16-02-2025

9610ZJA801239240027 JA

PART 1 : PHYSICS

SECTION-I (i)

1) In the figure shown, mass on both side at t = 0 is M kg. Water in left side leaks out at constant
rate µ kg/s. Velocity of water flow in downward direction relative to cylindrical tank is Ve m/s.

Acceleration at any time t will be:

(A)

(B)

(C)

(D)

2) If surface tension of liquid air boundary is Sℓa, surface tension of solid air surface is SSa and
surface tension of liquid solid boundary is SℓS, the contact angle θ is given by

(A)

(B)

(C)

(D) None of these

3) Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The
expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate

of fractional change in density is constant. The velocity v of any point on the surface of the
expanding sphere is proportional to :

(A) R3
(B)

(C) R
(D) R2/3

4) A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational
acceleration. On an inclined plane inside the rocket, making an angle θ with the horizontal, a point
object of mass m is kept. The minimum coefficient of friction μmin between the mass and the inclined
surface such that the mass does not move is:

(A) 2 tan θ
(B) 3 tan θ
(C) tan θ
(D) tan 2θ

SECTION-I (ii)

1) A small hole of area a is at the bottom of a container of area A. The liquid is filled up to height h

from base. As liquid comes out then

(A)
level of liquid in container falls at rate of
magnitude of acceleration of top surface of liquid is
(B)

acceleration of top surface of liquid is

(C)

(D) None of these

2) A homogeneous disk of radius r and mass m is mounted on an axle OG of length L and negligible
mass The axle is pivoted at the fixed point O, and the disk is constrained to roll on a horizontal floor.
Knowing that the 'disk' rotates counterclockwise at the rate about the axis OG,
 Choose the correct option(s)

(A)
The angular velocity of the disk
The angular velocity of the disc
(B)

The X component of angular momentum of disc about 'O' is

(C)

(D) The Z component of angular momentum of disc about 'O' is zero

3) Two independent harmonic oscillators of equal mass are oscillating about the origin with angular
frequencies ω1 and ω2 and have total energies E1 and E2, respectively. The variations of their

momenta p with positions x are shown in the figures. If and , then the correct
equation(s) is(are) :-

(A)

(B)

(C)

(D)

4) A particle of mass m is at rest in a train moving with constant velocity with respect to ground.
Now the particle is accelerated by a constant force F0 acting along the direction of motion of train
for time t0. A girl in the train and a boy on the ground measure the work done by this force. Which of
the following are incorrect?

(A) Both will measure the same work

(B) Boy will measure higher value than the girl
(C) Girl will measure higher value than the boy
(D) Data are insufficient for the measurement of work done by the force F0

5) A spherical surface separates air & medium for which μ = 1.615 for violet and μ = 1.600 for red
color. A paraxial beam parallel to optical axis is incident on the surface as shown. The distance

between point of convergence for violet and red color is Δf.

(A) Δf = 0.40 cm
(B) Point of convergence for red is closer to optical centre than that for violet.
(C) Point of convergence for violet is closer to optical centre than that for red.
(D) Δf = 0.84 cm.

6) Two metallic sphere A and B are made of same material and have got identical surface finish. The
mass of sphere A is four times that of B. Both the spheres are heated to the same temperature and
placed in a room having lower temperature but thermally insulated from each other.

(A) The ratio of heat loss of A to that of B is 24/3.

(B) The ratio of heat loss of A to that of B is 22/3.
(C) The ratio of the initial rate of cooling of A to that of B is 2–2/3.
(D) The ratio of the initial rate of cooling of A to that of B is 2–4/3.

SECTION-II (i)

Common Content for Question No. 1 to 2

The two wires shown in figure are made of the same material which has a breaking stress of 8 × 108
N/m2. The area of the cross–section of the upper wire is 0.006 cm2 and that of the lower wire is
0.003 cm2. The mass m1 = 10 kg, m2 = 20 kg and the hanger is light.

