27-07-2025

3010CJA101001250005 JM

PART-1 : PHYSICS

SECTION-I

1)

A person can throw a ball to a maximum height 40 m find the maximum distance to which person
can throw the stone :

(A) 50 m
(B) 80 m
(C) 75 m
(D) 25 m

2) At a height 0.4 m from the ground, the velocity of a projectile in vector form is : .
The angle of projection is :- (g = 10m/s2)

(A) 45°
(B) 60°
(C) 30°
(D) tan–1 (3/4)

3) From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 m/s at an
angle of elevation of 30°, then the ratio of the total time taken by the ball to hit the ground to reach
at same height is :- (Take g = 10 m/sec2)

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

4) A particle is projected horizontally with a speed of m/s, from some height at t = 0. At what

time will its velocity make 60° angle with the initial velocity.

(A) 1 sec
(B) 2 sec
(C) 1.5 sec
(D) 2.5 sec

5) To a man walking at the rate of 3 km/h the rain appears to fall vertically. When he increases his
speed to 6 km/h it appears to meet him at an angle 45° with vertical. Find the speed of rain :-

(A)
(B)

(C)

(D)

6) A ball is dropped vertically on ground. An observer sitting on a train that moves towards right
with constant velocity observes the ball. The trajectory observed is:

(A)

(B)

(C)

(D)

7) A person, reaches a point directly opposite on the other bank of a flowing river, while swimming
at a speed of 5 m/s at an angle of 120° with the flow. The speed of the flow must be

(A) 2.5 m/s

(B) 3 m/s
(C) 4 m/s
(D) 1.5 m/s

8) A coin is released inside a lift at a height of 2m from the floor of the lift. The height of the lift is
10m. The lift is moving with an acceleration of 11 m/s2 downwards. The time after which the coin
will strike with the lift is :

(A) 4 s
(B) 2 s
(C)

(D)

9) A swimmer swims in still water at a speed = 5 km/hr. He enters a 200 m wide river, having river
flow speed = 4 km/hr at point A and proceeds to swim at an angle of 127° with the river flow
direction. Another point B is located directly across A on the other side. The swimmer lands on the
other bank at a point C, from which he walks the distance CB with a speed = 3 km/hr. The total time
in which he reaches from A to B is

(A) 5 minutes
(B) 4 minutes
(C) 3 minutes
(D) None of these

10)

A ball is horizontlly projected with a speed 'u' from the top of a plane inclined at an angle 45º with
the horizontal. How far from the point of projection will the ball strike the plane?

(A)

(B)

(C)

(D)

11) Two trains A & B 100 km apart are travelling towards each other on different tracks with
starting speed of 50 km/h for both. The train A accelerates at 20 km/h2 and the train B retards at the
rate 20 km/h2. The distance covered by the train A when they cross each other is :-

(A) 45 km
(B) 55 km
(C) 65 km
(D) 60 km

12)

Two particles A and B are moving in horizontal plane as shown in figure at t = 0, then time after
which A will catch to B will be-
(A) 4 s
(B) 5 s
(C) 2 s
(D) 8 s

13) The graph shows the variation of velocity of a rocket with time. Then, the maximum height
attained by the rocket is :-

(A) 1.1 km
(B) 5 km
(C) 55 km
(D) none of these

14) A body is released from the top of a tower of height H metres. It takes t time to reach the
ground. Where

is the body time after the release

(A)
At metres from ground

(B)
At metres from ground

(C)
At metres from the ground

(D)
At metres from the ground

15) A stone is thrown vertically upward with an initial velocity u from the top of a tower, reaches the
ground with a velocity 3u. The height of the tower is:

(A)

(B)

(C)

(D)
16) Block A of mass 4 kg is to be kept at rest against a smooth vertical wall by applying a force F as

shown in figure. The force required is :- (g = 10 m/s2)

(A) 40 N
(B) 20 N
(C) 10 N
(D) 15 N

17) In the two cases respectively, the ratio of contact force between the blocks will be :-

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

18) For the arrangement shown in figure the tension in the string is :-

(A) mg/2
(B) 3mg/2
(C) mg
(D) 2 mg

19) For the figure shown what will be the tension in the rope connecting blocks of mass m and 5 m :-

(A)

(B)
(C)

(D) Zero

20) In the given figure by what acceleration the boy must go up so that 100 kg block remains
stationary on the wedge? The wedge is fixed and friction is absent everywhere (g = 10m/s2) :-

