JEE MAINS Answers & Solutions for 2019 

JEE (MAIN)-2019 (Online) Phase-2

(Physics, Chemistry and Mathematics)

1. Determine the charge on the capacitor in the following circuit:

[image]

(1) 200μC200μC (2) 60μC60μC (3) 10μC10μC (4) 2μC2μC

Answer (1)

Sol. At steady state current through capacitor is zero.

VC=V1VC?=V1? V1=56×V0×39V1?=65?×V0?×93?

[image]

V1=5×72×36×9=20VV1?=6×95×72×3?=20V

Q1=CV1Q1?=CV1? =200μC=200μC

1. The magnetic field of a plane electromagnetic wave is given by :

B?=B?0i?[cos?(kz−ωt)+B1j?cos?(kz+ωt)B=B0?i[cos(kz−ωt)+B1?j?cos(kz+ωt)

where B0=3×10−5TB0?=3×10−5T and B1=2×10−6TB1?=2×10−6T

The rms value of the force experienced by a stationary charge Q=10−4CQ=10−4C at z=0z=0 is closest to:

(1) 0.6N0.6N (2) 0.9N0.9N (3) 3×10−2N3×10−2N (4) 0.1N0.1N

Answer (1)

Sol.?E?1?=CB1Sol.?E1??=CB1? ?E?2?=CB2?E2??=CB2? AlsoE?1⊥E?2AlsoE1?⊥E2?

Fnet=θ2E12+E22Fnet?=2?θ?E12?+E22?? =10−42×3×108×30×10−6=2?10−4?×3×108×30×10−6 =90×108×10−102=2?90×108×10−10? =0.6N=0.6N

1. A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θθ , where θθ is the angle by which it has rotated, is given as kθ2kθ2 . If its moment of inertia is 1 then the angular acceleration of the disc is:

(1)2k1θ(2)k1θ(1)12k?θ(2)1k?θ (3)k2lθ(4)k4lθ(3)2lk?θ(4)4lk?θ

Answer (1)

Sol.τ=dEdθSol.τ=dθdE? 2k0=lα2k0=lα α=2k0lα=l2k0?

1. The stream of a river is flowing with a speed of 2km/h2km/h . A swimmer can swim at a speed of 4km/h4km/h . What should be the direction of the swimmer with respect to the flow of the river to cross the river straight?

(1)60?(2)90?(1)60?(2)90? (3)150?(4)120?(3)150?(4)120?

Answer (4)

Sol. Draw velocity diagram

[image]

sin?θ=vrvsr=12sinθ=vsr?vr??=21?θ=30?θ=30??=90+θ=120??=90+θ=120?

1. A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its (1n)th(n1?)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

 

Answer (3)

Sol. m1=Mnm1?=nM?

U1=−Mn×gL2nU1?=n−M?×g2nL?

1. A signal Acosot is transmitted using v0v0? sinot as carrier wave. The correct amplitude modulated (AM) signal is:

v0sin?ω0t+A2sin?(ω0−ω)t+A2sin?(ω0+ω)tv0+A)cos?ωtsin?ω0tv0sin?ω0t+Acos?ωtv0sin?[ω0(1+0.01Asin?ωt)](1)v0?sinω0?t+2A?sin(ω0?−ω)t+2A?sin(ω0?+ω)tv0?+A)cosωtsinω0?tv0?sinω0?t+Acosωtv0?sin[ω0?(1+0.01Asinωt)]?(1)

Answer (1)

Sol.A=(v0+Acos?ωt)sin?ω0t=v0sin?(ω0t)+A2[sin?(ω0−ω)t+sin?(ω0+ω)t](2)Sol.A=(v0?+Acosωt)sinω0?t=v0?sin(ω0?t)+2A?[sin(ω0?−ω)t+sin(ω0?+ω)t]?(2)

1. The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane : (i) a ring of radius R, (ii) a solid cylinder of radius R22R? and (iii) a solid sphere of radius R44R? If, in each case, the speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is :

(1)14:15:20(2)10:15:7(3)4:3:2(4)2:3:4(3)(1)14:15:20(3)4:3:2?(2)10:15:7(4)2:3:4?(3)

Answer (Bonus)

Sol. mgh=12lpω2Sol. mgh=21?lp?ω2For ring→h1=1(2mr2)2v2mgv2(r)2=v2gFor ring→h1?=21(2mr2)?mgv2?(r)2v2?=gv2?For cylinder→h2=12(32mr2)v2(r)2=3v24gFor cylinder→h2?=21?(23?mr2)(r)2v2?=4g3v2?For sphere→h3=12×75m(r)2gv2r2=7v210gFor sphere→h3?=21?×g57?m(r)2?r2v2?=10g7v2?h1:h2:h3h1?:h2?:h3?2:32:14102:23?:1014?20:15:1420:15:14

1. A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is

(E is mid- point of arm CD)

[image]

 

Answer (4)

Sol.R4=R8,R2=7R8Sol.R4?=8R?,R2?=87R?1Req=8R+87RReq?1?=R8?+7R8?Req=7R64Req?=647R?

1. In the density measurement of a cube, the mass and edge length are measured as (10.00 ±0.10)±0.10) kg and (0.10±0.01)(0.10±0.01) m, respectively. The error in the measurement of density is

 

Answer (Bonus)

Sol. ρ=M/V=10(0.1)3=10,000kg/m3ρ=M/V=(0.1)310?=10,000kg/m3

dρρ=[dMM+dVV]ρdρ?=[MdM?+VdV?]dρ10,000=0.110+0.030.110,000dρ?=100.1?+0.10.03?dρ=3100kg/m3dρ=3100kg/m3

1. A rectangular coil (Dimension 5 cm × 2.5 cm) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45° about Z-axis, then the torque on the coil is

(1) 0.55 Nm
(2) 0.27 Nm
(3) 0.42 Nm
(4) 0.38 Nm

Answer (2)

Sol. τ=M×Bτ=M×B

τ=100×3×5×2.5×10−4×1×12τ=100×3×5×2.5×10−4×1×2?1?=0.27Nm=0.27Nm

1. A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 116161? th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is :

(1)2T114(2)4T115(1)2T141??(2)4T151?? (3)4T114(4)2T110(3)4T141??(4)2T101??

Answer (2)

Sol. T=2πlgeffT=2πgeff?l??

T′T=geffgeffTT′?=geff?geff???geff=g−g16=1516ggeff?=g−16g?=1615?gT′T=1615TT′?=1516??

1. The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to

(1) 1/L
(2) L
(3) 1/L²
(4) L²

Answer (1)

Sol. B=μ0niB=μ0?ni

?=μ0ni(nL)A?=μ0?ni(nL)A

(Inductance) L′=μ0n2LAL′=μ0?n2LA [η=NL][η=LN?]

L′=μ0NA(NL)L′=μ0?NA(LN?)L′∝1LL′∝L1?

1. For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be:

(1) 10051005? m/s
(2) 100 m/s
(3) 805805? m/s
(4) 80 m/s

Answer (1)

Sol. V∝TV∝T?

