01-06-2025

9610ZJA801896250002 JA

PART 1 : PHYSICS

SECTION-I (i)

1) The river flow varies with perpendicular distance from point A as then the drifting of the
man if he wants to cross the river in minimum time, will be- (if velocity of man is vm and h is the
width of river)

(A)

(B)

(C)

(D)

2) A ball is held at a fixed height from a smooth incline plane 'B' of inclination φ = 60°. Now, a
smooth platform 'P' of variable length is placed below the ball and the ball is released from rest. The
ball slides on the platform and touches the incline 'B' in the shortest time? Find the angle of platform

'P' with vertical (in degrees).

(A) 0
(B) 90
(C) 30
(D) 45

3) Two particles P and Q move in a straight line towards each other at initial velocities v1 and v2 and
with constant accelerations a1 and a2 directed against the corresponding velocities at the initial
instant. What must be the maximum initial separation between them for which they meet during the
motion?

(A)

(B)
(C)

(D) None of these

4) A particle of mass m is initially situated at the point P inside a hemispherical surface of radius r as
shown in figure. A horizontal acceleration of magnitude a0 is suddenly produced on the particle in
horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the

sphere again is

(A)

(B)

(C)

(D) None of these

SECTION-I (ii)

1) Suppose two particles, 1 and 2, are projected in vertical plane simultaneously. Their angles of
projection are 30° and θ, respectively, with the horizontal. Let they collide after a time t in air. Then

(A) θ = sin–1 (4/5) and they will have same speed just before the collision.
(B) θ = sin–1 (4/5) and they will have different speed just before the collision
(C)
(D) It is possible that the particles collide when both of them are at their highest point.

2) In a situation, a board is moving with a constant velocity v with respect to earth, while a man A
and man B are running with a velocity 2v with respect to earth and both men are running from the
opposite ends of the board at the same time, as shown. Length of the board is L. If they meet after

time T, then
 Value of T is
(A)

Value of T is
(B)

(C) Displacement of man B with respect to board in time T is 3L/4

(D) Displacement of man A with respect to board in time T is L/4

3) A curved section of a road is banked for a speed v. If there is no friction between road and tyres of
the car, then:

(A) car is more likely to slip at speeds higher than v than speeds lower than v
(B) car cannot remain in static equilibrium on the curved section
(C) car will not slip when moving with speed v
(D) none of the above

4) Ship A is located 4 km north and 3 km east of ship B. Ship A has a velocity of 20 kmh–1 towards
the south and ship B is moving at 40 kmh–1 in a direction 37° north of east. X and Y-axes are along
east and north directions, respectively. Then which of the options is correct ?

(A) Velocity of A relative to B is

Position of A relative to B as a function of time is given by
(B)

Velocity of A relative to B is
(C)

(D) Position of A relative to B as a function of time is given by

5) The motion of a body is given by the equation where v is speed in m/s and t in
second. If body was at rest at t = 0

(A) The terminal speed is 2.0 m/s

(B) The speed varies with the time as v = 2(1 – e–3t) m/s
(C) The speed is 0.1 m/s when the acceleration is half the initial value
(D) The magnitude of the initial acceleration is 6.0 m/s2

6) The coordinates of a particle moving in a plane are given by x(t) = acos(pt) and y(t) = bsin(pt)
where a, b (<a) and p are positive constants of appropriate dimensions. Then :

(A) The path of the particle is an ellipse.

(B) The velocity and acceleration of the particle are normal to each other at t = π/2p
(C) The acceleration of the particle is always directed towards a focus.

(D)
The distance travelled by the particle in time interval t = 0 to t = is a

7) A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point ‘C’ starting from ‘A’. he
swims with a speed of 5 km/hr, at an angle θ, w.r.t. the river. If AB = BC = 400 m. Then :-

(A) the value of θ is 53°.

(B) time taken by the man is 6 min.
(C) time taken by the man is 8 min.
(D) the value of θ is 45°.

8) A man is standing on a road and observes that rain is falling at angle 45° with the vertical. The
man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the
start of the motion, it appear to him the rain is still falling at angle 45° with the vertical, with speed
m/s. Motion of the man is in the same vertical plane in which the rain is falling. Then which of
the following statement(s) are true :

(A) It is not possible

(B) Speed of the rain relative to the ground is 2 m/s
(C) Speed of the man when he finds rain to be falling at angle 45° with the vertical, is 4 m/s.
The man has travelled a distance 16 m on the road by the time he again finds rain to be falling
(D)
at angle 45°.