1) For m1 = 10 kg, m2 = 36 kg, find the maximum load (in kg) that can be put on the hanger without
breaking any wire -

2) Find the maximum load (in kg) that can be put on the hunger without breaking any wire.
Common Content for Question No. 3 to 4

A particle is moving on given curve path. At certain instant its velocity and acceleration shown in
figure.

3) Radius of curvature (in m) at point P is :-

4) Rate of change of speed (in m/s2) of particle at P is :-

Common Content for Question No. 5 to 6

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a
speed . The projectile at the highest point of its trajectory splits into two equal parts. One part
falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the
splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity
g = 10 m/s2.

5) The value of t is _________.

6) The value of x is _________.

SECTION-II (ii)

1) A large tank is filled with water (density = 103kg/m3). A small hole is made at a depth 10m below
water surface. The range of water issuing out of the hole is R on ground. What extra pressure (in
atm) must be applied on the water surface so that the range becomes 2R (take 1 atm = 105Pa and g

= 10m/s2)

2) The emissivity of tungsten is approximately 0.35. A tungsten sphere 1 cm in radius is suspended

within a large evacuated enclosure whose walls are at 300 K. What power input is required to
maintain the sphere at a temperature of 3000 K if heat conduction along the supports is neglected ?

Express your answer in kW after rounding off the nearest integer. Take σ = × 10–8 S.I. units and π
= .

3) In a cyclic thermodynamic process an ideal gas is heated at constant volume while its internal
energy is increased by 350 J. It next expands adiabatically until it returns to its original pressure.
Finally, the gas is returned isobarically to its original state when 60.0 J of work are performed on the
gas while its internal energy decreases by 200 J. What is the change in internal energy during the
adiabatic expansion (in S.I. unit)? Assume the gas is ideal with cp = 20.8 J/mol/K and cv = 12.5
J/mol/K. If your answer is X. Give the value of (X + 500).

PART 2 : CHEMISTRY

SECTION-I (i)

1) Identify correct nucleophilicity order

(A) i > ii > iii

(B) ii > iii > i
(C) i > iii > ii
(D) iii > i > ii

2) Correct order of stability for given intermediates ?

(A) II > I > III

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

3) Two elements A and B form two different sets of compounds A2B3 and A3B4. 2.02 kg of A2B3 has
1.806 × 1025 atoms of B and 1.44 kg of A3B4 has 9.03 × 1024 atoms of A. What will be gram molecular
mass (GMM) of another compound AB formed from combination of A and B.

(A) 75
(B) 86
(C) 94
(D) 97
4) For 1 mole substance variation of absolute entropy with temperature is given by following graph.
What is molar enthalpy of vapourisation

(A) 2 kJ
(B) 5 kJ
(C) 20 kJ
(D) 10 kJ

SECTION-I (ii)

1) The incorrect order of ionization energies of F-, Cl-, F and Cl is :

(A) Cl < F < Cl- < F-

(B) Cl- < F- < Cl < F
(C) F- < Cl- < Cl < F
(D) Cl- < Cl < F- < F

2) For which of the following conditions will favour forward reaction ? (Reactant
absorb heat to form the products)

(A) Increase in temperature

(B) Adding more PCl5
(C) Adding more PCl3
(D) Decrease in pressure

3) Which of the following do not exist ?

(A) He2
(B) HeH+
(C) Be2
(D) HeH−

4) Which of the following is/are correct statement(s).

(A) Increasing covalent character : NaCl < MgCl2 < AlCl3
(B) Increasing covalent character : LiF < LiCl < LiBr < LiI.
(C) Increasing polarizability : F– < Cl– < Br – < I–
(D) Decreasing ionic nature : MCl3 > MCl2 > MCl

5)
Mark out the incorrect statement:

(A)

(B)

Both H3PO4 and are more acidic than

(C)

(D) Only is amphiprotic anion in the solution.

