26-01-2025

1001CJA10T-AS250003 JA

PART-1 : PHYSICS

SECTION-I (i)

1) Car B overtakes another car A at a relative speed of 40 ms–1. How fast will the image of car B
appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from
the car A?

(A) 4 ms–1
(B) 0.2 ms–1
(C) 40 ms–1
(D) 0.1 ms–1

2) A small disc is performing angular oscillation with a wire of torsional constant k, with an angular
amplitude θ0. When it passes through half of its amplitude, a sharp angular impulse is provided, so
that its, angular velocity increases to 1.5 times of the angular velocity of same position. Find its new
angular amplitude :-

(A)

(B)

(C)

(D)

3) ABC is a smooth horizontal wire frame. PQ is a wire of mass m, which can slide on frame. There is
a film of a liquid of surface tension T covering the triangular part. Wire PQ is given a velocity v0
when it is at a distance x0 from B. Find v0 if wire PQ is just able to reach upto a distance 2x0 from B.

(A)
(B)

(C)

(D)

4) In the figure shown S1S2 are ideal springs each of spring constant 'k'. A is a block of mass 'm'
placed on a smooth horizontal surface. PO is a massless rod that can oscillate about an axis
perpendicular to the plane of the paper and passing through 'O'. The time period of small oscillation

of the block is

(A)

(B)

(C)

(D)

SECTION-I (ii)

1) A ray of light is incident along a vector on a plane mirror lying in y–z plane. The unit
vector along the reflected ray can be

(A)

(B)

(C)

(D)

2) In an arrangement shown in figure an object is placed in front of multiple slabes of different

refractive indices and width. If object is in air then choose the correct alternative (s).
(A) If observer is in air, object will appear to him at distance 39 cm.
(B) If observer is in medium of refractive index 1.5, object will appear to him at distance 37.5 cm.
(C) If observer is in medium of refractive index 1.5, object will appear to him at distance 48.5 cm.
(D) No matter in what medium observer is, object will always appear close to him.

3) A uniform plank of mass m, free to move in horizontal direction, is placed on the top of a solid
cylinder of mass 2m and radius R. The plank is attached with one end of a massless spring of force
constant k and the other end of spring is fixed with a wall as shown in fig. The plank is slightly
displaced towards spring and released. There is no slipping any where then :-

Total energy of the system, when compression is spring is x and velocity of plank is v, is
(A)
.

(B)
Angular frequency of oscillation is .

(C)
Time period of oscillation is .

(D)
Angular frequency of oscillation is .

4) For a concave mirror, graph of square of magnification and image distance from pole is given for

real object. Choose the correct statement(s).

(A) This graph is hyperbola.

(B) length a must be equal to length b.
(C) radius of concave mirror is 60 cm
(D) the length c will be half of length a

5) A hemisphere made of a transparent material of variable refractive index has radius 12 cm and

centre at origin O. The refractive index of the material of the sphere varies as , where x is
in cm. A ray of light is incident at the point O at an angle θ0 with the x-axis and comes out through a

point P on its curved surface. Given then :-

(A) Path of the ray inside the sphere is parabolic.

(B) Path of the ray inside the sphere is circular.
(C) x-coordinate of point P is 3 cm.
(D) x-coordinate of point P is 1.50 cm.

6) An object is approaching at thin convex lens of focal length 0.3 m with speed u = 0.01 m/s. Object
is at a distance of 0.4 m from the lens. At the instant shown in figure.

(A) The magnitude of rate of change of position of image is 0.09 m/s.

(B) The magnitude of rate of change of position of image is 0.18 m/s.
(C) The magnitude of rate of change of lateral magnification of image is 0.3 /s.
(D) The magnitude of rate of change of lateral magnification of image is 0.6 /s.

SECTION-II (i)

Common Content for Question No. 1 to 2

There are two spheres of transparent materials. Sphere A has refractive index of and
refractive index of sphere B is not known to us (consider it as µ2). Radii of each sphere is 50 cm.
Both the spheres are cut so that the section of cut is 30 cm from center. Later the smaller parts
struck together as shown. A laser beam
from point S (source) is incident on the curved surface of the part of sphere A at point P. The length
of beam from S to P is 50 cm. Also the angle made by beam with the principal axis is θ = 37°

. The beam finally emerges from the curved surface of the part of sphere B. The beam
follows a total deviation of 53°.

1) If the image of first refraction is formed at a distance D cm from the surface of A, find the value

of .

