
IIT JEE Advanced 2015 Paper 1 (Code 8)
Test Name: IIT JEE Advanced 2015 Paper 1 (Code 8)
Question 1:
Consider a hydrogen atom with its electron in the n^{th} orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of n is (hc = 1242 eV nm)
Solutions:
Energy of electron in nth orbit, En=13.6n2eVEnergy of incident photon, Ep=hcλ=1242 eV nm90 nm=13.8 eVThe kinetic energy of ejected electron, K = 10.4 eV By conservation of energy, we know that
EpEn=K
∴ 13.8 eV 13.6 eV n2=10.4 eV ⇒n2=13.63.4=4 ⇒n=2[[VIDEO:13854]]
Question 2:
A bullet is fired vertically upwards with velocity v from the surface of a spherical planet . When it reaches its maximum height, its acceleration due to the planet’s gravity is 1/4^{th} of its value at the surface of the planet. If the escape velocity from the planet is
vesc=vN, then the value of N is (ignore energy loss due to atmosphere)
Solutions:Let R be the radius of planet and h be the maximum height to which the bullet reaches when thrown upwards from its surface.
Given that,
gh=14gswhere gh=acceleration due to gravity at height hgs=acceleration due to gravity on the surface of planet⇒GMh2=14GMR2⇒h=2RApplying conservation of mechanical energy at the two points, we get
GMmR+12mv2=GMmh+0 ∵velocity at maximum height will be zero⇒12mv2=GMm2R ∵h=2R⇒v2=GMR⇒v=GMR …..(1)The escape velocity from the planet is given by
vesc=2GMRFrom (1), we getvesc=v2⇒N=2[[VIDEO:13855]]
Question 3:
Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speed v_{1} and v_{2}, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed v_{1} = 3 m/s, then v_{2} in m/s is (g = 10 m/s^{2})
Solutions:Given that, both the discs reach their respective final points B and D with the same speed. So, their final kinetic energies are equal. Applying the conservation of energy:
3212mv12+mg×30=3212mv22+mg×27⇒v22=v12+4g=9+40=49⇒v2=7 m s1[[VIDEO:13856]]
Question 4:
Two spherical stars A and B emit blackbody radiation. This radius of A is 400 times that of B and A emits 10^{4} times the power emitted from B. The ratio
λAλBof their wavelengths λ_{A} and λ_{B} at which the peaks occur in their respective radiation curves is
Solutions:Let A_{A} and T_{A} be the area and temperature of star A and A_{B }and T_{B} be the area and temperature of star B.
Given that,
R_{A} = 400 R_{B}
where R_{A} and R_{B} are the radius of star A and B respectively.
Power emitted from star A (P_{A}) is given by
PA=eσAATA4 …..1where σ = Stefan’s constant
Similarly, power emitted from star B (P_{B}) is given by
PB=eσABTB4 …..2Dividing equation (1) by (2), we get
PAPB=AATA4ABTB4=4πRA24πRB2TA4TB4⇒TBTA=RARB·PBPA14From the Wein’s law, we haveλT=constant⇒λATA=λBTB ∴λAλB=TBTA=400 110414=2[[VIDEO:13857]]
Question 5:
A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of nT years, then the value of n is
Solutions:We know that Power generated ∝ Activity
Let electric power available = A_{0}
Total power requirement, A=12.5100×A0⇒A=18A0From the relation
AA0=12nWhere n = number of half life
⇒18=12n⇒n=3[[VIDEO:13858]]
Question 6:
A Young’s double slit interference arrangement with slits S_{1} and S_{2} is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x^{2} = p^{2}m^{2} λ^{2} – d^{2}, where λ is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is
Solutions:
Path length S2O = μx2+d2 where O is point of interference on the water surfacePath length S1O = x2+d2Path difference, ∆x=S2OS1OFor constructive interference, Δx=mλ⇒x2+d2 μ1=mλ Given, μ=4/3 ⇒x2=9m2λ2d2On comparing, p =3. [[VIDEO:13859]]
Question 7:
Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M_{1}. When the setup is kept in a medium of refractive index 7/6, the magnification becomes M_{2}. Then magnitude
M2M1is
Solutions:For mirror in air,
1v115=110⇒1v=115110=130⇒v=30Magnification, M=vu=2For lens in air, we have
1v120=110⇒v=20 cmMagnification, M’=vu=1Net magnification in air, M1=MM’=2Image formation by concave mirror is unaffected by presence of medium. So, magnification due to it, M will remain same. Focal length of lens will change when kept in other medium.
f’f=μ1μμ’1where μ is the refractive index of lens materialμ’ is the refractive index of medium⇒f’f=1.511.57/61⇒f’=74f=704 cmSo, calculating image distance again for the lens, we get
1v120=470⇒v=140 cmMagnification, M’=vu=14020=7Net magnification, M2=MM’=14∴M2M1=142=7
Question 8:
An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the yaxis in the yz plane at
z=32a(see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the xy plane with its centre at the origin is
λLnε0(ε_{0} = permittivity of free space), then the value of n is
Solutions:Consider a cylindrical gaussian surface with line charge as axis passing through its centre, such that rectangular surface ABCD are at edges of the cylinder.
To calculate the flux through the cylindrical surface AB, calculate the angle subtended by the line charge to line AD or BC.
From the given figure,
tanθ=a223a⇒θ=300From gauss law, total flux through cylindrical surface =
Qε0Flux through surface ABCD will be ,
6003600×Qε0=16×Qε0∴ Total flux through ABCD will be
Q6ε0=λL6ε0Hence, the value of n is 6. [[VIDEO:13860]]
Question 9:
Two identical glass rods S_{1} and S_{2} (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S_{1} on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S_{2}. The distance d is
Option 1:  60 cm 
Option 2:  70 cm 
Option 3:  80 cm 
Option 4:  90 cm 
Solutions:
On surface of glass S2, refraction takes place from air to glass rod.μ2vμ1u=μ2μ1R⇒μ2∞1u=μ2110⇒u=20 cm
as distance between S_{1}_{ }and S_{2} is_{ }d, then distance of I from S_{1}_{ }will be (d20) now for refraction from S_{1} glass to air:
μ1v’μ2u’=μ1μ2Rv’=d20 and u’=50⇒1d20μ250⇒1μ210⇒1d20+μ250=μ2110Substituting μ2=1.5, we get1d20+3100=120⇒d=70 cmHence, the correct option is (B).