SECTION-II

1) A cannon fires a projectile as shown in figure. The dashed line shows the trajectory in the absence
of gravity and the solid line shows the trajectory in the presence of gravity. Particle will be at A,B

and C at t = 1s, t = 2s and t = 3s respectively. Find the value of in meters if g = 10ms–2.

2) A small boat is travelling downstream with a speed of 5 m/s relative to river flowing with speed

. It encounters a log of wood floating on the river at a distance of a from itself. It turns by an
angle of 37° and just crosses the edge of the log after 8 sec. If the log was fixed and other data
remain same, what would be the time (in sec.) taken by it to just cross the edge of the log.
3) A body A is dropped from a height h1 = 80 m above the ground. At the same time another body B
is fired from the top of a tower of height h2 = 20 m at an angle α to the horizontal as shown. If the

two collide in mid air, find the value of tanα.

4) In an action film, hero is supposed to throw a grenade from his car, which is going with 90 km/h,

to his enemy's car, which is going at . The enemy's car is 1.4 m in front of the hero's
when he let go off the grenade. If the hero throws the grenade so that its initial velocity relative to
him is at an angle of 45° above the horizontal, what should be the magnitude of the initial velocity
(in m/s) of grenade relative to the hero ? The cars are both traveling in the same direction on a level
road. Ignore air resistance.

5) A boat travels across a river from a point A to a point B of opposite bank along the line AB forming
angle 60° with the bank. A flag on the pole (mounted on the boat) flutters at an angle 60° with the
line AB as shown in figure. If velocity of wind ( ) is perpendicular to the flow of river, find

the speed (in m/s) of boat with respect to the bank.

6) Both the cars are accelerating towards right and their initial velocity is also in rightward direction
as shown in figure. Maximum separation between car A and car B is 50α. The value of α is
 PART 2 : CHEMISTRY

SECTION-I (i)

1) The graph between |ψ|2 and r(radial distance) is shown below. This represents :-

(A) 3s orbital
(B) 1s orbital
(C) 2p orbital
(D) 2s orbital

2) Uncertainty in position is twice the uncertainty in momentum, Uncertainty in velocity is :

(A)

(B)

(C)

(D)

3) Which of the following order is CORRECT for indicate property ?

(A) C > Si > Ge > Sn > Pb (Ionization energy)

(B) S > O > Se > Te (Electron gain enthalpy)
(C) SiO2 > Al2O3 > MgO (Basic character)
(D) Cl2O7 > N2O5 > P4O10 (acidic character)

4) Decreasing order of electron affinity of given elements with following valence shell electronic
configuration P→ 2s22p4 , Q→ 3s23p4 , R→ 3s23p3 , S→ 2s22p3
(A) P > Q > R > S
(B) S > R > Q > P
(C) Q > P > S > R
(D) Q > P > R > S

SECTION-I (ii)

1) Which statements are correct -

(A) O < C < S < Se — Atomic size

(B) Na < Al < Mg < Si – IE1
(C) SrO<MgO<Cs2O<K2O — Basic character
(Difference in electron gain enthalpies of Ne & Cl) < (Difference in electron gain enthalpies of F
(D)
& Ar)

2) Which of the following statements is/are correct.

(A) Zeff of B is more than that of Be.

(B) Screening caused by ‘s’ electrons is more as compared to those by ‘f’ electrons
(C) Atomic size of Zr and Hf are almost similar (atomic no. Zr → 40, Hf → 72)
(D) Order of size Sc < Y < La (atomic no. Sc → 21, Y → 39, La → 57)

3) Which of the following order is correct :

(A) P < Si < Be < Mg < Na (Metallic character)

(B) Mg+2 < Na+ < F¯ < O2– (Ionic radius)
(C) Li < B < Be < C < N < O (2nd ionization energy)
(D) Li+ < Na+ < K+ < Rb+ < Cs+ (Ionic mobility)

4) Which of the following order(s) is / are CORRECT :

(A) Li < Be < B < C (IE1)

(B) HF < HCl < HBr < HI (Bond length)
(C) Na2O < MgO < Al2O3 < SiO2 < P2O5 (Acidic)
(D) Li+(g) < Na+(g) < K+(g) < Cs+(g) (Ionic radius)

