
JEE Advanced 2016 Paper 1 (Code6)
Test Name: JEE Advanced 2016 Paper 1 (Code6)
Question 1:
A parallel beam of light is incident from air at an angle α on the side PQ of a right angled triangular prism of refractive index
n=2. Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45°. The angle θ of the prism is
Option 1:  15° 
Option 2:  22.5° 
Option 3:  30° 
Option 4:  45° 
Solutions:The path of the beam can be drawn as follows:
Using Snell’s law, we have
sinαsinβ=n1⇒sin45°=2sinβ⇒sinβ=12∴β=30°The angle of incident of light beam on the face PR for which light undergoes total internal reflection isâ€‹ given by
sinic=1n=12⇒ic=450From the figure, we can see that
ic=θ+β=45°⇒θ=45°30°=15°Hence, the correct answer is option A.
Question 2:
In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light and the corresponding stopping potential (V_{0}) are given below:
λ (μm) V_{0}(Volt) 
0.3 2.0 0.4 1.0 0.5 0.4 
Given that c = 3 × 10^{8} ms^{–1 }and e = 1.6 × 10^{–19} C, Planck’s constant (in units of J s) found from such an experiment is
Option 1:  6.0 × 10^{–34} 
Option 2:  6.4 × 10^{–34} 
Option 3:  6.6 × 10^{–34} 
Option 4:  6.8 × 10^{–34} 
Solutions:From Einstein’s equation of photoelectric effect, we have
hcλϕ=eV0 …..(i)
For different wavelengths of light, we can write (i) as
hcλ1ϕ=eV1⇒hc0.3×106ϕ=1.6×1019×2 …..(ii)hcλ2ϕ=eV2⇒hc0.4×106ϕ=1.6×1019×1 …..(iii)hcλ3ϕ=eV3⇒hc0.5×106ϕ=1.6×1019×0.4 …..(iv)Subtracting (iii) from (ii), we get
hc10610.310.4=1.6×101921⇒h=6.4×1034 (Substituting for c=3×108 m/s)Similarly, subtracting (iv) from (iii), we get
hc10610.410.5=1.6×101910.4⇒h=6.4×1034Hence, the correct answer is option B.
Question 3:
A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30°C and the entire stored 120 litres of water is initially cooled to 10°C. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is
(Specific heat of water is 4.2 kJ kg^{–1} K^{–1} and the density of water is 1000 kg m^{–3})
Option 1:  1600 
Option 2:  2067 
Option 3:  2533 
Option 4:  3933 
Solutions:Given, Capacity of water cooler = 120 litres Total mass of water in the cooler = 120 kg Heat generated by external device = 3 kW = 3000 W
Consider that T be the temperature of the coolder at any time t. So, the rate of the cooling is
msdTdt=3000P⇒∫1030dT=3000Pms∫03dt⇒3010=(3000P)×3×3600120×4.2×103⇒3000P=20×120×423×36=28003⇒P=300028003≈2067Hence, the correct answer is option B.
Question 4:
An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:As we know the electric field due to infinite line charge is given by ,
E=λ2πεrCurrent density,
j=qAt=σλ2πεr …(i)
Differentiating the above equation w.r.t t, we get
dqdt=jA=j2πrl⇒dλldt=λ2πεr×σ2πrl⇒dλdt=λε×σ⇒dλλ=σε⇒λ=λ0eσεt⇒j=j0eσεtj(t) decays exponentially, thus graph A best suits the above result.
Hence, the correct answer is option A.
Question 5:
A uniform wooden stick of mass 1.6 kg and length l rests in an inclined manner on a smooth, vertical wall of height h(< l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30° with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/l and the frictional force f at the bottom of the stick are (g = 10 m s^{–2})
Option 1:  hl=316, f=1633N 
Option 2:  hl=316, f=1633N 
Option 3:  hl=3316, f=833N 
Option 4:  hl=3316, f=1633N 
Solutions:The freebody diagram of the figure is shown below:
The force f_{r} is the frictional force due to the rough surface.
Balancing the vertical forces, we have
Nsin30°+N=mg⇒N(12+1)=mg⇒N=23mgSimilarly,
Ncos30°fr=0⇒fr=32N=32×23mg=13×1.6×10=163∴fr==1633 NBalancing the torque about point A,
Nhsin60=mgl2cos60⇒2mg3×2h3=mg×l2×12⇒hl=3316Hence, the correct answer is option D.
Question 6:
The position vector
r→of a particle of mass m is given by the following equation
r→t=αt3i^+βt2j^, where α =
103m s^{–3}, β = 5 m s^{–2} and m = 0.1 kg. At t = 1 s, which of the following statement(s) is(are) true about the particle?
Option 1:  The velocity
v→ is given by v→=10i^+10j^ m s1 
Option 2:  The angular momentum
L→with respect to the origin is given by L→=5/3 k^N m s 
Option 3:  The force
F→is given by F→=i^+2j^ N 
Option 4:  The torque
τ→with respect to the origin is given by τ→=20/3 k^ N m 
Solutions:Given, Position vector
r→t=αt3i^+βt2j^. α =
103m s^{–3}, β = 5 m s^{–2} m = 0.1 kg
Velocity vector,
v→=drdt=3at2i^+2βtj^At t = 1 s,
v→=10i^+10j^Angular momentum is given as L=mr→×v→L=m2αβt4k^+3αβt4k^L=mαβt4(k^)at t = 1 s
L→=53k^a→=dv→dt=6αti^+2βj^At t =1 seca→=20i^+10j^F→=ma→=2i^+j^τ=dLdt=dmαβt4dtk^τ=4mαβt3k^at t =1 sτ=203k^Hence, the correct answers are options A,B and D.
Question 7:
A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n_{1} and n_{2} (> n_{1}), as shown in the figure. A ray of light is incident with angle θ_{i} from medium 1 and emerges in medium 2 with refraction and θ_{f} with a lateral displacement
l.