V1V2=T1T2V2?V1??=T2?T1???200V2=400500V2?200?=500400??V2=20054=1005m/sV2?=20045??=1005?m/s

1. A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done?

(1) 2.55×10−6J2.55×10−6J
(2) 6.25×10−6J6.25×10−6J
(3) 3.75×10−6J3.75×10−6J
(4) 2.16×10−6J2.16×10−6J

Answer (3)

Sol. U=12q2CU=21?Cq2?

Ui=12525=2.5μJUi?=21?552?=2.5μJUf=12522=6.25μJUf?=21?252?=6.25μJW=Uf−Ui=3.75μJW=Uf?−Ui?=3.75μJ

1. A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance '3a' from the centre will be:

(1) GM9a29a2GM?
(2) 2GM9a29a22GM?
(3) GM3a23a2GM?
(4) 2GM3a23a22GM?

Answer (3)

Sol. E=GM(3a)2+2GM(3a)2E=(3a)2GM?+(3a)22GM?

E=GM3a2E=3a2GM?

1. A moving coil galvanometer has resistance 50 ΩΩ and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 k ΩΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:

(1) 10 V
(2) 20 V
(3) 15 V
(4) 40 V

Answer (2)

Sol. V=Ig(Rs+Rg)V=Ig?(Rs?+Rg?)

=4×10−3[5050]=4×10−3[5050]=20V=20V

1. The pressure wave, P=0.01sin?[1000t−3x]Nm−2P=0.01sin[1000t−3x]Nm−2 corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0?C0?C . On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms−1336 ms−1 . Approximate value of T is:

(1) 4?C4?C
(2) 12?C12?C
(3) 15?C15?C
(4) 11?C11?C

Answer (1)

Sol. V=10003m/sV=31000?m/s

V∝TV∝T?dVV=12dTTVdV?=21?TdT?8×33×1000=12dT2733×10008×3?=21?273dT?dT=273×2×81000=4.36?CdT=1000273×2×8?=4.36?C

1. Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

(1) 889.2 nm
(2) 488.9 nm
(3) 388.9 nm
(4) 642.7 nm

Answer (2)

Sol. 1λ1=R[122−132]λ1?1?=R[221?−321?]

\frac{1}{\lambda_2} = R \left[\frac{1}{2^2} - \frac{1}{4^2}\right]\) \[\frac{5\lambda_1}{36} = \frac{12\lambda_2}{4 \times 16}

λ2=5×660×6436×12λ2?=36×125×660×64?=489nm=489nm

1. A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

(1) 1.5 kg
(2) 1.8 kg
(3) 1.0 kg
(4) 1.2 kg

Answer (4)

Sol. 2V=2V4+m2V22V=42V?+m2?V2?

V=V2−V4V=V2?−4V?3V2=m2(V+V4)23V?=m2?(V+4V?)m2=65kgm2?=56?kg

1. An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v?v?, m is its mass and kBkB? is Boltzmann constant, then its temperature will be :

(1) mv?25kB5kB?mv?2?

(2) mv?26kB6kB?mv?2?

(3) mv?27kB7kB?mv?2?

(4) mv?23kB3kB?mv?2?

Answer (4)

Sol. v?=3RTMv?=M3RT??

v?=3RTmNAv?=mNA?3RT??v?=3kBTmv?=m3kB?T??

1. A system of three charges are placed as shown in the figure :

[image]

If D >> d, the potential energy of the system is best given by :

(1) 14π?0[−q2d−qQdD2]4π?0?1?[−dq2?−D2qQd?]

(2) 14π?0[q2d+qQdD2]4π?0?1?[dq2?+D2qQd?]

(3) 14π?0[−q2d+2qQdD2]4π?0?1?[−dq2?+D22qQd?]

(4) 14π?0[−q2d−qQd2D2]4π?0?1?[−dq2?−2D2qQd?]

Answer (4)

Sol. U=−Kq2d+QVU=−dKq2?+QV

V=−KPD2=−KqdD2V=−D2KP?=−D2Kqd?U=−Kq2d−KQqdD2U=−dKq2?−D2KQqd?

1. A rigid square loop of side 'a' and carrying current I2I2? is lying on a horizontal surface near a long current I1I1? carrying wire in the same plane as shown in figure. The net force on the loop due to the wire will be :

[image]

(1) Repulsive and equal to μ0I1I24π4πμ0?I1?I2??

(2) Repulsive and equal to μ0I1I22π2πμ0?I1?I2??

(3) Zero

(4) Attractive and equal to μ0I1I23π3πμ0?I1?I2??

Answer (1)

Sol. F=I2a(B1−B2)F=I2?a(B1?−B2?)

B1=μ0I12πaB1?=2πaμ0?I1??B2=μ0I14πaB2?=4πaμ0?I1??F=μ0I1I24πF=4πμ0?I1?I2??

1. A string is clamped at both the ends and it is vibrating in its 4th4th harmonic. The equation of the stationary wave is Y=0.3Y=0.3 sin (0.157x)(0.157x) cos (200πt)(200πt) . The length of the string is : (All quantities are in SI units)

(1) 60 m60 m

(2) 20 m20 m

(3) 40 m40 m

(4) 80 m80 m

Answer (4)

Sol. λ=0.157λ=0.157

=3.1420=π20=203.14?=20π?

In 4th4th harmonic

[image]

∴2πλ=π/20∴λ2π?=π/20λ=40 mλ=40 m∴L=4λ2=80m∴L=24λ?=80m

1. An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ1 kΩ load resistance. Signal voltage of 10 mV10 mV is applied across the base-emitter. This produces a 3 mA3 mA change in the collector current and 15 μA15 μA change in the base current of the amplifier. The input resistance and voltage gain are :

(1) 0.33 kΩ,1.50.33 kΩ,1.5

(2) 0.33 kΩ,3000.33 kΩ,300

(3) 0.67 kΩ,2000.67 kΩ,200

(4) 0.67 kΩ,3000.67 kΩ,300

Answer (4)

1. The electric field of light wave is given as E?=10−3cos?(2πx5×10−7−2π×6×1014t)x^NCE=10−3cos(5×10−72πx?−2π×6×1014t)x^CN? . This light falls on a metal plate of work function 2eV2eV . The stopping potential of the photoelectrons is : Given, E(in eV)=12375λ(in A)E(in eV)=λ(in A)12375?

(1) 0.48V0.48V
(2) 2.48V2.48V
(3) 0.72V0.72V
(4) 2.0V2.0V

Answer (1)

Sol. λ=5×10−7mλ=5×10−7m

ν=3×1085×10−7=6×1014Hzν=5×10−73×108?=6×1014HzE=123755000=2.475eVE=500012375?=2.475eVKmax=E−?Kmax?=E−?=0.48eV=0.48eV

1. A ball is thrown vertically up (taken as ++ z-axis) from the ground. The correct momentum- height (p- h) diagram is:

[image]

Answer (4)

Sol. V=V02−2ghV=V02?−2gh?