(A) 2 m/s2
(B) 4 m/s2
(C) 6 m/s2
(D) 8 m/s2

SECTION-II

1) Blocks A and B of masses 2 kg and 1 kg respectively are pushed up a frictionless slope by a 21 N

force applied parallel to the slope as shown in the figure. Find the magnitude of the force of normal

reaction between A and B.[g = 10 m/s2]

2) Figure represents a painter in a crate which hangs alongside a building. When this painter of
mass 100 kg pulls the rope, the force exerted by him on the floor of the crate is 450 N. If the mass of

the crate is 25 kg, the acceleration of the painter will be (Take g = 10 m/s2):-

3) A uniform rope AB of mass 1 kg connects two blocks of mass 1 kg and 3 kg as shown. A pulling
force of 20 N is applied on 3 kg as shown, then find the tension at end B of the rope (in N)?
4) Two monkeys of masses 10 kg and 8 kg are moving along a vertical light rope, the former
climbing up with an acceleration of 2 m/s2, while the latter coming down with a uniform velocity of

2m/s. Find tension in the rope at the fixed support. (in N)

5) The spring shown in the figure has a natural length of 1 m. What is the initial acceleration of the

block when released (in m/s2)?

PART-2 : CHEMISTRY

SECTION-I

1) Which of the following sample contains maximum no. of atoms :-

(A) 1 mg of O2
(B) 1 mg of N2
(C) 1 mg of Na
(D) 1 mg of water

2) Arrange the following in order of increasing Masses (Atomic mass : N= 14, Fe = 56, O = 16) :-
(i) 1 Mole of oxygen molecule
(ii) 1 Molecule of oxygen
(iii) 10–10 g of iron
(iv) 1 g-atom of nitrogen

(A) i < ii < iii < iv

(B) i < iii < iv < ii
(C) iii < iv < ii < i
(D) ii < iii < iv < i

3) If in urea [CO(NH2)2] there are 20 g -atoms of nitrogen present then the mass of urea will be-

(A) 600 g
(B) 60 g
(C) 6 kg
(D) 120 g

4) In the different experiments a-particles, proton, deuteron and neutron are projected towards gold
nucleus with the same kinetic energy. The distance of closest approach will be minimum for

(A) α-particle
(B) proton
(C) deuteron
(D) neutron

5) In a certain electronic transition in the hydrogen atoms from an initial state (A) to a final state (B),
the difference in the orbit radius is 24 times the first Bohr radius. Identify the transition :

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

6) Amongst the following statements, that which was not proposed by Dalton was :

all the atoms of a given element have identical properties including identical mass. Atoms of
(A)
different elements differ in mass.
chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in
(B)
a chemical reaction.
when gases combine or reproduced in a chemical reaction they do so in a simple ratio by
(C)
volume provided all gases are at the same T & P.
(D) matter consists of indivisible atoms.

7) The radius of the second Bohr orbit of Li2+ ion, in terms of the Bohr radius,

(A)

(B)
(C)

(D)

8) The minimum energy that must be possessed by photons in order to produce the photoelectric
effect with platinum metal is:
[Given: The threshold frequency of platinum is 1.3 × 1015 s–1 and h = 6.6 × 10–34 J s.]

(A) 3.21 × 10–14 J

(B) 6.24 × 10–16 J
(C) 8.58 × 10–19 J
(D) 9.76 × 10–20 J

9) P is the probability of find the 1s electron of hydrogen atom in a spherical shell of infinitesimal
thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr2dr. The qualitative
sketch of the dependence of P on r is

(A)

(B)

(C)
(D)

10) Which of the following orbital does not make sense?

(A) 3d
(B) 3f
(C) 5p
(D) 7s

11) In which of the following compounds cation and anion size ratio is minimum ?

(A) CsF
(B) LiI
(C) LiF
(D) CsI

12) Which of the following is not isoelectronic series ?

