[tabs title=”JEE MAINS EXAM Previous Year Paper” type=”centered”]
[tab title=”Physics”]
PHYSICS
SECTION A
Q 1. A current through a wire depends on time as i = α0t + βt2 where α0 = 20 A/s and β = 8 As-2. Find the charge crossed through a section of the wire in 15 s.
(A) 2100 C
(B) 260 C
(C) 2250 C
(D) 11250 C
Answer: (D)
Q 2. Each side of a box made of metal sheet in cubic shape is ‘a’ at room temperature ‘T’, the coefficient of linear expansion of the metal sheet is ‘α’. The metal sheet is heated uniformly, by a small temperature ΔT, so that its new temperature is T+ΔT. Calculate the increase in the volume of the metal box:
(A) 4/3 πa3 α ΔT
(B) 4πa3 α ΔT
(C) 3a3 α ΔT
(D) 4a3 α ΔT
Answer: (C)
Q 3. Given below are two statements:
Statement-I: Two photons having equal linear momenta have equal wavelengths.
Statement-II: If the wavelength of a photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
(A) Statement-I is false but Statement-II is true
(B) Both Statement-I and Statement-II are true
(C) Both Statement-I and Statement-II are false
(D) Statement-I is true but Statement-II is false
Answer: (D)
Q 4. A cube of side ‘a’ has point charges +Q located at each of its vertices except at the origin where the charge is –Q. The electric field at the centre of cube is:
Answer: (C)
Q 5. If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?
Answer: (A)
Q 6. In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:
Answer: (A)
Q 7. Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be
Answer: (A)
Q 8. In Young’s double-slit experiment, the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
(A) 4: 1
(B) 2: 1
(C) 3: 1
(D) 1: 4
Answer: (A)
Q 9. Consider two satellites S1 and S2 with periods of revolution 1 hr and 8 hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is
(A) 8: 1
(B) 1: 8
(C) 2: 1
(D) 1: 4
Answer: (A)
Q 10. If an emitter current is changed by 4mA, the collector current changes by 3.5 mA. The value of β will be:
(A) 7
(B) 0.875
(C) 0.5
(D) 3.5
Answer: (A)
Q 11. n moles of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes:
A→ B: Isothermal expansion at temperature T so that the volume is doubled from V1 to V2 and pressure changes from P1 to P2.
B → C: Isobaric compression at pressure P2 to initial volume V1.
C → A: Isochoric change leading to change of pressure from P2 to P1.
Total work done in the complete cycle ABCA is –
(A) 0
(B) nRT(ln2 + 1/2)
(C) nRTln2
(D) nRT (ln2 – 1/2)
Answer: (D)
Q 12. The work done by a gas molecule in an isolated system is given by, 
(A) [M0LT0]
(B) [M2LT2]
(C) [MLT–2]
(D) [ML2T–2]
Answer: (C)
Q 13. Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be
(A) 2: 1
(B) 1: 4
(C) 4: 1
(D) 1: 2
Answer: (B)
Q 14. Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as:
I1 = M.I. of thin circular ring about its diameter,
I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3 = M.I. of solid cylinder about its axis and
I4 = M.I. of solid sphere about its diameter.
Then:
(A) I1 = I2 = I3< I4
(B) I1 + I2 = I3 + 5/2 I4
(C) I1 + I3< I2 + I4
(D) I1 = I2 = I3> I4
Answer: (D)
Q 15. A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is
(A) 3.6V
(B) 10.0V
(C) 5.6V
(D) 2.0V
Answer: (C)
Q 16. The focal length f is related to the radius of curvature r of the spherical convex mirror by:
(A) f = r
(B) f = – ½ r
(C) f = +½ r
(D) f = – r
Answer: (C)
Q 17. If Y, K and η are the values of Young’s modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.
Answer: (A)
Q 18. In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.