Direction of velocity changes at top most point

1. If 'M' is the mass of water that rises in a capillary tube of radius 'r', then mass of water which will rise in a capillary tube of radius '2r' is

(1) 2 M
(2) M
(3) 4 M
(4) M22M?

Answer (1)

Sol. h∝1/rh∝1/r

M∝πr2hM∝πr2hM∝rM∝r

1. The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index μμ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light used is λλ , t will be

[image]

(1) 2nDλa(μ−1)a(μ−1)2nDλ?

(2) 2Dλa(μ−1)a(μ−1)2Dλ?

(3) nDλa(μ−1)a(μ−1)nDλ?

(4) Dλa(μ−1)a(μ−1)Dλ?

Answer (Bonus)

Sol. (μ−1)t=nλ(μ−1)t=nλ

t=nλμ−1t=μ−1nλ?

1. C60C60? , an allotrope of carbon contains

(1) 16 hexagons and 16 pentagons
(2) 18 hexagons and 14 pentagons
(3) 20 hexagons and 12 pentagons
(4) 12 hexagons and 20 pentagons

Answer (3)

Sol. Fullerene C60C60? contains 20 six membered rings and 12 five membered rings.

1. The major product of the following reaction is

CH3C≡CH→(ii)DI(i)DC1(1equiv.)→CH3?C≡CH(i)DC1(1equiv.)(ii)DI?→

(1)CH3C(I)(CI)CHD2(2)CH3CD(I)CHD(CI)(1)CH3?C(I)(CI)CHD2?(2)CH3?CD(I)CHD(CI) (3)CH3CD(CI)CHD(I)(4)CH3CD2CH(CI)(I)(3)CH3?CD(CI)CHD(I)(4)CH3?CD2?CH(CI)(I)

Answer (1)

Sol. CH3−C=C−H→1eqDC1CH3−C=CHCH3?−C=C−HDC11eq?CH3?−C=CH

[image]

Both addition follow Markownikov's rule.

1. The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is

Zn(s)+Cu2+(aq)?Zn2+(aq)+Cu(s),Zn(s)+Cu2+(aq)?Zn2+(aq)+Cu(s),E?=2Vat298KE?=2Vat298K

(Faraday's constant, F=96000Cmol−1F=96000Cmol−1

(1) 192

(2) 384

(3) -384

(4) -192

Answer (3)

Sol. ΔG?=−nFEcell?ΔG?=−nFEcell??

=−2×(96000)×2V=−2×(96000)×2V=−384kJ/mole=−384kJ/mole

1. The number of water molecule(s) not coordinated to copper ion directly in CuSO4⋅5H2OCuSO4?⋅5H2?O is

(1) 4
(2) 1
(3) 2
(4) 3

Answer (2)

Sol. In CuSO4⋅5H2OCuSO4?⋅5H2?O four H2OH2?O molecules are directly coordinated to the central metal ion while one H2OH2?O molecule is hydrogen bonded.

1. Magnesium powder burns in air to give

(1) Mg(NO3)2Mg(NO3?)2? and Mg3N2Mg3?N2?
(2) MgOMgO and Mg(NO3)2Mg(NO3?)2?
(3) MgOMgO and Mg3N2Mg3?N2?
(4) MgOMgO only

Answer (3)

Sol. Mg burn in air and produces a mixture of nitride and oxide.

So, Mg3N2Mg3?N2? and MgOMgO are formed.

1. The one that will show optical activity is (en = ethane-1,2-diamine)

[image]

Answer (3)

1. The major product of the following reaction is

CH3CH=CHCO2CH3→LiAlH4?CH3?CH=CHCO2?CH3?LiAlH4???CH3CH2CH2CH2OHCH3?CH2?CH2?CH2?OHCH3CH2CH2CHOCH3?CH2?CH2?CHOCH3CH2CH2CO2CH3CH3?CH2?CH2?CO2?CH3?CH3CH=CHCH2OHCH3?CH=CHCH2?OH

Answer (4)

Sol. LiAlH4LiAlH4? reduces esters to alcohols but does not reduce C=CC=C

[image]

1. For any given series of spectral lines of atomic hydrogen, let Δv‾=v‾max−v‾minΔv=vmax?−vmin? be the difference in maximum and minimum frequencies in cm−1cm−1 . The ratio Δv‾Lyman/Δv‾BalmerΔvLyman?/ΔvBalmer? is

(1)9:4(2)27:5(1)9:4(2)27:5 (3)4:1(4)5:4(3)4:1(4)5:4

Answer (1)

Sol. v‾∝ΔEv∝ΔE

For H- atom

v‾=R[1n12−1n22]v=R[n12?1?−n22?1?]

For Lyman series,

v‾(max)∝13.6(1−1∞)v(max)∝13.6(1−∞1?)v‾(min)∝13.6(1−14)v(min)∝13.6(1−41?)∴v‾max−v‾min∝13.6(14)∴vmax?−vmin?∝13.6(41?)

For Balmer series,

v‾(max)∝13.6(14−1∞)v(max)∝13.6(41?−∞1?)v‾(min)∝13.6(14−19)v(min)∝13.6(41?−91?)∴v‾min−v‾min∝13.6(19)∴vmin?−vmin?∝13.6(91?)vLymanvBalmer=94vBalmer?vLyman??=49?

1. Among the following, the set of parameters that represents path functions, is

(A) q+wq+w (B) q

(C) w
(D) H- TS

(1) (A), (B) and (C)
(2) (B) and (C)

(3) (B), (C) and (D)
(4) (A) and (D)

Answer (2)

1. The major product of the following reaction is

[image]

Answer (2)

Sol. Alkaline KMnO4KMnO4? converts RR with a benzyllic hydrogen into benzoic acid.

[image]

1. The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is

[image]

(1) D<B<A<CD<B<A<C (2) D<A<C<BD<A<C<B (3) B<C<A<DB<C<A<D (4) A<B<C<DA<B<C<D

Answer (2)

[image]

1. The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are

[image]

Answer (3)

Sol. Graph- (i) : In[Reactant] vs time is linear Hence, 1st order Graph- (ii) : [Reactant] vs time is linear Hence, zero order

1. The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01MBaCl20.01MBaCl2? in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L−1L−1 ) in solution is

(1)16×10−4(2)4×10−4(1)16×10−4(2)4×10−4 (3)6×10−2(4)4×10−2(3)6×10−2(4)4×10−2

Answer (3)

∴2[XY]=4×(0.01)×3∴2[XY]=4×(0.01)×3 [XY]=0.06[XY]=0.06 =6×10−2molL=6×10−2Lmol?

1. Consider the van der Waals constants, a and b, for the following gases.

<table>GasArNeKrXea/(atm dm6 mol-2)1.30.25.14.1b/(10-2 dm3 mol-1)3.21.71.05.0</table>

Which gas is expected to have the highest critical temperature?

(1) Ne
(2) Kr
(3) Xe
(4) Ar

Answer (2)

Sol. Critical temperature = 8a27Rb27Rb8a?