6) The incorrect set of quantum numbers of unpaired electron of chlorine is

(A) 3, 1, 1, +1/2
(B) 3, 2, -2, +1/2
(C) 4, 2, -1, -1/2
(D) 4, 1, 0, 0

SECTION-II (i)

Common Content for Question No. 1 to 2

(1) (2) (3) (4)

(5) (6) (7)

(8) (9)
From the above compounds answer the following questions :

1) Number of heterocyclic aromatic compound in the above molecules

2) Number of aromatic compounds in the above compounds :

Common Content for Question No. 3 to 4

24 gm pure sample of magnesium is burned in air to form magnesium oxide and magnesium nitride.
When products are treated with excess of H2O, 3.4 gm of gaseous NH3 is generated according to
given reactions.
Mg + O2 MgO
Mg + N2 Mg3N2
Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3
MgO + H2O Mg(OH)2

3)

Calculate the amount of magnesium oxide (in gm) in products when magnesium is burnt in air :-

4) Calculate the amount of Mg(OH)2 (in gm) produced in above reactions.

Common Content for Question No. 5 to 6

In photo electric experiments, photon with definite energy hv collides with an electron close to the
surface of metal target cathode and transfers its energy to it. The relation is
hv = hv0 + kinetic energy
where v0 = Threshold frequency associated with the ejection of electron.
h = plank's constant (6.6 ×10–34 J. sec)

5) The work function of some metals is listed below

The number of metals in the list which will show photoelectric effect when light of 6000 Å
wavelength falls on the metal is

6) If one photon has 25 eV energy and work function of material is 7 eV then the value of stopping
potential required to stop photo current will be –

SECTION-II (ii)
1)
How many statement are true?
a) (I) and (III) are modest Bronsted bases whereas (II) is not
b) In (III) Na is more basic than Nb
c) When (II) is protonated in the presence of a strong acid, protonation occurs at C-2
d) All the nitrogen present in (I), (II), and (III) is 2 hybridised

2) If ZI is the number of electrons in g sub shell, ZII is the azimuthal quantum number of an atomic
orbital of g subshell, ZIII is the minimum principal quantum number of an atomic orbital of g sub
shell, ZIV is the number of angular nodes in an atomic orbital of the g sub shell and ZV is the number
of elements expected in the g block of the periodic table, then the value of ZI + ZII + ZIII + ZIV + ZV is

'Y' find the value of .

3) The percent yield for the following reaction carried out in carbon tetrachloride (CCI4) solution is
80% (the remaining amount of reactants are left over unreacted in the reaction)
Br2 + CI2 → 2BrCI
If the amount of BrCI formed from the reaction of 0.025 mol Br2 and 0.025 mol CI2 is 'x'. If the
amount of Br2 left unreacted is 'y'. Find x/y.

PART 3 : MATHEMATICS

SECTION-I (i)

1) In the expansion of (1 + x)2 (1 + y)3 (1 + z)4 ( 1 + ω)5, the sum of the coefficient of the terms of
degree 12 is:

(A) 61
(B) 71
(C) 81
(D) 91

2) The remainder when (2021)2022 + (2022)2021 is divided by 7 is

(A) 0
(B) 1
(C) 2
(D) 6

3) cot 12° cot 102° + cot 102° cot 66° + cot 66° cot 12° is
(A)
(B) 1
(C)
(D) 2

4) Number of solutions to the equation in is

(A) 2
(B) 3
(C) 4
(D) 5

SECTION-I (ii)

1) Let G be a circle of radius R > 0. Let G1, G2,…,Gn be n circles of equal radius r > 0. Suppose each
of the n circles G1, G2,…,Gn touches the circle G externally. Also, for I = 1, 2,…, n-1, the circle Gi
touches Gi+1 externally, and Gn touches G1 externally. Then, which of the following statements is/are
TRUE?

(A) If n = 4, then ( -1) r < R

(B) If n = 5, then r < R
(C) If n = 8, then ( -1) r < R
(D) If n = 12, then ( + 1) r > R

2) Point M moved on the circle . Then it broke away from it and moving along
a tangent to the circle, cuts the x-axis at the point (-2, 0). The coordinates of a point on the circle at
which the moving point broke away is :

(A)

(B)

(C)
(D)

3) TP and TQ are tangents to parabola y2 = 4x and normals at P and Q intersect at a point R

on the curve. The locus of the centre of the circle circumscribing ΔTPQ is a parabola whose

(A) vertex is (1, 0).