The transition A, B and C respectively represents:
(A) The series limit of Lyman series, third member of Balmer series and second member of Paschen series
(B) The first member of the Lyman series, third member of Balmer series and second member of Paschen series
(C) The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series
(D) The series limit of Lyman series, second member of Balmer series and second member of Paschen series
Answer: (A)
Q 19. Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is
Answer: (A)
Q 20. Match List-I with List-II
| List – I | List – II |
| (A) Isothermal | (i) Pressure constant |
| (B) Isochoric | (ii) Temperature constant |
| (C) Adiabatic | (iii) Volume constant |
| (D) Isobaric | (iv) Heat content is constant |
Choose the correct answer from the options given below:
(A) (A) – (ii), (B) – (iv), (C) – (iii), (D) – (i)
(B) (A) – (ii), (B) – (iii), (C) – (iv), (D) – (i)
(C) (A) – (i), (B) – (iii), (C) – (ii), (D) – (iv)
(D) (A) – (iii), (B) – (ii), (C) – (i), (D) – (iv)
Answer: (B)
SECTION-B
Q 1. In connection with the circuit drawn below, the value of current flowing through the 2kΩ resistor is _______ × 10–4 A.
Answer: 25
In zener diode there will be o change in current after 5V
Zener diode breakdown
⇒ i = 5 / 2 × 103
⇒ i = 2.5 × 10–3 A
⇒ i = 25 × 10–4 A
Q 2. An inclined plane is bent in such a way that the vertical cross-section is given by y = x2 / 4 where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction μ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _____ cm.
Answer: 25
Given,
y = x2 / 4
μ = 0.5
Condition for block will not slip downward
mg sin θ = μmg cos θ
⇒ tan θ = μ
And we know that
⇒ tanθ = dv / dx
⇒ dv / dx = μ ⇒ x/2 = 0.5 [y = x2 / 4 dy / dx = x / 2]
⇒ x = 1,
put x = 1 in equation y = x2/4
⇒ y = (1)2 / 4 ⇒ y = ¼ ⇒ y = 0.25
y = 25 cm
Q 3. An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of the light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by 30° in clockwise direction, the intensity of emerging light will be________ Lumens.
Answer: 75
Given: I0 = 100 lumens, θ = 30o
Inet = I0 cos2θ
Inet = 100 × (√3/2)2 = 100 × 3 / 4
Inet = 75 lumens
Q 4. The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be ______ N.
[g = 10 ms-2]
Answer: 25
Given: μs = 0.2
m = 0.5 kg
g = 10 m/s2
We know that
fs = μsN and …. (1)
To keep the block adhere to the wall
Here, N = F … (2)
fs = mg …. (3)
From equation (1), (2), and (3), we get
⇒mg = μs F
⇒ F = mg / μs ⇒ F = 0.5 × 10 / 0.2
F = 25 N (for all values of F greater than or equal to 25 N this case is possible)
Q 5. A hydraulic press can lift 100 kg when a mass ‘m’ is placed on the smaller piston. It can lift _______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ‘m’ on the smaller piston.
Answer: 25600
Atmospheric pressure P0 will be acting on both the limbs of the hydraulic lift.
Applying pascal’s law for the same liquid level
⇒ P0 + mg / A1 = Po + (100)g / A2
⇒ mg / A1 = (100)g / A2 ⇒ m / 100 = A1/ A2 …(1)
Diameter of piston on side of 100 kg is increased by 4 times so new area = 16A2
Diameter of piston on side of (m) kg is decreasing
A1 = A1 / 16
(In order to increasing weight lifting capacity, diameter of smaller piston must be reduced)
Again, mg / (A1/16) = M’g / 16A2 ⇒ 256m / M’ = A1/ A2
From equation (1) = 256m / M’ = m / 100 ⇒ M’ = 25600 kg
Q 6. An audio signal υm = 20sin2π(1500t) amplitude modulates a carrier υc =80 sin 2π (100,000t). The value of percent modulation is ________.
Answer: 25
We know that, modulation index = Am / Ac
From given equations, Am = 20 and Ac = 80
Percentage modulation index = Am / Ac × 100
⇒ 20 / 80 × 100 = 25%
The value of percentage modulation index is
= 25
Q 7. A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of the secondary coil are 24, then the number of turns of the primary are ______.
Answer: 440
Given,
Primary voltage, Vp = 220 V
Secondary voltage, vs = 12 V
No. of turns in secondary coil is Ns = 24
No. of turns in primary coil, Np = ?
We know that for a transformer
⇒ Np / Ns = Vp / Vs
⇒ Np = Vp × Ns / Vs = 220 × 24 / 12
⇒ Np = 440
Q 8. A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30° with the original direction. The ratio of velocities of the balls after collision is x : y, where x is _________.
Answer: 1
Momentum is conserved just before and just after the collision in both x-y direction.