Question 10:
A conductor (shown in the figure) carrying constant current I is kept in the xy plane in a uniform magnetic field
B→. If F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is(are)
Option 1:  If
B→is along z^, F ∝ (L + R) 
Option 2:  If
B→is along x^, F = 0 
Option 3:  If
B→is along y^, F ∝ (L + R) 
Option 4:  If
B→is along z^, F = 0 
Solutions:The force on a current carrying wire is calculated by following formula:
∫dF→=i∫dl→×B→When
B→is along
z^the magnetic force on the wire is given by
F→=i[2 (L+R)i^×B→]=i[2 (L+R)B](j^)⇒F ∝ (L + R)
When
B→is along
x^the magnetic force on the wire is given by
F→=i[2 (L+R)i^×B→]⇒F = 0
When
B→is along
y^the magnetic force on the wire is given by
F→=i[2 (L+R)i^×B→]=i[2 (L+R)B](k^)⇒F ∝ (L + R)
Hence, the correct options are (A), (B), (C).
Question 11:
A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is(are)
Option 1:  The average energy per mole of the gas mixture is 2RT. 
Option 2:  The ratio of speed of sound in the gas mixture to that in helium gas is
6/5. 
Option 3:  The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2. 
Option 4:  The ratio of the rms speed of helium atoms to that of hydrogen molecules is
1/2. 
Solutions:Total energy of the mixture of hydrogen and helium is:
3RT2+5RT2=4RTAverage energy per moles will be 4RT2=2RTFor ratio of speed of sound in gas mixture to that in helium mixture:
Vsound =γRTMVsound, mixVsound, helium=γmixRTMmixγHeRTMHe⇒γmixMHeMmixγHeNow, γmix=1×52+1×721×32+1×52=32Mmix=1×2+1×42=3therefore, ⇒γmixMHeMmixγHe=65For the ratio of the rms speed of helium atoms to that of hydrogen molecules â€‹
vrms=3RTM ∴vrms ∝1MLet the rms velocity for helium and hydrogen be vHe and vH2 respectvely.vHe vH2=1412=12Hence, the correct options are (A), (B) and (D).
Question 12:
In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10^{–8} Ω m and 1.0 × 10^{–7} Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is
Option 1:  247564μΩ 
Option 2:  187564μΩ 
Option 3:  187549μΩ 
Option 4:  2475132μΩ 
Solutions:The resistances due to aluminium bar and iron hole are calculated as under:
RAl=ρℓA=2.7×108×50×103494×106=3×105 ΩRFe=1.0×107×50×1034×106=125×105The two resistances will be in parallel. So,Req=RAl  RFe=3×105×125×1053×105+125×105 =3×125128×105×106 μΩ = 187564μΩHence, the correct option is (B). [[VIDEO:13861]]
Question 13:
For photoelectric effect with incident photon wavelength λ, the stopping potential is V_{0}. Identify the correct variation(s) of V_{0} with λ and 1/λ.
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Stopping potential is given by
hcλ=eV0+φV0=hceλϕeAs, V0 ∝ 1λThe graphs given in (A) and (C) represent correct relations.
Hence, the correct options are (A) and (C). [[VIDEO:13862]]
Question 14:
Consider a vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
Option 1:  If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. 
Option 2:  If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm. 
Option 3:  If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm. 
Option 4:  If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm. 
Solutions:For vernier callipers,
1 division on main scale=18cm=0.125 cm5 divisions of the vernier scale=4 divisions of the main scale = 4×0.125=0.5 cm∴ 1 division of vernier scale=0.55=0.1 cmLeast count of vernier = 1 main scale division − 1 vernier scale division = 0.125 − 0.1 = 0.025 cm
Least count of screw gauge =
pitch100When the pitch of the screw gauge is twice the least count of vernier callipers Least count of the screw gauge =
0.05100 cm=0.5100 mm=0.005 mmWhen least count of linear scale of screw gauge is twice the least count of vernier callipers
Least count of linear scale = 2 âœ• 0.025 cm = 0.05 cm Then pitch of the screw gauge will be = 2 âœ• 0.05 cm = 0.1 cm = 1 mm ∴ Least Count of screw gauge =
1 mm100=0.01 mmHence, the correct options are (B) and (C).
Question 15:
Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)
Option 1:  M∝c 
Option 2:  M∝G 
Option 3:  L∝h 
Option 4:  L∝G 
Solutions:Let L ∝ Kh^{x} c^{y} G^{z}
Here, x, y, z are the powers of h, c and G respectively. ∴ L ∝ (ML^{2}T^{−1})^{x} (LT^{−1})^{y} (M^{−1}L^{3}T^{−2})^{z}
⇒ L ∝ M^{x−z} L^{2x}^{+y+3z} T^{−x−y−2z}
∴ x − z = 0 …..(2) 2x + y + 3z = 1 …..(3) x + y + 2z = 0 …..(4)
On solving, we get
x=12, y=32, z=12 Substituting in 1 we getL=Kh12c32G12∴ Option (C) and (D) are correct.
Similarly Let M ∝ h^{a} c^{b} G^{c}
Here, a, b, and c are the powers of h, c and G respectively ∴ M ∝ (ML^{2}T^{−1})^{x} (LT^{−1})^{y} (M^{−1}L^{3}T^{−2})^{z} ∴ M ∝ M^{x−z} L^{2}^{x+y+}^{3}^{z} T^{−x−y−2z}
∴ x − z = 1 …..(5) 2x + y + 3z = 0 …..(6) x + y + 2z = 0 …..(7)
Solving the above equation, we get
x=12, z=12, y=12∴ M∝h, M∝c ∴ Option (A) is correct.Hence, the correct options are (A), (C), and (D).
Question 16:
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω_{1} and ω_{2} and have total energies E_{1} and E_{2}, respectively. Then variation of their momenta p with position x are shown in the figures. If
ab=n2and
aR=n, then the correct equation(s) is(are)
Option 1:  E_{1}ω_{1} = E_{2}ω_{2} 
Option 2:  ω2ω1=n2 
Option 3:  ω_{1}ω_{2} = n_{2} 
Option 4:  E1ω1=E2ω2 
Solutions:Let m be the mass of each of the independent harmonic oscillators. From the figure, maximum momentum for first harmonic oscillator is b and maximum momentum of the second harmonic oscillator is R. Energy of first harmonic oscillator (E_{1}) is given by
E1=12mω12a2=b22m ⇒ab=1mω1=n2 …..1Energy of second harmonic oscillator E2 is given byE2=12mω22R2=R22m⇒1mω2=1 …..2 From 1 and 2 ω2ω1=n2On dividing the energies of two harmonic oscillator, we getE1E2=ω1ω22 aR2=1n2·ω1ω2·n2⇒E1ω1=E2ω2Hence, the correct options are (B) and (D).