2) If the beam after emerging from B meets the principal axis at I, the total length of beam (in cm)

from S to I will be given by . Find the value of Q.

Common Content for Question No. 3 to 4

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of
curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered

and placed on a horizontal surface.

3) Find x (in cm) where should a pin be placed on the optic axis such that its image is formed at the
same place ?

4) If the concave part is filled with water of refractive index 4/3, find the distance (in cm) through
which the pin should be moved so that the image of the pin again coincides with the pin ?

Common Content for Question No. 5 to 6

In the given figure, the block is attached with a system of three ideal springs A, B, C. The block is
displaced slightly by a distance 'x' from its equilibrium position vertically downwards and released.
5) The ratio of deformation of spring B to spring C is

6) The time period of small vertical oscillations of block is . Find a

SECTION-II (ii)

1) A solid sphere (radius = R) rolls without slipping in a cylindrical trough (radius = 5R) as shown in
figure. For small displacements from equilibrium perpendicular to the length of the trough, the

sphere executes simple harmonic motion with a period T = then value of n is

2) A spherical soap bubble has internal pressure P0 and radius r0 and is in equilibrium in an

enclosure with pressure . The enclosure is gradually evacuated. Assuming temperature and

surface tension of soap bubble to be fixed find the value of of soap bubble.

3) A ray of light is incident parallel to BC at a height h = 3.0 cm from BC. Find the height (in cm)
above BC at which the emergent ray leaves the surface AC. If is given that and length BC =
20 cm. Take tan15º = 0.25.

PART-2 : CHEMISTRY

SECTION-I (i)

1) In a photo emissive cell with exciting wavelength λ, the fastest electron has a speed V. If the

exciting wavelength is changed to , the speed of the fastest emitted electrons will be :
(Assume work function of metal is too small than energy of radiation)

(A)

(B)

(C)
Less than

(D)
Greater than

2) The equilibrium constant for the reaction CO(g) + H2O(g) ⇌ CO2(g) + H2(g) is 5. How many moles
of CO2 must be added to 10 litre container already containing 3 moles each of CO and H2O to make
0.2 M equilibrium concentration of CO ?

(A) 15
(B) 19
(C) 5
(D) 20

3) For the oxidation-reduction reaction ;

K2Cr2O7 + XH2SO4 + YSO2 → K2SO4 + Cr2(SO4)3 + ZH2O
the values X, Y and Z are respectively :

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

4) The number of 'two center – two electron' and 'three center – two electron' bond in [Al(BH4)3] are
respectively :

(A) twelve and zero

(B) twelve and three
(C) Six and six
(D) Nine and three

SECTION-I (ii)

1) Identify reagents for following reaction sequence -

(A) conc.HI

(B) conc. H2SO4/

(C) conc. H3PO4/

(D) conc. HBr

2) Choose the correct statement(s) :

(A) 'TC' of CH4 gas is greater than H2 gas.

(B) Vander Waal's constant 'b' of CH4 gas is greater than H2 gas.
(C) Compressibility factor (Z) of CH4 gas is greater than H2 gas
(D) At their respective Boyle's temperatures both CH4 and H2 gases ideally in low pressure region.

3) For gases A and B plots between [d = density of gas (g/L), P = pressure (atm)] and P at 300 K
temperature was obtained as :

Choose correct statements. [Given : R = 0.082 L atm K–1 mol–1].

(A) Gas A shown non-ideal behaviour

(B) Molar mass of gas B is 81.18 g mol–1

(C)
The plot of vs Temperature for ideal gas should have a positive slope.
(D) Gas B shows ideal behaviour.

4) Select CORRECT order against property given :

(A) S–2 > Cl– > K+ > Ca+2 (Ionic radius)
(B) B > Tℓ > Ga > Aℓ > In (Ionization energy)
(C) Ne > Ar ≃ Kr > Xe > He (ΔHeg)

(D) (Ionic mobility)

5) Which of the following option is/are CORRECT about NO2 and ClO2?

(A) Both are paramagnetic species.

(B) Both compounds dimerise readily.
(C) Both have sp2 hybridisation of central atom.
(D) Both are polar and planar.

6) Select the CORRECT option :

(A) BeCl2 > MgCl2 > CaCl2 > SrCl2 > BaCl2 : Covalent character
MF > MCℓ > MBr > MI (Melting point)
(B)
(where M = Li-------Cs)
(C) CO < CO2 < SO2 < SO3 : Acidic strength
(D) LiCl > CsCl > RbCl > NaCl > KCl : (Water Solubility)

SECTION-II (i)

Common Content for Question No. 1 to 2

Carbocation can rearrange to more stable carbocation via 1,2-shift.