In y-direction,
pi = 0
pf = mv1 sin30o – mv2 sin30o
Pf = m × ½ v1 – m × ½ v2
pi = pf, so
= mv1 / 2 – mv2 / 2 = 0
⇒ mv1 / 2 = mv2 / 2 ⇒ v1 = v2
v1 / v2 = 1
Q 9. An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is _______ × 107 m/s.
Answer: 15
Given: f = 5 GHz
εr =2
μr = 2
Velocity of wave ⇒ v = c / n ….(1)
Where, n = √μrεr and c = speed of light = 3 × 108 m/s
n = √2 × 2 = 2
put the value of n in we get
⇒ v = 3 × 108 / 2 = 15 × 107 m/s
⇒ X × 107 = 15 × 107
X = 15
Q 10. A resonance circuit having inductance and resistance 2 × 10–4 H and 6.28 Ω respectively oscillates at 10 MHz frequency. The value of the quality factor of this resonator is________. [π = 3.14]
Answer: 2000
Given: R = 6.28 Ω
f = 10 MHz
L = 2 × 10-4 Henry
We know that quality factor Q is given by
⇒ Q = XL / R = ωL / R
also, ω = 2πf, so
⇒ Q = 2πfL / R
⇒ Q = 2π × 10 × 106 × 2 × 10-4 / 6.28 = 2000
Q = 2000
[/tab]
[tab title=”Chemistry”]
CHEMISTRY
SECTION-A
Q 1. Which reagent (A) is used for the following given conversion?
(A) Cu / ∆ / high pressure
(B) Molybdenum Oxide
(C) Manganese Acetate
(D) Potassium Permanganate
Answer: (B)
Q 2. S-1: Colourless cupric metaborate is converted into cuprous metaborate in a luminous flame.
S-2: Cuprous metaborate is formed by reacting copper sulphate with boric anhydride heated in non luminous flame.
(A) S-1 is true and S-2 is false
(B) S-1 is false and S-2 is true
(C) Both are true
(D) Both are false
Answer: (D)
Q 3. Find A and B.
Answer: (B)
Q 4. EoM2+/ M has a positive value for which of the following elements of 3d transition series?
(A) Cu
(B) Zn
(C) Cr
(D) Co
Answer: (A)
Q 5. What is the reason for the formation of a meta product in the following reaction?
(A) Aniline is ortho/para directing
(B) Aniline is meta directing
(C) In acidic medium, aniline is converted into anilinium ion, which is ortho/para directing
(D) In acidic medium, aniline is converted into anilinium ion which is meta directing
Answer: (D)
Q 6. Identify X, Y, Z in the given reaction sequence.
(A) X = Na[Al(OH)4] ; Y = CO2 ; Z = Al2O3.xH2O
(B) X = Na[Al(OH)4] ; Y = SO2 ; Z = Al2O3.xH2O
(C) X = Al(OH)3 ; Y = CO2 ; Z = Al2O3
(D) X = Al(OH)3 ; Y = SO2 ; Z = Al2O3
Answer: (A)
Q 7. The missing reagent P is:
Answer: (A)
Q 8. Arrange Mg, Al, Si, P and S in the correct order of their ionisation potentials.
Answer: P > S > Si > Mg > Al
Q 9. Which force is responsible for the stacking of the α-helix structure of protein? Hydrogen (1H, 2H, 3H) is ________.