Question 17:
A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing through its centre O with two point masses each of mass
M8at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is
89ωand one of the masses is at a distance of
35Rfrom O. At this instant the distance of the other mass from O is
Option 1:  23R 
Option 2:  13R 
Option 3:  35R 
Option 4:  45R 
Solutions:Let us consider that the other mass is at x distance from centre O. Applying conservation of angular momentum about the axis, we get
MR2ω=MR2+M835R2+M8x289ω⇒MR2ω=MR289ω+M935R2ω+M9x2ω⇒9MR2ω=8MR2ω+M35R2ω+Mx2ω⇒9R2=8R2+35R2+x2⇒x2=R235R2⇒x=45RHence, the correct option is (D). [[VIDEO:13863]]
Question 18:
The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density λ are kept parallel to each other. In their resulting electric field, point charges q and −q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statement is
Option 1:  Both charges execute simple harmonic motion. 
Option 2:  Both charges will continue moving in the direction of their displacement. 
Option 3:  Charge +q executes simple harmonic motion while charge −q continues moving in the direction of its displacement. 
Option 4:  Charge −q executes simple harmonic motion while charge +q continues moving in the direction of its displacement 
Solutions:When +q is displaced towards the +ve xdirection (towards right), it experiences a repulsive force in opposite direction (towards left). Thus repulsive force dominates in this and resulting force is towards
–xdirection (towards left)
The restoring force is given by:
F=q2kλdx2kλd+x=2k2xqλd2x2F≅4kλqd2x ∵d2x2≈d2 as x is very smallThis equation represents a simple harmonic motion as
F∝xNow, in case of charge
q , when it is displaced towards +x direction (towards right), it gets attracted and â€‹continues moving in the direction of its displacement.
Hence, the correct answer is option C.
Question 19:
Match the nuclear processes given in column I with the appropriate option(s) in column (II).
Column I  Column II  
(A)  Nuclear fusion  (P)  Absorption on thermal neutrons by
U92235 
(B)  Fission in a nuclear reactor  (Q)  Co2760nucleus 
(C)  β−decay  (R)  Energy production in stars via hydrogen conversion to helium 
(D)  γray emission  (S)  Heavy water 
(T)  Neutrino emission 
Question 20:
A particle of unit mass is moving along the xaxis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U_{0} are constants). Match the potential energies in column I to the corresponding statement(s) in column II.
Column I  Column II  
(A)  U1x=U021xa22  (P)  The force acting on the particle is zero at x = a. 
(B)  U2x=U02xa2  (Q)  The force acting on the particle is zero at x = 0. 
(C)  U3x=U02xa2 expxa2  (R)  The force acting on the particle is zero at x = −a 
(D)  U4x=U02xa13xa3  (S)  The particle experiences an attractive force towards x = 0 in the region x < a. 
(T)  The particle with total energy
U04can oscillation about the point x = −a. 
Solutions:
(A) U1=U021xa22⇒Force, F=dU1dx=U02·21xa2×2xa2⇒F=U021xa2×2xa2 Now, F=0 if x=0 and1xa2=0⇒xa2=1⇒xa=±1⇒ x=±a∴A→P, Q, R, T (B) U2x=U02a2·x2F=dU2dx=U02a2·2x⇒F=U0a2·x Now, F=0 if x=0∴ (B) →Q, SC U3x=U02xa2·exa2F=dU3dx=U02xa2·exa2·2xa2+exa2·U02·2xa2⇒F=U02·2xa2xa2·exa2+exa2=U0·xa21xa2·exa2=U0a2·x·exa2 1xa2Now, F=0 at x=0, x=±a. Also F is negative for x<a∴(C)→P, Q, R, S D U4x=U02xa13xa3F=dU4dx=U021a13·3x2a3⇒F=U02a1x2a2Now, F=0 at x=±a. Also F is negative for x<a.∴(D) →P, R, T[[VIDEO:13866]]
Question 21:
The number of resonance structures for N is
Solutions:
The resonance structures of N can be represented as
Thus, the species N has nine resonance structures.
Question 22:
The total number of lone pairs of electron in N_{2}O_{3} is
Solutions:
The total number of lone pairs of electrons in N_{2}O_{3} is 8.
Question 23:
For the octahedral complexes of Fe^{3}^{+} in SCN^{−} (thiocyanato−S) and in CN^{−} ligand environments, the difference between the spinonly magnetic moments in Bohr magnetons (when approximated to the nearest integer) is [Atomic number of Fe = 26]
Solutions:Electronic configuration of Fe^{3}^{+}: [Ar] 3d^{5} CN^{−} is a strong field ligand which causes the pairing of electrons in the dorbitals of metal, whereas, SCN^{−} is a weak field ligand.
Now, valence shell electronic configuration of Fe^{3}^{+} in [Fe(CN)_{6}]^{3−} is given as:
Number of unpaired electrons, n = 1 Magnetic moment,
μ = nn+2 = 11+2 = 3B.M.
Now, valence shell electronic configuration of Fe^{3}^{+} in [Fe(SCN)_{6}]^{3−} is given as
Number of unpaired electrons, n = 5 Magnetic moment,
μ = nn+2 = 55+2 = 35B.M.
Difference in magnetic moment values =
35 – 3 = 4 B.M.
Question 24:
Among the triatomic molecules/ions,
BeCl2, N3, N2O, NO2+, O3, SCI2, ICI2, I3and XeF_{2} the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the dorbital (s) is [Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54]
Solutions:Among the given triatomic molecules/ions, the species having linear geometry with hybridisation not involving the contribution from the d orbital(s), are BeCl_{2},
N3, N_{2}O and
NO2+.
They all involve sp hybridisation.
Thus, the total number of such molecules is 4.
Question 25:
Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H^{−} is
Solutions:Order of energy for H^{−} is decided by the (n + l) rule.
Electronic configuration of H^{−} = 1s^{2} 2s^{0} 2p^{0}
The second excited state of H^{−} is 2p, which has a degeneracy of 3.
Question 26:
All the energy released from the reaction X → Y, âˆ†_{r}G^{o} = −193 kJ mol^{−1} is used for oxidizing M^{+} as M^{+} → M^{3}^{+} + 2e^{−}, E^{o} = −0.25 V. Under standard conditions, the number of moles of M^{+} oxidized when one mole of X is converted to Y is [F = 96500 C mol^{−1}]
Solutions:Conversion of one mole of X to Y
X → Y ΔrGo = 193 kJ mol1Now, for the reaction
M+ → M3+ + 2 e Eo = 0.25 Vthe standard Gibbs free energy change is given as
ΔGo = nFEo⇒ ΔGo = 2 × 96500 × 0.25 = 48.25 kJ mol1Thus, 48.25 kJ mol^{−1} energy oxidise one mole of M^{+}. So, 193 kJ mol^{−1} energy will oxidise ⇒
19348.25 = 4moles of M^{+}.