Which of the following statement(s) is(are) true?
Option 1:  l is independent of n_{2} 
Option 2:  n_{1}sinθ_{i} = n_{2}sinθ_{f} 
Option 3:  l is dependent on n(z) 
Option 4:  n_{1}sinθ_{i} = (n_{2} – n_{1})sinθ_{f} 
Solutions:The lateral displacement l depends on the refractive index of the glass slab n(z) and not on n_{2}, hence options A and C are true.
From Snell’s law, we have
n1sinθi=n(z)sinθ=n2sinθf⇒n1sinθi=n2sinθfHence, option B is true and D are false.
Hence, the correct answers are options A, B and C.
Question 8:
A planoconvex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true?
Option 1:  The refractive index of the lens is 2.5. 
Option 2:  The radius of curvature of the convex surface is 45 cm. 
Option 3:  The faint image is erect and real. 
Option 4:  The focal length of the lens is 20 cm. 
Solutions: Given: Object distance = −30 cm, As per the formula,
1f=μ1R …..(i)
As per question, magnification of the image, m = 2 Thus, the image distance will be v = −mu = 60 cm,
Using lens formula,
1f=160130=120⇒f=20 cmFrom equation (i)
μ1R=120 …..(ii)
Now lets consider the reflection. As per question, the source of the lens behaves as a convex mirror. For the real object, image formed is virtual. According to the question,
u =
30 cm v = +10 cm
2R=110130 ∵f=R2⇒R=30 cmUsing equation (ii),
μ130=120⇒μ=2.5Hence, the correct answers are options A and D.
Question 9:
Highly excited states for hydrogenlike atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true?
Option 1:  Relative change in the radii of two consecutive orbitals does not depend on Z 
Option 2:  Relative change in the radii of two consecutive orbitals varies as 1/n 
Option 3:  Relative change in the energy of two consecutive orbitals varies as 1/n^{3} 
Option 4:  Relative change in the angular momenta of two consecutive orbitals varies as 1/n 
Solutions:According to Bohr, the radius of a Hydrgenlike atom is given as
a=n2Zkwhere k is a constant. Since relative change in radius is given as
∆aa=(n+1)2n2Z×n2Z=1+2nn2=1n2+2n≈2nHence, A and B are true.
Energy of Bohr’s atom is
E=13.6Zn2⇒∆EE=n21(n+1)21n2This clearly does not vary as
1n3. Hence, option C is false.
Also, by Bohr’s quantisation condition L = nh. Hence, relative change in L varies as
1n.
Hence the correct answers are option A, B and D.
Question 10:
A lengthscale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (k_{B}), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for l is(are) dimensionally correct?
Option 1:  l=nq2εkBT 
Option 2:  l=εkBTnq2 
Option 3:  l=q2εn2/3kBT 
Option 4:  l=q2εn1/3kBT 
Solutions:We know that, dimension of permittivity (ε) of a dielectric material:
ε=AT2MLT2L2=M1L3A2T4Boltzmann constant (k_{B}):
kB=ML2T2θ1Absolute temperature (T):
T=θNumber per unit volume (n):
n=L3Charge (q)
q=AT
∴εkBTnq2=L1T2A2L3T2A2=Lq2εn13kBT=A2T2L3T2A2×L3=LHence, the correct answers are options B and D.
Question 11:
Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/h along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let v_{P}, v_{Q} and v_{R} be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 m/s. Which of the following statement(s) is(are) true regarding the sound heard by the person?
Option 1:  The plot below represents schematically the variation of beat frequency with time 
Option 2:  v_{P} + v_{R} = 2v_{Q} 
Option 3:  The plot below represents schematically the variation of beat frequency with time 
Option 4:  The rate of change in beat frequency is maximum when the car passes through Q. 
Solutions:Consider the given figure:
The frequency heard by the car due to source M:
fM=118vs+v0cosθvsThe frequency heard by the car due to source N:
fN=121vs+v0cosθvsThen frequency of beat is
n=(121119)1+v0cosθvs=31+v0cosθvsAs θ increases, cosθ decreases and hence n decreases. Since θ decreases as the car goes from the P to Q, it can be deduced that option A is true.
Since P and R are at equal distance in the opposite ends, the angle formed by them will be θ and −θ. Thus, the beat frequency at this position are
vP=31+v0cosθvs;vR=31v0cosθvsAt Q, θ = 90°
∴vQ=3(10)=3⇒vQ=vP+vR2The rate of change of beat is
dndθ=3v0vssinθThis is maximum at θ = 90°, which is at Q.
Hence, the correct answers are options A, B and D.
Question 12:
A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 A s^{–1}. Which of the following statement(s) is(are) true?
Option 1:  The induced current in the wire is in opposite direction to the current along the hypotenuse. 
Option 2:  The magnitude of induced emf in the wire is
μ0πvolt. 
Option 3:  There is a repulsive force between the wire and the loop. 
Option 4:  If the loop is rotated at a constant angular speed about the wire, an additional emf of
μ0πvolt is induced in the wire. 
Solutions:Given, Current in the rectangular loop is increasing at the rate of 10 A/s. Thus, current is induced in the wire parallel to the hypotenuse of the triangle.
Consider a case in which current I is flowing from the infinite long wire.
So, the flux through the shaded area is δϕ=μ0I2πr2r.δr=μ0Iπdr ⇒ϕ=μ0Iπ∫0ldr=μ0IlπThus, mutual induction is found to be
M=ϕI=μ0lπWhen the current flows through the triangle, the flux associated with the wire ϕ=μ0lπIFrom Faraday’s law, we have,
e=dϕdt ⇒e=μ0lπdϕdtSubstituting l = 10 cm and rate of change of current as 10 A/s, we find
e=μ0π VHence, option B is correct.