(B)
foot of directrix is

(C)
length of latus-rectum is
(D)
focus is

4) A parabola touching both the coordinate axes lies in the first quadrant and has its focus at (3, 4)

(A) Equation of tangent at vertex is

(B) Equation of directrix is
Latus rectum of the parabola is
(C)

(D) Equation of axis is

5) Let from the point with abscissa 25, two tangents are drawn to the ellipse 24x2 + 25y2 = 600 with

foci at S1 and S2. The points of contact of tangents are A and B. If the distance of A from S1 is
units, then

(A)
The distance of A from S2 is units
(B) The distance of B from S1 is 5 units
(C) The distance of B from S2 is 5 units

(D)
The distance of A from the directrix corresponding to S2 is units.

6) In the given figure, the radii of the fixed circle-1 and the fixed circle-2 are 3 and 10 respectively
and the distance between their centres is 5. The moving circle-3 is externally tangent to the circle-1
and internally tangent to the circle-2. The path traced by the centre of the circle-3 is a conic. then

choose the correct option(s).

(A)
eccentricity =

(B)
eccentricity =
(C) length of major axis is 13
(D) if the conic is closed curve then its area is 39π

SECTION-II (i)

Common Content for Question No. 1 to 2

If ƒ(x) = (x – 1)(x – 3)(x – 5)(x – 7) + 15 = (x – α)(x – β)(ax2 + bx + c) ∀ x ∈ R (where α, β, a, b, c are
rational number), then

1) The sum of roots of ƒ (x) = 0 is -

2) is -

Common Content for Question No. 3 to 4

Consider two different A.P.'s with common difference 'D' and 'd' respectively such that sum of three
consecutive terms of each A.P. is 15 and D – d = 1, d > 0. If A and B denote the value of product of

these three terms of each A.P. respectively such that .

3) is equal to-

4) Value of |A – B| is-

Common Content for Question No. 5 to 6

A variable straight line 'L' is drawn through O(0, 0) to meet this lines L1 : y – x – 10 = 0 and L2 : y – x
– 20 = 0 at the points A and B respectively.

5) A point P is taken on 'L' such that , then locus of P is (y – x)2 = a. Then a

is

6) A point P is taken on 'L' such that then the locus of P is 3x – 3y + k = 0. Then k is

SECTION-II (ii)

1) Number of divisors of 1024 × (225)2 × (25)7 which are of the form 3k + 1 (where k is a whole
number) is equal to -

2) Number of ways in which 6 distinct objects can be kept into two identical boxes so that no box
remains empty is

3) The number of values x lying in the interval satisfying the equation

1 + cos10x cos6x = 2cos2 8x + sin2 8x is :
 ANSWER KEYS

PART 1 : PHYSICS

SECTION-I (i)

Q. 1 2 3 4
A. A C C C

SECTION-I (ii)

Q. 5 6 7 8 9 10
A. A,B A,C,D B,D A,C A,C A,C

SECTION-II (i)

Q. 11 12 13 14 15 16
A. 2.00 14.00 5.00 8.65 to 8.67 0.50 7.50

SECTION-II (ii)

Q. 17 18 19
A. 3 2 350

PART 2 : CHEMISTRY

SECTION-I (i)

Q. 20 21 22 23
A. A B B C

SECTION-I (ii)

Q. 24 25 26 27 28 29
A. A,B,D A,B,D A,C,D A,B,C B,D B,C,D

SECTION-II (i)

Q. 30 31 32 33 34 35
A. 4.00 6.00 28.00 58.00 0.00 18.00

SECTION-II (ii)

Q. 36 37 38
A. 3 7 10

PART 3 : MATHEMATICS
 SECTION-I (i)

Q. 39 40 41 42
A. D A B A

SECTION-I (ii)

Q. 43 44 45 46 47 48
A. C,D B,C A,B,D A,C B,C,D A,C,D

SECTION-II (i)

Q. 49 50 51 52 53 54
A. 16.00 4.00 2.00 15.00 80.00 40.00

SECTION-II (ii)