Hence, the 4 moles of M^{+} oxidised when one mole of X is converted to Y.
Question 27:
If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride−ammonia complex (which behaves as a strong electrolyte) is −0.0558 °C, the number of chloride(s) in the coordination sphere of the complex is [K_{f} of water = 1.86 K kg mol^{−1}]
Solutions:Concentration of aqueous solution of cobalt (III) chlorideammonia complex, m = 0.01 m Freezing point, T_{f} = −0.0558 ^{o}C = 272.9442 K ⇒ Depression in freezing point, ΔT_{f} = 273 − (272.9442) = 0.0558 K K_{f} of water = 1.86 K Kg mol^{−1}
Now, ΔT_{f} = iK_{f}m
⇒0.0558 = i × 1.86 × 0.01⇒ i = 3
Thus, if the given complex behaves as a strong electrolyte, then the number of particles given by the complex upon dissociation in the aqueous solution is 3.
Thus, the complex has the molecular formula [Co(NH_{3})_{5}Cl]Cl_{2}.
Hence, the number of chloride in the coordination sphere of the complex is 1. [[VIDEO:13903]]
Question 28:
The total number of stereoisomers that can exist for M is
Solutions:Number of stereoisomers = 2^{n} Here, n is the number of stereocentres. Thus, the total number of stereoisomers for the given compound is four as there are two stereocenters in the molecule as shown below:
These four stereoisomers will include geometrical as well as optical isomers, but this compound is unable to show geometrical isomerism. Therefore, only two optical isomers are possible for the given compound.
Question 29:
The correct statement (s) about Cr^{2}^{+} and Mn^{3}^{+} is (are) [Atomic numbers of Cr = 24 and Mn = 25]
Option 1:  Cr^{2}^{+} is a reducing agent 
Option 2:  Mn^{3}^{+} is an oxidizing agent 
Option 3:  Both Cr^{2}^{+} and Mn^{3}^{+} exhibit d^{4} electronic configuration 
Option 4:  When Cr^{2}^{+} is used as a reducing agent, the chromium ion attains d^{5} electronic configuration 
Solutions:The electronic configurations of Cr^{2}^{+} and Mn^{3}^{+} ions are given below: Cr^{2}^{+} = [Ar] 3d^{4} Mn^{3}^{+} = [Ar] 3d^{4} So, both the ions show d^{4} electronic configuration.
Cr is reducing as its configuration changes from d^{4} to d^{3}, the latter having a stable halffilled t_{2}g level.
Mn^{3+} acts as a strong oxidising agent as it needs only one electron to attain the stable halffilled d^{5} electronic configuration.
Thus, the correct statements are A, B and C.
Question 30:
Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is (are)
Option 1:  Impure Cu strip is used as cathode 
Option 2:  Acidified aqueous CuSO_{4} is used as electrolyte 
Option 3:  Pure Cu deposits at cathode 
Option 4:  Impurities settle as anode−mud 
Solutions:During refining of blister copper to obtain pure copper by the process of electrolytic refining, the impure copper is used as anode and acidified copper sulphate solution is used as an electrolyte. In this process, pure copper from anode is deposited on the strip of pure copper which is used as cathode, while the impurities settle down as anode mud.
The correct answer is B, C and D.
Question 31:
Fe^{3}^{+} is reduced to Fe^{2}^{+} by using
Option 1:  H_{2}O_{2} in presence of NaOH 
Option 2:  Na_{2}O_{2} in water 
Option 3:  H_{2}O_{2} in presence of H_{2}SO_{4} 
Option 4:  Na_{2}O_{2} in presence of H_{2}SO_{4} 
Solutions:Na_{2}O_{2} in water furnishes H_{2}O_{2} and NaOH which reduces Fe^{3}^{+} to Fe^{2}^{+} as shown in the reactions below: Na_{2}O_{2} + 2 H_{2}O → H_{2}O_{2} + 2 NaOH 2 Fe^{3}^{+} + H_{2}O_{2} + 2 OH^{−} → 2 Fe^{2}^{+} + 2 H_{2}O + O_{2}
Hence, the correct options are A and B.
Question 32:
The % yield of ammonia as a function of time in the reaction
N_{2} (g) + 3H_{2} (g) â‡Œ 2NH_{3} (g), ΔH < 0
at (P, T_{1}) is given below.
If this reaction is conducted at (P, T_{2}), with T_{2} > T_{1}, the % yield of ammonia as a function of time is represented by
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:The given reaction is an exothermic reaction. When an exothermic reaction is carried out at high temperatures; then at equilibrium; according to Le Chatelier’s principle, the reaction will proceed in backward direction. Thus, on increasing the temperature, the percentage yield of ammonia will be given by the following graph:
Hence, the correct answer is option B.
Question 33:
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are
Option 1:  12,18 
Option 2:  1,14 
Option 3:  12,12 
Option 4:  14,18 
Solutions:Oxygen atoms are arranged in ccp array. Number of oxygen atoms per unit cell = 4 Number of octahedral voids per unit cell = 4 Number of tetrahedral voids per unit cell = 8 Number of occupied octahedral voids per unit cell = 4m Number of occupied tetrahedral voids per unit cell = 8n
(4m
×3) + (8n
×2) + (4
×(−2)) = 0 12m + 16n − 8 = 0 3m + 4n − 2 = 0 3m + 4n = 2
This relation is satisfied by values
12,18. Hence, the correct answer is option A.
Question 34:
Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is(are)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:(A)
(B)
(C)
(D)
Hence, the correct options are B and D.
Question 35:
The major product of the following reaction is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Question 36:
In the following reaction, the major product is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Question 37:
The structure of D(+)glucose is
The structure of L(–)glucose is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions: L(–)glucose
Hence, the correct option is A.
Question 38:
The major product of the reaction is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Question 39:
Match the anionic species given in Column I that are present in the ore(s) given in Column II.
Column I  Column II  
(A)  Carbonate  (P)  Siderite 
(B)  Sulphide  (Q)  Malachite 
(C)  Hydroxide  (R)  Bauxite 
(D)  Oxide  (S)  Calamine 
(T)  Argentite 
Question 40:
Match the thermodynamic processes given under Column I with the expressions given under Column II.
Column I  Column II  
(A)  Freezing of water at 273 K and 1 atm  (P)  q = 0 
(B)  Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions  (Q)  w = 0 
(C)  Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container  (R)  ΔSsys < 0 
(D)  Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm  (S)  ΔU = 0 
(T)  ΔG = 0 
Solutions:
A → (R) and (T)
H_{2}O (â„“) â‡Œ H_{2}O(s)
ΔG = 0 and ΔS_{sys} < 0
(B) → (P), (Q), (S)
Free expansion i.e. w = 0, ΔU = 0, q = 0
(C) → (P), (Q), (S)
q = 0, ΔU = 0, w = 0
(D) → (P), (Q), (S) and (T)
Question 41:
Let the curve C be the mirror image of the parabola y^{2} = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = –5, then the distance between A and B is
Solutions:Let P(t^{2}, 2t) be a point on the curve y^{2} = 4x and Q(h, k) be its mirror image with respect to x + y + 4 = 0 on C.