(A) H-bond
(B) Ionic bond
(C) Covalent bond
(D) Van der Waals forces
Answer: (A)
Q 10. The composition of gun metal is:
(A) Cu, Zn, Sn
(B) Al, Mg, Mn, Cu
(C) Cu, Ni, Fe
(D) Cu, Sn, Fe
Answer: (A)
Q 11. The gas evolved due to anaerobic degradation of vegetation causes:
(A) Global warming and cancer
(A) Global warming and cancer
(B) Acid rain
(C) Ozone hole
(D) Metal corrosion
Answer: (A)
Q 12. The slope of the straight line given in the following diagram for adsorption is:
(A) 1/n (0 to 1)
(B) 1/n (0.1 to 0.5)
(C) log n
(D) log (1/n)
Answer: (A)
Q 13. Match the following:
| (i) Caprolactam | (a) Neoprene |
| (ii) Acrylonitrile | (b) Buna N |
| (iii) 2-chlorobuta-1,3-diene | (c) Nylon – 6 |
| (iv) 2-Methylbuta-1,3-diene | (d) Natural rubber |
(A) (i) →(b) , (ii) → (c) , (iii) → (a) , (iv) → (d)
(B) (i) → (a) , (ii) → (c) , (iii) → (b) , (iv) → (d)
(C) (i) → (c) , (ii) → (b) , (iii) → (a) , (iv) → (d)
(D) (i) → (c) , (ii) → (a) , (iii) → (b) , (iv) → (d)
Answer: (C)
Q 14. In the given reactions,
1) I2 + H2O2 + 2OH– → 2I– + 2H2O + O2
2) H2O2 + HOCl → Cl– + H3O+ + O2
(A) H2O2 acts as an oxidising agent in both the reactions
(B) H2O2 acts as a reducing agent in both the reactions
(C) H2O2 acts as an oxidising agent in reaction (1) and as a reducing agent in reaction (2)
(D) H2O2 acts as a reducing agent in reaction (1) and as an oxidizing agent in reaction (2)
Answer: (B)
Q 15. What is the major product of the following reaction
Answer: (A)
Q 16. Which of the following ores are concentrated by cyanide of group 1st element?
(A) Sphalerite
(B) Malachite
(C) Calamine
(D) Siderite
Answer: (A)
Q 17. What is the major product of the following reaction?
Answer: (C)
Q 18. Which of the following pairs are isostructural
a) TiCl4, SiCl4
b) SO2-3, CrO2-3
c) NH3, NO–3
d) ClF3, BCl3
(A) B, C
(B) A, C
(C) A, B
(D) A, D
Answer: (C)
Q 19. Identify the major product:
Answer: (B)
Q 20. The products A and B are:
Answer: (A)
Section B
Q 1. Cu2+ + NH3 ⇌ [Cu(NH3)]2+ K1 = 104
[Cu(NH3)]2+ + NH3 ⇌ [Cu(NH3)]2+ K2 = 1.58 × 103
[Cu(NH3)2]2+ + NH3 ⇌ [Cu(NH3)3]2+ K3 = 5 × 102
[Cu(NH3)3]2+ + NH3 ⇌ [Cu(NH3)4]2+ K4 = 102
If the dissociation constant of [Cu(NH3)4]2+ is X × 1012.
Determine X.
Answer: 1.26
Overall reaction constant (β):
β = K1× K2 × K3 × K4
= 104 (1.58 ×103) × 5 × 102 × 102 = 7.9 × 1011
Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ ; β = 7.9 × 1011
So, the dissociation constant (KDisso.) will be:
1/β = 1/ 7.9 × 1011
KDisso = 1 / β = 1.26 × 10-12
Hence, the value of X = 1.26
Q 2. What is the coordination number in Body Centered Cubic (BCC) arrangement of identical particles?
Answer: 8
The easiest way is to look at the atom at the body center. It lies at the center of the body diagonal and touches the eight corner atoms. So, coordination number = 8.
Q 3. Cl2(g)⇌ 2Cl(g)
For the given reaction at equilibrium, moles of Cl2(g) is equal to the moles of Cl(g) and the equilibrium pressure is 1atm. If Kp of this reaction is x ×10–1, find x.
Answer: 5
According to the question: Cl2 ⇌ 2Cl [Ptotal = 1 atm].
Moles of Cl2 = Moles of Cl (at equilibrium).
Given: Kp = x ×10-1
nCl2 = nCl.
Therefore, P Cl2 = P Cl
Hence, P Cl2 = P Cl = 0.5 atm
kp= (P Cl)2/ P Cl2 = 0.5 = 5 x 10-1
So, x = 5
Q 4. Among the following compounds, how many are amphoteric in nature?
Be(OH)2, BeO, Ba(OH)2, Sr(OH)2
Answer: 2
The oxide and hydroxide of Be are amphoteri(C) Hence BeO and Be(OH)2 is amphoteri(C) Ba(OH)2 and Sr(OH)2 are basic.
Q 5. S8 + bOH– → cS2- + sS2O2-3 + H2O. Find the value of c.
Answer: 4
Let us look at the half reactions i.e. oxidation and reduction separately.
