18-08-2025

4602CJA105021250010 JA

PART-A-PHYSICS

SECTION-I(i)

1) Two points A & B on a disc have velocities v1 & v2 at some moment. Their directions make angles
60° and 30° respectively with the line of separation as shown in figure. The angular velocity of disc

is :

(A)

(B)

(C)

(D)

2) A disc of radius R has a light pole fixed perpendicular to the disc at the circumference which has a
pendulum of length R attached to its other end as shown in figure. The disc is rotated with a
constant angular velocity ω. The string is making an angle 30° with the rod. Then the angular

velocity ω of disc is :

(A)

(B)

(C)

(D)
3) A block of mass 2kg is placed on wedge having an angle 37° with horizontal. Wedge rotates about
an axis as shown with angular velocity 5 rad/s. If block just start accelerating upward w.r.t. wedge
with acceleration of 1 m/s2 at the given instant find friction force acting on block. (Distance of block

is 1m at the given instant.)

(A) 16 N
(B) 8 N
(C) 26 N
(D) 38 N

4) A particle initially at rest starts moving from point A on the surface of a fixed smooth hemisphere
of radius r as shown. The particle looses its contact with hemisphere at point B. C is centre of the

hemisphere. The equation relating α and β is :-

(A) 3 sin α = 2 cos β

(B) 2 sin α = 3 cos β
(C) 3 sin β = 2 cos α
(D) 2 sin β = 3 cos α

5) A skier plans to ski a smooth fixed hemisphere of radius R. He starts from rest from a curved
smooth surface of height (R/4). The angle θ at which he leaves the hemisphere is :-

(A) cos–1 (2/3)

(B)

(C)

(D)

6) ACB is a smooth quater circular path of radius R. F4 is always directed from particle’s position to
point B. Work done by F4 is :

(A) FR
(B)

(C)

(D)

SECTION-I(ii)

1) A particle of mass m is kept on a wedge of mass M as shown in the figure, system is released from
rest. Horizontal part (AB) of wedge is rough with coefficient of friction µ and it is long enough.
Horizontal surface on which wedge is kept is smooth then choose the correct statement/s.

(A) Final velocity of particle is zero

(B) Final velocity of wedge is zero
(C) Displacement of the particle with respect to wedge on horizontal part of it is h/µ
(D) Total work done by contact force applied by particle on wedge is positive

2) A pendulum bob of mass m connected with a string of length ℓ lies in vertical plane. It is given
just sufficient velocity so as to complete vertical circle. Mark the CORRECT statements :-

(A) Net acceleration at B has magnitude

(B) Net workdone by all the forces as bob moves from A to B is –mgℓ
(C) Net acceleration at B has magnitude
Taking top most point as reference total mechanical energy is
(D)

3) A small package of weight w is projected into a vertical return loop at A with a velocity v. The
package travels without friction along a circle of radius r and is deposited on a horizontal surface at

C. For each of the two loops shown :-

In figure (a) the smallest velocity for which the package will reach the horizontal surface at C is
(A)
given by
In figure (b) the smallest velocity for which the package will reach the horizontal surface at C is
(B)
given by
(C) In figure (a) force exerted by the loop on the package as it passes point B is given by 3 mg.
(D) In figure (b) force exerted by the loop on the package as it passes point B is given by 2 mg.

4) Initially spring is compressed by 0.5 m and surface is smooth. Choose the correct option(s) :-

(A) Initial potential energy in spring is 12.5 J.

(B) Kinetic energy of block A, when spring is at natural length for first time is 12.5 J.
(C) Work done by all forces on B is zero till spring attains it's natural length for first time.
(D) None of these.

5) Surface of inclined wedge is frictionless and block does not slips on inclined wedge, initial velocity

of system is zero. Mass of wedge is m (g = 10 m/s2)

(A) Rate of change of momentum of system (block + wedge) is equal to (m + 10) .

(B) a = 7.5 m/s2
(C) Kinetic energy of block is increasing with rate 562.5t, where t is time in second.
(D) Work done by normal on the wedge is negative.

6) A body of mass 'm' is moving slowly up the rough hill from point A to point B as shown in figure by
a force which is acting tangential to surface at each point on the hill. Work done by this force is :
(A) Independent of shape of trajectory.
(B) Independent of horizontal component of displacement.
(C) Depends on coefficient of friction between the surfaces.
(D) Depends on the mass of the body.

