AIEEE 2009 — 26 Apr Offline Set A (Previous Year Paper)

JEE Main · 2009 · with answers and explanations

questions
86
Total marks
416
Duration
180 minutes
Negative marking
−2

Subjects: Physics · Chemistry · Mathematics

Question 1

Physics · 4 marks · −1

Statement-1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero

  1. A.Statement-1 is true, Statement-2 is false
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  3. C.Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: B

Explanation: Electrostatic force is conservative; path independence is equivalent to zero work round any closed loop.

Question 2

Physics · 4 marks · −1

The above is a plot of binding energy per nucleon EbE_b, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions: (i) A+B→C+ϵA + B \rightarrow C + \epsilon (ii) C→A+B+ϵC \rightarrow A + B + \epsilon (iii) D+E→F+ϵD + E \rightarrow F + \epsilon and (iv) F→D+E+ϵF \rightarrow D + E + \epsilon where ϵ\epsilon is the energy released? In which reactions is ϵ\epsilon positive?

Figure for question 2
  1. A.(i) and (iv)
  2. B.(i) and (iii)
  3. C.(ii) and (iv)
  4. D.(ii) and (iii)

Correct answer: A

Explanation: Energy is released when products have higher binding energy per nucleon: fusion of light A, B into C, and fission of heavy F into D, E.

Question 3

Physics · 8 marks · −2

A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. The current (I) in the resistor (R) can be shown by :

Figure for question 3
  1. A.Graph (1): full sinusoid
  2. B.Graph (2): full-wave rectified humps
  3. C.Graph (3): half-wave rectified (alternate humps)
  4. D.Graph (4): constant DC

Correct answer: C

Explanation: A single diode passes only one half-cycle: half-wave rectification.

Question 4

Physics · 8 marks · −2

The logic circuit shown below has the input waveforms 'A' and 'B' as shown. Pick out the correct output waveform.

Figure for question 4
  1. A.Waveform (1)
  2. B.Waveform (2)
  3. C.Waveform (3)
  4. D.Waveform (4)

Correct answer: A

Explanation: Both inputs are inverted and fed to a NOR: Y=Aˉ+Bˉ‾=ABY=\overline{\bar A+\bar B}=AB, high only when both A and B are high.

Question 5

Physics · 8 marks · −2

If x, v and a denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period T, then, which of the following does not change with time?

  1. A.a2T2+4π2v2a^2T^2 + 4\pi^2v^2
  2. B.aTx\frac{aT}{x}
  3. C.aT+2πvaT + 2\pi v
  4. D.aTv\frac{aT}{v}

Correct answer: B

Explanation: a=−ω2xa=-\omega^2x, so aTx=−4π2T\frac{aT}{x}=-\frac{4\pi^2}{T}, constant.

Question 6

Physics · 8 marks · −2

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45∘45^\circ with the x-axis meets the experimental curve at P. The coordinates of P will be

  1. A.(2f, 2f)
  2. B.(f2,f2)\left(\frac{f}{2}, \frac{f}{2}\right)
  3. C.(f, f)
  4. D.(4f, 4f)

Correct answer: A

Explanation: ∣u∣=∣v∣|u|=|v| on the lens equation gives 2f2f.

Question 7

Physics · 8 marks · −2

A thin uniform rod of length ℓ\ell and mass m is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is ω\omega. Its centre of mass rises to a maximum height of

  1. A.13ℓ2ω2g\frac{1}{3}\frac{\ell^2\omega^2}{g}
  2. B.16ℓωg\frac{1}{6}\frac{\ell\omega}{g}
  3. C.12ℓ2ω2g\frac{1}{2}\frac{\ell^2\omega^2}{g}
  4. D.16ℓ2ω2g\frac{1}{6}\frac{\ell^2\omega^2}{g}

Correct answer: D

Explanation: 12⋅mℓ23ω2=mgh⇒h=ℓ2ω26g\frac12\cdot\frac{m\ell^2}{3}\omega^2=mgh\Rightarrow h=\frac{\ell^2\omega^2}{6g}.