Oxidation:
S8 → S2O2-3
S8 + 24OH– → 4S2O2-3 + 12H2O +16e- ….(1)
Reduction:
S8 → S2-
S8 + 16e- → 8S2- ….(2)
Adding both the reactions (1) and (2),
2S8 + 24OH– → 4S2O2-3 + 8S2- + 12H2O
Dividing the whole equation by 2,
S8 + 12OH– → 2S2O2-3 + 4S2- + 6H2O
So, c = 4
Q 6. 4.5 g of a solute having molar mass of 90 g/mol is dissolved in water to make a 250 mL solution. Calculate the molarity of the solution.
Answer: 0.2
WB (given weight of solute) = 4.5 g
MB (Molar mass of solute) = 90 g/mol
VS (Volume of solution) = 250 mL
= 250 / 1000 L = 1/4 L
Molarity (M) = nB / VS(L) = WB / MB.VS(L) = 4.5 / (90 ×1/4) = 0.2 molar
Q 7. Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is 3.3× 10-4 sec-1.
Answer: 1518
= 30 x 103 x 0.506 = 1518 sec
Q 8. 9.45g of CH2ClCOOH is dissolved in 500 mL of H2O solution and the depression in freezing point of the solution is 0.5°C. Find the percentage dissociation.
(Kf)H2O = 1.86K kg mole-1.
Answer: 34.4%
CH2Cl COOH CH2ClCOO– + H+
Initial 100
Dissociated α α α
Left (1-α) α α
i = final moles / initial moles = 1 – α + α + α / 1 = 1 + α
ΔTf = i × Kf × m
0.5 = (1 + α) × 1.86 × m
Molality (m) = nB / WA (kg) = WBMB x WA (kg) = 9.45 × 1000 / 94.5 × 500 = 0.2
Here, A is for solute and B is for solvent
WA=WH2O = 500g (Density of H2O = 1g/mL)
0.5 = (1+ α) ×1.86 × 0.2
α = 0.344
Percentage of dissociation = 34.4%
Q 9. For a chemical reaction, Keq is 100 at 300K, the value of ΔGo is –xR Joule at 1 atm pressure. Find the value of x. (Use ln 10 = 2.3)
Answer: 1382
Given: Keq= 100 at 300K and ΔGo = –xR Joule
ΔGo = -RTln(Keq) = -2.303RTlog(Keq) = -1381.8R
Therefore, x = 1381.8 or 1382
Q 10. The mass of Li3+ is 8.33 times the mass of a proton. If Li3+ and proton are accelerated through the same potential difference, then the ratio of de Broglie’s wavelength of Li3+ to proton is x ×10–1. Find x
Answer: 2
m (Li3+) = 8.33 × mp+ (given)
Debroglie’s wavelength (λ) = h / p
KE = ½ mv2
Multiplying by m on both sides
m. KE = ½ mv2
2m. KE = (mv)2 = p2
P = √2m. KE
Also KE= q × V
So, P = 2mq.V
Comparing with x × 10 -1 = 2 × 10 -1
x = 2
[/tab]
[tab title=”Maths”]
MATHS
SECTION A
Q 1: The statement among the following that is a tautology is:
(A) A∧(A∨B)
(B) B→[A∧(A→B)]
(C) A∨(A∧B)
(D) [A∧(A→B)]→B
Answer: (D)
A∧ (~ A∨B)→B
= [(A∧~A)∨(A∧B)]→ B
= (A∧ B)→ B
= ~ (A∧B)∨B
= t
Q 2: Let f :R→R be defined as f(x) = 2x-1 and g:R – {1} →R be defined as g(x) = (x-½)/(x-1).
Then the composition function f(g(x)) is:
(A) Both one-one and onto
(B) onto but not one-one
(C) Neither one-one nor onto
(D) one-one but not onto
Answer: (D)
f(g(x)) = 2g(x) – 1
= 2(x – 1/2)/(x – 1) – 1
= x/(x – 1)
f(g(x)) = 1 + 1/(x – 1)
∴ one-one, into
Q 3: If f:R→ R is a function defined by f(x) = [x – 1] cos ((2x – 1)/2)𝜋 , where [.] denotes the greatest integer function, then f is:
(A) discontinuous only at x = 1
(B) discontinuous at all integral values of x except at x = 1
(C) continuous only at x = 1
(D) continuous for every real x
Answer: (D)
Doubtful points are x = n, n∈I
f(n) = 0
Hence continuous.