5) If n + =5, then

(A) Number of possible orbitals are 4

Maximum number of electrons that can be filled with ms = +1/2 is 6 in an atom with above
(B)
condition
(C) for iron (Z = 26) atom, number of electron with above condition must be 1 only
(D) For Germanium (Z = 32), number of electron with above condition can be 5

6) 1st excitation potential for the H-like (hypothetical) sample is 24 V. Then :

(A) Ionisation energy of the sample is 36 eV
(B) Ionisation energy of the sample is 32 eV
(C) Binding energy of 3rd excited state is 2 eV

(D)
2nd excitation potential of the sample is V

7)

Choose the correct relation on the basis of Bohr's theory -

(A)
velocity of electron ∝

(B)
frequency of revolution ∝
(C) radius of orbit ∝ n2Z

(D)
force on electron ∝

8) Hydrogen atoms in a particular excited state ‘n’, when all returned to ground state, 6 different
photons are emitted. Which of the following is/are incorrect.

(A) Out of 6 different photons only 2 photons have wavelength equal to that of visible light.
If highest energy photon emitted from the above sample is incedent on the metal plate having
(B)
work function 8 eV, AKE of liberated photo-electron may be equal to or less than 4.75 eV.
(C) Total number of radial nodes in all the orbitals of nth shell is 24.
(D) Total number of angular nodes in all the orbitals in (n-1)th shell is 23.

SECTION-II

1) An α-particle of K.E. 5.4 MeV is projected towards Cr-nucleus (Z = 24). What is its distance of
closest approach (in 10–14 m) ? (e = 1.6 × 10–19C)

2) An electromagnetic radiation makes 5000 waves in 20 cm. The frequency of radiation (in 1012 Hz)
is.

3) Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain

metal (work function = 2.5 eV). The maximum kinetic energy of the photo-electrons will be :

4) Work function of Ag metal is 7.52 × 10–19J. When Ag is exposed to UV light, the photoelectrons
ejected. The maximum kinetic energy of ejected photoelectrons is 6.97 × 10–19 J. What is value of
minimum potential (in volt) at which photocurrent becomes zero ?

5) The EN of an atom ‘y’ on Pauling scale is 3.06. Then EN of y on Alfred Rochow’s scale is ____
6) If bond length of F2 = 1.44Å, Bond length of H2 = 0.74Å. Find out the bond length of H – F ? (EN
of F is 4.0, EN of H is 2.1)

PART 3 : MATHEMATICS

SECTION-I (i)

1) If ‘x’ satisfies the inequality

|x – 1| + |x – 2| + | x – 3| ≥ 6, then exhaustive set of ‘x’ is :

(A) only 1 or 4
(B) x ≤ 0
(C) x ≥ 0
(D) x ≤ 0 or x ≥ 4

2) If α, β are the roots of the equation x2 – x – 1 = 0 and An = αn + βn, then An+2 + An–2 =

(A) 3An+1
(B) 3An–1
(C) An+1 + An–1
(D) 3An

3) If λ be the ratio of the roots of the quadratic equation in x, 3m2x2+m(m–4)x+2 = 0, then the least

value of m for which , is

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

4) Each set Xr contains 5 elements and each set Yr contains 2 elements and . If
each element of S belong to exactly 10 of the Xr's and to exactly 4 of the Yr's, then n is

(A) 10
(B) 20
(C) 100
(D) 50

SECTION-I (ii)

1) For two sets A and B if n(A) = 7, n(B) = 13 and n(A ∩ B) = 5, then the correct statement is
(A) n(A ∪ B) = 15
(B) n(A – B) = 6
(C) n(A × B) = 91
(D) n{(A ∪ B) × (A ∩ B)} = 75

2) If A, B be any two sets, then (A ∪ B)' is equal to __________ (Here U is universal set).

(A) A' ∩ B'

(B) U – (A ∪ B)
(C) ((A – B) ∪ (B – A))'
(D) A' ∩ (U – B)

3) In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product
C. If 14 people liked products A and B. 12 people liked products C and A, 14 people liked products B
and C and 8 liked all the three products. Which of the following are correct?

(A) 11 people like C only

(B) 6 people like B only
(C) 15 people like exactly two products
(D) 20 people like exactly one product

4) If the quadratic equation x2 + 2ax + b(a – 1) = 0 has real roots for all real values of a, and b is an
integer, then which of the following is correct ?

