∞)
(D) x ∈ R

4) If , then value of xyz2 is equal to

(A) 0
(B) 1
(C) (a + b + c)
(D) 2

5) Let f(x) = x2 – 2ax + a – 2 and g(x) = . If the complete set of real values of 'a'
for which ∀ x ∈ R is (k1, k2) then the value of (10k1 + 3k2) is equal to
[Note : [k] denotes greatest integer less than or equal to k.]

(A) 35
(B) 27
(C) 20
(D) 8

6) The sum of roots of the equation,

x + 1 – 2log2(3 + 2x) + 2log4(10 – 2–x) = 0 is

(A) log214
(B) log212
(C) log213
(D) log211

7) It ax + b sec(tan–1 x) = c and ay + b sec(tan–1y) = c, then is equal to

(A)

(B)

(C)

(D)

8) Solution set of the inequality is

(A) (0, 4)
(B) (–∞, 1)
(C) (4, ∞)
(D) (2, 4)

9) Number of possible ordered pair(s) (x, y) satisfying the equation cot–1x = –ey is

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

10) The number of value(s) of n such that logn(31443) is an integer is (n ∈ N)

(A) 1
(B) 5
(C) 8
(D) 13

11) The number of solutions of the equation,

for x ∈ [–1, 1], and [x] denotes the greatest integer less than or equal to x, is

(A) 2
(B) 0
(C) 4
(D) Infinite

12) The number of solution(s) of the equation is

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

13) The sum of the infinite series

is

(A)

(B)

(C)

(D) None of these

14) Number of integers satisfying inequality,

is

(A) 5
(B) 6
(C) 7
(D) 8
15) A value of x satisfying the equation sin (cot–1(1 + x)) = cos(tan–1 x), is

(A)

(B) –1
(C) 0

(D)

16) Solve for ‘x’

(A)
( – ∞, ]

(B)

(C)

(D)

17) is equal to

(A)

(B)

(C)

(D)

18) if

(A)
(B)
(C)
(D) x > 1

19) Number of real solution(s) of the equation sin–1(sinx) = 1 – |x| is

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

20) is equal to

(A)

(B)

(C)

(D)

SECTION-II

1) Let x1, x2, x3 are the solutions of the equation , then

is

2) If are the positive roots of the equation . If the value

of is equal to k, then is

3) For positive real numbers x and y, let ƒ(x, y) = . If the sum of the solutions of the equation

4096ƒ(ƒ(x, x), x) = x13 can be expressed as (where m, n are coprime numbers), then (m – 10n) is

4) If , then the value of λ is

5) Let , then the value of , 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 A D C B A D B D A C D C C D C C C A A

SECTION-II

Q. 21 22 23 24 25
A. 4 1 4 5 2

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

SECTION-II

Q. 46 47 48 49 50
A. 3 5 5 6 3

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

SECTION-II

Q. 71 72 73 74 75
A. 1 3 1 6 7
 SOLUTIONS

PART 1 : PHYSICS

1)

2)

5R = 4 + 4R
R = 4Ω

3) i = i1 + i2

= neAV + n(2e)A

= neAV

4) V = Eeq = = 14 volt

5)

If bulbs are identical then their resistances are equal.

Power P = I2R

current through A, B, is IB = IA =
If emf of battery is V0

PA = PB =

Current through C, IC =

Current through E, D IE = ID =

PC = IC2R =
PE = PD =
Power in C is maximum so maximum brightness.

6) V = ε – ir
ε = 10V

r = 5Ω

imix =

7)

Within the plates electric fields due to charges Q1 and Q2 are

E1 = and E2 =
As these fields are in opposite directions and Q1 > Q2, the net electric field within the plates is

E = E1 – E2 =

Hence V = Ed = (Q1 – Q2) =

8) Let . Equivalent circuit

Alternative Method : Let charge distribution on plates as shown :

Potential of 1 and 4 is same

 y = 2x

9)
q = CV = 2 × 9 = 18 μC

10) When switch is open =

qi = Ceq. V = × 200 = 1500 µC

When switch is closed Ceq. = 30 µF
qf = 30 × 200 = 6000 µC
Charge flow through AB = qf – qi = 4500 µC = 4.5 × 10–3 C

11)

VAB = ; VBC =

12) 12 µF and 6µF are in series and again are in parallel with 4µF. Therefore resultant of these

three will be
This equivalent system is series with 1µF, its equivalent capacitance.

= ....(1)
Equivalent of 8µF, 2µF and

2µF = ....(2)
(1) and (2) are in parallel and are in series with C.

and Ceq = 1 =

⇒ +C= C ⇒ C–C= ⇒ C= × = µF

13)
80 = ⇒ t = 4 sec
10 × 4 + v × 4 = 60 ⇒ v = 5 m/s

14)

t2 = ⇒ x = a + bt2
⇒ constant accelerated motion

15)

V1 = u1 – gt
V2 = u2 – gt
Now (Tf is time of flight)
So, till
V2 – V1 = (u2 – u1) = constant then after 1 will be on ground and 2 will falling down.