SECTION-III

1) The block of mass m is attached to a frame by spring of force constant k, entire system moves
horizontally without any frictions anywhere. The frame and block are at rest with x = x0, the relaxed
length of spring. If frame is given a constant acceleration a0. Determine the maximum velocity (in
m/s) of block relative to frame. (Given : a0 = 2 m/s2, m = 4 kg, k = 4 N/m)

2) A bob is attached to a string of length ℓ = . The string is held horizontal and the bob is
released from this position. At the time when the centripetal acceleration becomes equal in
magnitude to the tangential acceleration, find the speed (in m/s) of the bob. (g = 10 m/s2)

3) A 2 kg block is gently pushed from rest at A and it slides down along the fixed smooth circular
surface as shown in figure. If the attached spring has a force constant k = 20 N/m. What is
unstretched length of spring (in m) so that it does not allow the block to leave the surface until angle

with the vertical is θ = 60°. (Take g = 10 m/s2)

4) Two blocks of masses m & 2m are attached to a spring and is kept on the floor as shown. The
minimum compression in the spring from its natural length so that the block of mass 2m leaves

contact with the surfaces is N . Fill “N” is OMR sheet :-

5) A particle of mass 'm' connected to a string of length 'l' then find out the tension in the string
when the particle is at its bottom most point. The particle is projected horizontally with a speed
from the bottom most point. Find Z if the tension is Zmg.

PART-B-CHEMISTRY

SECTION-I(i)

1) Choose the correct option for free expansion of an ideal gas under adiabatic condition from the
following :

(A) q = 0, ΔT ≠ 0, w = 0
(B) q = 0, ΔT < 0, w ≠ 0
(C) q ≠ 0, ΔT = 0, w = 0
(D) q = 0, ΔT = 0, w = 0

2) A system absorbs 100 kJ heat in the process shown in the figure. What is ΔU for the system?

(A) −50 kJ
(B) +50 kJ
(C) +150 kJ
(D) −150 kJ

3) Two moles of helium gas undergoes a cyclic process as shown in the figure. Assuming ideal
behaviour of gas, the magnitude of net work done by the gas in this cyclic process is :-
(A) 0
(B) 100R ln2
(C) 100R ln4
(D) 200R ln4

4) Enthalpy of CH4 + O2 CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x
and y respectively. Then which relation is correct :

(A) x > y
(B) x < y
(C) x = y
(D)

5) Consider the reaction equilibrium

On the basis of Le-Chatelier's principle, the condition favourable for the forward reaction is -

(A) Lowering the temperature and increasing the pressure

(B) Any value of temperature as well as pressure
(C) Lowering of temperature as well as pressure
(D) Increasing temperature as well as pressure

6) The equilibrium constant for a reaction A + B C + D is 1 × 10–2 at 298 K and is 2 at 273 K. The
chemical process resulting in the formation of C and D is :

(A) exothermic
(B) endothermic
(C) unpredictable
(D) there is no relationship between ΔH and K

SECTION-I(ii)

1) Which of the following is/are correct about chemical equilibrium :-

(A) Equilibrium condition is most stable condition under given conditions

(B) Equilibrium can be achieved from both reactant as well as product side
(C) Catalyst does not affect the equilibrium constant & equilibrium composition
(D) For any given reaction equilibrium constant depends on temperature only
2) The vapour density of N2O4 at a certain temperature is 23. What is the % dissociation of N2O4 at
this temperature :

(A) 53.3%
(B) 100%
(C) 106.6%
(D) 50%

3) P-V plot for two gases (assuming ideal) during adiabatic processes are given in the figure. Plot A

and plot B should correspond respectively to :

(A) He and H2
(B) H2 and He
(C) SO3 and CO2
(D) N2 and Ar

4) An amount of 4 moles of an ideal monoatomic gas expands adiabatically and reversibly by which
its temperature decreases from 47°C to 17°C. Which of the following is/are true?

(A) q = 0
(B) ΔH = 0
(C) ΔH = –600 cal
(D) ΔU = –360 cal

5) ΔS is positive in which of the following reactions/processes ?