Question 8

Physics · 8 marks · −2

Let P(r)=QπR4rP(r) = \frac{Q}{\pi R^4}r be the charge density distribution for a solid sphere of radius R and total charge Q. For a point 'p' inside the sphere at distance r1r_1 from the centre of the sphere, the magnitude of electric field is

  1. A.0
  2. B.Q4πϵ0r12\frac{Q}{4\pi\epsilon_0r_1^2}
  3. C.Qr124πϵ0R4\frac{Qr_1^2}{4\pi\epsilon_0R^4}
  4. D.Qr123πϵ0R4\frac{Qr_1^2}{3\pi\epsilon_0R^4}

Correct answer: C

Explanation: Enclosed charge ∫0r1QrπR44πr2dr=Qr14R4\int_0^{r_1}\frac{Qr}{\pi R^4}4\pi r^2dr=\frac{Qr_1^4}{R^4}; E=Qr124πϵ0R4E=\frac{Qr_1^2}{4\pi\epsilon_0R^4}.

Question 9

Physics · 4 marks · −1

The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

  1. A.2→12 \to 1
  2. B.3→23 \to 2
  3. C.4→24 \to 2
  4. D.5→45 \to 4

Correct answer: D

Explanation: Need a smaller energy gap than 4→34\to3: only 5→45\to4.

Question 10

Physics · 4 marks · −1

One kg of a diatomic gas is at a pressure of 8×1048\times10^4 N/m2^2. The density of the gas is 4 kg/m3^3. What is the energy of the gas due to its thermal motion?

  1. A.3×1043\times10^4 J
  2. B.5×1045\times10^4 J
  3. C.6×1046\times10^4 J
  4. D.7×1047\times10^4 J

Correct answer: B

Explanation: V=14V=\frac14 m3^3; U=52PV=52⋅8×104⋅0.25=5×104U=\frac52PV=\frac52\cdot8\times10^4\cdot0.25=5\times10^4 J.

Question 11

Physics · 4 marks · −1

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30∘30^\circ. Another straight thin wire with steady current I1I_1 flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is

Figure for question 11
  1. A.zero
  2. B.μ0I(b−a)24ab\frac{\mu_0I(b-a)}{24ab}
  3. C.μ0I4π[b−aab]\frac{\mu_0I}{4\pi}\left[\frac{b-a}{ab}\right]
  4. D.μ0I4π[2(b−a)+π3(a+b)]\frac{\mu_0I}{4\pi}\left[2(b-a) + \frac{\pi}{3}(a+b)\right]

Correct answer: B

Explanation: Radial segments give no field at O; arcs give μ0I4ππ6(1a−1b)=μ0I(b−a)24ab\frac{\mu_0I}{4\pi}\frac{\pi}{6}\left(\frac1a-\frac1b\right)=\frac{\mu_0I(b-a)}{24ab}.

Question 12

Physics · 4 marks · −1

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30∘30^\circ. Another straight thin wire with steady current I1I_1 flowing out of the plane of the paper is kept at the origin. Due to the presence of the current I1I_1 at the origin

Figure for question 12
  1. A.The forces on AB and DC are zero
  2. B.The forces on AD and BC are zero
  3. C.The magnitude of the net force on the loop is given by μ0II14π[2(b−a)+π3(a+b)]\frac{\mu_0II_1}{4\pi}\left[2(b-a) + \frac{\pi}{3}(a+b)\right]
  4. D.The magnitude of the net force on the loop is given by μ0II124ab(b−a)\frac{\mu_0II_1}{24ab}(b-a)

Correct answer: B

Explanation: The field of I1I_1 is tangential, parallel to the arcs AD and BC, so they feel no force; the radial segments do.

Question 13

Physics · 4 marks · −1

A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th4^{th} bright fringe of the unknown light. From this data, the wavelength of the unknown light is

  1. A.393.4 nm
  2. B.885.0 nm
  3. C.442.5 nm
  4. D.776.8 nm

Correct answer: C

Explanation: 3⋅590=4λ⇒λ=442.53\cdot590=4\lambda\Rightarrow\lambda=442.5 nm.

Question 14

Physics · 4 marks · −1

Two points P and Q are maintained at the potentials of 10V and -4V respectively. The work done in moving 100 electrons from P to Q is

  1. A.−19×10−17-19\times10^{-17} J
  2. B.9.60×10−179.60\times10^{-17} J
  3. C.−2.24×10−16-2.24\times10^{-16} J
  4. D.2.24×10−162.24\times10^{-16} J

Correct answer: D

Explanation: W=q(VQ−VP)=(−100e)(−14)=2.24×10−16W=q(V_Q-V_P)=(-100e)(-14)=2.24\times10^{-16} J.