Q 4: If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the ordinate of the point which divides PQ internally in the ratio 1:2 is:
(A) –2t3
(B) –t3
(C) 0
(D) 2t3
Answer: (A)
Equation of tangent at P(t, t3)
(y – t3) = 3t2(x – t) ⋯(1)
Now solve the above equation with
y = x3 ⋯(2)
By (1) & (2)
x3 – t3 = 3t2 (x – t)
x2 + xt + t2 = 3t2
x2 + xt – 2t2 = 0
⇒(x – t)(x + 2t) = 0
⇒x = – 2t
⇒Q(-2t, -8t3)
Ordinate of required point = (2t3 + (-8t)3)/3
= -2t3
Q 5: The value of – 15C1 + 2. 15C2 – 3.15C3+ ….-15.15C1 + 14C1+ 14C3 + 14C5 + …14C11 is
(A) 214
(B) 213 – 13
(C) 216 – 1
(D) 213 – 14
Answer: (D)
= (14C1+ 14C3 + 14C5 + …14C11 + 14C13) – 14C13
= 213 – 14
∴ S1 + S2 = 213 – 14
Q 6: An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:
(A) 3/16
(B) 1/2
(C) 5/16
(D) 1/32
Answer: (B)
P(odd no. twice) = P(even no. thrice)
nC2 (1/2)n = nC3 (1/2)n
⇒ n = 5
Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)
= 5C2 (1/2)5 + 5C2 (1/2)5 + 5C2 (1/2)5
= 16/25
= 1/2
Q 7: Let p and q be two positive number such that p + q = 2 and p4 + q4 = 272. Then p and q are roots of the equation:
(A) x2 – 2x + 2 = 0
(B) x2 – 2x + 8 = 0
(C) x2 – 2x + 136 = 0
(D) x2 – 2x + 16 = 0
Answer: (D)
(p2 + q2)2 – 2p2q2 = 272
((p + q)2 – 2pq)2 – 2p2q2 = 272
⇒16 – 16pq + 2p2q2 = 272
(pq)2 – 8pq –128 = 0
⇒pq = (8±24)/2 = 16, – 8
⇒pq = 16
Now
x2 – (p + q)x + pq = 0
x2 – 2x + 16 = 0
Q 8: The area (in sq.units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is:
(A) 24π + 3√3
(B) 12π + 3√3
(C) 12π – 3√3
(D) 24π – 3√3
Answer: (D)
The curves intersect at point (3, ± 3√3)
Required area
Q 9: If ∫(cos x -sin x)/√(8-sin 2x) dx = a sin-1(sin x + cos x)/b + c where c is a constant of integration, then the ordered pair (a, b) is equal to:
(A) (1, –3)
(B) (1, 3)
(C) (–1, 3)
(D) (3, 1)
Answer: (B)
Put sin x + cos x = t ⇒1 + sin 2x = t2
(cos x -sin x ) dx = dt
⇒I = ∫dt/√(8-(t2-1))
= ∫dt/(9-t2)
= sin-1 (t/3) + c = sin-1(sin x + cos x)/3 + c
⇒ a = 1 and b = 3
Q 10: The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is:
(A) x = a
(B) x = 0
(C) x = -a/2
(D) x = a/2
Answer: (B)
h = (at2+a)/2, k = (2at+0)/2
⇒ t2 =(2h-a)/a and t=k/a
⇒ k2/a2 =(2h-a)/a
⇒ Locus of (h, k) is y2 = a (2x – a)
⇒ y2 = 2a(x- a/2)
Its directrix is x – a/2 = -a/2
⇒ x = 0
Q 11: A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts of this line on the coordinate axes is 1/4. Three stones A, B and C are placed at the points (1,1),(2,2), and (4,4) respectively. Then which of these stones is/are on the path of the man?