So, species with greatest value of abba? has greatest value of critical temperature i.e. Kr.

1. The major product of the following reaction is

[image]

Answer (2)

[image]

More stable product due to conjugation.

1. Among the following, the molecule expected to be stabilized by anion formation is

 

Answer (3)

Sol. C2C2? has s- p mixing and the HOMO is π2px=π2pyπ2px?=π2py? and LUMO is σ2p2σ2p2? . So, the extra electron will occupy bonding molecular orbital and this will lead to an increase in bond order.

C2C2? has more bond order than C2C2?

1. Excessive release of CO2CO2? into the atmosphere results in

(1) Depletion of ozone
(2) Polar vortex
(3) Formation of smog
(4) Global warming

Answer (4)

Sol. CO2CO2? causes global warming.

1. For a reaction,

N2(g)+3H2(g)→2NH3(g);N2?(g)+3H2?(g)→2NH3?(g);

Identify dihydrogen (H2)(H2?) as a limiting reagent in the following reaction mixtures.

(1) 35g35g of N2+8gN2?+8g of H2H2? (2) 28g28g of N2+6gN2?+6g of H2H2? (3) 56g56g of N2+10gN2?+10g of H2H2? (4) 14g14g of N2+4gN2?+4g of H2H2?

Answer (3)

Sol. 28gN228gN2? react with 6gH26gH2?

N2+3H2?2NH3N2?+3H2??2NH3? 1mol3mol6g1mol3mol6g

For 56g56g of N2N2? 12g12g of H2H2? is required.

1. Match the catalysts (Column I) with products (Column II).

Column I Column II Catalyst Product

(A) V2O5V2?O5? (i) Polyethylene

(B) TiCl4/Al(Me)3TiCl4?/Al(Me)3? (ii) Ethanal

(C) PdCl2PdCl2? (iii) H2SO4H2?SO4?

(D) Iron Oxide (iv) NH3NH3?

(1) (A)-(iv); (B)-(iii); (C)-(ii); (D)-(i)

(2) (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)

(3) (A)-(iii); (B)-(i); (C)-(ii); (D)-(iv)

(4) (A)-(ii); (B)-(iii); (C)-(i); (D)-(iv)

Answer (3)

Sol. (A) V2O5V2?O5? →→ Preparation of H2SO4H2?SO4? in contacts process

(B) TiCl4+Al(Me)3→TiCl4?+Al(Me)3?→ Polyethylene (Ziegler- Natta catalyst)

(C) PdCl2PdCl2? →→ Ethanal (Wacker's process)

(D) Iron oxide →NH3→NH3? in Haber's process

1. The element having greatest difference between its first and second ionization energies, is

(1) K
(2) Sc
(3) Ca
(4) Ba

Answer (1)

Sol. Alkali metals have high difference in the first ionisation and the second ionisation energy as they achieve stable noble gas configuration after first ionisation.

1. The correct order of the oxidation states of nitrogen in NO, N2ON2?O N2O2N2?O2? and N2O3N2?O3? is

(1) NO2<NO<N2O3<N2O(1) NO2?<NO<N2?O3?<N2?O (2) N2O<NO<N2O3<NO2(2) N2?O<NO<N2?O3?<NO2? (3) NO2<N2O3<NO<N2O(3) NO2?<N2?O3?<NO<N2?O (4) N2O<N2O3<NO<N2O2(4) N2?O<N2?O3?<NO<N2?O2?

Answer (2)

Sol. (oxide) (oxidation state)

NO+1NO+1 NO+2NO+2 NO3+3NO3?+3 NO2+4NO2?+4 SO2NO2<NO<N2O3<NO2SO2?NO2?<NO<N2?O3?<NO2?

1. Which of the following statements is not true about sucrose?

(1) The glycosidic linkage is present between C1C1? of αα -glucose and C1C1? of ββ -fructose
(2) On hydrolysis, it produces glucose and fructose
(3) It is a non-reducing sugar
(4) It is also named as invert sugar

Answer (1)

Sol. Sucrose contains glycosidic link between C1C1? of αα - D glucose and C2C2? of ββ - D- Fructose.

C12H22O11+H2O?Glucose+FructoseC12?H22?O11?+H2?O?Glucose+Fructose

1. The ore that contains the metal in the form of fluoride is

(1) malachite
(2) sphalerite
(3) magnetite
(4) cryolite

Answer (4)

Sol. Magnetite Fe3O4Fe3?O4?

Sphalerite ZnS

Cryolite Na3AlF6Na3?AlF6?

Malachite CuCO3⋅Cu(OH)2CuCO3?⋅Cu(OH)2?

1. The correct IUPAC name of the following compound is

[image]

(1) 3-chloro-4-methyl-1-nitrobenzene
(2) 5-chloro-4-methyl-1-nitrobenzene
(3) 2-methyl-5-nitro-1-chlorobenzene
(4) 2-chloro-1-methyl-4-nitrobenzene

Answer (4)

[image]

All Groups attached are to be treated as substituents and lowest set of locant rule is followed.

2-Chloro- 1-methyl- 4-nitrobenzene

1. Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0?C0?C . This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is

[image]

Answer (2)

Sol. In acidic medium aniline is more reactive than phenol that's why electrophilic aromatic substitution of Ph−N2Ph−N2? takes place with aniline

[image]

1. The aerosol is a kind of colloid in which

(1) solid is dispersed in gas
(2) gas is dispersed in solid
(3) liquid is dispersed in water
(4) gas is dispersed in liquid

Answer (1)

1. The degenerate orbitals of [Cr(H2O)6]3+[Cr(H2?O)6?]3+ are

(1) dxxdxx? and dyzdyz?
(2) dx2−y2dx2−y2? and dxydxy?
(3) dz2dz2? and dxzdxz?
(4) dyzdyz? and dz2dz2?

Answer (1)

Sol. Cr3+Cr3+ has d3d3 configuration and forms an octahedral inner orbitals complex. The set of degenerate orbitals are (dxy,dyz(dxy?,dyz? and dxz)dxz?) and (dx2−y2(dx2−y2? and dz2)dz2?) .

1. The major product of the following reaction is

(1) KOH (alc.) (2) Free radical polymerisation Cl Cl (1) (2) (3) (4)

[image]

[image]

1. The organic compound that gives following qualitative analysis is

[image]

[image]

Answer (2)

[image]

1. Let p, q ∈ R. If 2 - √3 is a root of the quadratic equation, x² + px + q = 0, then:

(1) q² - 4p - 16 = 0
(2) p² - 4q + 12 = 0
(3) p² - 4q - 12 = 0
(4) q² + 4p + 14 = 0

Answer (3)

Sol. p, q are rational numbers

2+√3 in the other root

Now, p=−4p=−4 q=1q=1

⇒p2−4q−12=16−4−12⇒p2−4q−12=16−4−12=0=0

Note:- (Erratum) p, q, should be given as rational numbers instead of real numbers

1. If the line, x−12=y+13=z−242x−1?=3y+1?=4z−2? meets the plane, x+2y+3z=15x+2y+3z=15 at a point P, then the distance of P from the origin is :

(1)25(2)92(1)25?(2)29? (3)72(4)52(3)27?(4)25??