Question 15

Physics · 4 marks · −1

The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is (hc = 1240 eV nm)

  1. A.3.09 eV
  2. B.1.41 eV
  3. C.151 eV
  4. D.1.68 eV

Correct answer: B

Explanation: 3.10−1.68=1.423.10-1.68=1.42 eV.

Question 16

Physics · 4 marks · −1

A particle has an initial velocity 3i^+4j^3\hat{i} + 4\hat{j} and an acceleration of 0.4i^+0.3j^0.4\hat{i} + 0.3\hat{j}. Its speed after 10 s is

  1. A.10 units
  2. B.727\sqrt{2} units
  3. C.7 units
  4. D.8.5 units

Correct answer: B

Explanation: v⃗=7i^+7j^\vec v=7\hat i+7\hat j.

Question 17

Physics · 4 marks · −1

A motor cycle starts from rest and accelerates along a straight path at 2 m/s2^2. At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (speed of sound = 330 ms−1^{-1})

  1. A.49 m
  2. B.98 m
  3. C.147 m
  4. D.196 m

Correct answer: B

Explanation: vo=0.06⋅330=19.8v_o=0.06\cdot330=19.8 m/s; s=v22a=98s=\frac{v^2}{2a}=98 m.

Question 18

Physics · 4 marks · −1

Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as function of time will be

Figure for question 18
  1. A.Pair (1)
  2. B.Pair (2)
  3. C.Pair (3)
  4. D.Pair (4)

Correct answer: C

Explanation: v decreases linearly from 0 to −v1-v_1, jumps to +v1+v_1 at impact, and repeats (sawtooth between ±v1\pm v_1); height is a series of parabolic arches.

Question 19

Physics · 4 marks · −1

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q/q equals

  1. A.−22-2\sqrt{2}
  2. B.−1-1
  3. C.1
  4. D.−12-\frac{1}{\sqrt{2}}

Correct answer: A

Explanation: 2kQqa2+kQ22a2=0⇒Qq=−22\sqrt2\frac{kQq}{a^2}+\frac{kQ^2}{2a^2}=0\Rightarrow\frac Qq=-2\sqrt2.

Question 20

Physics · 4 marks · −1

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature θ\theta along the length x of the bar from its hot end is best described by which of the following figure.

Figure for question 20
  1. A.Graph (1)
  2. B.Graph (2)
  3. C.Graph (3)
  4. D.Graph (4)

Correct answer: B

Explanation: Uniform bar in steady state: θ\theta decreases linearly with x.

Question 21

Physics · 4 marks · −1

A transparent solid cylindrical rod has a refractive index of 23\frac{2}{\sqrt{3}}. It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure. The incident angle θ\theta for which the light ray grazes along the wall of the rod is

Figure for question 21
  1. A.sin⁡−1(12)\sin^{-1}\left(\frac{1}{2}\right)
  2. B.sin⁡−1(32)\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)
  3. C.sin⁡−1(23)\sin^{-1}\left(\frac{2}{\sqrt{3}}\right)
  4. D.sin⁡−1(13)\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)

Correct answer: D

Explanation: Critical angle: sin⁡C=32\sin C=\frac{\sqrt3}{2}; r=90∘−Cr=90^\circ-C, sin⁡r=12\sin r=\frac12; sin⁡θ=23⋅12\sin\theta=\frac{2}{\sqrt3}\cdot\frac12.

Question 22

Physics · 4 marks · −1

Three sound waves of equal amplitudes have frequencies (ν−1)(\nu - 1), ν\nu, (ν+1)(\nu + 1). They superpose to give beats. The number of beats produced per second will be

  1. A.4
  2. B.3
  3. C.2
  4. D.1

Correct answer: C

Explanation: The resultant amplitude A[1+2cos⁡(2πt)]A[1+2\cos(2\pi t)] has two maxima per second (one full and one secondary), conventionally counted as 2 beats per second (difference between extreme frequencies).