Q. 55 56 57
A. 60 31 0
 SOLUTIONS

PART 1 : PHYSICS

1)

....(1)
Mg – T = Ma ....(2)

2)
Ssℓ + Sℓa cosθ = Ssa

3)

Density of sphere is

Since, = constant

⇒ = constant (Let K)

Velocity v of any point on the surface of sphere is equal to (rate of change of radius of
outer layer).
4)
Free body diagram is shown in the figure.
Psuedo force on the object is 2 mg in the downward direction as seen from the rocket frame of
reference.
Normal force in the direction perpendicular to the inclined plane.
N = 3 mg cos(θ)
Maximum frictional force in the direction along the plane,
Fr = mmN
∴ Fr = 3 mm mg cos(q)
This should balance the force on the block along the downward direction of plane.
3 mg sin(θ) = 3mm mg cos(q)
mm = tan(θ)

5) Applying Bernoulli’s equation between P and Q

.....(i)
From equation of continuity,

Acceleration of top layer

6)

7)

8) Displacement would be different for both observers.

9)

= 0.40 cm
10) m ∝ V

⇒
e & σ is same for both

11) Maximum possible tension in upper wire= 480 N

Maximum possible tension in lower wire= 240 N

12)

Maximum possible tension in upper wire= 480 N

Maximum possible tension in lower wire= 240 N

13)

14) Rate of charge of speed = at

at = 10 cos 30º =

15)
Range R = =5m

Time of flight T = = 1 sec.

Time of motion of one part falling vertically downwards is = 0.5 sec =

⇒ Time of motion of another part,

t = = 0.5 sec.
From momentum conservation
⇒ Pi = Pf
2m × 5 = m × v
v = 10 m/s

Displacement of other part in 0.5 sec in horizontal direction =

= 10 × 0.5 = 5 m = R
Total distance of second part from point 'O' is,

x=
x = 7.5 m
⇒ t = 0.5 sec.
16)

Range R = =5m

Time of flight T = = 1 sec.

Time of motion of one part falling vertically downwards is = 0.5 sec =

⇒ Time of motion of another part,

t = = 0.5 sec.
From momentum conservation
⇒ Pi = Pf
2m × 5 = m × v
v = 10 m/s

Displacement of other part in 0.5 sec in horizontal direction =

= 10 × 0.5 = 5 m = R
Total distance of second part from point 'O' is,

x=
x = 7.5 m
⇒ t = 0.5 sec.

17) Range doubles when velocity of efflux is doubled. Velocity double when new height is 4
times of initial height. Therefore pressure becomes 4 times or pressure increase by 3atm

18) P = σ A e (30004 – 3004) = 2kW

 19)
U2 = U1 + 350
U3 = U1 + 200
U23 = U3 – U2 = –150

PART 2 : CHEMISTRY

20)

21) Fact

22) Moles of B in A2B3 =

∴ Moles of A2B3 = 10

Moles of A in A3B4 =
∴ Moles of A3B4 = 5

Molecular weight of A2B3 =

Molecular weight of A3B4 =

Let us assume x gm and y gm be the molar atomic weights of A & B respectively. Therefore :
2x + 3y = 202 ...(i)
3x + 4y = 288 ...(ii)
Solving eq. we get x = 56 & y = 30
GMM = GAM of A + GAM of B
= 56 + 30 = 86
23)
OA — Heating of solid.
AB — Melting of solid.
BC — Heating of liquid.
CD — Vapourisation of liquid.
For CD

24) F > Cl IE
– –
F < Cl IE
F + e– → F–
Cl + e– → Cl–
energy releases in Cl– is more than F– than IE of Cl– is more than F– , needed more energy to
form Cl– .
F > F– IE
–
Cl > Cl IE

25)

Reaction is:-

As per Le chatlier's principle

Increase in temperature,
decrease in pressure,
adding more PCl5 will favour forward reaction.

26)

Bond order is zero for

He2, Be2 and HeH−
27) According to Fajan's rule.

28)

(A) Statement (1) is correct.