Then,
ht21=k2t1=2t2+2t+42⇒ht2=k2t=t22t4⇒h=2t+4 and k=t2+4If C intersects the line y = −5, then k = −5.
⇒t2+4=5⇒t=±1∴ h=2, k=5 and h=6, k=5So, the points A and B are (−2, −5) and (−6, −5).
⇒ AB = 4 (Using Distance Formula)
Therefore, the distance between A and B is 4 units. [[VIDEO:13790]]
Question 42:
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is
Solutions:Let the fair coin be tossed n number of times.
Given:
P(at least two heads)
≥0.96
⇒1
P(no head)
P(one head)
≥0.96
⇒1nC012012nnC11212n1≥0.96 ⇒112nn12n≥0.96 ⇒0.04≥n+112n ⇒125≥n+12n ⇒25≤2nn+1This inequality holds for
n≥8.
Therefore, the minimum number of times a fair coin needs to be tossed is 8.
Question 43:
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of
mnis
Solutions:Keeping all 5 five girls together and considering it as one entity, we have 6 entities (5 boys and a group of 5 girls) which can be arranged in 6! ways. But, the group of 5 girls can also arrange among themselves in the group in 5! ways. So, the number of ways in which 5 boys and 5 girls can stand in a queue such that all the girls stand consecutively in the queue is 6! × 5!. ∴ n = 6! × 5!
Now, 4 girls which stand in the queue consecutively can be selected out of 5 girls in
C45ways. Keeping all 4 five girls together and considering it as one entity, we have 6 entities (5 boys and a group of 4 girls) which can be arranged in 6! ways. But, the group of 4 girls can also arrange among themselves in the group in 4! ways. The remaining girl can be arranged in 5 ways. So, the number of ways in which 5 boys and 5 girls can stand in a queue such that exactly 4 girls stand consecutively in the queue is
C45× 6! × 4! × 5. ∴ m =
C45× 6! × 4! × 5
Thus,
mn=C45×6!×4!×56!×5!=5
Therefore, the value of
mnis 5.
Question 44:
If the normals of the parabola y^{2} = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)^{2} + (y + 2)^{2} = r^{2}, then the value of r^{2} is
Solutions:The end points of the latus rectum of the parabola are (1, ±2).
The equation of the normal to the parabola at the point (1, 2) is
y2=22x1 yy1=y12axx1⇒x+y3=0Similarly, the equation of the normal to the parabola at the point(1, −2) is
xy3=0.
The centre of the given circle is (3, −2).
The normals to the parabola drawn at the end points of its latus rectum are tangents to the given circle.
So, the lengths of perpendicular drawn from the centre of the circle to the normals is equal to the radius of the circle.
∴32312+12=r=3+2312+12⇒r2=2Therefore, the value of r^{2} is 2.
Question 45:
Let f : â„ → â„ be a function defined by
fx=x,x≤20,x>2, where [x] is the greatest integer less than or equal to x. If
I=∫12xf x22+fx+1dx,then the value of (4I – 1) is
Solutions:
I=∫12xfx22+fx+1dx⇒I=∫10xfx22+fx+1dx+∫01xfx22+fx+1dx+∫12xfx22+fx+1dx+∫22xfx22+fx+1dx ….. 1We have
fx=x,x≤20,x>2So, fx2=x2=0,1≤x<11,1≤x<2and fx2=0∀x>2 ∴I=∫10x·02+0dx+∫01x·02+1dx+∫12x·12+0dx+∫22x·02+0dx Using 1⇒I=14×212⇒I=1421⇒I=14∴ 4I1=0Therefore, the value of (4I − 1) is 0.
Question 46:
A cylindrical container is to be made from certain solid material with the following constraints. It has a fixed inner volume of V mm^{3}, has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is a radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of
V250πis
Solutions:Given:
Fixed inner volume,
V=πr2h …..iLet the volume of the material used to make the container be v and the height of inner cylindrical part be h.
v=πr+222+πr+22hπr2h⇒v=2πr+22+πh4r+4⇒v=2πr+22+4πhr+1⇒v=2πr+22+4r+1Vr2 Using i⇒v=2πr+22+4V1r+1r2⇒dvdr=4πr+2+4V1r22r3For maxima or minima,
dvdr=0 ⇒4πr+2+4V1r22r3=0⇒4π12+4V102103=0 Given: r=10 mm⇒π1212103V=0⇒V=103π⇒V250π=4Therefore, if the volume of the material used to make the container is minimum when the inner radius of the container is 10 mm, then the value of
V250πis 4. [[VIDEO:13791]]
Question 47:
Let F(x) =
∫xx2+π62cos^{2} t dt for all x ∈
ℝand f :
0, 12 → [0, ∞)be a continuous function. For a
∈0, 12, if F'(a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x = a, then f(0) is
Solutions:We have
F(x) =
∫xx2+π61+cos2tdt =txx2+π6+sin2t2xx2+π6=x2x+π6+12sin2x2+π3sin2x∴ F’x=2x1+12cos2x2+π34x2cos2xNow,∫0a fxdx=F’a+2⇒∫0a fxdx=2a1+12cos2a2+π34a2cos2a+2Differentiating both sides w.r.t. a, we getfa=2+124cos2a2+π34asin2a2+π34a+4sin2aSubstituting a=0, we getf0=2+124×120+0=3Therefore, f(0) is 3. [[VIDEO:13792]]
Question 48:
The number of distinct solutions of the equation :
54cos2 2x + cos4 x + sin4 x + cos6 x + sin6 x = 2in the interval 10, 2π is
Solutions:
54cos22x+cos4x+sin4x+cos6x+sin6x=2 ⇒54cos22x+cos2x+sin2x22cos2xsin2x+cos3x+sin3x3cos2xsin2x=2 ⇒54cos22x+112sin22x+134sin22x=2 ⇒54cos22x54sin22x=0 ⇒54cos4x=0 ⇒cos4x=0Now,
x∈0,2π ⇒4x∈0,8π x=π8,3π8,5π8,7π8,9π8,11π8,13π8,15π8Therefore, the number of distinct solutions of the given equation is 8. [[VIDEO:13793]]
Question 49:
Let y(x) be a solution of the differential equation (1 + e^{x}) y’ + ye^{x} = 1. If y(0) = 2, then which of the following statements is (are) true?