(A) B only
(B) A only
(C) the three
(D) C only
Answer: (A)
x/a + y/b = 1
h/a + k/b = 1 ⋯⋯(1)
and (1/a + 1/b)/2 = 1/4
∴1/a + 1/b = 1/2 ⋯⋯(2)
(From (1) and (2))
Line passes through fixed point B(2, 2)
Q 12: The function f(x) = (4×3- 3×2)/6 – 2sin x + (2x – 1)cos x:
(A) increases in [1/2, ∞)
(B) decreases (-∞, 1/2]
(C) increases in (-∞, 1/2]
(D) decreases [1/2, ∞)
Answer: (A)
f’(x) = (2x – 1) (x – sin x )
⇒ f’x ≥ 0 in x∈(-∞, 0] ⋃ [1/2, ∞)
and f’x ≤ 0 in x∈(0, ½)
Q 13: The distance of the point (1, 1, 9) from the point of intersection of the line (x – 3)/1 = (y – 4)/2 = (z – 5)/2 and the plane x + y + z = 17 is:
(A) 38
(B) 19√2
(C) 2√19
(D) √38
Answer: (D)
(x – 3)/1 = (y – 4)/2 = (z – 5)/2 = λ
x = λ + 3, y = 2λ+ 4, z = 2λ+5
Which lies on given plane hence
⇒ λ+3+2λ +4+2λ+5 = 17
⇒ λ = 5/5 = 1
Hence, point of intersection is Q (4, 6, 7)
∴ Required distance =PQ
= √(9+25+4)
= √38
Q 14:
(A) 2/3
(B) 0
(C) 1/15
(D) 3/2
Answer: (A)
Q 15: Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:
(A) 25
(B) 20√3
(C) 30
(D) 25√3
Answer: (D)
tan θ = h/75 = 75/3h
h2 = 752/3
h = 25√3m
Q 16: A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed is:
(A) 560
(B) 1050
(C) 1625
(D) 575
Answer: (C)
(2I,4F)+ (3I,6F) + (4I,8F)
= 6C2 8C4 + 6C3 8C6 + 6C4 8C8
= 15 × 70 + 20 × 28 + 15 × 1
= 1050 + 560 + 15 = 1625
Q 17: The equation of the plane passing through the point (1,2,–3) and perpendicular to the planes 3x + y – 2z = 5 and 2x – 5y – z = 7, is:
(A) 3x – 10y – 2z + 11 = 0
(B) 6x – 5y – 2z – 2 = 0
(C) 11x + y + 17z + 38 = 0
(D) 6x – 5y + 2z + 10 = 0
Answer: (C)
Q 18: The population P = P(t) at time ‘t’ of a certain species follows the differential equation dP/dt = 0.5P – 450. If P(0) = 850, then the time at which population becomes zero is:
(A) ½ loge 18
(B) 2loge 18
(C) loge 9
(D) loge 18
Answer: (B)
dp/dt = (p-900)/2
ln |900| – ln |50| = t/2
t/2 = ln |18|
⇒ t = 2ln 18
Q 19: If e^{(cos2x + cos4x + cos6x + …..∞ )loge2 satisfies the equation t2 – 9t + 8 = 0, then the value of 2sin x/(sin x + √3cos x) 0< x<π/2 is:
(A) 3/2
(B) 2√3
(C) 1/2
(D) √3
Answer: (C)
0 < x < π/2 ⇒ cotx = √3
⇒(2 sinx)/(sinx+√3 cosx )
= 2/(1+√3 cot x)
= 1/2
Q 20: The system of linear equations
3x – 2y – kz = 10
2x – 4y – 2z = 6
x + 2y – z = 5m
is inconsistent if :
(A) k = 3, m = 4/5
(B) k ≠ 3,m∈R
(C) k ≠ 3, m ≠ 4/5
(D) k = 3, m ≠ 4/5
Answer: (D)
⇒ 3(4 + 4) + 2(–2 + 2) -k(4 + 4) = 0
⇒ k = 3
⇒ 10(4 + 4) +2(-6 + 10m) -3(12 + 20m) ≠ 0
⇒ m ≠ 4/5
⇒ 3(-6 + 10m) – 10(- 2 + 2) – 3(10m – 6) ≠ 0
⇒ 0
⇒3(-20m – 12) +2(10m – 6) +10(4 + 4) ≠ 0
⇒ m ≠ 4/5
Section B
Q 1:
Solution:
Answer: 1
= tan 𝜋/4
= 1
Q 2: If 
Answer: 3
Q 3: If one of the diameters of the circle x2+y2 – 2x – 6y + 6 = 0 is a chord of another circle ‘C’ whose center is at (2,1), then its radius is ________