1)
The given carbocation undergoes how many rearrangements to form the final stable carbocation?

2) How many of the given carbocations can undergo rearrangement

(i) (ii) (iii) (iv)

(v) (vi) (vii)

(viii) (ix) (x)

Common Content for Question No. 3 to 4

(major)

3) Calculate the total number of chiral centers in 'y'?

4) Total number of possible structurally isomeric products (z) are :

Common Content for Question No. 5 to 6

5) Number of filled σg electrons = x

Number of filled πu electrons = y
Number of filled electrons in HOMO = z
Find (x + y + z).

6) How many of the following species have fractional bond order & diamagnetic in nature.
, , NO , CN– , , , F2
 SECTION-II (ii)

1) Total number of stereo isomers possible of the given compound is

2)

How many of the following will give white ppt with AgNO3 solution?
(a) CH2 = CH – Cl (b) CH2 = CH – CH2 – Cl

(c) (d)

(e) (f) CH3 – CH2 – O – CH2 – Br

(g) (h)

3) A vessel of uniform cross-section of length 500 cm as shown in figure is divided in two parts by a
weightless and frictionless piston. One part contains 5 moles of He(g) and other part contains 2
moles of H2(g) and 4 moles of O2(g), at the same temperature and pressure. A reaction takes place in
other part and finally vessel is cooled to 300 K and 1 atm. What is the final length (in mm) of He
compartment ? (Assume volume of piston and volume of H2O(ℓ) formed are negligible).

PART-3 : MATHEMATICS

SECTION-I (i)

1) The value of

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

2) If a ∈ and the equation –3(x – [x])2 + 2(x – [x]) + a2 = 0 (where [x] deontes the greatest integer
≤ x) has no integral solution, then all possible values of a lie in the interval :
(A) (–1, 0) ∪ (0, 1)
(B) (1, 2)
(C) (–2, –1)
(D) (–∞, –2) ∪ (2, ∞)

3) Let f(n) be defined for all whole numbers n with f(a + b) = f(a) + f(b) – 2f (ab) and f(1) = 1.
Then the value of f(0) + f(1) + f(2) + ... + f(2025) is

(A) 505
(B) 0
(C) 1012
(D) 1013

4) Let α, β, γ be the roots of x3 – 3x2 + 2x + 4 = 0 and y =

, then the value of y at x = 2 is equal to -

(A) 1
(B) 2
(C) 4
(D) 8

SECTION-I (ii)

1) The maximum value of the function defined by is , then integral

value of x satisfying the inequality denote greatest integer function, is

(A) 1
(B) 2
(C) 5
(D) 6

2) Let x denote the total number of one-one function from a set A with 3 elements to a set B with 4
elements and y denote the total number of one-one function from the set A to the set A × B. Then

(A) x + y = 1344
(B) x + y = 1354
(C) y = 55x
(D) 2y = 55x

3) Which of the following pair of functions are identical?

(A) f(x) = sgn(x2 –3x + 4) and g(x) = e[{x}]

(B)
f(x) = and g(x) = tanx
(C) f(x) = ℓn(1 + x) + ℓn(1–x) and g(x) = ℓn(1 – x2)

(D)
and

4) If graph of a function f(x) which is defined in [–1, 4] is shown in the adjacent figure then identify

the correct statement(s).

(A) domain of f(|x| – 1) is [–5, 5]

(B) range of f(|x| + 1) is [0, 2]
(C) range of f(–|x|) is [–1, 0]
(D) domain of f(|x|) is [–3, 3]

5) Let f, g and h be three functions defined as follows :

f(x) = , g(x) = 9 + x2 and h(x) = –x2 – 3x + k.

Identify which of the following statement(s) is(are) correct ?