(A) Cl–, P3–, Ar

(B) N3–, Ne, Mg2+
(C) B3+, He, Li+
(D) N3–, S2–, Cl–

13) According to Slater rule which of the following has the highest screening constant for last
electron :-

(A) Fluorine (F)

(B) Oxygen (O)
(C) Carbon (C)
(D) Nitrogen (N)

14) Which of the following orders for electron affinity is/are correct ?
(a) S > O < Se (b) Cl > F
(c) S > O (d) O > S
(e) N > P (f) C > B

(A) a, b, c, e
(B) a, b, c, f
(C) b, c, d, e
(D) b, c, f

15) The successive I.E. for an element X are as follows

X X+ X+2 X+3
If I.E.1 and I.E.3 values are 30 kJ / mole and 50 kJ / mole respectively, then the value of I.E.2 is
_________ kJ/mole :

(A) 29
(B) 30
(C) 50
(D) 40

16) Which of the following pairs of elements have almost similar atomic radii :-

(A) Zr, Hf
(B) Mo, W
(C) Co, Ni
(D) All

17) The nomenclature of ICl is iodine chloride because

(A) Size of I < Size of Cl

(B) Atomic number of I > Atomic number of Cl
(C) E.N. of I < E.N. of Cl
(D) E. A. of I < E. A. of Cl

18) Which of the compound is least soluble in water

(A) AgF
(B) AgCl
(C) AgBr
(D) AgI

19) Ionic conductances of hydrated M+ ions are in the order –

(A) Li+ (aq) > Na+ (aq) > K+ (aq) > Rb+ (aq) > Cs+ (aq)
(B) Li+ (aq) > Na+ (aq) < K+ (aq) < Rb+ (aq) < Cs+ (aq)
(C) Li+ (aq) > Na+ (aq) > K+ (aq) > Rb+ (aq) < Cs+ (aq)
(D) Li+ (aq) < Na+ (aq) < K+ (aq) < Rb+ (aq) < Cs+ (aq)

20) When two atoms combine to form a molecule:-

(A) Energy is released

(B) Energy is absorbed
(C) Energy is neither released nor absorbed
(D) Energy may either released or absorbed
 SECTION-II

1)

3.0 litre of water are added to 2.0 litre of 5 M HCl. What is the molarity of HCl (in M) in resultant
solutions

2) The number of electrons for Zn+2 cation that have the value of azimuthal quantum number equals
to zero is

3) The work function (ϕ) of some metals is listed below. The number of metals which will show
photoelectric effect when light of 300 nm wavelength falls on the metal is :

4)

Ratio of number of protons to neutrons in 3.011 × 1022 ions of D3O+ is -

5) Among F, O, S, Al, Zn, He and N the number of elements having electron affinity less than Cl.

PART-3 : MATHEMATICS

SECTION-I

1) If , then number of values x is

(A) 1
(B) 2
(C) 3
(D) None of these

2) Number of solutions of equation in [0,2π] which are solutions of equation

(A) 0
(B) 1
(C) 2
(D) 4

3) If , then the greatest positive solution of 1 + sin4 x = cos2 3x is -

(A) π
(B) 2π

(C)

(D) none of these

4) The possible values of θ ∈ (0, π) such that sin (θ) + sin (4θ) + sin(7θ) = 0 are:

(A)

(B)

(C)

(D)

5) If z = 5 + 2i, then the modulus of z, |z|, is :

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

6) If (x + iy)(2 – 3i) = 4 + i, then the values of x and y are :

(A) x = 2/13, y = 11/13

(B) x = –2/13, y = 11/13
(C) x = 2/13, y = –11/13
(D) x = –2/13, y = –11/13

7)

Roots of the equation

(x – a)2 + (x – b)2 + (x – c)2 + 4(a + b + c)x = 0 are (where a ≠ b ≠ c & a,b,c ∈ R)

(A) always real
(B) imaginary
(C) real & equal
(D) can't say

8) The quadratic equation with rational coefficients whose one root is tan15º, is-

(A) x2 – 2x + 2 = 0
(B) x2 – 4x + 2 = 0
(C) x2 – 4x + 1 = 0
(D) x2 – 3x + 2 = 0
9)

Graph of y = ax2 + bx + c is given in the adjacent figure, then which of the following is correct -

(where α,β are the zeros of ax2 + bx + c)

(A) α + β > 0, αβ < 0
(B) b < 0, D < 0
(C) b > 0, D > 0
(D) α + β < 0, αβ < 0

10) If the equation (p2 – 3p + 2)x2 + (p – 1)2x + (p2 – 4p + 3) = 0 is satisfied for more than two
different values of x, then p is equal to -

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

11) If 3α2 – 6α + 5 = 0 & 3β2 – 6β + 5 = 0, α ≠ β, then value of is -

(A)

(B)

(C)

(D)

12) If α & β are the roots of the quadratic equation x2 – (k – 2)x – k + 1 = 0, then minimum value of α2
+ β2 is -