Ka values decreases successively, since it is difficult to remove H⊕ ion from an anion than a
neutral compound. Similarly, it is difficult to remove H⊕ ion from an ionian than and anion.
Hence
(B) Statement (2) is incorrect.

This is valid only when H3PO4 during titration is completely converted to

This is again valid when starting with H2PO4 and is completely converted to HPO4 2-
(C) Statement (3) is correct.

29)

unpaired electron is present in 3p orbital = 3, 1, 1, +1/2

32)
nNH3 = = 0.2 mol,

nMg = = 1 mol
2Mg + O2 —→ 2Mg
0.7 mol 0.7 mol
⇒ 0.7 × 40 gm= 28 gm
3Mg + N2 —→ Mg3N2
3 × 0.1 0.1
= 0.3 mol
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

0.3 mol 0.2 mol

= 0.3 × 58
= 17.4 gm

33) nNH3 = = 0.2 mol,

nMg = = 1 mol
2Mg + O2 —→ 2Mg
0.7 mol 0.7 mol
⇒ 0.7 × 40 gm= 28 gm
3Mg + N2 —→ Mg3N2
3 × 0.1 0.1
= 0.3 mol
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

0.3 mol 0.2 mol

= 0.3 × 58
= 17.4 gm

34) = = 2.06 eV
W0 should be less than hv.

35) K.E. = hv – hv0 = (25 – 7) eV = 18 eV

0
V =
36)
But (II) is not protonated, hence the statement (a) is true.

In
So, Nb is more basic due to the presence of LP e's.

Statement (s) :
Statement (c ) is true
2
Statement (d): Due to resonsnce all the N in I,II, and III is hybridised and hence the
statement is true
So, statement a, c, d are true.

37)

ZI =18, ZII = 4, ZIII = 5, ZIV = 4, ZV = 18

g sub shell is present in n=5 principal energy level.
Minimum principal quantum number of an atomic orbital of g sub shell is 5. (ZIII = 5)
Azimuthal quantum number of an atomic orbital of g sub shell is 4. Because l=0 to n–1. (ZII =
4)
Number of angular nodes in an atomic orbital of g sub shell are 4. Because number of angular
nodes =l. (ZIV = 4)
Total 18 elements are expected in g block of the periodic table. (ZV = 18)
Number of electrons in g sub shell are 18. Because number of electrons in a sub shell
= 2(2l+1).(ZI=18)

38)

Br2 + Cl2 → 2BrCI

1 mol 1mol 2mol
0.025 0.025 0.050 mol
Theoretical yield of BrCI = 0.050 mol

actual yield is = = 0.04 mole

Br2 left unreacted = 0.025 – 0.02 = 0.005 mol.
PART 3 : MATHEMATICS

39)

(1 + x)2 (1 + y)3 (1 + z)4 (1 + w)5

General term = 2Ca 3Cb 4Cd 5Ce xa+b+d+e

or

40)

(2021)2022 + (2022)2021
= (2023 – 2)2022 + (2023 – 1)2021
= + 22022 + –1
=

∴ Given number is divisible by 7 hence remainder is zero

41)

42)

43)
(A) n = 4, R + r = r

(B) n = 5,
R+r<2r ⇒r>R

(C) n = 8,
R+r>

(D) n = 12,

R+r=

44)

Slopes of PA and PB are tan

where

,
= (6, 4)

45)
Parabola y2 = 4ax
Let normal at P (t1) Q (t2) meet at parabola at R(t3)
So T (a (t1 t2), a (t1 + t2))
R (a (t1 + t2)2, –2a (t1 + t2))
& t1t2 = 2
PTQR is cyclic quadrilateral
so centre
2x = at1t2 + a (t1+ t2)2

2x = 2a + a

2x = 2a +
2ax – 2a2 = 4y2

∵Qa=1

vertex (1, 0)

foot of directrix

length of latusrectum =

focus
46)