(A) 2C(s) + O2(g) → 2CO(g)

(B) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
(C) Br2(l) → Br2(g)

(D)

6) When 0.5mL of a 1.0MHCl solution is mixed with 5.0mL of a 0.1M NaOH solution, temperature of
solution increases by 2°C. Which of the following(s) can be predicated accurately from this
observation?

(A) If 10mL of same HCl is mixed with 10mL of same NaOH , temperature rise will be 4°C
(B) If 10mL of 0.05 M HCl is mixed with 10mL of 0.5M NaCl, the temperature rise will be 2°C
If 5mL of 0.1 M HCl is mixed with 5 mL 0.1 M NH3 solution, the temperature rise will be less
(C)
than 2°C
If 5 mL of 0.1 M CH3COOH is mixed with 5 mL 0.1 M NaOH, the temperature rise will be
(D)
less than 2°C
 SECTION-III

1) If the equilibrium constant of the reaction is 0.25, the equilibrium

constant of the reaction will be ________.

2) The reaction PCl5(g) PCl3(g) + Cl2(g) is at equilibrium. Now suddenly the volume is doubled.
The system is now allowed to attain a new equilibrium state. How many of the following are lesser at
the new equilibrium as compared to the old equilibrium ? Assume the entire process to be
isothermal.
(a) Moles of PCl5
(b) Moles of PCl3
(c) Moles of Cl2
(d) Concentration of PCl5
(e) Concentration of PCl3
(f) Concentration of Cl2
(g) Rate of the forward reaction
(h) Rate of the backward reaction
(i) equilibrium constant

3) When a system is taken from A to C through path ABC, 10 J of heat flows to the system and 4 J of

work is done by the system. How much heat flows into the system in path
ADC, if the work done by the system is 3 J ?

4) how many of following are intensive properties?

Temperature, Volume, Heat capacity, mole, entropy, enthalpy, internal energy, molarity, mass,
specific heat capacity :-

5) One mole of ideal monoatomic gas undergoes expansion along a straight line on P–V curve from
intial state (3 L, 8 atm) to final state B (7.5 L, 2 atm).Calculate the magnitude of q for the above
process in L atm.

PART-C-MATHEMATICS

SECTION-I(i)

1) Let such that –3 < f(x) < 2 ∀ x ∈ R and complete range of a is (p, s), then the
value of (s–p) is

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

2) If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest,
then a ratio of lengths of the sides of this triangle is

(A) 5 : 9 : 13
(B) 5 : 6 : 7
(C) 4 : 5 : 6
(D) 3 : 4 : 5

3) If the angle A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the
lengths of the sides opposite to A, B and C respectively, then the value of expression

is

(A)

(B)

(C) 1
(D)

4) If the equation 4x2 –4(5x + 1) + p2 = 0 has one root equals to two more than the other, then the
value of p is equal to

(A)

(B)
(C) 5 or –1
(D) 4 or –3

5) If roots of the equation ax2 + bx + c = 0 are α,β. Then is equal

to

(A) a2
(B) c2

(C)

(D)
6) If α, β, γ, δ are the roots of 5x4 – 4x3 + 3x2 – 7x + 1 = 0, then the value of |(1 – α)(1 – β)(1 – γ)(1 –
δ)| is

(A)

(B)

(C)

(D) 1

SECTION-I(ii)

1) Given that x2 – 3x – 4 < 0, < 0 and 2sin2x + 3sinx – 2 > 0. Then x can be

(A)

(B)

(C) π

(D)

2) For a quadratic plynomial f(x)=4x2 − 8kx + k, the statements which hold good are ?

(A) There is only one integral k for which f(x) is non-negative ∀ x ∈ R

(B) for k < 0 the number zero lies between the zeros of the polynomial.

(C)
f(x) = 0 has two distinct solutions in (0,1) for
(D) Minimum value of f(x) is k(1 + 12k) ∀ k ∈ R

3) The value of x satisfying the equation 22x – 8 × 2x = – 12 is

(A)

(B)

(C)

(D) 1

4) If x, y, z ∈ R+ and A, B, C are angles of a triangle such that x2 + y2 + z2 + 2xycosAcosB = 2xzcosB

+ 2yzcosA + 2xysinAsinB, then choose the correct option(s)

(A) z = xcosB + ycosA

(B)

(C) z2 = x2 + y2 + 2xycosC
(D) x + y > z

5)

If , where a, b, c, k are real positive

numbers then value of 'k' can be

(A) 1

(B)

(C)

(D) 2

6) If f(x) = x2 + bx + c and f(2 + t) = f(2–t) for all real numbers t, then which of the following is true
?