(A) 1
(B) 2
(C) 5/4
(D) 3/4

13) The roots of quadratic equation x2 – 12x + t = 0 are positive and one of them is square of the
other, if roots are α and β with α > β then

(A) 3α + β – t = 5
(B) 2α + 3β – t = 0
(C) α + β – t = 4
(D) α + 2β – t = 0

14) If equations 3ax2 + 4bx + c = 0 (a, b, c ∈ R) and 5x2 + 2x + 1 = 0 have a common root, then the

value of (b ≠ 0) is -

(A)

(B)

(C)

(D)

15) If α and β are the roots of equation 27x2 – 4x + 64 = 0, then the value of is -

(A)

(B)

(C)

(D)

16) The curve of the quadratic expression y = ax2 + bx + c is shown in the figure and α, β be the
2
roots of the equation ax + bx + c = 0, then correct option is (D is discriminant)

(A) a < 0, b < 0, c < 0, D < 0, α + β > 0, αβ > 0

(B) a < 0, b > 0, c < 0, D < 0, α + β > 0, αβ > 0
(C) a < 0, b > 0, c > 0, D < 0, α + β > 0, αβ < 0
(D) a < 0, b > 0, c > 0, D < 0, α + β > 0, αβ > 0

17) If the equation has roots equal in magnitude but opposite in sign, then m is
equal to-
(A)

(B)

(C)

(D) None of these

18) The expression a2x2 + bx + 1 will be positive for all x ∈ R if-

(A) b2 > 4a2

(B) b2 < 4a2
(C) 4b2 > a2
(D) 4b2 < a2

19) Solve the inequation

(A)

(B)

(C)

(D)

20) Find the value of m for which the expression can take all real values for

(A) m ∈ (4, 6)
(B) m ∈ [4, 6]
(C) m ∈ (4, 8)
(D) m ∈ (4, 6]

SECTION-II

1) If sinθ and cosθ are the roots of the quadratic equation ax2 + bx + c = 0 (ac ≠ 0) then value of

is equal to –

2) If the equations x2 - 11x + a = 0 and x2 - 14x + 2a = 0 must have a common root and a ≠ 0, then
‘a’ is
3) In (0, 6π), find the number of solutions of the equation tanθ + tan 2θ + tan 3θ = tanθ.tan2θ.tan3θ

4) If cos 2θ + 9 sin 2θ – 6 sin θ + 54 cos θ = 1 then the value of 100 tan2 θ + 9 tan θ is equal to
............

5) The positive integer value of n > 3 satisfying the equation 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 C A B A C A A B D D B C C B A B C A C

SECTION-II

Q. 21 22 23 24 25
A. 7 4 4 200 4

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

SECTION-II

Q. 46 47 48 49 50
A. 2 6 4 1 7

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

SECTION-II

Q. 71 72 73 74 75
A. 2 24 17 8181 7
 SOLUTIONS

PART-1 : PHYSICS

1)

Answer. : 80 m
Answer. : 80 m .

2)

Answer. : 30° .
Answer. : 30° .

3)

Answer. : 2 : 1
Answer. : 2 : 1

4)

Answer. : 2 sec
Answer. : 2 sec .

5)

x=3

6) For observe in train relative to it

Uinitial = In horizontal direction toward left
Acceleration = vertically downward
7)
For zero dift.

8) For the coin w.r.t. lift, relative vel = 0

acceleration = 11 – 10 = 1 m/s2 upwards
Therefore, it hits the lift after travelling for 8 m.

10)

Answer. : . .

11)

Answer. : 60 km.

12)

To catch (vy)A = (vy)B

⇒ vsin37° = 30

⇒v= = 50 m/s

Time taken = = 5 s.

13) Maximum height = × 110 × 1000 = 55 km

14)

Answer. : At metres from the ground

15) Let h be the height of the tower.

Using v2 – u2 = 2 as, we get;
(–3u)2 – u2 = 2(–g)(–h)
Here, u = u, a = – g, s = –h and v = –3u (upward direction +ve)
∴ 9u2 – u2 = 2gh or h = 4u2/g

16)

Answer. : 40 N .

17) a=

N1 = 2Ma =

a=

N2 = Ma =

18)

Let retardation provided by wooden block is 'a' and initial velocity of body is 'ν'. Using
3rd equation of motion,

⇒
Further,

Where, x = 1cm is distance travelled by body after 3cm distance and before coming to rest.