From Lenz law, it is know that the force between the wire and the loop is repulsive in nature. Hence, option C is correct.
Hence, the correct answers are options B and C.
Question 13:
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits blackbody radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to nonuniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
Option 1:  The temperature distribution over the filament is uniform. 
Option 2:  The resistance over small sections of the filament decreases with time. 
Option 3:  The filament emits more light at higher band of frequencies before it breaks up. 
Option 4:  The filament consumes less electrical power towards the end of the life of the bulb. 
Solutions:Resistance of a segment of tungsten is inversely proportional to its area of crosssection. And for a given current, heat produced in the segment of tungsten is directly proportional to the resistance of the segment. Towards the end of the life of filament, filament will become thinner. Therefore, the resistance of the filament will increase and so consumed power will be less, so it will emit less light. The filament emits more light at higher band of frequencies before it breaks up because at this point resistance of the tungsten is maximum and heat produced by the constant current is also maximum. The distribution of the temperature along the length of the tungsten will be non uniform. At the position where temperature is maximum, filament will break up.
Hence, the correct answers are option C and D.
Question 14:
The inductor L_{1} (inductance 1 mH, internal resistance 3 Ω) and L_{2} (inductance 2 mH, internal resistance 4 Ω), and a resistor R (resistance 12 Ω) are all connected in parallel across a 5 V battery. The circuit is switched on at time t = 0. The ratio of the maximum to the minimum current (I_{max} /I_{min}) drawn from the battery isâ€‹
Solutions:Given that the two inductors and one resistance are connected in parallel across a battery of 5 V.
At time t = 0, current in the inductors is zero. Thus using Kirchoff’s voltage law, we find Current through the resistance =
512A=IminAfter And at t =
∞, current in the inductors gets stabilised and behave like resistors. Thus, we find
I1=53AI2=54Awhere, current I_{1} and I_{2} corresponds to inductor L_{1} and L_{2}.
∴Imax=I1+I2+Imin=512+53+54=512×8⇒ImaxImin=512×8×125=8Hence, the required ratio is 8.
Question 15:
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has scale that displays log_{2}(P/P_{0}), where P_{0} is a constant. When the metal surface is at a temperature of 487 °C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 °C?
Solutions:We know that the power radiated from the metal is directly proportional to the fourth power of temperature. Given T_{1} = 487 °C = (487 + 273) K = 760 K Hence,
P1∝(760)4According to the question,
log2P1P0=1⇒P1P0=2⇒P0=P12For temperature T_{2} = 2767 + 273 = 3040 K
P2∝(3040)4∴log2P2P0=log22P2P1=log22×30407604=log2(29)=9Hence, the value displayed by the sensor is 9.
Question 16:
A hydrogen atom in its ground state is irradiated by light of wavelength 970 Å. Taking
hce= 1.237 × 10^{−6} eV m and the ground state energy of hydrogen atom as −13.6 eV, the number of lines present in the emission spectrum is
Solutions:Given: Wavelength of light, λ = 970 Å Energy of photon
=hcλ=1.237×106970×1010=1237970×10 eV=12.75 eVThe ground state energy of hydrogen atom =
13.6 eV
Therefore, on absorption of this photon, its energy changes energy to =
13.6 + 12.75 =
0.85 eV This is the energy of the 4th excited state.
∴Number of lines in the spectra = ^{4}C_{2} = 6
Hence, the number of lines in emission spectra is 6.
Question 17:
Consider two solid spheres P and Q each of density 8 gm cm^{–3} and diameters 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm^{–3} and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm^{–3} and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is
Solutions:The formula for calculating terminal speed v_{T} is
vT=2r2(ρσ)g9η …..(i) where,
ρ be the density of the material of the sphere and
σ be the density of the medium
For sphere P, (i) can be written as
vT,P=2(0.5)2(80.8)g9×3 …..(ii)For sphere Q, (i) can be written as
vT,Q=2(0.25)2(81.6)g9×2 …..(iii)Dividing (ii) by (iii), we get
vT,PvT,Q=2(0.5)2(80.8)g9×3×9×22(0.25)2(81.6)g⇒vT,PvT,Q=(0.5)2×7.23×2(0.25)2×6.4⇒vT,PvT,Q=3Hence, the correct value of the ratio is 3.
Question 18:
The isotope
B512having a mass 12.014 u undergoes β–decay to
C612.
C612has an excited state of the nucleus
C612*at 4.041 MeV above its ground state. If
B512decays to
C612*, the maximum kinetic energy of β–particle in units of MeV is (1 u = 931.5 MeV/c^{2}), where c is the speed of light in vacuum)
Solutions:The reaction is given as
B512→C*612+e+νKinetic energy of the β–particle is
K=(mBmC*)c2∆mc2 = (mBmC)c2∆mc2 =(12.01412)×931.54.041 =9 MeVHence, the maximum kinetic energy of the β–particle is 9 MeV.
Question 19:
P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr^{2}dr. The qualitative sketch of the dependence of P on r is
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
For the 1s orbital of hydrogen atom, with increase in distance from the nucleus, the probability of finding the electron increases upto a point, after which it begins to decrease.
The qualitative sketch of the dependence of P on r is
Hence, the correct answer is option C.
Question 20:
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (âˆ†S_{surr}) in J K^{−1} is (1 L atm = 101.3 J)
Option 1:  5.763 
Option 2:  1.013 
Option 3:  −1.013 
Option 4:  −5.763 
Solutions:
Given:
Temperature (T) = 300 K Pressure (p) = 3.0 atm
According to the first law of thermodynamics,
∆U = q + wFor an isothermal process,∆U = 0⇒ q = w
Now,
∆SSurr = qSurrT = qSysT =wSysT⇒∆SSurr = pextVf – ViT = 321300 = 0.01 L atm K1⇒∆SSurr = 1.013 J/K
Hence, the correct answer is option C.