(A) f(1) < f(2) < f(4)

(B) f(2) < f(1) < f(4)
(C) f(2) < f(4) < f(1)
(D) f(2.1) < f(1.5) < f(3)

SECTION-III

1) For positive integer n, let Sn denotes the minimum value of the sum where a1,
a2,....,an are positive real numbers whose sum is 17. If there exist a unique positive integer n for

which Sn is also an integer, then is

2)

In a ΔABC with usual notations, if b2sin2C + c2sin2B = 12, then area of triangle will be

3) Find the largest natural number 'a' for which the maximum value of f (x) = a –1 + 2x – x2 is
smaller than the minimum value of g(x) = x2 – 2ax + 10 – 2a.

4) α, β, γ, δ are the root of the equation x4 – 3x + 7 = 0 then the sum of digits of (α4 + 4) (β4 + 4)(γ4 +
4)(δ4 + 4) is equal to
5) If α, β, γ are roots of equation x3 – 2x2 – 1 = 0 and Tn = αn + βn + γn then value of is equal
to ....
 ANSWER KEYS

PART-A-PHYSICS

SECTION-I(i)

Q. 1 2 3 4 5 6
A. D D C C C B

SECTION-I(ii)

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

SECTION-III

Q. 13 14 15 16 17
A. 2 4 1 4 4

PART-B-CHEMISTRY

SECTION-I(i)

Q. 18 19 20 21 22 23
A. D A C B A A

SECTION-I(ii)

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

SECTION-III

Q. 30 31 32 33 34
A. 4 6 9 3 9

PART-C-MATHEMATICS

SECTION-I(i)

Q. 35 36 37 38 39 40
A. C C D B C B

SECTION-I(ii)

Q. 41 42 43 44 45 46
A. A,B,D A,B,C A,D A,B,D A,C,D B,D
 SECTION-III

Q. 47 48 49 50 51
A. 6 3 1 9 2
 SOLUTIONS

PART-A-PHYSICS

1)

For rigid body separation between two point remains same.

v1 cos60° = v2 cos30°

= ⇒ v1 = v2

ωdisc = = = = =

ωdisc =

2)

3)
50cos37 – 20sin37 – fr = 2 × 1
40 – 12 – fr = 2
fr = 26 N

4) Let v be the speed of particle at B, just when it is about to loose contact.

From application of Newton's second law to the particle normal to the spherical surface.
 = mg sin β .......... (1)
Applying conservation of energy as the block moves from A to B.

mv2 = mg (r cos α – r sin β) ....... (2)

Solving 1 and 2 we get
3 sin β = 2 cos α

5) Conserving energy between points A & B,

mg =
Also EFD at point B will give,

6) Work done by displacement in the direction of force F1 = F1 R

Work done by F2 = F2 X dispalcement in the directon of force F2 = F2 R

Work done by F3 =
Workd done by F4 = F4 × displacement in the direction of F4 =

7)

(A) The surface is rough and long enough so, final velocity of particle is zero.
(B) Inertial must zero momentum is zero so final momentum the displacement of particle.
(C) It wedge is at rest mgh = mmgx ⇒ x n/m
(D) Contact force ⊥ displacement so, WD = 0.

8)

(A)

(B)

(C) a=
(D)

9)

1st case (a)

At 'C' mg = ⇒ vC =
Every conservative

vA =

EC ⇒
vB =

NB =
In case 'b' to reach at C, vC = 0

⇒
vA =

10) Ans. (A,B,C)

Wall = ΔKE

11)

w.r.t. plank So 10gsin37° = 10acos37°

= 7.5 m/s2

12)

By work energy theorem (WET)

WF = Wg = Wfr = ΔKE
WF – µmgx – mgh = 0 – 0
 ∴ WF = mgh + µmgx

i.e. independent of shape of trajectory here WF depends on x (horizontal component of AB),

coefficient of friction, mass of the body.