Answer (2)

Sol. Let point on line is p(2r+1,3r−1,4r+2)p(2r+1,3r−1,4r+2)

It lies on the plane x+2y+3z=15x+2y+3z=15

∴2r+1+6r−2+12r+6=15∴2r+1+6r−2+12r+6=15⇒r=12⇒r=21?∴P=(2,12,4)∴P=(2,21?,4)∴OP=4+14+16=814=92∴OP=4+41?+16?=481??=29?

1. If a tangent to the circle x2+y2=1x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is :

(1)x2+y2−16x2y2=0(1)x2+y2−16x2y2=0 (2)x2+y2−2x2y2=0(2)x2+y2−2x2y2=0 (3)x2+y2−4x2y2=0(3)x2+y2−4x2y2=0 (4)x2+y2−2xy=0(4)x2+y2−2xy=0

Answer (3)

Sol. Let any tangent to circle x2+y2=1x2+y2=1 is

xcos?θ+ysin?θ=1xcosθ+ysinθ=1∴P(1cos?θ,0);Q(0,1sin?θ)∴P(cosθ1?,0);Q(0,sinθ1?)∴Mid−pointofPQletM(12cos?θ,12sin?θ)=(h,k)∴Mid−pointofPQletM(2cosθ1?,2sinθ1?)=(h,k)⇒cos?θ=12h;sin?θ=12k⇒cosθ=2h1?;sinθ=2k1?

On squaring and adding

1h2+1k2=4⇒x2+y2=4x2y2h21?+k21?=4⇒x2+y2=4x2y2

1. If f(x) is a non-zero polynomial of degree four, having local extreme points at x=−1,0,1x=−1,0,1 ; then the set

S={x∈R:f(x)=f(0)}S={x∈R:f(x)=f(0)}

contains exactly :

(1) Four irrational numbers

(2) Four rational numbers

(3) Two irrational and one rational number

(4) Two irrational and two rational numbers

Answer (3)

Sol. f(x)=A(x+1)x(x−1)f(x)=A(x+1)x(x−1)

A(x3−x)A(x3−x)⇒f(x)=A(x44−x22)+C⇒f(x)=A(4x4?−2x2?)+C

Now f(0)=Cf(0)=C

∴f(x)=f(0)⇒A(x44−x22)=0∴f(x)=f(0)⇒A(4x4?−2x2?)=0 ⇒x22(x22−1)=0⇒2x2?(2x2?−1)=0 ⇒x=0,0,−2,2⇒x=0,0,−2?,2?

∴S={0,−2,2}∴S={0,−2?,2?}∴S={0,−2,2}∴S={0,−2?,2?}

1. If the standard deviation of the numbers −1,0,−1,0, 1, k is 55? where k>0k>0 , then k is equal to :

(1)6(2)26(3)2103(4)453(4)(1)(3)?6?2310???(2)(4)?26?435???(4)

Answer (2)

Sol. Mean of given observation =k4=4k?

∴σ2=5(given)…(i)∴σ2=5(given)…(i)Alsoσ2=(k4+1)2+(k4)2+(k4−1)2+(3k4)24…(ii)Alsoσ2=4(4k?+1)2+(4k?)2+(4k?−1)2+(43k?)2?…(ii)∴ from (i) and (ii)∴ from (i) and (ii)∴12k2164+2⇒12k216=18∴416?12k2?+2⇒1612k2?=18⇒k2=24⇒k=26⇒k2=24⇒k=26?

1. If the fourth term in the Binomial expansion of (2x+xlog?ax)6(x>0)(x2?+xloga?x)6(x>0) is 20×8720×87 , then a value of xx is :

(1) 8383
(2) 8
(3) 8−28−2
(4) 8282

Answer (4)

Sol.T4=20×87=6C3(2x)3×(xlog?ax)3Sol.T4?=20×87=6C3?(x2?)3×(xloga?x)3⇒8×20×(xlog?axx)3=20×87⇒8×20×(xxloga?x?)3=20×87⇒(xlog?axx)3=(82)3⇒(xxloga?x?)3=(82)3⇒xlog?axx=64Takelog?aboth side⇒xxloga?x?=64Takeloga?both side⇒(log?ax)2−(log?ax)=2⇒(loga?x)2−(loga?x)=2⇒log?ax=−1orlog?ax=2⇒loga?x=−1orloga?x=2⇒x=18orx=82⇒x=81?orx=82

1. The integral ∫sec?2/3xcos?sec?4/3x dx∫sec2/3xcossec4/3x dx is equal to :

(Here C is a constant of integration)

(1)−3cot?−1/3x+C(1)−3cot−1/3x+C(2)−3tan?−1/3x+C(2)−3tan−1/3x+C(3)−34tan?−4/3x+C(3)−43?tan−4/3x+C(4)3tan?−1/3x+C(4)3tan−1/3x+C

Answer (2)

Sol.I=∫sec?2x⋅cos?sec?3x dxSol.I=∫sec2x⋅cossec3x dxI=∫sec?2x dxtan?3x Puttan?x=tI=∫tan3xsec2x dx? Puttanx=t⇒sec?2x dx= dt⇒sec2x dx= dt⇒I=t33(−13)+C⇒I=−3(tan?x)−13+C⇒I=(3−1?)3t3??+C⇒I=−3(tanx)3−1?+C

1. If one end of a focal chord of the parabola, y2=16xy2=16x is at (1, 4), then the length of this focal chord is :

(1) 24

(2) 20

(3) 22

(4) 25

Answer (4)

Sol. ?y2=16x⇒a=4?y2=16x⇒a=4

One end of focal chord (1, 4) ∴2∴2 at =4=4

⇒t=12⇒t=21?

Length of focal chord=a(t+1t)2Length of focal chord=a(t+t1?)2  =4×(2+12)2 =4×(2+21?)2  =25 =25

1. If the function f defined on (π6,π3)(6π?,3π?) by

[image]

is continuous, then k is equal to :

(1)1(2)12(3)12(4)2(3)(1)1(3)2?1??(2)21?(4)2?(3)

Answer (2)

lim?x→π42cos?x−1cot?x−1=k∴ByLhspitariurele.limx→4π??cotx−12?cosx−1?=k∴ByLhspitariurele. lim?x→π42sin?xcos?ec2x=k12limx→4π??cosec2x2?sinx?=k21?