(A) Number of integers in the range of f(x) is 8

(B) Number of integral values of k for which h(f(x)) > 0 and h(g(x)) < 0 ∀ x ∈ R is 20
(C) Number of integral values of k for which h(f(x)) > 0 and h(g(x)) < 0 ∀ x ∈ R is 19
(D) Maximum value of g(f(x)) is 73

6) If α, β, γ are the roots of x3 – x2 + ax + b = 0 and β, γ, δ are the roots of x3 – 4x2 + mx + n = 0. If α,

β, γ and δ are in AP with common difference d then

(A) a = m
(B) a = m – 5
(C) n = b – a – 2
(D) b = m + n – 3

SECTION-II (i)

Common Content for Question No. 1 to 2

The number of solution of the equation ƒ(x) = g(x) is the number of points of intersection of the
graphs y = ƒ(x) and y = g(x).
1) Number of roots of the equation |x2 – 2x – 5| = λ, where 0 < λ < 6 is -

2) Number of roots of the equation |sin 2x| = cosx in -

Common Content for Question No. 3 to 4

Let ƒ(x) = |x – 1| + 2|x – 2| + 3|x – 3| and g(x) = 4|x – 4| + 5|x – 5|.

3) Minimum value of ƒ(x) + g(x) is -

4) Number of solutions of the equation ƒ(x) = g(x).

Common Content for Question No. 5 to 6

Consider the expression ƒ(x) = x2 – 2|x| (k – 1) + 7(2k – 9).

5) If 'k' is a single digit natural number then number of possible values of 'k' for which ƒ(x) = 0 has 4
real solutions -

6) Number of value/s of 'k' so that ƒ(x) = 0 has exactly 3 real solutions -

SECTION-II (ii)

1) Let m and n be the numbers of real roots of the quadratic equations x2 – 12x + [x] + 31 = 0 and x2
– 5|x + 2| – 4 = 0 respectively, where [x] denotes the greatest integer ≤ x. Then m2 + mn + n2 is
equal to ____.

2) The number of integral values of m, for which the roots of will lie between
–2 and 4 is

3) If three different polynomials x2 + ax + b, x2 + x + ab and ax2 + x + b have exactly one common

zero, where a, b are non-zero real numbers, then value of a + 2b is -
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-II (i)

Q. 11 12 13 14 15 16
A. 0.00 124.00 15.00 1.15 to 1.16 4.00 5.50

SECTION-II (ii)

Q. 17 18 19
A. 28 3 1

PART-2 : CHEMISTRY

SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-II (i)

Q. 30 31 32 33 34 35
A. 4.00 7.00 3.00 3.00 12.00 0.00

SECTION-II (ii)

Q. 36 37 38
A. 4 3 3125

PART-3 : MATHEMATICS
 SECTION-I (i)

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

SECTION-I (ii)

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

SECTION-II (i)

Q. 49 50 51 52 53 54
A. 4.00 4.00 15.00 2.00 4.00 1.00

SECTION-II (ii)

Q. 55 56 57
A. 9 3 0
 SOLUTIONS

PART-1 : PHYSICS

1)
Mirror used is convex mirror (rear-view mirror)

Given,

∴ Car will appear to move with speed 0.1 m/s.

Hence option (4)

2)

This problem can be solved by analogy with linear SHM.

V2 = w2 (A2 – x2)

Here, x = A/2

Hence

3)

(Along -ve x)
from work energy theorem
5) According to law of reflection Here so

6)
when observer in air

when observer in medium

d' = 48.5 cm

7) At the instant when velocity of plank is v velocity of CM. of cylinder is 0 and angular velocity

of cylinder is ω. E =

I= = mR2 =

For free oscillation =0

⇒a=

⇒ ω2 = and t = =

8) = graph will be a parabola

m = 0 at focus ⇒ f = 30
b =1 = m at v = 0 = a as mag at 2f

at v = 15 m =

9)

Let at any point A of trajectory of ray the tangent to the path of ray makes an angle θ with x-
axis. From Snell’s law
 , so the path of the ray is circular.
At

10)

u = 40 cm, F = 30 cm, cm/s

v = 120 cm
Velocity of image,

cm/s

/s

11) Rays are marginal

By figure

by shell's law

1 sin 74 =

So light becomes parallel to principle axis hence

D=∞

12)

For next surface let angle of emergence is 'e' and by symmetry angle of incidence = 37°
Total deviation = 53° = (i - r1)+(e - r2)
r1 = r2 = 37° i = 74°
53° = 74° - 37°+ e - 37°
e = 53°
Total length - SP + PQ + QI
SP = 50 cm
PQ = 20 cm

QI =

Total length
13)
The optical arrangement is equivalent to the concave mirror of focal length F given by

where fx is the focal length of the lens without silvering and fm is the focal length of the mirror.

fg = 60 cm

fg = = = 10 cm

For the image to be formed at the place of the object

X = R = 2F = 7.5 × 2 = 15 cm

14) When the concave part is filled with water of refractive index 4/3, the optical arrangement
is equivalent to concave mirror of focal length F such that

fw = 180 cm
fg = 60 cm (calculated earlier)

X1 = R = 2F =
Δx = 15.0 – 13.85 = 1.15 cm

15) Let elongation in spring A, B and C be x1, x2 and x3 respectively.