13)

Maximum velocity will be at elongation

Now from WET

v=

v=

14)
v2 = 2gℓ cos θ
at = g sin θ = 2g cos θ
tan θ = 2

= 4 m/sec

15) When the block has fallen by 60°,

Applying NLM along the radial direction

Extension in spring = 0.5 m ⇒ Natural length = (R–x) = 1m

16) For 2m to leave contact

kx = 2mg

x=
 for m : – mgh +
–2mgh + kh2 = kh2 + kx2 + 2khx + mgx

x=

17)

T – mg =

T = 4mg
∵ Given T = zmg/z mg = 4mg

PART-B-CHEMISTRY

18) During free expansion of an ideal gas under adiabatic condition q = 0, ΔT = 0, w = 0.

19)

W=–1× = –1.5 bar. m3 = kJ

= –150 kJ and q = +100 kJ

∴ ΔU = q + w = –50 kJ

20) wAB = –nRT . In = – 2 × R × 300 × In

= +600R . In 2
wBC = – nR(T2 – T1) = – 2 × R × (400 – 300)
= – 200 R

wCD = –nRT In = – 2 × R × 400 × In

= – 800 R . In 2
wDA = –nR(T2 –T1) = –2 × R × (300 – 400)
= +200R
∴ wnet = wAB + wBC + wCD + wDA = – 200 R . In 2
= –100 R In 4

21)

CH4 + 2O2 → CO2 + 2H2O, ∆H = x

X−Y<O
X<Y

22) Question Explanation:

Favourable conditions for the forward reaction.

Concept:
This question is based on Le-chatelier's principle.

Solution:

for forward reaction

Them.↓ because of exothermic reaction
pressure ↑ because of lower gaseous moles toward product side.

Final Answer:
The correct option is (1).

23) For exothermic reaction equilibrium constant with decrease in temp.

24)

Question Explanation:

Asking properties of equilibrium and equilibrium constant.

Concept:

This question is based on properties of equilibrium.

Solution:

(1) Equilibrium condition is stable and it doesn't change until we change pressure, volume,
temperature etc.
(2) Equilibrium can be achieved from both directions.
(3) Catalyst only affect time required to attain equilibrium but doesn't change equilibrium
constant and equilibrium composition.
(4) Equilibrium constant depends on temperature for a given equilibrium.
Final Answer:

The correct options are (1, 2, 3 & 4).

25) = 1 + (n – 1) α

=1+α

⇒1+a= =2
⇒ α = 1 or 100%

26) (B) PV = constant for isothermal proces

PVγ = constant for adiabatic process so more
value of γ, more decrease in pressure as volume increases.

27)

q = 0, ,
ΔU = nCvΔT = –360 cal
ΔH = nCpΔT = –600 cal

28) Concept of Entropy Change (ΔS) :

Entropy (ΔS) is a measure of disorded or randomness. If a reaction increases disorded, ΔS is
positive.

Analyzing Each Reaction :

1. C(s) + O2(g) → 2CO(g)

• Solid (C) turns into gas (CO)
• Gases have higher entropy than solids.
• ΔS > 0 (Positive Entropy Change)

2. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

• Gases (NH3, O2) react to form liquid (H2O)
• Liquids have lower entropy than gases
• ΔS is negative

3. Br2(l) – Br2(g)
• Liquid turning into gas
• Gas has more disorder than liquid
• ΔS > 0 (Positive Entropy Change)

4. A2(g) (latm) → A2(g) (0.1 atm)

• Same gas but at lower pressure
• Lower pressure means more volume & randomness
• ΔS > 0 (Positive Entropy Change)
 Correct answer is option (1, 3 and 4).

29) Weak acid and weak base gives less heat in neutralisation hence smaller temperature rise
than in case of strong acids and bases if all other conditions are same

30)

31) Application of Le-chatelier principle

32) will be same as u is a state function.

33)

Temperature, molarity, specific heat capacity are intensive properties

34) Work = 22.5 L atm

ΔUAB = –13.5 L atm
qAB = 9 L atm

PART-C-MATHEMATICS

35)
(1) x2 – ax – 2 < 2(x2 + x + 1) ⇒ a ∈(–6,2)
(2) x2 – ax – 2 > –3x2 – 3x – 3 ⇒ a ∈ (–1,7)
⇒ a ∈ (–1,2)

36) a < b < c are in A.P.