(A) Greatest value of b is 4

(B) No such value of b is possible
(C) Least value of b is 0
(D) b can be any integer

5) Consider the equation :

and S be the solution set of above

equation, then

(A) Sum of absolute values of real solution(s) of given equation is a prime number
(B) equation has more than two real solutions

(C)

(D)
(where α, β ∈ S and α > β)

6) The adjoining figure shows the graph of y = ax2 + bx + c. Then

(A) a > 0
(B) b > 0
(C) c > 0
(D) b2 < 4ac

7) If x2 – 3x + 2 is a factor of x4 – px2 + q, then

(A) equation x4 – px2 + q = 0 has four distinct real roots

(B) equation x4 – px2 + q = 0 has two real and two imaginary roots
(C) p = – 5, q = – 4
(D) p = 5, q = 4

8) If one root of the equation 4x2 + 2x – 1=0 is ‘α’, then

α can be equal to
(A)

(B)
α can be equal to
(C) other root is 4α3 – 3α
(D) other root is 4α3 + 3α

SECTION-II

1) If α, β are roots of equation 3x2 – 2x + 1 = 0, then value of .

2) If (x4 + 2x2 + 1) – (b + 4) (x4 + x2) + (b + 3)x4 = 0 has atleast one real solution then least value of

is

3) Find the sum of all possible integral solutions of equation

||x2 – 6x + 5| – |2x2 – 3x + 1|| = 3|x2 – 3x + 2|

4) Let α and β be the roots of equation x2 − ax + b = 0 (where a, b are constants). If

then the value of

5) If A and B are two sets such that n(A) = 150,

n(B) = 250 and n(A ∪ B) = 300. Then is

6) How many ordered pairs of integers which satisfy the equation ?

 ANSWER KEYS

PART 1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

Q. 5 6 7 8 9 10 11 12
A. B,C,D A,C,D B,C A,B A,B,D A,B A,B C,D

SECTION-II

Q. 13 14 15 16 17 18
A. 5.00 6.00 4.00 7.00 5.00 3.00

PART 2 : CHEMISTRY

SECTION-I (i)

Q. 19 20 21 22
A. D C D D

SECTION-I (ii)

Q. 23 24 25 26 27 28 29 30
A. A,B A,B,C,D A,B,D B,C,D A,D B,C,D A,B,D C,D

SECTION-II

Q. 31 32 33 34 35 36
A. 1.28 7.50 0.05 4.33 to 4.37 2.32 0.90 to 0.93

PART 3 : MATHEMATICS

SECTION-I (i)

Q. 37 38 39 40
A. D D B B

SECTION-I (ii)

Q. 41 42 43 44 45 46 47 48
A. A,C,D A,B,D A,B,D A,C A,C,D B,C A,D A,C
 SECTION-II

Q. 49 50 51 52 53 54
A. 2.00 0.25 15.00 1.00 1.00 7.00
 SOLUTIONS

PART 1 : PHYSICS

1)

dα = vrdt =vr

dx = vr

⇒ dx = dy ⇒ dx =

⇒ ⇒

2)
aball = gsin(90 – θ) = gcosθ

tmin for θ = 30°

3)

For maximum

=
OR
Relative velocity = v1 + v2,
Relative acceleration According to question the relative distance over which the

relative velocity vanishes must be longer than the initial separation. So

4) Let the particle touches the sphere at the point A.

Let

In

or

But

5)

If they collide, their vertical component of velocities should be same

6)

At the plane VAB = 4V

7) There is no friction between road and tyres of car so that car cannot remain in static
equilibrium on curved section. Whenever speed of car is greater than or less than v car will
slip.

8)
VA = 20(– )
VB = 40 cos 37 + 40 sin 37

= 40 × + 40 ×
= 32 + 24

= –20 – ( )
= –32 – 44 k/h
At time = 0
=( )
at time t

= (3 – 32t) + (4 – 44t)

9)

n6 – n(6 – 3v) = 3t

v = 2(1 – e–3t)
at t = ∞, v = 2 m/s
initial acceleration, a = 6
a=3
6 – 3v = 3

10) (A) x(t) = a cos pt y(t) = b sin pt

(B)

ax = –ap2 cos pt
ay = –bp2 sin pt
= a2p3 sin pt cos pt – b2p3 sin pt cos pt = 0
 so, sin2pt = 0 at t =
(D) at t=0, x(0) = a, y(0) = 0

At

So, distance travelled is

(C) The acceleration is directed towards the center of the ellipse and not towards the focus.
This can be calculated by the expression of acceleration.