1. All the points in the set

S={α+iα−i:α∈R}(i=−1)S={α−iα+i?:α∈R}(i=−1?)

lie on a :

(1) Straight line whose slope is 1
(2) Circle whose radius is 22?
(3) Circle whose radius is 1
(4) Straight line whose slope is -1

Answer (3)

S:Σ:ΣS=α+iα−iLetS=x+iyS:Σ:ΣS=α−iα+i?LetS=x+iy⇒x+iy=(α+i)2α2+1 (by rationalisation)⇒x+iy=α2+1(α+i)2? (by rationalisation)⇒x+iy=(α2−1)α2+1+i(2α)α2+1 (On comparing both sides)⇒x+iy=α2+1(α2−1)?+α2+1i(2α)? (On comparing both sides)⇒x=α2−1α2+1…(i)y=2αα2+1…(i)⇒x=α2+1α2−1?…(i)y=α2+12α?…(i)

By squaring and adding

⇒x2+y2=1⇒x2+y2=1

1. Let α?=3i?+j?α=3i+j? and β?=2i?−j?+3k?β?=2i−j?+3k .If β?=β?1−β?2β?=β?1?−β?2? where β?1β?1? is parallel to α?α and β?2β?2? is perpendicular to α?α ,then β?1×β?2β?1?×β?2? is equal to :

\begin{array}{l l l}{{(\mathrm{1})\frac{1}{2}(3\vec{\mathrm{i}}-\vec{\mathrm{j}}\vec{\mathrm{j}}+5\vec{\mathrm{k}})}&{{(\mathrm{2})\frac{1}{2}(-3\vec{\mathrm{i}}+\vec{\mathrm{j}}\vec{\mathrm{j}}+5\vec{\mathrm{k}})}}\\ {{(\mathrm{3})\ -3\vec{\mathrm{i}}+\vec{\mathrm{j}}\vec{\mathrm{j}}+5\vec{\mathrm{k}})}&{{(\mathrm{4})\ 3\vec{\mathrm{i}}-\vec{\mathrm{j}}\vec{\mathrm{j}}-5\vec{\mathrm{k}}}}\end{array} \quad (3)

Answer (2)

\begin{array}{r l} & {\mathrm{Sol.}\vec{\beta} = \vec{\beta}_{1} - \vec{\beta}_{2}\qquad \dots (\mathrm{i})}\\ & {\therefore \quad \vec{\beta}_{2}\cdot \vec{\alpha} = 0}\\ & {\mathrm{and}\mathrm{L e t}\vec{\beta}_{1} = \lambda \vec{\alpha}}\\ & {\vec{\alpha}\cdot \vec{\beta} = \vec{\alpha}\cdot \vec{\beta}_{1} - \vec{\alpha}\cdot \vec{\beta}_{2}}\\ & {\Rightarrow 5 = \lambda \alpha^{2}}\\ & {\Rightarrow 5 = \lambda \times 10}\\ & {\Rightarrow \lambda = \frac{1}{2}}\\ & {\therefore \quad \frac{\vec{\beta}_{1}}{\vec{\beta}_{1}} = \frac{\vec{\alpha}}{\vec{\alpha}}}\\ & {\mathrm{Cross~product~with~}\vec{\beta}_{1}\mathrm{~in~equation~}(\mathrm{i})}\\ & {\Rightarrow \vec{\beta}\times \vec{\beta}_{1} = -\vec{\beta}_{2}\times \vec{\beta}_{1}}\\ & {\Rightarrow \frac{\vec{\beta}\times\vec{\beta}_{1}}{\vec{\beta}\times\vec{\beta}_{1}} = \frac{(\vec{\beta}\times\vec{\alpha})}{2}}\\ & {\Rightarrow \vec{\beta}_{1}\times \vec{\beta}_{2} = \frac{1}{2}\left|\begin{array}{l l}{\vec{\mathrm{i}} \vec{\mathrm{j}} \vec{\mathrm{k}}\\ {2} - 1 3\\ {3 1 0}\end{array}\right|}\\ & {\qquad = \frac{1}{2}\left[-3\vec{\mathrm{i}} -\vec{\mathrm{j}} (-9) + \vec{\mathrm{k}}(5)\right]}\\ & {\qquad = \frac{1}{2}\left[-3\vec{\mathrm{i}} +9\vec{\mathrm{j}} +5\vec{\mathrm{k}}\right]} \end{array} \quad (3)

Cross product with β?1β?1? in equation (i)

β?×β?1=−β?2×β?1β?×β?1?=−β?2?×β?1?⇒β?×β?1β?×β?1=(β?×α?)2⇒β?×β?1?β?×β?1??=2(β?×α)?

\Rightarrow \vec{\beta}_{1}\times \vec{\beta}_{2} = \frac{1}{2}\left|\begin{array}{l l}{\vec{\mathrm{i}} \vec{\mathrm{j}} \vec{\mathrm{k}} 2−132−13 3 1 0}\end{array}\right|

=12[−3i?−j?(−9)+k?(5)]=21?[−3i−j?(−9)+k(5)]=12[−3i?+9j?+5k?]=21?[−3i+9j?+5k]

1. Let ∑k=110f(a+k)=16(210−1)∑k=110?f(a+k)=16(210−1) , where the function f satisfies f(x+y)=f(x)f(x+y)=f(x) f(y)f(y) for all natural numbers xx ,y and f(1)=2f(1)=2 .Then the natural number 'a' is :

(1) 2

(2) 3

(3) 16

(4) 4

Answer (2)

Sol. f(x+y)=f(x)⋅f(y)f(x+y)=f(x)⋅f(y)

∴Letf(x)=bx∴Letf(x)=bx∴f(1)=2∴f(1)=2∴b′=2∴b′=2⇒f(x)=2x⇒f(x)=2xNow,∑k=1102a+k=16(210−1)Now,k=1∑10?2a+k=16(210−1)⇒2a∑k=1102k=16(210−1)⇒2ak=1∑10?2k=16(210−1)⇒2a×((210)−1)×2(2−1)=16×(210−1)⇒2a×(2−1)((210)−1)×2?=16×(210−1)⇒2a=8⇒2a=8⇒a=3⇒a=3

1. If [11120101][1301]…[1n−101]=[17801],[10?11?10?21?][10?31?]…[10?n−11?]=[10?781?], then the inverse of [n01][n0?1?] is :

(1) [10131][113?01?]
(2) [1−1301][10?−131?]
(3) [1−1201][10?−121?]
(4) [10121][112?01?]

Answer (2)

Sol. [11120101][1301][1401]…[1n−101]=[17801][10?11?10?21?][10?31?][10?41?]…[10?n−11?]=[10?781?]

⇒[11+2+3+?+(n−1)01]=[17801]⇒[10?1+2+3+?+(n−1)1?]=[10?781?]⇒(n−1)n2=78⇒2(n−1)n?=78⇒n=13⇒n=13

Now, inverse of [11301]=[1−1301][10?131?]=[10?−131?]

1. Let S={0∈[−2π,2π];2cos?20+3sin?0=0}S={0∈[−2π,2π];2cos20+3sin0=0} . Then the sum of the elements of SS is :

(1) ππ
(2) 2π2π
(3) 13π6613π?
(4) 5π335π?