∠C = 2∠A (Given)

put a = b – λ, c = b + λ, λ > 0
⇒ required ratio = 4 : 5 : 6

37) Given : A, B, C are in A.P.

⇒ 2B = A + C ...(1)
∵ A + B + C = 180° ...(2)
from (1) & (2) ⇒ 3B = 180° ⇒ B = 60°

from sine rule

= 2(sinAcosC + cosAsinC)

E = 2sin(A + C) = 2sinB =

38) Let roots are α, α + 2

4λ2 – 20x – 4 + P2 = 0
2a + 2 = 5 ⇒ a – 3/2

α(α + 2) =

2
P = 25 P=±5

39) ax2 + bx + c = 0
x(ax + b) = –c

ax + b =

40) 5x4 – 4x3 + 3x2 – 7x + 1 = 5(x – α)(x – β)(x – γ)(x – δ)

put x = 1

41)

We have,
x2 – 3x – 4 < 0 –1 < x < 4 …(i)
Also,
 …(ii)
2
and 2 sin x + 3 sinx – 2 > 0
(sinx + 2) (2 sinx – 1) > 0

[ (sin x + 2) > 0]

…(iii)

42) A. f(x) ≥ 0 ∀ x ∈ R D≤0 K (4K − 1) ≤ 0 .

B. K < 0 product of roots < 0
C. for roots ∈ (0, 1), D > 0 K(4K − 1) > 0,

f(0) > 0 K > 0, and f(1) > 0

D. Min. value

43) (2x)2 – 8 × 2x = – 12
putting 2x = t
t2 – 8t + 12 = 0
t = 6, t = 2
2x = 6, 2x = 2
x = log2 6, x = 1

44) x2(cos2B + sin2B) + y2(cos2A + sin2A) + z2 + 2xycosAcosB – 2xzcosB – 2yzcosA –

2xysinAsinB = 0
⇒ (xcosB + ycosA – z)2 + (xsinB – ysinA)2 = 0

⇒ z = xcosB + ycosA &

⇒ x, y, z are proportional to a, b & c respectively.

45) Let log a = x, log b = y, log c = z

given (x – y)2 + (y – z)2 + (z – x)2 + x2 + y2 + z2 – x – y – z + k

(x – y)2 + (y – z)2 + (z – x)2 +

46) ƒ(x) = x2 + bx + c
Leading cofficient +ve parabola upward
ƒ(2+t) = ƒ(2–t) ⇒ Symmatric about '2'
Diagram
Now parabola have its minimum at x = 2 clearly seen by graph, point which is away from the
vertex have greater value so
47) Interpret the problem geometrically consider n right triangle joined at their vertices with
bases a1, a2, a3,.......,an and heights 1,3,.....,2n–1. The sum of their hypotenusses is the value of
Sn. The minimum value of Sn, then is the length of the straight line. Connecting the bottom
vertex of first right triangle & the top vertex of the last right triangle, so

∴
Now 172 + n4 = m2 (m ∈ I)
(m – n2) (m + n) = 289.1
∴ n2 = 144 ∴ n = 12

48)

b2sin2C + c2sin2B

= 12 ⇒

⇒
⇒ 4Δ = 12 ⇒ Δ=3

49) Let V1 and V2 be the vertices of the parabola f(x) and g(x) respectively.
We have V1 ≡ (1, a) and V2 ≡ (a, – a2 + 10 – 2a)

Now, a < – a2 + 10 – 2a
⇒ a2 + 3a – 10 < 0
∴ (a + 5) (a – 2) < 0
⇒–5<a<2
∴ Largest natural number a = 1 Ans.

50) α4 – 3α + 7 = 0
α4 + 4 = 3α – 3
α4 + 4 = 3(α – 1)
Q1(α – 1)(β – 1)(γ – 1)(δ – 1)
x4 – 3x + 7 = (x – α)(x – β)(x – γ)(x – δ)
Put x = 1
1 – 3 + 7 = (1 – α)(1 – β)(1 – γ)(1 – δ)
5 = (α – 1)(β – 1)
(Q1 )(α – 1)(β – 1)(γ – 1)(δ – 1) = Q1 × 5
= 405

51) x3 – 2x2 – 1 = 0 or α3 – 1 = 2α2

So =

= =2