Question 23

Physics · 4 marks · −1

The height at which the acceleration due to gravity becomes g9\frac{g}{9} (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is

  1. A.2R
  2. B.R2\frac{R}{\sqrt{2}}
  3. C.R2\frac{R}{2}
  4. D.2R\sqrt{2}R

Correct answer: A

Explanation: (RR+h)2=19⇒h=2R\left(\frac{R}{R+h}\right)^2=\frac19\Rightarrow h=2R.

Question 24

Physics · 4 marks · −1

Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire-2 has cross-sectional area 3A. If the length of wire 1 increases by Δx\Delta x on applying force F, how much force is needed to stretch wire 2 by the same amount?

  1. A.F
  2. B.4F
  3. C.6F
  4. D.9F

Correct answer: D

Explanation: ℓ2=ℓ13\ell_2=\frac{\ell_1}{3}; F2=F3AAℓ1ℓ2=9FF_2=F\frac{3A}{A}\frac{\ell_1}{\ell_2}=9F.

Question 25

Physics · 4 marks · −1

In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (= 0.5∘0.5^\circ), then the least count of the instrument is

  1. A.one minute
  2. B.half minute
  3. C.one degree
  4. D.half degree

Correct answer: A

Explanation: LC =0.5∘30=1′=\frac{0.5^\circ}{30}=1'.

Question 26

Physics · 4 marks · −1

An inductor of inductance L = 400 mH and resistors of resistances R1=2 ΩR_1 = 2\,\Omega and R2=2 ΩR_2 = 2\,\Omega are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

Figure for question 26
  1. A.6e−5t6e^{-5t} V
  2. B.12te−3t\frac{12}{t}e^{-3t} V
  3. C.6(1−e−t/0.2)6\left(1 - e^{-t/0.2}\right) V
  4. D.12e−5t12e^{-5t} V

Correct answer: D

Explanation: The L-R2R_2 branch sits directly across 12 V: VL=12e−tR2/L=12e−5tV_L=12e^{-tR_2/L}=12e^{-5t} V.

Question 27

Physics · 4 marks · −1

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram. The net work done on the gas in the cycle ABCDA is

Figure for question 27
  1. A.Zero
  2. B.276 R
  3. C.1076 R
  4. D.1904 R

Correct answer: B

Explanation: By gas: 400R+1000Rln⁡2−400R−600Rln⁡2=400Rln⁡2≈276R400R+1000R\ln2-400R-600R\ln2=400R\ln2\approx276R (magnitude).

Question 28

Chemistry · 4 marks · −1

Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect ?

  1. A.Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.
  2. B.The ionic sizes of Ln (III) decrease in general with increasing atomic number.
  3. C.Ln (III) compounds are generally colourless.
  4. D.Ln (III) hydroxides are mainly basic in character.

Correct answer: C

Explanation: Most Ln3+Ln^{3+} compounds are coloured (f-f transitions).

Question 29

Chemistry · 4 marks · −1

A liquid was mixed with ethanol and a drop of concentrated H2SO4H_2SO_4 was added. A compound with a fruity smell was formed. The liquid was :

  1. A.CH3OHCH_3OH
  2. B.HCHO
  3. C.CH3COCH3CH_3COCH_3
  4. D.CH3COOHCH_3COOH

Correct answer: D

Explanation: Esterification gives ethyl acetate.

Question 30

Chemistry · 4 marks · −1

Arrange the carbanions, (CH3)3Cˉ(CH_3)_3\bar{C}, CˉCl3\bar{C}Cl_3, (CH3)2CˉH(CH_3)_2\bar{C}H, C6H5CˉH2C_6H_5\bar{C}H_2, in order of their decreasing stability :

  1. A.C6H5CˉH2>CˉCl3>(CH3)3Cˉ>(CH3)2CˉHC_6H_5\bar{C}H_2 > \bar{C}Cl_3 > (CH_3)_3\bar{C} > (CH_3)_2\bar{C}H
  2. B.(CH3)2CˉH>CˉCl3>C6H5CˉH2>(CH3)3Cˉ(CH_3)_2\bar{C}H > \bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_3\bar{C}
  3. C.CˉCl3>C6H5CˉH2>(CH3)2CˉH>(CH3)3Cˉ\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}
  4. D.(CH3)3Cˉ>(CH3)2CˉH>C6H5CˉH2>CˉCl3(CH_3)_3\bar{C} > (CH_3)_2\bar{C}H > C_6H_5\bar{C}H_2 > \bar{C}Cl_3

Correct answer: C

Explanation: −I-I of three Cl (and d-orbital delocalisation) > benzylic resonance > secondary > tertiary (alkyl groups destabilise).