Option 1:  y(−4) = 0 
Option 2:  y(−2) = 0 
Option 3:  y(x) has a critical point in the interval (−1, 0) 
Option 4:  y(x) has no critical point in the interval (−1, 0) 
Solutions:The given differential equation is
1+exy’+yex=1⇒y’+ex1+exy=11+exThis is a linear differential equation.
Integrating factor =
e∫ex1+exdx
=eloge1+ex=1+exTherefore, the solution of the given equation is given by
y1+ex=∫11+ex.1+exdx+C
⇒y1+ex=x+CGiven: y0=2∴21+1=C⇒C=4So,
y=4+x1+ex∴ y4=0y2=2x2+1For critical points, dydx=0⇒1+exx+4ex1+ex2=0⇒13exxex=0⇒3+xex=1⇒3+x=ex
From the graph we can see that the critical point lies in the interval (−1, 0).
Hence, the correct options are A and C.
Question 50:
Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is represented by the differential equation Py” + Qy’ + 1 = 0, where P, Q are functions of x, y and y’
here y’ = dydx, y” = d2ydx2, then which of the following statements is (are) true?
Option 1:  P = y + x 
Option 2:  P = y − x 
Option 3:  P + Q = 1 − x + y + y’ + (y’)^{2} 
Option 4:  P − Q = x + y − y’ − (y’)^{2} 
Solutions:Let the equation of the circle be
xh2+yh2=r2 …..1
On differentiating (1) with respect to x, we get
xh+yhdydx=0 …..2⇒xh+ydydxhdydx=0 ⇒h=x+ydydx1+dydx …..3Differentiating (2) with respect to x, we get
1+yhd2ydx2+dydx2=0Substituting the value of h from (3), we get
1+yx+ydydx1+dydxd2ydx2+dydx2=0⇒1+yx1+dydxd2ydx2+dydx2=0⇒1+dydx+yxd2ydx2+1+dydxdydx2=0⇒1+y’+yxy”+1+y’y’2=0⇒yxy”+1+y’+y’2y’+1=0∴P=yx, Q=1+y’+y’2So, P + Q =
1x+y+y’+y’2Hence, the correct options are B and C. [[VIDEO:13794]]
Question 51:
Let g:
ℝ → ℝbe a differentiable function with g(0) = 0, g’ (0) = 0 and g'(1) ≠ 0. Let
fx = x x gx,x ≠00,x = 0and h(x) = e^{x} for all x ∈
ℝ. Let (f ° h) (x) denote f(h(x)) and (h ° f) (x) denote h (f(x)). Then which of the following is (are) true ?
Option 1:  f is differentiable at x = 0 
Option 2:  h is differentiable at x = 0 
Option 3:  f ° h is differentiable at x = 0 
Option 4:  h ° f is differentiable at x = 0 
Solutions:We have
g(0) = 0, g‘(0) = 0 and g‘(1) ≠ 0
Consider option (A).
fx = gx,x>0gx,x<00,x=0 f’0+=limh→0fhf0h =limh→0ghg0h =g’0 =0f’0=limh→0fhf0h =limh→0gh+g0h =g’0 =0So, the function f(x) is differentiable at x = 0.
Consider option (B).
It is given that
hx=ex.
∴hx = ex,x>0ex,x<01,x=0 h’0+=limx→0hxh0x =limx→0ex1x =1h’0=limx→0hxh0x =limx→0ex1x =1So, h(x) is not differentiable at x = 0.
Consider option (C).
fohx=fhx=ghx ∵hx>0foh’0+=limx→0ghxgh0x =limx→0gexg1x =g’1foh’0=limx→0ghxgh0x =limx→0gexg1x =g’1So, (foh)(x) is not differentiable at x = 0.
Consider option (D).
hofx=hfxhof’0+=limx→0hfxhf0x =limx→0hgxh0x = limx→0hgx1x =0 hof’0=limx→0hfxhf0x =limx→0hgx1x =0So, hof(x) is differentiable at x = 0.
Hence, the correct options are A and D. [[VIDEO:13795]]
Question 52:
Let
fx=sin π6sin π2sin xfor all x ∈ â„ and
gx=π2sin xfor all x ∈ â„. Let (f âˆ˜ g) (x) denote f(g(x)) and (g âˆ˜ f) (x) denote g(f(x)). Then which of the following is(are) true?
Option 1:  Range of f is 12, 12 
Option 2:  Range of f∘ g is 12, 12 
Option 3:  limx→0 fxgx=π6 
Option 4:  There is an x ∈ â„ such that (g âˆ˜ f) (x) = 1 
Solutions:
fx=sinπ6sinπ2sinx
Let π2sinx=θ ⇒θ∈π2, π2So, fx=sinπ6sinθLet π6sinθ=Φ⇒Φ∈π6, π6∴ fx=sinΦ⇒fx∈12, 12Now,fogx=sinπ6sinπ2sinπ2sinx Similarly,
fogx∈12, 12
limx→0f(x)g(x)=sinπ6sinπ2sinxπ6sinπ2sinx×π6sinπ2sinxπ2sinx=π6Now, gf(x)=π2sinsinπ6sinπ2sinxπ2sin12≤gf(x)≤π2sin120.75≤gf(x)≤0.75So, there is no x ∈ R such that (gof)(x) = 1.
Hence, the correct options are A, B and C. [[VIDEO:13796]]
Question 53:
Let âˆ†PQR be a triangle. Let
a→=QR→, b→=RP→ and c→=PQ→. If a→=12, b→=43 and b→·c→=24, then which of the following is(are) true?
Option 1:  c→22a→=12 
Option 2:  c→22+a→=30 
Option 3:  a→×b→+c→×a→=483 
Option 4:  a→·b→=72 
Solutions:In
∆PQR,
a→=QR→, b→=RP→ and c→=PQ→. So,
a→+b→+c→=0⇒b→+c→=a→⇒b→+c→.b→+c→=a→.a→⇒b→2+c→2+2b→·c→=a→2⇒c→2=1444848⇒c→2=48⇒c→=43Thus,
∆PQRis an isosceles triangle.
Now,
c→22a→=48212=12Let
θbe the angle between
b→and
c→. Now,
b→·c→=24⇒b→c→cosθ=24⇒cosθ=12⇒θ=120° θ≠60°⇒∠PQR=∠QRP=30° ∵PQ→=RP→
a→+b→+c→=0⇒a→×b→+a→×c→=0⇒a→×b→=c→×a→∴a→×b→+c→×a→=2a→×b→=212×43×sin30°=483Also,
a→+b→+c→=0⇒a→+b→=c→⇒a→+b→.a→+b→=c→.c→⇒a→2+b→2+2a→·b→=c→⇒144+48+2a→·b→=48⇒a→·b→=72Hence, the correct options are A, C and D.