Answer (2)

Sol.:2cos?20+3sin?0=0Sol.:2cos20+3sin0=0⇒2sin?20−3sin?0−2=0⇒2sin20−3sin0−2=0⇒(2sin?0+1)(sin?0−2)=0⇒(2sin0+1)(sin0−2)=0⇒sin?0=−12;sin?02=2⇒sin0=−21?;2sin0?=2

[image]

Sum of all solutions in [−2π,2π][−2π,2π] is

Ψ=(π+π6)+(2π−π6)+(−π6+π6)=2πΨ=(π+6π?)+(2π−6π?)+(−6π?+6π?)=2π

1. Let f(x)=15−?x−10?f(x)=15−?x−10? ; x∈Rx∈R . Then the set of all values of xx , at which the function, g(x)=f(f(x))g(x)=f(f(x)) is not differentiable, is :

(1) (10, 15)

(2) (5, 10, 15, 20)

(3) (10)

(4) (5, 10, 15)

Answer (4)

Sol. Given f(x)=15−?(10−x)?f(x)=15−?(10−x)?

⇒f(f(x))=15−?f(10−x)−5?⇒f(f(x))=15−?f(10−x)−5?

Non- differentiable at points where

10−x=0 and ?10−x?=510−x=0 and ?10−x?=5⇒x=10 and x−10=±5⇒x=10 and x−10=±5⇒x=10 and x=15,5⇒x=10 and x=15,5

1. The area (in sq. units) of the region

A={(x,y):x2≤y≤x+2}is:A={(x,y):x2≤y≤x+2}is:(1)316(2)103(1)631?(2)310?(3)92(4)136(3)29?(4)613?

Answer (3)

Sol.

[image]

∴Required area=∫−12((x+2)−x2)dx∴Required area=∫−12?((x+2)−x2)dx=(x22+2x−x33)−12=(2x2?+2x−3x3?)−12?=(2+4−83)−(12−2+13)=(2+4−38?)−(21?−2+31?)=8−3−12=8−3−21?=5−12=92=5−21?=29?

1. If the line y=mx+73y=mx+73? is normal to the hyperbola x224−y218=124x2?−18y2?=1 , then a value of m is :

(1) 255?2?
(2) 355?3?
(3) 152215??
(4) 5225??

Answer (1)

Sol. mx−y+73mx−y+73? is normal to hyperbola

x224−y218=124x2?−18y2?=1

then 24m2−18(−1)2=(24+18)2(73)2m224?−(−1)218?=(73?)2(24+18)2?

⇒24m2−18=42×427×7×3⇒m224?−18=7×7×342×42?⇒m=25⇒m=5?2?

If lx+my+n=0lx+my+n=0 is a normal to x2a2−y2b2=1a2x2?−b2y2?=1,

then a2l2−b2m2=(a2+b2)2n2l2a2?−m2b2?=n2(a2+b2)2?

1. Let αα and ββ be the roots of the equation x2+x+1=0x2+x+1=0 . Then for y≠0y?=0 in R,

?y+1αβαy+β1β1y+α??y+1αβ?αy+β1?β1y+α??

is equal to :

(1) y(y2−1)y(y2−1)
(2) y3−1y3−1
(3) y(y2−3)y(y2−3)
(4) y3y3

Answer (4)

Sol. Let α=ωα=ω and β=ω2β=ω2 are roots of x2+x+1=0x2+x+1=0

?y+ω1ω21y+ω2ωω2ω1+y?operate c1→c1+c2+c3?y+ω1ω2?1y+ω2ω?ω2ω1+y??operate c1?→c1?+c2?+c3?=y?11ω21y+ω2ω1ω1+y?(By R2→R2−R1)=y?111?1y+ω2ω?ω2ω1+y??(By R2?→R2?−R1?)=y?11ω20y+ω2−1ω−ω20ω−11+y−ω2?=y?100?1y+ω2−1ω−1?ω2ω−ω21+y−ω2??=y((y+ω2−1)(1+y−ω2)−ω(ω−1)(1−ω))=y((y+ω2−1)(1+y−ω2)−ω(ω−1)(1−ω))=y(y2−(ω2−1)2)+yω(ω−1)2=y(y2−(ω2−1)2)+yω(ω−1)2=y3+y(ω−1)2(ω−(ω+1)2)=y3=y3+y(ω−1)2(ω−(ω+1)2)=y3

1. The value of cos?210?−cos?10?cos?50?+cos?250?cos210?−cos10?cos50?+cos250? is :

(1) 3443?
(2) 32(1+cos?20?)23?(1+cos20?)
(3) 3223?
(4) 34+cos?20?43?+cos20?

Answer (1)

Sol. (1+cos?20?2)+(1+cos?100?2)−12(2cos?10?cos?50?)(21+cos20??)+(21+cos100??)−21?(2cos10?cos50?)

=1+12(cos?20?+cos?100?)−12[cos?60?+cos?40?]=1+21?(cos20?+cos100?)−21?[cos60?+cos40?]=(1−14)+12[cos?20?+cos?100?−cos?40?]=(1−41?)+21?[cos20?+cos100?−cos40?]=34+12[2cos?60?×cos?40?−cos?40?]=43?+21?[2cos60?×cos40?−cos40?]=34=43?

1. A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then :

(1) m = n = 68
(2) m + n = 68
(3) m = n = 78
(4) n = m - 8

Answer (3)

1. Slope of a line passing through P(2,3)P(2,3) and intersecting the line, x+y=7x+y=7 at a distance of 4 units from PP , is :

(1) 7−17+17?+17?−1?
(2) 1−71+71+7?1−7??
(3) 5−15+15?+15?−1?
(4) 1−51+51+5?1−5??

Answer (2)

Sol. Point at 4 units from P(2,3)P(2,3) will be

A(4cos?θ+2,4sin?θ+3)will satisfy x+y=7A(4cosθ+2,4sinθ+3)will satisfy x+y=7⇒cos?θ+sin?θ=12on squaring⇒cosθ+sinθ=21?on squaring⇒sin?2θ=−34sin?2θ⇒2tan?θ1+tan?2θ=−34⇒sin2θsin2θ=−43??⇒1+tan2θ2tanθ?=−43?⇒3tan?2θ+8tan?θ+3=0⇒3tan2θ+8tanθ+3=0⇒tan?θ=−8±276(lgnoring−ve sign)⇒tanθ=6−8±27??(lgnoring−ve sign)⇒tan?θ=−8+276=1−71+7⇒tanθ=6−8+27??=1+7?1−7??

1. The value of ∫0π/2sin?3xsin?x+cos?xdx∫0π/2?sinx+cosxsin3x?dx is :

(1) π−244π−2?
(2) π−288π−2?
(3) π−144π−1?
(4) π−122π−1?

Answer (3)

Sol.I=∫0π2sin?3xdxsin?x+cos?xSol.I=∫02π??sinx+cosxsin3xdx?⇒I=∫0π2cos?3xdxsin?x+cos?x⇒I=∫02π??sinx+cosxcos3xdx?⇒2I=∫0π2(1−12sin?(2x))dx⇒2I=∫02π??(1−21?sin(2x))dx⇒I=12[x+14cos?2x]0π2⇒I=21?[x+41?cos2x]02π??I=12(π−12)=π−14I=21?(2π−1?)=4π−1?