Considering spring forces and constant relations
x2 = 4x3 ....(i)
x2 = 2x1 ....(ii)
and x1 + 2x2 + x3 = x ....(iii)

; ;

Also,
16) Let elongation in spring A, B and C be x1, x2 and x3 respectively. Considering spring forces
and constant relations
x2 = 4x3 ....(i)
x2 = 2x1 ....(ii)
and x1 + 2x2 + x3 = x ....(iii)

; ;

Also,

17) The kinetic energy of the ball is K = , where Ω is the rotation rate of the ball
about its centre of mass. Since the centre of the ball moves along a circle of radius 4R, its

displacement from equilibrium is s = (4R)θ and its speed is v = . Also, since the

ball rolls without slipping. So Ω =

The kinetic energy is then

K=
When the ball has an angular displacement θ, its centre is distance h = 4R(1 – cosθ) higher
than when at the equilibrium position. Thus, the potential energy is Ug = mgh = 4mgR(1 –

cosθ). For small angles, .

2
Hence Ug ≈ 2mgRθ , and the total energy is

E = K + Ug = .

Since E = constant in time,

This reduces to , or .
With the angular acceleration equal to a negative constant times the angular position, this is in

the defining form of a simple harmonic motion equation with ω = .

 The period of the simple harmonic motion is then T =

18)

Finally Poutside = 0 ⇒ Pinside =

Now P1V1 = P2V2

19)

sin 45° = µ sin r ⇒ r = 30°

From ΔBQR and ΔCSR

i = 45° + r = 75°

So, ⇒ MR = 12 cm
So, BR = 15 cm and RC = 5 cm
So, h' = 1 cm

PART-2 : CHEMISTRY

20) ∵ (K.E.)max = –w

........ (1)

∴ ........ (2)
From (2) / (1)
21) Let x mole of CO2 is added

3 mole 3 mole x mole +

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
2 2 x+1 1

Keq. = =5
x = 19

22) Theoretical.

25) TC of CH4 = 190 K

TC of H2 = 33 K
b of CH4 = 0.043 lit./mole
b of H2 = 0.027 lit. /mole

26) For ideal gas

PM = dRT

M= = 3.3 × 0.032 × 300 = 81.18

30)

31)

(i), (ii), (iii), (iv), (vi), (vii), (ix)

 38) Moles of O2 remaining after reaction = 4 – 1 = 3
V ∝ n is due to same T and P and each compartment after reaction
∴ Length of He compartment

× 500 = 312.5 cm.

PART-3 : MATHEMATICS

39) from (I) ∩ (II) ∩ (III)

A – (A ∩ (B ∪ C))

40) Given equation is

–3(x – [x])2 + 2 (x – [x]) + a2 = 0
⇒ a2 = 3. {x}2 – 2{x}

= 3. (∵ {x} ≠ 0 ⇒ a2 ≠ 0)
2
⇒ a ∈ (0, 1)
⇒ a ∈ (–1, 0) ∪ (0, 1).

41) Putting a = b = 0 f(0) = 0

Putting a = n, b = 1 f(n + 1) = f(n) + 1 – 2 f(n)
f(n + 1) + f(n) = 1
now
Therefore, period of function is 2
f(3) = f(1) = 1; f(4) = f(2) = f(0) = 0

f(0) + f(1) + f(2) + ... + f(2025) = 0 + 1 + 0 + 1 + 0 + 1 ...

42) On solving,
Also x3 – 3x2 + 2x + 4
= (x – α) (x – β) (x – γ).
At x = 2,
43)
at both are equal to

So
and

Now

44) Number of elements in A = 3

Number of elements in B = 4
Number of elements in A × B = 12
Number of one one function ⇒ 4 × 3 × 2
From set A to set B = 24
Number of one one function
From set A to set A × B ⇒ 12 × 11 × 10 = 1320

45) (A) f(x) = sgn(x2 – 3x + 4)

Now

⇒ f(x) = 1
⇒ g(x) = e[{x}]
Here, {x}∈[0, 1)
⇒ [{x}] = 0
0
⇒ g(x) = e = 1
Here, Domain, Range and graph of f and g are same. So, f and g are identical.