Question 31

Chemistry · 4 marks · −1

The alkene that exhibits geometrical isomerism is :

  1. A.propene
  2. B.2-methyl propene
  3. C.2-butene
  4. D.2- methyl -2- butene

Correct answer: C

Explanation: Each carbon of 2-butene carries H and CH3CH_3.

Question 32

Chemistry · 4 marks · −1

In which of the following arrangements, the sequence is not strictly according to the property written against it ?

  1. A.CO2<SiO2<SnO2<PbO2CO_2 < SiO_2 < SnO_2 < PbO_2 : increasing oxidising power
  2. B.HF < HCl < HBr < HI : increasing acid strength
  3. C.NH3<PH3<AsH3<SbH3NH_3 < PH_3 < AsH_3 < SbH_3 : increasing basic strength
  4. D.B < C < O < N : increasing first ionization enthalpy

Correct answer: C

Explanation: Basic strength decreases from NH3NH_3 to SbH3SbH_3.

Question 33

Chemistry · 4 marks · −1

The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :

  1. A.benzoic acid
  2. B.salicylaldehyde
  3. C.salicylic acid
  4. D.phthalic acid

Correct answer: C

Explanation: Kolbe-Schmitt reaction.

Question 34

Chemistry · 4 marks · −1

Which of the following statements is incorrect regarding physissorptions ?

  1. A.It occurs because of vander Waal's forces.
  2. B.More easily liquefiable gases are adsorbed readily.
  3. C.Under high pressure it results into multi molecular layer on adsorbent surface.
  4. D.Enthalpy of adsorption (ΔHadsorption\Delta H_{adsorption}) is low and positive.

Correct answer: D

Explanation: Adsorption is exothermic; ΔH\Delta H is negative.

Question 35

Chemistry · 4 marks · −1

Which of the following on heating with aqueous KOH, produces acetaldehyde ?

  1. A.CH3COClCH_3COCl
  2. B.CH3CH2ClCH_3CH_2Cl
  3. C.CH2ClCH2ClCH_2ClCH_2Cl
  4. D.CH3CHCl2CH_3CHCl_2

Correct answer: D

Explanation: Gem-dihalide hydrolyses to the gem-diol, which loses water to CH3CHOCH_3CHO.

Question 36

Chemistry · 4 marks · −1

In an atom, an electron is moving with a speed of 600m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h=6.6×10−34h = 6.6\times10^{-34} kg m2^2s−1^{-1}, mass of electron =9.1×10−31= 9.1\times10^{-31} kg)

  1. A.1.52×10−41.52\times10^{-4} m
  2. B.5.10×10−35.10\times10^{-3} m
  3. C.1.92×10−31.92\times10^{-3} m
  4. D.3.84×10−33.84\times10^{-3} m

Correct answer: C

Explanation: Δv=0.03\Delta v=0.03 m/s; Δx=h4πmΔv≈1.92×10−3\Delta x=\frac{h}{4\pi m\Delta v}\approx1.92\times10^{-3} m.

Question 37

Chemistry · 8 marks · −2

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l). At 298K standard Gibb's energies of formation for CH3OH(l)CH_3OH(l), H2O(l)H_2O(l) and CO2(g)CO_2(g) are −166.2-166.2, −237.2-237.2 and −394.4-394.4 kJ mol−1^{-1} respectively. If standard enthalpy of combustion of methanol is −726-726 kJ mol−1^{-1}, efficiency of the fuel cell will be

  1. A.80 %
  2. B.87%
  3. C.90%
  4. D.97%

Correct answer: D

Explanation: ΔG=−394.4−474.4+166.2=−702.6\Delta G=-394.4-474.4+166.2=-702.6; efficiency 702.6726=97%\frac{702.6}{726}=97\%.

Question 38

Chemistry · 8 marks · −2

Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively :

  1. A.200 and 300
  2. B.300 and 400
  3. C.400 and 600
  4. D.500 and 600

Correct answer: C

Explanation: PX+3PY=2200P_X+3P_Y=2200, PX+4PY=2800P_X+4P_Y=2800: PY=600P_Y=600, PX=400P_X=400.