16)

Since acceleration of both is g downward hence relative acceleration will be zero and relative
velocity will remains constant & path will be straight line as seen by other.

17)

xf = – 15 + Area

= – 15 + 2 +
= –15 + 10 = – 5m

18)

u = 25
Velocity becomes zero at 2.5 sec.
So distance travelled in 3rd second
= |S2.5 – S2| × 2

S2.5 = 25 × 2.5 – × 10 × (2.5)2

= 31.25
S2 = 25 × 2 – 5 × 22 = 50 – 20 = 30
So distance=(31.25–30)×2=1.25×2 = 2.5m

19)

Displacement = Area of v-t graph with time axis.

20) Question Explanation:
We need to find height of fall when a body travels half of its total path in the last second of its
fall from rest under gravity.

Concept:
This question is based on motion under gravity.

Solution:

∴ h = 10 (2n – 1) ... (i)

h = 5n2 ... (ii)

21)

Velocity is perpendicular to initial velocity after time

22)
From A to B

23) Number the resistors, starting with the last element in the chain. As current of 1 A flows
through the first resistor, a current of 1 A has to flow through the second one as well, thus
there is a potential different (p.d) of 1 V across each resistor. As a consequence, the p.d.
across the third resistor is (1 + 1) = 2V, and the current flowing through it must be 2A. The
current flowing through the next resistor is (1 + 2) = 3A. The current in the fifth resistor can
 be determined using the p.d. (2+3) = 5 V across the resistors with currents of 2 and 3 A,
respectively, flowing through them and so on, as shown in figure.

24)

=18m/s

-18=32-10t
10t=50
t=5s

25)

PART 2 : CHEMISTRY

26) ⇒

YA =

y = C + m.x

27) = xsolute
Po – 80 = 0.2 Po
Po = 100 torr

28) For different solvents, Kf will be different.

29) ΔTb = Kb × m
0.15 = 0.5 × m
m = 0.3
∵ due to dilution
m1v1 = m2v2
0.3 × v = m2 × 2V
m2 = 0.15
ΔTf = Kf × m2
= 1.8 × 0.15
= 0.27°C
so F.P. of solution = –0.27°C

30)

31) Ethanol Methanol

(P) (B)
YB = ?

YB =

YB = = 0.513

32) Only 4 lines in visible region (Balmer)

6 → 2, 5 → 2, 4 → 2, 3 → 2

33)

V1 = 200

V2 = 50
If we are concern about e– then mass and charge (e) are same
so ⇒ =1:2

34) Isotopes = Same atomic no.

Isotones = Same no. of neutrons
Isodiaphers = Same (N–P)
Isobars = Same mass no.

35) Energy of single e– species depends upon 'n' only

So energy of 3s = 3p = 3d

36)

A. Question Explanation:

The question asks you to identify which of the given metal displacement reactions are
spontaneous, based on the reactivity series of metals.

B. Concept: Electrochemical series

C. Solution -

A. Reactivity Series of Metals: A list of metals arranged in order of their reactivity. A more
reactive metal can displace a less reactive metal from its salt solution.
B. Displacement Reaction: A reaction in which a more reactive metal displaces a less reactive
metal from its salt solution.
C. Electrochemical Series: The electrochemical series is a more precise list, but for this
question, a general understanding of metal reactivity is sufficient.
D. General reactivity order from most to least reactive. K, Na, Ca, Mg, Al, Zn, Fe, Pb, H, Cu, Ag,
Au.

D. Final Answer:

Let's analyze each statement:

A. (a) Copper can displace iron from FeSO4 solution: Copper is less reactive than iron, so
this is incorrect.
B. (b) Iron can displace copper from CuSO4 solution: Iron is more reactive than copper, so
this is correct.
C. (c) Silver can displace copper from CuSO4 solution: Silver is less reactive than copper, so
this is incorrect.
D. (d) Iron can displace silver from AgNO3 solution: Iron is more reactive than silver, so this
is correct.

Therefore, the correct statements are (b) and (d).

The correct answer is 3. b and d.

E. Question Level - Medium

 37)

At 300 K

E = E° –

E = 0.29 –

0.29 + 0.06 ⇒ 0.35 V

38) + 2e– → 2Hg ...(1)

+ –
2e → Hg ...(2)
(1) – (2)
→ Hg + Hg+2 E° = 0.78 – 0.85
⇒ – 0.07 V

– 0.07 = 0.03 log K

–2.3 = log K
K = 10–2.3
⇒ 10–3 × 10+0.7 = 5 × 10–3

39) k=

2.768 × 10–3 =

constant = 82.4 × 2.768 × 10–3 = 0.2281

40) Anode :

Cathode :
________________________________

________________________________
 = 0.118 volt

41) (i)

(ii)

(iii)
(size almost same due to transition contraction)
Al < Ga (IE)
(iv)

42) General outermost shell e– configuration of d-block is ns1–2 (n–1)d1-10

43) A cation is smaller than it's parent atom.