Question 21:
Among [Ni(CO)_{4}], [NiCl_{4}]^{2−}, [Co(NH_{3})_{4}Cl_{2}]Cl, Na_{3}[CoF_{6}], Na_{2}O_{2} and CsO_{2}, the total number of paramagnetic compounds is
Option 1:  2 
Option 2:  3 
Option 3:  4 
Option 4:  5 
Solutions:
Complex  Hybridisation of central metal arom/ion  Valence d electron configuration of central metal atom/ion  Number of unpaired electrons in central atom  Magnetic nature 
[Ni(CO)_{4}]  sp^{3}  3d^{10}  0  Diamagnetic 
[NiCl_{4}]^{2−}  sp^{3}  3d^{8}  2  Paramagnetic 
[Co(NH_{3})_{4}Cl_{2}]Cl  d^{2}sp^{3}  3d^{6}  0  Diamagnetic 
Na_{3}[CoF_{6}]  sp^{3}  3d^{6}  4  Paramagnetic 
Na_{2}O_{2} has
O22anion that is diamagnetic in nature.
CsO_{2} has
O2anion having one unpaired electron in one of the molecular orbitals. So, CsO_{2} is paramagnetic in nature.
Thus, among the given compounds, [NiCl_{4}]^{2−}, Na_{3}[CoF_{6}] and CsO_{2} are paramagnetic in nature.
Hence, the correct answer is option B.
Question 22:
The increasing order of atomic radii of the following Group 13 elements is
Option 1:  Al < Ga < In < Tl 
Option 2:  Ga < Al < In < Tl 
Option 3:  Al < In < Ga < Tl 
Option 4:  Al < Ga < Tl < In 
Solutions:
In general, the atomic radii of elements increase on moving down a group. But in case of group 13 elements, due to the poor shielding effect exhibited by the 3d electrons, the atomic radius of Ga is less than that of Al.
So, the correct increasing order of atomic radii of the given group 13 elements is
Ga < Al < In < Tl
Hence, the correct answer is option B.
Question 23:
On complete hydrogenation, natural rubber produces
Option 1:  ethylene−propylene copolymer 
Option 2:  vulcanised rubber 
Option 3:  polypropylene 
Option 4:  polybutylene 
Solutions:
Natural rubber is a linear polymer of isoprene units. Upon complete hydrogenation, it produces ethylene−propylene copolymer.
Hence, the correct answer is option A.
Question 24:
The product(s) of the following reaction sequence is (are)
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The given reaction sequence can be completely written as
Thus, the product of the given reaction sequence is 1,4dibromobenzene.
Hence, the correct answer is option B.
Question 25:
The correct statement(s) about the following reaction sequence is(are)
Cumene C9H12→ii) H3O+i) O2P→CHCl3/NaOHQmajor+RminorQ→PhCH2BrNaOHS
Option 1:  R is steam volatile 
Option 2:  Q gives dark violet coloration with 1% aqueous FeCl_{3} solution 
Option 3:  S gives yellow precipitate with 2, 4−dinitrophenylhydrazine 
Option 4:  S gives dark violet coloration with 1% aqueous FeCl_{3} solution 
Solutions:
The given reaction sequence can be completely written as
Compound R is not steam volatile. Rather, due to greater proximity of −OH group to −CHO group leading to intramolecular hydrogen bonding, compound Q is steam volatile.
Compounds having presence of phenolic group, give dark violet coloration with 1% aqueous FeCl_{3} solution. Thus, compound Q gives dark violet coloration with 1% aqueous FeCl_{3} solution, while compound S does not.
Due to the presence of a carbonyl group (−CO−), compounds S gives yellow precipitate with 2, 4−dinitrophenylhydrazine.
Hence, the correct answers are options B and C.
Question 26:
The crystalline form of borax has
Option 1:  tetranuclear [B_{4}O_{5}(OH)_{4}]^{2−} unit 
Option 2:  all boron atoms in the same plane 
Option 3:  equal number of sp^{2} and sp^{3} hybridized boron atoms 
Option 4:  one terminal hydroxide per boron atom 
Solutions:
Among the given statements regarding the crystalline form of borax, statements given in options A, C and D are correct.
Crystalline structure of borax has tetranuclear [B_{4}O_{5}(OH)_{4}]^{2−} unit, in which only two B atoms lie in the same plane.
Two of the B atoms are sp^{2} hybridised and the other two sp^{3} hybridised. Also, each B atom has a terminal −OH group bonded to it.
Hence, the correct answers are options A, C and D.
Question 27:
The reagent(s) that can selectively precipitate S^{2−} from a mixture of S^{2−} and
SO42in aqueous solution is (are)
Option 1:  CuCl_{2} 
Option 2:  BaCl_{2} 
Option 3:  Pb(OOCCH_{3})_{2} 
Option 4:  Na_{2}[Fe(CN)_{5}NO] 
Solutions:
The reactions that the given reagents can undergo with a mixture of S^{2−} and
SO42in aqueous solution are as illustrated below.
a) CuCl_{2}
CuCl2 →S2 CuS↓Black ppt.CuCl2 →SO42 CuSO4Blue solution
b) BaCl_{2}
BaCl2 →S2 BaSSolubleBaCl2 →SO42 BaSO4↓White ppt.
c) Pb(OOCCH_{3})_{2}
PbOOCCH32 →S2 PbS↓Black ppt.PbOOCCH32 →SO42 PbSO4↓White ppt.
d) Na_{2}[Fe(CN)_{5}NO]
Na2FeCN5NO →S2 FeCN5NOS4Violet colour solutionNa2FeCN5NO →SO42 No reaction
Thus, from the above reactions, it is clear that among the given reagents, only CuCl_{2} can selectively precipitate S^{2−} from a mixture of S^{2−} and
SO42in aqueous solution.