Question 39

Chemistry · 8 marks · −2

The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301) :

  1. A.230.3 minutes
  2. B.23.03 minutes
  3. C.46.06 minutes
  4. D.460.6 minutes

Correct answer: C

Explanation: k=0.1k=0.1 min−1^{-1}; t=2.3030.1log⁡100=46.06t=\frac{2.303}{0.1}\log100=46.06 min.

Question 40

Chemistry · 8 marks · −2

Given : EFe3+/Fe∘=−0.036E^\circ_{Fe^{3+}/Fe} = -0.036 V, EFe2+/Fe∘=−0.439E^\circ_{Fe^{2+}/Fe} = -0.439 V. The value of standard electrode potential for the change, Fe3+(aq)+e−→Fe2+(aq)Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq) will be :

  1. A.−0.072-0.072 V
  2. B.0.385 V
  3. C.0.770 V
  4. D.−0.270-0.270 V

Correct answer: C

Explanation: E∘=3(−0.036)−2(−0.439)=0.770E^\circ=3(-0.036)-2(-0.439)=0.770 V.

Question 41

Chemistry · 8 marks · −2

On the basis of the following thermochemical data : (ΔfG∘H(aq)+=0\Delta_fG^\circ H^+_{(aq)} = 0) H2O(l)→H+(aq)+OH−(aq)H_2O(l) \rightarrow H^+(aq) + OH^-(aq); ΔH=57.32\Delta H = 57.32 kJ; H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l); ΔH=−286.20\Delta H = -286.20 kJ. The value of enthalpy of formation of OH−OH^- ion at 25∘25^\circC is :

  1. A.−22.88-22.88 kJ
  2. B.−228.88-228.88 kJ
  3. C.+228.88+228.88 kJ
  4. D.−343.52-343.52 kJ

Correct answer: B

Explanation: Adding: ΔfH(OH−)=57.32−286.20=−228.88\Delta_fH(OH^-)=57.32-286.20=-228.88 kJ.

Question 42

Chemistry · 8 marks · −2

Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ?

  1. A.108 pm
  2. B.127 pm
  3. C.157 pm
  4. D.181 pm

Correct answer: B

Explanation: r=2a4=127r=\frac{\sqrt2a}{4}=127 pm.

Question 43

Chemistry · 4 marks · −1

Which of the following has an optical isomer ?

  1. A.[Co(NH3)3Cl]+[Co(NH_3)_3Cl]^+
  2. B.[Co(en)(NH3)2]2+[Co(en)(NH_3)_2]^{2+}
  3. C.[Co(H2O)4(en)]3+[Co(H_2O)_4(en)]^{3+}
  4. D.[Co(en)2(NH3)2]3+[Co(en)_2(NH_3)_2]^{3+}

Correct answer: D

Explanation: cis-[M(AA)2X2][M(AA)_2X_2] is chiral.

Question 44

Chemistry · 4 marks · −1

Solid Ba(NO3)2Ba(NO_3)_2 is gradually dissolved in a 1.0×10−41.0\times10^{-4} M Na2CO3Na_2CO_3 solution. At what concentration of Ba2+Ba^{2+} will a precipitate begin to form ? (KspK_{sp} for BaCO3=5.1×10−9BaCO_3 = 5.1\times10^{-9}).

  1. A.4.1×10−54.1\times10^{-5} M
  2. B.5.1×10−55.1\times10^{-5} M
  3. C.8.1×10−88.1\times10^{-8} M
  4. D.8.1×10−78.1\times10^{-7} M

Correct answer: B

Explanation: [Ba2+]=5.1×10−910−4[Ba^{2+}]=\frac{5.1\times10^{-9}}{10^{-4}}.

Question 45

Chemistry · 4 marks · −1

Which one of the following reactions of Xenon compounds is not feasible ?

  1. A.XeO3+6HF→XeF6+3H2OXeO_3 + 6HF \rightarrow XeF_6 + 3H_2O
  2. B.3XeF4+6H2O→2Xe+XeO3+12HF+1.5O23XeF_4 + 6H_2O \rightarrow 2Xe + XeO_3 + 12HF + 1.5O_2
  3. C.2XeF2+2H2O→2Xe+4HF+O22XeF_2 + 2H_2O \rightarrow 2Xe + 4HF + O_2
  4. D.XeF6+RbF→Rb[XeF7]XeF_6 + RbF \rightarrow Rb[XeF_7]

Correct answer: A

Explanation: The reverse (hydrolysis of XeF6XeF_6) is what occurs; XeO3XeO_3 is not fluorinated by HF.