19)

For block of mass 'm'.

mg – T = ma

T = m(g – a) = m
20)
T = 100 × 10 × sin 53°
T = 800 N
T – 500 = 50a
a = 6 m/s2

21)

Answer (7)

22)
ANSWR
equation for painter & crate system.
2T – 125 g = 125 a .... (1)
equation for painter
Np + T –100 g = 100 a (Np = 450 N)
T – 550 = 100 a ..... (2)
from eqation (1) & (2)
a = 2 m/sec2

23)

Answer (4)
A uniform rope AB of mass 1 kg connects two blocks of mass 1 kg and 3 kg as shown. A pulling
force of 20 N is applied on 3 kg as shown, then find the tension at end B of the rope (in N)?

24)

Answer (200)
 25)

Answer (4)

PART-2 : CHEMISTRY

26)

No. of atoms ⇒

27) (i) 1 Mol of oxygen = 32 g

(ii) 1 Molecule of oxygen =

(iii) 10–10 g of iron = 10–10 g
(iv) 1 g atom of nitrogen = 14 g

28) Moles of N-atom = 20 moles

Moles of urea = 10 moles
Moles of urea = = 600 gm

29) Neutron because it has zero charge.

30) Initial state (A) final state (B)

Put the value of in equation (1)

24 = ...(2)

,
will satisfy the above equation (2)

31)

The theory given by the Dalton was for the atoms and its properties, it does not depend on
volume. C option is wrong.

32)
 (n = 2, z = 3)

33)

W = hv°
= 6.6 × 10–34 × 1.3 × 1015
= 8.58 × 10–19 J

34) ρ = Ψ × 4πr2dr
∴ 1s orbital
at r = 0 ρ = 0
∴ (B)

35)

Ans-(B)
3f orbital does not exist according to quantum theory In 3f orbital, l = 3, and according to
quantum theory, the quantum number (l) should be less than n. So due to this reason 3f orbital
does not exist.

36) is lowest when cation is small but anion is large.

37) N3– is not isoelectronic with S2– and Cl–

38) Fluorine (F) has maximum no. of inner electrons.

39) Most of the 3rd period elements have higher E.A. as compared to corresponding 2nd period
element. Elements which have s2 or p3-configuration in outermost shell have low E.A. as
compared to its next and previous element in periodic table
40) IE1 = 30 kJ/mole IE3 = 50 kJ/mole so IE2 should be between 30 and 50 so correct ons is 40
kJ/mole

41)
, ,

42) ICl → Iodine chloride

EN of I < EN of Cl

43) Ionic compound are move soluble in water

AgF AgCl AgBr AgI Ionic char.

44) Conductivity

45)

Energy is released

46)

Ans : (2)

M1V1 = M2V2
2 × 5 = M2 × 5
M2 = 2M

47)

48) 4 will show photoelectric effect

E = ϕ + K.E

4.13 = ϕ + K.E.
(Li) 4.1 – 2.4 = K.E (Mg) 4.1 – 3.7 = K.E.
1.7. = K.E. 0.4 = K.E
(Na) 4.1 – 2.3 = K.E (Cu) 4.1 – 4.8 = K.E.
1.8 = K.E. –0.7 = K.E.
(K) 4.1 – 2.2 = K.E. (Ag) 4.1 – 4.3 = K.E.
 1.9 = K.E. 0.2 = K.E.

49)

50) Explanation : elements having less EA than 'Cl' is asked

Concept : EA

Solution/Explanation/Calculation :
'Cl' has max EA in P.T
∴ All elements has less EA than Cl.
correct answer := 7.

PART-3 : MATHEMATICS

51)

Answer. : 2
Answer. : 2.

52)
θ ∈ [0,2π]
adding both the equations

which is possible when

⇒
None of them satisfies the given equations
⇒ No solutions.

53) sinx = 0 and cos3x = ±1

54) sinθ + sin4θ + sin7θ = 0 2sin cos + sin4θ = 0

⇒ sin4θ[2cos3θ + 1] = 0
⇒ sin4θ = 0 ⇒ 4θ = 0, π, 2π, 3π, 4π

⇒ θ=0, ,π
but 0 and π are not included. and 2cos3θ + 1 = 0

⇒ cos3θ =
⇒ 3θ =

⇒ θ=

but (0, π)

So, θ =

56)

real part is zero

x=nπ ±

57) (x – a)2 + (x – b)2 + (x – c)2 + 4(a + b + c)x = 0

⇒ (x + a)2 + (x + b)2 + (x + c)2 = 0
⇒ if a ≠ b ≠ c then we have imaginary roots.