(B)

=
=
= |tanx|
g(x) = tanx
Clearly f and g are not identical functions.
(C) f(x) = ℓn(1 + x) +ℓn(1 – x)
= ℓ(1 + x).(1 – x) [∵ logA + logB = log(A × B)]
= ℓn(1 – x2)
⇒ g(x) = ℓn(1 – x20)
Here, Domain, Range and graph of f and g are same. So, f and g are identical.

(D)

Domain of

Domain of
Here, domain of f and g are different
So, they are not identical.

46) (A) – 1 ≤ |x| – 1 ≤ 4

⇒ 0 ≤ |x| ≤ 5 ⇒ –5 ≤ x ≤ 5
(B) f(|x| + 1)
Range of f(|x| + 1) is [0, 2]

(C) Range of f(–|x|) is [–1, 0]

(D) –1 ≤ |x| ≤ 4
⇒ x ∈ [–4, 4]

47) f(x) = ; g(x) = 9 + x2 ; h(x) = –x2 – 3x + k

(A) Range of f is (0, 8]
(C) h[f(x)] > 0 and h{g(x)} < 0
h(0) ≥ 0 ⇒ k ≥ 0
h(8) > 0 ⇒ –64 – 24 + k > 0 ⇒ k > 88
h(9) < 0 ⇒ –81 – 27 + k < 0 ⇒ k < 108
Number of integral values of k is 19
(D) Maximum value of g{f(x)} is g(8) = 64 + 9 = 73

48) ∵ α, β, γ, δ are in AP with common difference d, then

β = α + d, γ = α + 2d and δ = α + 3d ....(i)
Given, α, β, γ are the roots of x3 – x2 + ax + b = 0, then
α+β+γ=1 ....(ii)
αβ + βγ + γα = a ....(iii)
αβγ = –b ....(iv)
Also, β, γ, δ are the roots of x3 – 4x2 + mx + n = 0, then
β+γ+δ=4 ....(v)
βγ + γδ + δβ = m ....(vi)
βγδ = –n ....(vii)
From Eqs. (i) and (ii), we get
3α + 3d = 1 ....(viii)
and from Eqs. (i) and (v), we get
3α + 6d = 4 ....(ix)
From Eqs. (viii) and (ix), we get

d = 1, α =
Now, from Eq. (i), we get

β = , γ = and δ =
From Eqs. (iii), (iv), (vi) and (vii), we get

a= ,b= ,m= ,n=

∴ a = m – 5, n = b – a – 2 and b = m + n – 3

49)
∴ Number of roots = 4

50)
Number of roots = 4

51) ƒ(x) + g(x) = |x – 1| + 2|x – 2| + 3|x – 3| + 4|x – 4| + 5|x – 5|.

clearly minimum is achieved at x = 4 & (ƒ(x) + g(x))min = 15.

52) ƒ(x) = g(x) has two intersection points.

Plot the graphs of ƒ(x) & g(x)

53) ƒ(x) = (|x| – 7)(|x| – (2k – 9))

|x| = 7 ⇒ x = ±7
|x| = 2k – 9

for 4 real solutions 2k – 9 > 0 ⇒ k>

Also 2k – 9 ≠ 7
k≠8

54) For 3 real solutions, 2k – 9 = 0 ⇒ k=

55) x2 – 12 x + [x] + 31 = 0
x2 – 12x + 31 = –[x]
(x – 6)2 – 5 = –[x]
By graph

zero point of intersection, m = 0

x2 – 5 | x + 2 | – 4 = 0
case-I : x < –2
x2 + 5x + 6 =
x = – 3, – 2 (rejected)
case-II : x ≥ –2
x2 – 5x – 14 = 0
x = 7, – 2
No. of solution (n) = 3
So m2 + mn + n2 = 9

56) (1)

(2)
(3)
(4)
Considering all the cases, permissible integral value of m are

57) Let x = α be common zero then

α2 + aα + b = 0 .... (1)
2
α + α + ab = 0 .... (2)
2
aα + α + b = 0 .... (3)
 (1) – (2)
⇒ (α – 1) (α – b) = 0
(1) or (3) gives b2 + ab + b = 0
and (3) gives ab2 + 2b = 0
b ≠ 0 ⇒ ab = –2, if ab = –2, b2 + b – 2 = 0
b = 1 or –2, a = –2 or 1, a = –2, b = 1 and a = 1, b = –2
(does not satisfy the mentioned condition polynomial will be same).
a + 2b = 0.