11) Let he crosses river in in time 't' then we

have

And we have,
(VSR cosθ + VR)t = 0.4
⇒ (5cosθ + 1)t = 0.4
⇒ 5 cosθ + 1 = 5 sinθ

⇒ sinθ – cosθ =
⇒ θ = 53°
And, t = 6 min.

12) vr = vy + vx

vr/m = vr – vm
8 = v2 + (v – t/2)2 ...(i)

tan 45° =
v – t/2 = –v
2v = t/2 ⇒ v = t/4
now from equ. (i)
t=8
vm = 4 m/s

13)

y1 = gt12 = × 10 × 12 = 5m

y2 = gt22 = × 10 × 22 = 20m

y3 = g t32 = × 10 × 32 = 45 m; = 5m

14)

ℓ = 54 m
d = 36 m

15 cos θ + 7 = 20 sin θ
225 cos2θ + 49 + 210 cos θ = 400 – 400 cos2θ
625 cos2θ + 210 cos θ – 351 = 0

15)
 16)
In frame of enemy
For vertical motion

0
⇒v =7

17)
Flag on the boat flies in the direction of wind relative to boat, so β is the angle make by
relative velocity of the wind relative to the boat.
, vboat, earth =

Putting
v = 5 m/s

18) vA = vB (for maximum separation)

(10 + 2t) = 20 + t
(t = 10)

PART 2 : CHEMISTRY
19)
node = 1 and, from graph, ℓ = 0
n–ℓ–1=1 i.e.,
n–1=1

20)

21)

Acidic nature of oxide EN of central atom

22)

Element = O S P N

Electron affinity of
∴ Q>P>R>S

23)

Basic character of oxide, increases on moving down the group. Acidic character of oxides
increases left to right in a period.

24)

A long period Zeff increases & along group it remains constant.

Also screening by s electrons is more as compared to other because of closeness.
25) Theoritical

26) Down the group ⇒ metallic character increases.

L → R in period ⇒ Non metallic character increases

27)

If n + =5,
m = +1 or -1

n=4 l=1 4p

n = 3 l = 2 3d
no. of posible orbitals = 4

28)

1st excitation potential = 10.2 Z2 = 24 V

∴ Z2 = 24/10.2

∴ IE = 13.6 Z2 = = 32 eV.

Binding energy of 3rd excited state = 0.85 Z2 = = 2eV.

2nd excitation potential of sample = 12.09 Z2 = = V.

29)

(A) v = 2.18 × 106 × ⇒v∝ or v ∝

(B) f = or f = ∝ f∝

(C) r ∝ n2 / Z [T ∝ ] F=

F∝ F∝
So ans (A,B,D)

30) Value of n is 4
For n = 4 to n = 1 total 6 spectral lines are observed
(4 → 3, 3 → 2, 2 → 1, 4 → 2, 4 → 1, 3 → 1)
the highest energy photon along them is 4 → 1

For = 3 + 3(2) + 5(1) + 0 = 14

For = 0 + 3(1) + 5(2) = 13

31)

K.E. of a = 5.4 × 106 × 1.6 × 10–19 Joule.

i.e. r =

=
= 128 × 10–16 m
= 1.28 × 10–14 m

32) Given → 5000 waves in 20 cm

⇒ = number of waves per unit length

33)

En = – eV; E2 = –

E4 = – eV/atom
ΔE = E4 – E2 = 2.55 eV
Absorbed energy = work function of metal + K.E.
2.55 = 2.5 + K.E. ;
K.E. = 0.05 eV

34) KEmax = eVs

Vs = 4.35 volt
 35) XP = XAR + 0.744
⇒ XAR = 3.06 – 0.744
= 2.316 (2.31 or 2.32)

36) According to Schomaker-Stevenson formula

rF = rF + rH – 0.09 (XF – XH)
∵ rF = 1.44/2 = 0.72Å, rH = 0.74/2 = 0.37Å
∴ dH–F = 0.72 + 0.37 – 0.09 (4.0 – 2.1)
= 1.09 – (0.09 × 1.9) = 1.09 – 0.171 = 0.919Å

PART 3 : MATHEMATICS

37) |x – 1| + |x – 2| +|x – 3| ≥ 6
∴ x ≤ 0 or x ≥ 4

Alternately, we can solve with options.