47) , a = 5, b = ,e=

So, the point is on the directrix = 25. So AB is a focal chord, also as

⇒ if . Also S1A + S2A = 2a = 10 = S1B + S2B

⇒ S2A = , S2B = 5. Distance from directrix = × distance from focus =

48)
cc2 = 10 – r
cc1 = 3 + r
cc1 + cc2 = 13 > c1c2
(PS1 + PS2 = 2A > S1S2)
locus is ellipse
so 2a = 13 and 2ae = 5

⇒
Now,
b2 = a2 (1 – e2)

b=6
A = πab = 39π

49) ƒ(x) = (x – 1)(x – 3)(x – 5)(x – 7) + 15

= (x – α)(x – β)(ax2 + bx + c)
(x – 1)(x – 7)(x – 3)(x – 5) + 15
(x2 – 8x + 7)(x2 – 8x + 15) + 15
Let x2 – 8x = t
⇒ (t + 7)(t + 15) + 15 ⇒ t + 22t + 120
⇒ (t + 12)(t + 10) ⇒ (x2 – 8x + 12)(x2 – 8x + 10)
⇒ (x – 6)(x – 2)(x2 – 8x + 10)
ƒ(x) = (x – 6)(x – 2)(x2 – 8x + 10)
(x – 6) (x – 2) (x2 – 8x + 10) = (x – α) (α – β) (ax2 + bx + c)
Sum of root = 16

50) ƒ(x) = (x – 1)(x – 3)(x – 5)(x – 7) + 15

= (x – α)(x – β)(ax2 + bx + c)
(x – 1)(x – 7)(x – 3)(x – 5) + 15
(x2 – 8x + 7)(x2 – 8x + 15) + 15
Let x2 – 8x = t
⇒ (1 + 7)(t + 15) + 15 ⇒ t + 22t + 120
⇒ (t + 12)(t + 10) ⇒ (x2 – 8x + 12)(x2 – 8x + 10)
⇒ (x – 6)(x – 2)(x2 – 8x + 10)
ƒ(x) = (x – 6)(x – 2)(x2 – 8x + 12)
(x – 6) (x – 2) (x2 – 8x + 10) = (x – α) (α – β) (ax2 + bx + c)

51) Let three consecutive terms are

a – D, a, a + D and b – d, b, b + d
⇒ sum is 15 ⇒ a = b = 5

⇒
⇒ d = –17, 1
as d > 0 so d = 1
D=d+1=2
A = 3 × 5 × 7 = 105
B = 4 × 5 × 6 = 120

52) Let three consecutive terms are

a – D, a, a + D and b – D, b, b + D
⇒ sum is 15 ⇒ a = b = 5

⇒
⇒ d = –17, 1
as d > 0 so d = 1
D=d+1=2
A = 3 × 5 × 7 = 105
B = 4 × 5 × 6 = 120
53) ∵

⇒
or 400 = 4(rsinθ – rcosθ)2 + (rsinθ – rcosθ)2 ∴ Locus of P is (y – x)2 = 80

54) ∵ ⇒

⇒ [from Eqs. (i) and (ii)]

or

or [∵ P ≡ (rcosθ, rsinθ)]
∴ Locus of P is 3x – 3y + 40 = 0

55) 1024 × (225)2 × (25)7 = 210 × 34 × 518

2 & 5 are of the form 3k + 2.
If two number of the form 3k + 2 are multiplied together than we get a number of the form 3k
+1
⇒ Divisor must contain even powers of 2 & 5 and it must not contain any power of 3.
⇒ Number of such divisors = 6 × 10 = 60
where number of ways to select power of 2 = 6
& number of ways to select power of 5 = 10

56)

T.N.O.W for distinct boxes 26

For identical boxes

= 32
No identical box remains
empty = 32 -1 = 31

57) We have 1 + cos10x cos6x = 2cos28x + sin28x

= cos28x + 1
Therefore

cos10x cos6x = cos28x =

= 2 cos28x + sin28x
2 cos10x cos6x = 1 + cos16x
cos16x + cos4x = 1 + cos16x
cos4x = 1
4x = 2nπ, n ∈ Z
x = nπ/2, n ∈ Z
Therefore for any value of n.

Therefore the number of values of x ∈ (π/9, 4π/9) is zero.