1. Let S be the set of all values of xx for which the tangent to the curve y=f(x)=x3−x2−2xy=f(x)=x3−x2−2x at (x,y)(x,y) is parallel to the line segment joining the points (1, f(1)f(1) ) and (−1,f(−1))(−1,f(−1)) , then S is equal to :

(1) {13,−1}{31?,−1}
(2) {−13,1}{−31?,1}
(3) {−13,−1}{−31?,−1}
(4) {13,1}{31?,1}

Answer (2)

Sol. y=f(x)=x3−x2−2xy=f(x)=x3−x2−2x

dydx=3x2−2x−2dxdy?=3x2−2x−2f(1)=1−1−2=−2,f(−1)=−1−1+2=0f(1)=1−1−2=−2,f(−1)=−1−1+2=0

According to question, 3x2−2x−23x2−2x−2

=f(1)−f(−1)1−(−1)=1−(−1)f(1)−f(−1)?⇒3x2−2x−2=−2−02⇒3x2−2x−2=2−2−0?⇒3x2−2x−1=0⇒3x2−2x−1=0⇒x=2±46=1,−13⇒x=62±4?=1,3−1?So,S={−13,1}So,S={3−1?,1}

1. A plane passing through the points (0, -1, 0) and (0, 0, 1) and making an angle π44π? with the plane y−z+5=0y−z+5=0 , also passes through the point :

(1) (2,1,4)(2?,1,4)
(2) (2,−1,4)(2?,−1,4)
(3) (−2,−1,−4)(−2?,−1,−4)
(4) (−2,1,−4)(−2?,1,−4)

Answer (1)

Sol. Let the required plane be xa+y−1+z1=1ax?+−1y?+1z?=1 given plane is y−z+5=0y−z+5=0

∴cos?π4=−1−11a2+1+12∴cos4π?=a21?+1+1?2?−1−1?⇒a2=12⇒a2=21?⇒1a=±2⇒a1?=±2?⇒±2x−y+z=1⇒±2?x−y+z=1∴(2,1,4) satisfies 2x−y+z=1∴(2?,1,4) satisfies 2?x−y+z=1

1. The solution of the differential equation xdydx+2y=x2(x≠0)xdxdy?+2y=x2(x?=0) with y(1)=1y(1)=1, is :

(1) y=45x3+15x2y=54?x3+5x21?
(2) y=x35+15x2y=5x3?+5x21?
(3) y=x24+34x2y=4x2?+4x23?
(4) y=34x2+14x2y=43?x2+4x21?

Answer (3)

Sol. dydx+2xy=xdxdy?+x2?y=x y(1)=1y(1)=1 (given)

I.F = e∫2xdx=x2e∫x2?dx=x2

y×x2=∫x3dxy×x2=∫x3dx⇒yx2=x44+c⇒yx2=4x4?+c∴at x=1;y=1∴at x=1;y=1⇒c=34⇒c=43?∴y=x24+34x2∴y=4x2?+4x23?

1. Let the sum of the first n terms of a non-constant A.P., a1,a2,a3,…a1?,a2?,a3?,… be 50n+n(n−7)2A50n+2n(n−7)?A, where A is a constant. If d is the common difference of this A.P., then the ordered pair (d, a_{50}) is equal to :

(1) (50, 50 + 46A)
(2) (A, 50 + 45A)
(3) (A, 50 + 46A)
(4) (50, 50 + 45A)

Answer (3)

Sol. ∴Sn=(50−7A2)n+n2×A2∴Sn?=(50−27A?)n+n2×2A?

∴Common difference=A2×2=A∴Common difference=2A?×2=Aa50=a1+49×da50?=a1?+49×d=(50−3A)+49A=(50−3A)+49A=50+46A=50+46A

So, (d,a50)=(A,50+46A)(d,a50?)=(A,50+46A)

1. For any two statements p and q, the negation of the expression p∨(¬p∧q)p∨(¬p∧q) is :

(1) ¬p∧¬q¬p∧¬q
(2) p∨¬qp∨¬q
(3) p∧qp∧q
(4) p↔qp↔q

Answer (1)

Sol. ¬(p∨(¬p∧q))=¬(¬p∧q)∧¬p¬(p∨(¬p∧q))=¬(¬p∧q)∧¬p

=(¬q∨p)∧¬p=(¬q∨p)∧¬p=¬p∧(p∨¬q)=¬p∧(p∨¬q)=(¬q∧¬p)∨(p∧¬p)=(¬q∧¬p)∨(p∧¬p)=(¬p∧¬q)=(¬p∧¬q)

1. If the tangent to the curve, y=x3+ax−by=x3+ax−b at the point (1, -5) is perpendicular to the line, −x+y+4=0−x+y+4=0, then which one of the following points lies on curve?

(1) (-2, 1)
(2) (2, -2)
(3) (2, -1)
(4) (-2, 2)

Answer (2)

Sol. f(x)=x3+ax−b⇒f′(x)=3x2+af(x)=x3+ax−b⇒f′(x)=3x2+a

f(1)=−5andf′(1)=3+af(1)=−5andf′(1)=3+a⇒1+a−b=−5⇒1+a−b=−5⇒a−b=−6…(1)⇒a−b=−6…(1)

Also slope of tangent P(1, -5) = -1 = f'(1)

∴3+a=−1∴3+a=−1⇒a=−4⇒a=−4b=2b=2

Equation of the curve is f(x)=x3−4x−2f(x)=x3−4x−2

(2,-2) lies on the curve

1. If the function f:R−{1,−1}→Af:R−{1,−1}→A defined by

f(x)=x21−x2,is surjective, then A is equal to:f(x)=1−x2x2?,is surjective, then A is equal to:(1)[0,∞)(2)R−{−1}(3)R−(−1,0)(4)R−[−1,0)(4)?(1)[0,∞)(2)R−{−1}(3)R−(−1,0)(4)R−[−1,0)?(4)

Answer (4)

Sol. f(x)=x21−x2f(x)=1−x2x2?

⇒f(−x)=x21−x2=f(x)⇒f(−x)=1−x2x2?=f(x)⇒f′(x)=2x(1−x2)2⇒f′(x)=(1−x2)22x?

f(x) increases in x E (0, 0)

Also f(0)=0f(0)=0

lim?x→±∞f(x)=−1x→±∞lim?f(x)=−1

and F(x)F(x) is even function

Set A→R−[−1,0)A→R−[−1,0)

Graph of function

[image]

1. Four persons can hit a target correctly with probabilities 12,13,1421?,31?,41? and 1881? respectively. If all hit at the target independently, then the probability that the target would be hit, is:

7321192251922532(4)327?1921??19225?3225??(4)

Answer (4)

Sol. PP (at least one) =1−P=1−P (none)

=1−12×23×34×78=1−21?×32?×43?×87?=1−732=2532=1−327?=3225?

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