Question 54:
Let X and Y be two arbitrary, 3 âœ• 3, nonzero, skewsymmetric matrices and Z be an arbitrary 3 âœ• 3, nonzero, symmetric matrix. Then which of the following matrices is(are) skew symmetric?
Option 1:  Y^{3}Z^{4} − Z^{4}Y^{3} 
Option 2:  X^{44} + Y^{44} 
Option 3:  X^{4}Z^{3} − Z^{3}X^{4} 
Option 4:  X^{23} + Y^{23} 
Solutions:If X and Y are skewsymmetric matrices and Z is symmetric matrix, then
XT=X, YT=Y and ZT=Z A Y3Z4Z4Y3T =Y3Z4TZ4Y3T =Z4T·Y3TZ4T·Y3T =Y3Z4Z4Y3Therefore,
Y3Z4Z4Y3is a symmetric matrix.
B X44+Y44T =X44T+Y44T =X44+Y44Therefore,
X44+Y44is a symmetric matrix.
C X4Z3Z3X4T =X4Z3TZ3X4T =Z3T·X4TX4T·Z3T =X4Z3Z3X4Therefore,
X4Z3Z3X4is a skewsymmetric matrix.
D X23+Y23T =X23T+Y23T =X23Y23 =X23+Y23Therefore,
X23+Y23is a skewsymmetric matrix.
Hence, the correct options are C and D.
Question 55:
Which of the following values of α satisfy the equation
1+α21+2α21+3α22+α22+2α22+3α23+α23+2α23+3α2=648α?
Option 1:  −4 
Option 2:  9 
Option 3:  −9 
Option 4:  4 
Solutions:
1+α21+2α21+3α22+α22+2α22+3α23+α23+2α23+3α2=648α R2→R2R1, R3→R3R21+α21+2α21+3α23+2α3+4α3+6α5+2α5+4α5+6α=648αR2→R2R31+α21+2α21+3α22225+2α5+4α5+6α=648αC2→C2C1, C3→C3C11+α2α2+3α4α1+2α2005+2α2α4α=648α ⇒24α·α2+3α2α·4α1+2α=648α⇒8α3=648α⇒8α3+648α=0⇒α=0, ±9Hence, the correct options are B and C. [[VIDEO:13797]]
Question 56:
In â„^{3}, consider the planes P_{1} : y = 0 and P_{2} : x + z = 1. Let P_{3} be a plane, different from P_{1} and P_{2}, which passes through the intersection of P_{1} and P_{2}. If the distance of the point (0, 1, 0) from P_{3} is 1 and the distance of a point (α, β, γ) from P_{3} is 2, then which of the following relations is (are) true?
Option 1:  2α + β + 2γ + 2 = 0 
Option 2:  2α − β + 2γ + 4 = 0 
Option 3:  2α + β − 2γ − 10 = 0 
Option 4:  2α − β + 2γ − 8 = 0 
Solutions:
Given:P1: y=0P2: x+z=1Let P3 be the plane which passes through the intersection of planes P1 and P2. ⇒P3: P2+λP1=0⇒x+z1+λy=0Now, the distance of the point (0, 1, 0) from P_{3} is 1.
⇒0+λ+011+1+λ2=1⇒λ2+12λ=2+λ2⇒2λ=1⇒λ=12So, the equation of plane is x+z1+12y=0i.e. P3: 2xy+2z2=0Also, the distance of the point
α, β, γfrom P_{3} is 1.
⇒2αβ+2γ222+1+22=2⇒2αβ+2γ2=±6⇒2αβ+2γ=4 or 8⇒2αβ+2γ+4=0 or 2αβ+2γ8=0Hence, the correct options are B and D. [[VIDEO:13798]]
Question 57:
In
ℝ3, Let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P_{1} : x + 2y – z + 1 = 0 and P_{2} : 2x – y + z – 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P_{1}. Which of the following points lie(s) on M?
Option 1:  0,56, 23 
Option 2:  16,13,16 
Option 3:  56, 0,16 
Option 4:  13, 0,23 
Solutions:
The given line is parallel to P1 and P2 that means its direction cosines will be perpendicular to normal of both the planes.L1⊥n1→ and L1⊥n2→ , where n1→ and n2→ are normal vectors perpendicular to the two planes P1 and P2 respectively.Thus, the vector parallel to the given line is n1→× n2→.n1→× n2→=i^j^k^121211=i ^3j^5k^So, the equation of the line is x01=y03=z05Hence, the general point on the line is λ, 3λ, 5λ.The coordinates of the foot of perpendicular drawn from the points on line L to the plane P1 are given by xλ1=y+3λ2=z+5λ1=1λ6λ+5λ+11+4+1xλ=y+3λ2=z+5λ1=16x=λ16y=3λ13z=5λ+16These equations are satisfied by the points given in options A and B.Hence, the correct options are (A) and (B). [[VIDEO:13799]]
Question 58:
Let P and Q be distinct points on the parabola y^{2} = 2x such that a circle with PQ as diameter passes through the vertex 0 of the parabola. If P lies in the first quadrant and the area of the triangle âˆ†OPQ is
32, then which of the following is(are) the coordinates of P?
Option 1:  4, 22 
Option 2:  9, 32 
Option 3:  14, 12 
Option 4:  1, 2 
Solutions:
The equation of the given parabola is y^{2} = 2x.
∴ 4a = 2
⇒ a =
12The parabola can be written parametrically as
x = at^{2}
y = 2at
So, the coordinates of any distinct points P and Q on the parabola are
t122, t1 and t222, t2, respectively.
PQ is the diameter of the circle.
We know that the angle in the semicircle is
90°.