58) x = tan15º =
⇒ ⇒

59) It is obvious from figure that roots are of opposite sign but –ve root has greater magnitude

60) Question Explanation:

In this question we have a quadratic equation where coefficients are involving a parameter p,
which have more than two distinct values of x, then we have to find the value of parameter p.

Concept:
This question is based on theory of equation.
A quadratic equation ax2 + bx + c = 0 will become an identity if it has more than two roots and
the required condition is a = b = c = 0

Solution:
(p2 – 3p + 2)x2 + (p – 1)2x + (p2 – 4p + 3) = 0
Since given quadratic equation has more than 2 distinct values of x then it will become an
identity.
⇒ Coefficient of x2 = 0 and coefficients of x = 0 and constant term = 0
⇒ p2 – 3p + 2 = 0 and (p – 1)2 = 0 and p2 – 4p + 3 = 0
⇒ (p – 1)(p – 2) = 0 ⇒p=1 ⇒ (p – 1)(p – 3) = 0
⇒ p = 1 or 2 ⇒ p = 1 or 3
∴ So

Final Answer:
The correct option is (1). 1

61)

Question Explanation:

The problem provides two identical equations satisfied by α and β respectively with α ≠ β. This
implies α and β are the distinct roots of the quadratic equation 3x2 – 6x + 5 = 0 and we have to

calculate the value of the expression .

Concept:

This question is based on relation between roots and coefficient of quadratic equation.
If f(α) = 0 & f(β) = 0 then we can say a and b are the roots of f(x) = 0

Solution:
Given: 3α2 – 6α + 5 = 0 and 3β2 – 6β + 5 = 0 with α ≠ β
So, α and β are the roots of quadratic equation
3x2 – 6x + 5 = 0

Sum of roots = (for f(x) = ax2 + bx + c)

Product of roots =

α+β= =2

Now,

Final Answer:

The correct option is (2).

62) α2 + β2 = (α + β)2 – 2αβ = (k – 2)2 – 2(–k + 1) = k2 – 2k + 2 = (k – 1)2 + 1
min. of α2 + β2 = 1 for k = 1

63) x2 – 12x + t = 0 roots are α, β

α + β = 12 αβ = t
α+β=9+3 9×3=t
⇒ α = 9, β = 3 t = 27

64) 5x2 + 2x + 1 = 0
D<0
∴ Both roots are imaginary

Hence : ⇒

65)

66) (1) a < 0

(2) D < 0

(3)

∴ b>0
(4) ƒ(0) = c < 0
c<0

(5) ⇒
α+β>0
(b) αβ > 0

67) x2(m + 1) – xb(m + 1) = ax(m – 1) – c(m – 1)

x2(m + 1) – x [bm + b + am – a] + c (n – 1) = 0
If root's equal in magnitude but opposite in sign then
b=0
∴ –[bm + b + am – a] = 0
m(a + b) = a – b

68) a2x2 + bx + 1 > 0

∵a>0&D<0
∴ b2 – 4a2 < 0
b2 < 4a2

69) Answer : (C)

70) Answer : m ∈ (4, 6)

71)

Answer. 2
Answer. : 2

72) Let be α common root,

∴

For common root to exist

a = 24

73) tanθ + tan 2θ + tan 3θ – tanθ tan 3θ tan 2θ = 0 ...(1)

⇒ tan(θ + 2θ + 3θ) = 0
⇒ tan 6θ = 0
⇒ 6θ = nπ

⇒θ=

74) 9 sin 2θ – 6 sin θ + 54 cos θ = 2 sin2 θ

18 sin θ cos θ – 6 sin θ + 54 cos θ = 2 sin2 θ
18 cos θ (sin θ + 3) = 2 sin2 θ + 6 sin θ
18 cos θ (sin θ + 3) = 2 sin θ (sin θ + 3)
⇒ (sin θ + 3) [18 cos θ – 2 sin θ] = 0
as sin θ = – 3 is not possible
⇒ 9 cos θ – sin θ = 0
⇒ tan θ = 9
⇒ 100 tan2 θ + 9 tan θ = 100 (81) + 81
⇒ 100 tan2 θ + 9 tan θ = 8100 + 81 = 8181

75) Answer : 7