Hence, the correct answer is option A.
Question 28:
A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation from linearity for atomic number, Z > 20. For an unstable nucleus having N/P ratio less than 1, the possible mode(s) of decay is (are)
Option 1:  β^{−} −decay (β emission) 
Option 2:  orbital or K−electron capture 
Option 3:  neutron emission 
Option 4:  β^{+} −decay (positron emission) 
Solutions:
For an unstable nucleus having N/P ratio less 1, the possible modes of decay are orbital or K−electron capture and β^{+} −decay (positron emission).
Via these modes of decay, the N/P ratio increases.
e01 + p11 → n10 Kelectron capturep11 → n10 + β+1 β+decay
Hence, the correct answers are options B and D.
Question 29:
Positive Tollen’s test is observed for
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:Positive Tollen’s test is given by aldehydes and αhydroxyketones. Thus, among the given alternatives, the compound given in option D does not exhibit positive Tollen’s test.
Hence, the correct answers are options A, B and C.
Question 30:
The compound(s) with two lone pairs of electrons on the central atom is(are)
Option 1:  BrF_{5} 
Option 2:  ClF_{3} 
Option 3:  XeF_{4} 
Option 4:  SF_{4} 
Solutions:
The structure of the given compounds are as follows:
From the given structures, it is clear that among the given compounds, ClF_{3} and XeF_{4} have two lone pairs of electrons on the respective central atoms.
Hence, the correct answers are options B and C.
Question 31:
According to the Arrhenius equation,
Option 1:  a high activation energy usually implies a fast reaction 
Option 2:  rate constant increases with increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy. 
Option 3:  higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant 
Option 4:  the pre−exponential factor is a measure of the rate at which collisions occur, irrespective of their energy. 
Solutions:
According to the Arrhenius equation, rate constant for a reaction is related to the activation energy as follows:
k=AeEaRT
If a reaction has a very high activation energy, it generally implies that the reaction is slow.
In general, rate constant for a reaction increases with increase in temperature. This is due to increase in the number of effective collisions.
The higher the magnitude of activation energy, the stronger is the temperature dependence of the rate constant.
The preexponential factor, A is a measure of the rate of occurrence of collisions.
Hence, the correct answers are options B, C and D.
Question 32:
In the following monobromination reaction, the number of possible chiral products is
Solutions:
The given monobromination reaction can be completely written as
Thus, it is clear that the number of possible chiral products in the given reaction is 5.
Question 33:
The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^{−3}. The ratio of the molecular weights of the solute and solvent,
MWsoluteMWsolvent, is
Solutions:
Given,
χsolute = 0.1⇒χsolvent = 1 – 0.1 = 0.9⇒χsoluteχsolvent = 0.10.9 = 19⇒ wsoluteMsolute×Msolventwsolvent = 19 …..i
Also,
wsolution = Density × Volume = wsolute + wsolventGiven, density = 2.0 g cm3⇒ wsolute + wsolvent = 2 × Vsolution⇒Vsolution = wsolute + wsolvent2 …..ii
It is given that molarity of the solution is same as its molality.
⇒nsoluteVsolution = nsolutewsolvent⇒wsolvent = Vsolution = wsolute + wsolvent2 Using ii⇒2wsolvent = wsolute + wsolvent⇒wsolvent = wsolute …..iii
Using equation (iii) in equation (i),
MsoluteMsolvent = 9
Thus, the value of the ratio of the molecular weights of the solute and solvent is 9.
Question 34:
The number of geometric isomers possible for the complex [CoL_{2}Cl_{2}]^{−} (L = H_{2}NCH_{2}CH_{2}O^{−}) is
Solutions:
In the complex [CoL_{2}Cl_{2}]^{−}, L is an unsymmetrical bidentate ligand. Thus, the given complex belongs to [M(AB)_{2}a_{2}] type.
There are five geometrical isomers possible for this complex, whose structures are given below.
Question 35:
In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is
Solutions:
The balanced ionic equation for the oxidation of thiosulphate ions by permanganate ions in a neutral or faintly alkaline solution is given as:
8 MnO4 + 3 S2O32 + H2O → 8 MnO2 + 6 SO42 + 2 OH
Thus, it is clear that when 8 moles of permanganate ions quantitatively oxidise thiosulphate ions, 6 moles of sulphate ions are produced.
Question 36:
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is
Solutions:
The diffusion coefficient of an ideal gas is directly proportional to the mean free path and mean speed of the gas.
Diffusion coefficient,
D ∝ λνmeanIn terms of temperature and pressure, diffusion coefficient can be related to mean free path and mean speed of the gas as
D ∝ TP×TM
For the given ideal gas, molar mass does not vary with change in temperature and pressure.
Case I:
D1 ∝ TTP
Case II:
D2 ∝ 4T4T2P
⇒ D2D1 = 4×2TT2PTTP = 41
Thus, the diffusion coefficient of the gas becomes 4 times.
Question 37:
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant T1) = 10 P (computer turns out to be defective given that it is produced in plant T2), where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2 is
Option 1:  3673 
Option 2:  4779 
Option 3:  7893 
Option 4:  7583 
Solutions:Let P(T_{1}) be the probability that the computer is produced by plant T_{1}. So, P(T_{1}) = 20% =
15Let P(T_{2}) be the probability that the computer is produced by plant T_{2}. So, P(T_{2}) = 80% =
45Let P(D) be the probability of defective computers produced. So, P(D) =
7100Let
PDT2 = x be the probability that the computer turns out to be defective given that it is produced in plant T_{2}. Given:
PDT1=10PDT2 = 10x Now,
P(T1)×PDT1+P(T2)×PDT2=7100
⇒15×10x+45×x=7100⇒x=140Let
PT2D be the probability that the computer produced from plant T_{2} is not defective.