∴∠POQ=90°So, OP ⊥ OQ
Slope of the line OP × Slope of the line OQ = −1
t212t22×t112t12=1⇒t2t1=t22t124⇒ t2t1=4Area of
△OPQ=32 (Given)
∴12t144+t12×t244+t22=32⇒12×4×t12+4t22+416=32 ∵t1t2=4⇒4×16+4t12+t22+1616=18⇒8+t12+t22=18⇒t12+t2210=0⇒t1410t12+16=0 ∵t2=4t1⇒t12=2, 8⇒t1=±2, ±22P lies in first quadrant. So, t1 can be 2 or 22.∴Coordinates of P are 12×2, 2 or 12×8, 22 i.e. 1, 2 or 4, 22.Hence, the correct options are A and D. [[VIDEO:13800]]
Question 59:
Column I  Column II  
(A)  In
ℝ2, if the magnitude of the projection vector of the vector αi^+βj^ on 3i^+j^ is 3 and if α=2+3β,then possible value(s) of αis (are) 
(P)  1 
(B)  Let a and b real numbers such that the function
fx=3ax22,x<1bx+a2, x≥1is differentiable for all x ∈ ℝ.Then possible value(s) of a is (are) 
(Q)  2 
(C)  Let
ω≠1be a complex cube root of unity. If 33ω+2ω24n+3 +2+3ω3ω24n+3+3+2ω+3ω24n+3=0,the possible value(s) of n is (are) 
(R)  3 
(D)  Let the harmonic mean of two positive real numbers a and b be 4. If q is a positive real number such that a, 5, q, b is an arithmetic progression, then the value(s) of lq − al is (are)  (S)  4 
(T)  5 
Solutions:(A)
Projection of the vector αi^ +βj^ on 3i^ +j^ =αi^ +βj^.3 i^ +j^32+1=3α+β2Projection of the vector αi^ +βj^ on 3i^ +j^=3 Given∴3α+β2=3⇒3α+β=23 …..1Now, α=2+3β Given⇒α3β=2 …..2Solving (1) and (2), we get
α=2and
β=0Thus, the possible value of
αis 2.
(A) → (P)
(B)
fx=3ax22, x <1bx+a2, x≥1 is differentiable for all x∈R. f’1+=limh→0 f1+hf1h =limh→0 b1+h+a2ba2h = bf’1=limh→0 f1hf1h =lim h→0 3a1h22ba2h =lim h→0 3a1+h22h2ba2h =lim h→0 3a3ah2+6ah2ba2h =lim h→0 3ah2+6ah3a2ba2h For f to be differntiable 3a2ba2 has to be 0, otherwise it will be not defined.So, lim h→0 3ah2+6ahh=6aAs f is differentiable,
f‘(1^{+}) = f‘(1^{−})
∴b=6a …..(1) Also, 3a2ba2=0⇒3a2+6aa2=0 Using 1⇒a23a+2=0⇒a=1, 2Thus, the possible values of
αare 1 and 2.
(B) → (P), (Q)
(C)
Let a=33ω+2ω2∴aω=3ω3ω2+2And aω2=3ω23+2ωSo,33ω+2ω24n+3+2+3ω3ω24n+3+3+2ω+3ω24n+3=0⇒a4n+3+aω4n+3+aω24n+3=0⇒a4n+31+ω4n+3+ω24n+3=0⇒1+ω4n+3+ω24n+3=0Therefore, 4n+3 should not be multiple of 3 or n should not be multiple of 3.Thus, the values of n are 1, 2, 4 and 5.
(C) → (P), (Q), (S), (T)
(D)
Harmonic mean of a and b=2aba+bHarmonic mean of a and b=4 Given⇒2aba+b=4⇒aba+b=2 …..1 It is given that a, 5, q, b are in AP.∴10=a+q …..22q=5+b ⇒q=5+b2Substituting the value of q in 2, we get10=a+5+b2⇒2a+b=15 ⇒b=152a …..3From 1 and 3, we geta152aa+152a=2⇒15a2a2=2a+30⇒2a217a+30=0⇒2a212a5a+30=0⇒2aa65a6=0⇒a62a5=0⇒a=6, 52∴q=4, 152⇒qa=2, 5, 32Thus, the values of
qaare 2 and 5.
(D) → (Q), (T)
Question 60:
Column I  Column II  
(A)  In a triangle âˆ†XYZ, let a, b and c be the lengths of the sides opposite to the angles X, Y and Z, respectively. If 2(a^{2} − b^{2}) = c^{2} and
λ=sin XYsin Zthen possible values of n sin Z for which cos(nπλ) = 0 is (are) 
(P)  1 
(B)  In a triangle âˆ†XYZ, let a, b and c be the lengths of the sides opposite to the angles X, Y and Z, respectively. If 1 + cos 2X − 2cos 2Y = 2 sin X sin y, then possible value(s) of
abis (are) 
(Q)  2 
(C)  If â„^{2}, let
3 i^+j^, i^+3 j^ and βi^+1βj^be the position vectors of X, Y and Z with respect to the origin O, respectively. If the distance of Z from the bisector of the acute angle of OX→ with OY→ is 32, then possible value(s) of β is (are) 
(R)  3 
(D)  Suppose that F(α) denotes the area of the
region bounded by x = 0, x = 2, y^{2} = 4x and
y = αx − 1 + αx − 2 + ax, where α ∈{0, 1}. Then the value(s) of F(α) +
832, when α = 0 and α = 1, is (are) 
(S)  5 
(T)  6 
Solutions:(A) In âˆ†XYZ, a, b and c be the length of the sides opposite to the angles X, Y and Z, respectively. It is given that 2(a^{2} − b^{2}) = c^{2} ∴ 2(sin^{2}X − sin^{2}Y) = sin^{2}Z (Uisng Sine Rule) ⇒ 2 sin(X + Y) sin(X − Y) = sin^{2}Z …..(1) As, X + Y + Z =
180°
⇒X+Y=180°Z
⇒sinX+Y=sin180°Z⇒sinX+Y=sinZ∴ 2sinZ sin(X − Y) sin^{2}Z
⇒sinXYsinZ=12Given: λ=sinXYsinZ⇒λ=12∴cosnπλ=cosnπ2=0 ⇒nπ2=2k+1π2⇒n=2k+1Thus, the possible values of n are 1, 3 and 5.
(A) → (P), (R), (S)
(B) In âˆ†XYZ, a, b and c be the length of the sides opposite to the angles X, Y and Z, respectively.
1+cos2X2cos2Y=2sinX sinY⇒1+12sin2X212sin2Y=2 sinXsinY⇒2sin2Ysin2X=sinXsinY⇒2sin2Xsin2Y=sinXsinYNow,
sinXsinY=abLet sinXsinY=ab=m∴ 2m2=m⇒m2+m2=0⇒m=1, 2Thus, the possible values of
abare 1 and −2.
(B) → (P)
(C)
The angle made by the bisector of acute angle of OX→ and OY→ is 45°.The equation of the acute angle of the bisector of OX→ and OY→ is x=y or xy=0.Distance of βi ^+1βj^ from xy = β1β2 =322β1=32β=±3+1⇒2β=4, 2⇒β=2, 1⇒β=2, 1Thus, the possible values of
βare 1 and 2.
(C) → (P), (Q)
(D)
For α=0, y=3.Now, F0 denotes the area of the region bounded by x=0, x=2, y2=4x and y=3.∴f0=∫0232xdx =6823∴ f0+823=6
For α=1,y=x1+x2+xF1=∫013x2xdx+∫12x+12xdx =5823⇒F1+823=5Thus, the possible values of
Fα+832are 5 and 6. (D) → (S), (T)