Solution:
Answer: 3
distance between (1,3) and (2,1) is √5
∴ (√5)2+(2)2= r2
⇒r = 3
Q 4: Let three vectors a, b and c be such that is coplanar with a and b, a.c=7 and is perpendicular to c, where 
Answer: 75
Q 5: The minimum value of α for which the equation 4/sinx +1/(1-sinx )=α has at least one solution in (0,π/2) is _______
Solution:
Answer: 9
f(x) = 4/sinx + 1/(1 – sinx )
Let sinx = t ∵x∈(0, π/2) ⇒ 0 < t < 1
f(t)= 4/t + 1/(1-t)
f'(t) = (-4)/t2 + 1/(1 – t)2 = 0
⇒(t2-4(1-t)2/t2(1-t)2 = 0
⇒t = 2/3
fmin at t = 2/3
αmin = f(2/3) =4/(2/3) + 1/(1 – 2/3)
= 6 + 3
= 9
Q 6: Let A = {n∈N∶ n is a 3-digit number}, B = {9k + 2∶ k∈N} and C = {9k + l∶ k∈N} for some l(0 < l < 9). If the sum of all the elements of the set A∩(B∪C) is 274×400, then l is equal to ___________
Answer: 5
3 digit number of the form 9K+2 are {101,109,⋯,992}
⇒ Sum equal to (100/2)(1093) = S1= 54650
Now 274 × 400 =S1+S2
⇒274 × 400 = (100/2) [101 + 992] + S2
⇒274 × 400 = 50 × 1093 + S2
⇒S2 = 109600 – 54650
∴S2 = 54950
S2 = 54950 = (100/2) [(99+l) + (990+l)]
⇒ 2l + 1089 = 1099
⇒ l = 5
Q 7: Let
where α∈R. Suppose Q = [qij ] is a matrix satisfying PQ = kI3 for some non-zero k∈R. If q23= – k/8 and |Q| = k2/2, then α2 + k2 is equal to ___
Answer: 17
Q 8: Let M be any 3×3 matrix with entries from the set {0,1,2}. The maximum number of such matrices, for which the sum of diagonal elements of MTM is seven, is __________
Answer: 540
⇒ a2 + b2 + c2 + d2 + e2 + f2 + g2 + h2 + i2 = 7
Case I∶ Seven (1’s) and two (0’s)
9C2 = 36
Case II∶ One (2) and three (1’s) and five (0’s)
9!/5!3! = 504
∴Total = 540
Q 9: If the least and the largest real values of α, for which the equation z + α|z-1| + 2 i = 0 (z∈C and i =√(-1)) has a solution, are p and q respectively; then 4(p2 + q2) is equal to _________
Answer: 10
x + iy + α√((x–1)2 +y2) + 2i=0
⇒y + 2 = 0 and x + α√((x-1)2+y2)=0
y = –2 & x2 = α2(x2 – 2x + 1 + 4)
α2 = x2/(x2– 2x + 5)
⇒ x2 (α2 – 1) – 2xα2 + 5α2 = 0
∵x∈R ⇒ D≥0
⇒ 4α4 – 4(α2 – 1)5α2 ≥ 0
⇒ α2 [4α2 – 20α2 + 20] ≥ 0
⇒ α2 [-16α2 + 20] ≥ 0
⇒ α2 [α2 – 5/4] ≤ 0
⇒ α2 ∈ [0, 5/4]
⇒ α ∈ [-√5/2, √5/2]
then 4[p2 + q2] = 4[5/4 + 5/4]
= 10
Q 10: Let Bi (i = 1,2,3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α – 2β)p = αβ and (β – 3γ)p= 2βγ (All the probabilities are assumed to lie in the interval (0, 1)). Then (P(B1))/(P(B3) )is equal to _____
Answer: 6
Let x,y,z be probability of B1, B2, B3 respectively.
⇒ x(1 – y) (1 – z) = α
⇒ y(1 – x)(1 – z) = β
⇒ z(1 – x)(1 – y ) = γ
⇒ (1 – x)(1 – y)(1 – z) = p
Now (α – 2β)p = αβ
⇒ (x(1–y)(1–z)-2y(1-x)(1–z)) (1–x)(1–y)(1–z) = xy(1–x)(1–y) (1–z)2
⇒ x+ xy – 2y = xy
∴x = 2y⋯(1)
Similarly, (β–3γ) p = 2βγ
⇒ y = 3z ⋯(2)
From (1) & (2)
⇒x = 6z
Hence x/z = P(B1)/P(B3) = 6
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