38) An + 2 + An – 2 = αn–2(α4 + 1) + βn–2 (β4 + 1)

= αn–2{(α + 1)2 + 1} + βn–2 {(β + 1)2 +1}
= αn–2(α2 + 2α + 2) + βn–2 (β2 + 2β + 2)
= αn–2(3α2) + βn–2(3β2) = 3An

39)

3m2x2 + m(m – 4) x + 2 = 0

, , α2 + β2 = αβ
2
(α + β) = 3αβ

⇒ =
2
(m – 4) = 18, m =

40) n(X1) = 5 = n(X2) = n(X3) = ... = n(X20)

Total elements ins S repeating as well non repeating

= 20 × 5 = 100 = 10k
S = {a1, a2, ... ak}
n(S) = K Let
⇒ K = 10
a1 → 10 times
a2 → 10 times

ak → 10 times
X1 = {a, b, c, d, e}
X2 = {a, p, q, r, s}

Repeatings + Non repeating

S=n×2
⇒ 4 × 10 = 2n
⇒ n = 20

41)

n(A) = 7 ; n(B) = 13
n(A – B) = n(A) – n(A ∩ B) = 7 – 5 = 2
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 7 + 13 – 5 = 15
n(A × B) = n(A) × n(B) = 7 × 13 = 91
n{(A ∪ B) × (A ∩ B)} = 15 × 5 = 75

42)
(A ∪ B)' = (1)
(A) A' = (1), (4)
B' = (1), (2)
(B) U – (A ∪ B) = (1)
(C) A – B = (2), B – A = (4)
((A – B) ∪ (B – A))' = (1), (3)
(D) A' = (1), (4)
A ∪ (U – B) = (1)

43)

Let, no. of people liking product A = n(A) = 21

no. of people liking product B = n(B) = 26
no. of people liking product C = n(C) = 29
given, n(A∩B)=14,n(B∩C)=14,n(A∩C)=12,n(A∩B∩C)=8.
⇒ no. of people liking C only = 11
no. of people liking B only = 6
no. of people liking exactly one product = 3 + 6 + 11 = 20
no. of people liking exactly two products = 4 + 6 + 6 = 16.

44) For real roots 4a2 – 4b(a – 1) ≥ 0

⇒ a2 – ba + b ≥ 0
Now for above inequality to hold for all values of a, b2 – 4b ≤ 0
or 0 ≤ b ≤ 4.

45) Given equation is x2 = 4 − 3x x = 1, −4

46) By seeing graph of ax2 + bx + c = y

c > 0 and b > 0
So ans is (B) and (C).

47) (x – 1) (x – 2) is a factor of x4 – px2 + q

⇒ x = 1, 2 are roots
⇒ 1 – p + q = 0 and 16 – 4p + q = 0
on solving, p = 5, q = 4
∴ equation : x4 – 5x2 + 4 = 0
⇒ (x2 – 4) (x2 –1) = 0
4 real and distinct roots.

48) 4x2 + 2x – 1 = 0

x=

If α = ⇒ α3 =

4α3 =

⇒ 4α3 – 3α =
49)
⇒

=2

50) (x2 + 1)2 – (b + 4)(x2)(x2 + 1) + (b + 3) (x2)2 = 0 clearly x ≠ 0

Let
Now, ℓ2 – (b + 4)ℓ + (b + 3) = 0
∴ ℓ2 – bℓ – 4ℓ + b + 3 = 0
∴ (ℓ2 – 4ℓ + 3) – b(ℓ – 1) = 0
∴ (ℓ – 3)(ℓ – 1) – b(ℓ – 1) = 0
∴ ℓ = 1 or ℓ = b + 3
but
Now, ℓ = b + 3, ℓ > 1
⇒ b+3>1
⇒ b>–2
⇒ b2 ≥ 0
⇒ b2 + 4 ≥ 4

∴ Least value of is 0.25

51) ||A| – |B|| = |A + B|

iff AB ≤ 0
⇒ (x2 – 6x + 5) (2x2 – 3x + 1) ≤ 0

⇒x∈

52) a10 – a.a9 + b.a8 = 0

53)

n(A) = 150, n(B) = 250, n(A ∪ B) = 300

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ 300 = 150 + 250 – n(A ∩ B)
⇒ n(A ∩ B) = 100
⇒ n(A – B) = 50
⇒ n(B – A) = 150

Then,

54)
⇒ 2(n+2m) –nm = 0
⇒ 2n + 4m –mn – 8 + 8 = 0
⇒ (m–2) (n–4) = 8
2 4
4 2
–2 –4 (rejected) m = 0, n = 0
–4 –2
1 8
–1 –8
8 1
–8 –1
Number of ordered pairs = 7