44) (Z = 89) d-block, 7th period

(Z = 115) p-block, 7th period

45)

Acidic character of non–metallic oxides increases along a period (from left to right) of periodic
table.
Cl2O7 > SO3 > P4O10

46) b,d,h are correct

47)
 1, 3, 4 , 5 , 7

48) |P.E.| = 2 × |T.E|

K.E = | –T.E.| = 13.6 ×

2 × 13.6 ×
(n + 1)2 = 36
n+1=6⇒n=5

49)

50) Concentration =0

= 0.03

PART 3 : MATHEMATICS

51)
⇒ log2021 2020 + log2021x – 1 = logx 2021 . log2021 2020

Suppose log2021x = t, then logx 2021 =

Then, log2021 2020 + t – 1 = log2021 2020

t . log2021 2020 + t2 – t – log2021 2020 = 0
⇒ (t – 1) log2021 (2020) + t(t – 1) = 0
⇒ (t – 1) (log20212020 + t) = 0
⇒ t = 1 or t = –log2021 2020
⇒ log2021x = 1 or log2021x = log2021 (2020)–1
⇒ x = 2021 or x =
Product of solutions

m + n = 2021 + 2020 = 4041

52) We have,
and

53) Reducing the fractions on the left hand side to a common denominator, we find

The numerator of the last expression is positive since we have

⇒ log2x(log2x – 1) > 0,
which is fulfilled for x > 2 and 0 < x < 1.

54) = k (Let)
2
⇒ logx + logy + logz
= k(a+b+c)–k(3a+5b+7c)+k(2a+4b+6c)=0
⇒ logxyz2 = 0
⇒ xyz2 = 1

55) We have g(x) = =2+

As, sin–1

∴ = – 2, – 1, 0, 1.
Range of g(x) = {0, 1, 2, 3} for <0∀x∈R
⇒ f(0) < 0 and f(3) < 0
Now, f(0) < 0 ⇒ a – 2 < 0 ⇒ a < 2
and f(3) < 0 ⇒ 9– 6a + a – 2 < 0

a>

∴ a∈ .

Hence, k1 = , k2 = 2
∴ (10k1 + 3k2) = 14 + 6 = 20. Ans.

56)

Product →

57)

Let tan–1x = α and tan–1y = β then a tan α + b sec α = c and a tan β + b sec β = c
Obviously a tanθ + b secθ = c has roots tan α and tan β
⇒ (a2 – b2) tan2θ – 2 ac tan θ + c2 – b2 = 0

∴ tanα + tanβ = and tanα tan β =

So

58)

59) LHS = +ve & RHS = –ve therefore solution is not possible.
60)

61) Given equation

Now, is defined if

⇒ ...........(1)

and is defined if

⇒ ....(2)
So, from (1) and (2) we can conclude

Case - I if

⇒ 0 + π = x2
⇒ x2 = π

but
⇒ No value of 'x'

Case - II if

⇒
⇒ x2 = π

but
⇒ No value of 'x'
So, number of solutions of the equation is zero.

62)
63) Let

64)

∴
∴

⇒ 6 integers

65) sin (cot–1 (1 + x)) = cos (tan–1 x)

x2 + 2 + 2x = x2 + 1
2x = –1

66)
or |1 – y| – 2 = |3 – y| where y =
i.e. |y – 1| – 2 = |y – 3|
case (i) y < 1
1–y–2=3–y
∴ NOT POSSIBLE
case (ii) 1 y < 3
y–1–2 =3–y
2y = 6 ⇒ y = 3
∴ NOT POSSIBLE
case (iii) y 3
y – 1 – 2 = y – 3 always true
∴ the solution set for y is [3 , ∞)
67)

=
Now, using the property of cosA cosB + sinA sin B →

68)
Since we get
So we have to solve
and simultaneously

Now,
So
and

which is the required solution.

69)

70)
71) RHS = x

⇒ x = 1 or (log2x)2 + 5(log2x) + 4 = 0

⇒ x1 = 1, &

⇒ x1 + x2 + x3

72)

73) = x13
⇒
⇒ Take log on base 2
⇒ log2x = 1, 3, –4

∴ ∴ Sum =

74)

the value of λ is '6'

75)
at which are
angles of a triangle.
So, x = tanA, y = tanB, z = –tanC.

= 8 – 1 = 7.