∴ PT2D=45×(1x)17100=45×394093100=7893Hence, the correct answer is option C.
Question 38:
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is
Option 1:  380 
Option 2:  320 
Option 3:  260 
Option 4:  95 
Solutions:
Given: Total number of girls = 6 Total number of boys = 4
Number of ways of selecting the team is the same as the number of ways of selecting either 0 boys (or 4 girls) or 1 boy (3 girls and 1 boy).
∴ Number of ways of selecting the team = Number of ways of selecting 4 girls × Number of ways of selecting a captain + Number of ways of selecting 3 girls and 1 boy × Number of ways of ways of selecting a captain
=C46×4 +C36×C14×4=15×4+20×4×4=60+320=380Hence, the correct answer is option A.
Question 39:
The least value of α ∈ â„ for which
4αx2+1x≥1,for all x > 0, is
Option 1:  164 
Option 2:  132 
Option 3:  127 
Option 4:  125 
Solutions:Since x > 0,
⇒4αx2+12x+12×3≥4αx2×12x×12×3 AM ≥GM⇒4αx2+1x≥3α3⇒3α3≥1⇒α≥127Thus, the least value of
αis
127.
Hence, the correct answer is option C.
Question 40:
Let
π6<θ<π12.Suppose α_{1} and β_{1} are the roots of the equation x^{2} – 2x sec θ + 1 = 0 and α_{2} and β_{2} are the roots of the equation x^{2} + 2x tan θ – 1 = 0. If α_{1} > β_{1} and α_{2} > β_{2}, then α_{1} + β_{2} equals
Option 1:  2(sec θ – tan θ) 
Option 2:  2 sec θ 
Option 3:  –2 tan θ 
Solutions: Given,
x22xsecθ+1=0. So, the roots of the given equation will be
x=secθ±tanθ.
Now,
α1>β1 ⇒
α1=secθtanθAlso,
x2+2xtanθ1=0So, the roots of the given equation will be
x=tanθ±secθ.
Now,
α2>β2 ⇒β2=tanθsecθ=(tanθ+secθ)
⇒α1+β2=secθtanθ+tanθsecθ=2tanθâ€‹Hence, the correct answer is option C.
Question 41:
Let
S=x∈π, π : x ≠0, ±π2.The sum of all distinct solutions of the equation
3sec x + cosec x + 2(tan x – cot x) = 0 in the set S is equal to
Option 1:  7π9 
Option 2:  2π9 
Option 4:  5π9 
Solutions:The given equation is
3secx+cosecx+2tanxcotx=0
⇒3×1cosx+1sinx+2sinxcosxcosxsinx=0⇒3sinx+cosx+2sin2xcos2x=0⇒3sinx+cosx2cos2x=0
⇒32sinx+12cosx=cos2x⇒cosxπ3=cos2x⇒2x=2nπ±xπ3⇒x=2nππ3,2nπ3+π9In
π,π,
x=π3,π9,7π9,5π9∴∑xi=0Thus, the sum of all distinct solutions of the given equation in the set S is zero.
Hence, the correct answer is option C.
Question 42:
Let f : (0, ∞) → â„ be a differentiable function such that
f’x=2fxxfor all x ∈ (0, ∞) and f(1) ≠ 1. Then
Option 1:  limx→0+ f’1x=1 
Option 2:  limx→0+ x f1x=2 
Option 3:  limx→0+ x2 f’x=0 
Option 4:  f(x) ≤ 2 for all x ∈ (0, 2) 
Solutions:
Given: f’x=2fxx⇒f’x+fxx=2This is a linear differential equation. I.F. =
e∫1xdx=elogx=xThe solution of this differential equation is given by
xfx=∫2xdx+C⇒xfx=x2+C⇒fx=x+CxSince f(1) ≠ 1, ⇒ C ≠ 0
Now,
f1x=1x+Cx⇒f’1x×1×2=1×2+C⇒f’1x=1Cx2⇒limx→0+f’1x=limx→0+1Cx2=1 xf1x=x+Cx2⇒limx→0+xf1x=limx→0+x+Cx2=0 f’x=1Cx2⇒x2f’x=x2C⇒limx→0+x2f’x=limx→0+x2C≠0 limx→0+fx=limx→0+x+Cx=∞ or ∞Hence, the correct answer is option A.
Question 43:
The circle C_{1} : x^{2} + y^{2} = 3, with centre at O, intersects the parabola x^{2} = 2y at the point P in the first quadrant. Let the tangent to the circle C_{1} at P touches other two circles C_{2} and C_{3} at R_{2} and R_{3}, respectively. Suppose C_{2} and C_{3 }have equal radii
23and centres Q_{2} and Q_{3}, respectively. If Q_{2} and Q_{3} lie on the yaxis, then
Option 1:  Q_{2}Q_{3} = 12 
Option 2:  R_{2}R_{3} =
46 
Option 3:  area of the triangle OR_{2}R_{3} is
62 
Option 4:  area of the triangle PQ_{2}Q_{3} is
42 
Solutions:
Given:
x2+y2=3 …..1×2=2y …..2Solving (1) and (2), we get
y2+2y3=0⇒y+3y1=0⇒y=3, 1In the first quadrant, y is positive.
⇒ y = 1
∴ x2=2⇒x=2So, the point of intersection of (1) and (2) is
P2, 1.
The equation of tangent to the circle (1) at
P2, 1is given by
2x+y3=0.
Let θ be the angle made by this tangent with the positive direction of the xaxis. ∴ Slope of the tangent =
tanθ=2⇒Slope of OP
=12=tanα sayNow,
∠OTP=α
In ∆TQ3R3,sinα=Q3R3TQ3⇒13=23TQ3⇒TQ3=6∴ Q2Q3=2TQ3=2×6=12Also,
tanα=Q3R3TR3⇒12=23TR3⇒TR3=26∴R2R3=2TR3=2×26=46OP = Length of perpendicular from
P2, 1to
2x+y3=0
⇒OP=0+032+1=3Area of the triangle OR_{2}R_{3} =
12×OP×R2R3=12×3×46=62Area of the triangle PQ_{2}Q_{3} =
12×Q2Q3×Distance of P from the x axis=12×12×2=62Hence, the correct answers are options A, B and C.
Question 44:
A solution curve of the differential equation (x^{2} + xy + 4x + 2y + 4)
dydxy2=0, x>0,passes through the point (1, 3). Then the solution curve
Option 1:  intersects y = x + 2 exactly at one point. 
Option 2:  intersects y = x + 2 exactly at two points 
Option 3:  intersects y = (x + 2)^{2} 
Option 4:  does NOT intersect y = (x + 3)^{2} 
Solutions:
The given differential equation=x2+xy+4x+2y+4dydxy2=0⇒x+22+yx+2=y2·dxdy⇒dxdy=x+22y2+x+2y
⇒1x+22dxdy=1y2+1yx+2∴1x+22dxdy1yx+2=1y2Putting 1x+2=t, we get1x+22dxdy=dtdyOn substituting, we get
dtdyty=1y2⇒dtdy+ty=1y2Now, I.F =e∫1ydy=yThus, the solution is given by
ty=C+∫y1y2dy⇒ty=Clogy∴1x+2·y=ClogyIt is given that the solution curve passes through the point (1, 3).
⇒1=Clog3⇒C=1+log3Substituting the value of C, we get
yx+2=1+log3logyFor
y=x+2,
logy=log3That is, it intersects
y=x+2 at exactly one point.
So, option A is correct.
For
y=x+22,
x+22x+2=1+log3y⇒x+1=log3y⇒y=3ex+1That is, it intersects
y=x+22.
So, option C is correct.
For y=x+32,x+32x+2=logx+323e⇒logx+323e=x+32x+2⇒x+32=3e.ex+2 +1x+2+2For x>0, x+2>2Now, LHS=x+32>9RHS=3e.ex+2 +1x+2+2<3e12+12+2=3.e72That is, it does not intersect
y=x+32.
So, option D is incorrect.
Hence, the correct answers are options A and C.
Question 45:
In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively, and 2s = x + y + z. If
sx4=sy3=sz2and area of incircle of the triangle XYZ is
8π3,then
Option 1:  area of the triangle XYZ is
66 
Option 2:  the radius of circumcircle of the triangle XYZ is
3566 
Option 3:  sin X2 sin Y2 sin Z2=435 
Option 4:  sin2 X+Y2=35 
Solutions:Given:
sx4=sy3=sz2⇒sx4=sy3=sz2=3sx+y+z9=s9⇒sx4=s9, sy3=s9, sz2=s9⇒x=5s9, y=2s3, z=7s9Now, Area of incircle =
πr2=8π3
⇒∆s2=r2=83⇒∆2=8s23⇒ssxsysz=8s23⇒s·4s9·s3·2s9=8s23⇒s=9 ∴∆=83×9=66 sq. unitsSo, option A is correct.
Now, the radius of the circumcircle is given by
R=xyz4∆=5s9·2s3·7s94×66=35246So, option B is incorrect.
Also,
sinX2sinY2sinZ2=r4R=834×35246=435So, option C is correct.
sin2X+Y2=cos2Z2=1+cosZ2=1+x2+y2z22xy2=35So, option D is correct.
Hence, the correct answers are options A, C and D.
Question 46:
Let RS be the diameter of the circle x^{2} + y^{2} = 1, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point(s)
Option 1:  13,13 
Option 2:  14,12 
Option 3:  13,13 
Option 4:  14,12 
Solutions: It is given that P is the tangent to the given circle. So, the equation of the tangent will be
xcosθ+ysinθ=1. Q is the meeting point of the two tangents passing through points P and S. So, the point of intersection of the tangents will be
Q1, 1cosθsinθ. Now, normal to the circle at P, intersecting a line drawn through Q, will be
y=tanθx. The equation of line QE parallel to line RS will be y1cosθsinθ=0⇒y=1cosθsinθ
⇒y=1xx2+y2yx2+y2=x2+y2xy⇒y2+x=x2+y2Now, point
13,13 and point 13,13satisfy the given equation of the locus through point E. Hence, the correct answers are options A and C.
Question 47:
Let
P=312 2 0 α35 0, where α ∈ â„. Suppose Q = [q_{ij}] is a matrix such that PQ = kI, where k ∈ â„, k ≠ 0 and I is the identity matrix of order 3. If
q23=k8and det
Q=k22,then
Option 1:  α = 0, k = 8 
Option 2:  4α – k + 8 = 0 
Option 3:  det (P adj (Q)) = 2^{9} 
Option 4:  det (Q adj (P)) = 2^{13} 
Solutions:
P=312 2 0 α35 0⇒P=30+5α+103α210=20+12αP1=120+12α5α10α3α63α+410122PQ=kI⇒P1kQ=I⇒P1=1kQNow,
q23=k8⇒k×3α+420+12α=k8⇒6α+8=5+3α⇒α=1
Q=k22⇒kP 1=k22⇒k3P=k22⇒k320+12α=k22⇒k=4∴ 4αk+8=0P=Q=8∴ P adj Q=PQ2=8×64=29Hence, the correct answers are options B and C.
Question 48:
Let f : â„ → â„, g : â„ → â„ and h : â„ → â„ be differentiable functions such that f(x) = x^{3} + 3x + 2, g(f(x)) = x and h (g(g(x))) = x for all x ∈ â„. Then
Option 1:  g′(2)=115 
Option 2:  h′(1) = 666 
Option 3:  h(0) = 16 
Option 4:  h(g(3)) = 36 
Solutions:Given:
fx=x3+3x+2 …..1∴ f’x=3×2+3 …..2
gfx=x ⇒g’fx×f’x=1Putting x=0, we get⇒g’f0×f’0=1⇒g’2×f’0=1 From (1), f0=2⇒g’2=1f’0=13 From (2), f’0=3So, option A is incorrect.
hggx=x⇒hggfx=fx⇒hgx=fx …..3 ∵gfx=x⇒hgfx=ffx⇒hx=ffx …..4⇒h’x=f’fx·f’xPutting x=1, we geth’1=f’f1·f’1 =f’6×6 [Using (1) and (2)] =111×6=666So, option B is correct.
Putting x=0 in 4, we geth0=ff0=f2=16So, option C is correct.
Putting x=3 in 3, hg3=38 From (1), f3=38So, option D is incorrect.
Hence, the correct answers are options B and C.
Question 49:
Consider a pyramid OPQRS located in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with O as origin, and OP and OR along the xaxis and the yaxis, respectively. The base OPQR of the pyramid is a square with OP = 3. The point S is directly above the midpoint T of diagonal OQ such that TS = 3. Then
Option 1:  the acute angle between OQ and OS is
π3 
Option 2:  the equation of the plane containing the triangle OQS is x – y = 0 
Option 3:  the length of the perpendicular from P to the plane containing the triangle OQS is
32 
Option 4:  the perpendicular distance from O to the straight line containing RS is
152 
Solutions: From the given figure we get,
OQ→=3i^+3j^ and OS→=32i^+32j^+3k^Now, the angle between OS and OQ will be
cosθ=3×32+3×3232+32322+322+32=13Thus, option (A) is incorrect.
Normal of the plane OQS will be
OQ→×OS→=±ijk33032323=±9i9jWe know that the equation of the plane passing through the origin will be
λ→·η→=0⇒xi+yj+zk·(9i9j)=0⇒xy=0Hence, the equation will be x – y = 0. So, option (B) is correct.
The perpendicular distance between the point P(3,0,0) and the plane containing OQS will be the distance between point P(3,0,0) and the equation x – y = 0.
3012+12=32So, option (C) is correct.
Equation of RS is
x0320=y3323=z030 The angle between RS and OR is
cosθ=0+332+0322+322+3232=16The projection of OS will be OT. So, OT =
ORsinθ=
3116=356=152Therefore, the perpendicular distance from O to the straight line containing RS is
152.
So, option (D) is also correct. Hence, the correct answers are options B, C and D.
Question 50:
Let
z=1+3i2, where
i=1 and r, s∈1, 2, 3. Let
P=zrz2sz2szrand I be the identity matrix of order 2. Then the total number of ordered pairs (r, s) for which P^{2} = −I is
Solutions:
z=1+3i2=ωP=zrz2sz2szr=ωrω2sω2sωrGiven:
P2=I⇒ωrω2sω2sωrωrω2sω2sωr=1001⇒ω2r+ω4sωr+2s1+1rωr+2s1+1rω2r+ω4s=1001⇒ω2r+ω4s=1 or ωr+2s1+1r=0⇒r=1 or 3When r = 1,
ω2+ω4s=1⇒ω4s=1ω2=ω⇒s=1When r = 3,
ω6+ω4s=1⇒ω4s=11=2This is not possible for any value of s. The ordered pair (r, s) for which P^{2} = −I is (1, 1). Thus, the total number of ordered pairs is 1.
Question 51:
Let m be the smallest positive integer such that the coefficient of x^{2} in the expansion of (1 + x)^{2} + (1 + x)^{3} + … + (1 + x)^{49} + (1 + mx)^{50} is (3n + 1) ^{51}C_{3} for some positive integer n. Then the value of n is
Solutions:Coefficient of
x2 in the given expansion
=C22+C23+C24+…+C249+C250m2=C33+C23+C24+…+C249+C250m2=C34+C24+…+C249+C250m2 ∵Crn+Cr1n=Crn+1 =C35+…+C249+C250m2 =C350+C250m2 =C350+C250m2+C250C250=C351+C250m21Given:
C351+C250m21=3n+1C351⇒C250m21=3n·C351⇒C250m21=3n·513·C250⇒n=m2151Thus, the minimum value for n is 5.
Question 52:
The total number of distinct x ∈ [0, 1] for which
∫0xt21+t4dt=2x1is
Solutions:Given:
∫0xt21+t4dt=2x1Let
∫0xt21+t4dt = f(t) ⇒f'(x) = x21+x4=2So, f’(x) will be positive. Thus, f(x) is increasing. At x = 0, f(x) = 0. At x = 1, f(1) = ∫01t21+t4dt<∫011.dt⇒0<∫01t21+t4dt<∫011.dtHence, there will be one solution in [0, 1].
Question 53:
The total number of distinct x ∈ â„ for which
xx21+x32x4x21+8x33x9x21+27×3=10is
Solutions:Given:
xx21+x32x4x21+8x33x9x21+27×3=10⇒x3111+x3241+8×3391+27×3=10⇒x3111241391+x61112483927=10 ⇒x3100221362+x61002263624=10⇒2×3+12×6=10⇒6×6+x35=0⇒6×6+6×35×35=0⇒6×35×3+1=0⇒x3=56 or x3=1Thus, the total number of distinct values of x is 2.
Question 54:
Let α, β âˆŠ â„ be such that
limx→0x2sinβxaxsinx=1. Then 6(α + β) equals
Solutions:Given:
limx→0x2sinβxaxsinx=1⇒limx→0x2βxβx33!+…αxxx33!+…=1⇒limx→0x3ββ3×23!+…α1x+x33!x55!+…=1As the limit of the above expression is finite,
∴α1=0 ⇒α=1Now, the given expression becomes
limx→0ββ33!x2+…13!x25!+…=1∴ β=13!=16∴ 6α+β=61+16=7