Question 46

Chemistry · 4 marks · −1

Using MO theory predict which of the following species has the shortest bond length ?

  1. A.O22+O_2^{2+}
  2. B.O2+O_2^+
  3. C.O2−O_2^-
  4. D.O22−O_2^{2-}

Correct answer: A

Explanation: Bond orders 3, 2.5, 1.5, 1.

Question 47

Chemistry · 4 marks · −1

In context with the transition elements, which of the following statements is incorrect ?

  1. A.In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.
  2. B.In the highest oxidation states, the transition metal show basic character and form cationic complexes.
  3. C.In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding.
  4. D.Once the d5d^5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases.

Correct answer: B

Explanation: Highest oxidation states are acidic (e.g. Mn2O7Mn_2O_7, CrO3CrO_3).

Question 48

Chemistry · 4 marks · −1

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×1031.0\times10^3 ms−1^{-1} (Mass of proton = 1.67×10−271.67\times10^{-27} kg and h=6.63×10−34h = 6.63\times10^{-34} Js) :

  1. A.0.032 nm
  2. B.0.40 nm
  3. C.2.5 nm
  4. D.14.0 nm

Correct answer: B

Explanation: λ=hmv≈0.40\lambda=\frac{h}{mv}\approx0.40 nm.

Question 49

Chemistry · 4 marks · −1

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

  1. A.The solution formed is an ideal solution
  2. B.The solution is non-ideal, showing +ve deviation from Raoult's law.
  3. C.The solution is non-ideal, showing -ve deviation from Raoult's law.
  4. D.n-heptane shows +ve deviation while ethanol shows -ve deviation from Raoult's law.

Correct answer: B

Explanation: Heptane breaks ethanol's H-bonds: weaker mixed interactions, positive deviation.

Question 50

Chemistry · 4 marks · −1

The number of stereoisomers possible for a compound of the molecular formula CH3−CH=CH−CH(OH)−MeCH_3-CH=CH-CH(OH)-Me is :

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

Correct answer: C

Explanation: One C=C (cis/trans) and one stereocentre: 4.

Question 51

Chemistry · 4 marks · −1

The IUPAC name of neopentane is

  1. A.2-methylbutane
  2. B.2, 2-dimethylpropane
  3. C.2-methylpropane
  4. D.2,2-dimethylbutane

Correct answer: B

Explanation: C(CH3)4C(CH_3)_4.

Question 52

Chemistry · 4 marks · −1

The set representing the correct order of ionic radius is :

  1. A.Li+>Be2+>Na+>Mg2+Li^+ > Be^{2+} > Na^+ > Mg^{2+}
  2. B.Na+>Li+>Mg2+>Be2+Na^+ > Li^+ > Mg^{2+} > Be^{2+}
  3. C.Li+>Na+>Mg2+>Be2+Li^+ > Na^+ > Mg^{2+} > Be^{2+}
  4. D.Mg2+>Be2+>Li+>Na+Mg^{2+} > Be^{2+} > Li^+ > Na^+

Correct answer: B

Explanation: Na+Na^+ 102, Li+Li^+ 76, Mg2+Mg^{2+} 72, Be2+Be^{2+} 31 pm.

Question 53

Chemistry · 4 marks · −1

The two functional groups present in a typical carbohydrate are :

  1. A.-OH and -COOH
  2. B.-CHO and -COOH
  3. C.>C=O and -OH
  4. D.-OH and -CHO

Correct answer: C

Explanation: Polyhydroxy aldehydes or ketones: carbonyl and hydroxyl.

Question 54

Chemistry · 4 marks · −1

The bond dissociation energy of B - F in BF3BF_3 is 646 kJ mol−1^{-1} whereas that of C-F in CF4CF_4 is 515 kJ mol−1^{-1}. The correct reason for higher B-F bond dissociation energy as compared to that of C-F is :

  1. A.smaller size of B-atom as compared to that of C- atom
  2. B.stronger σ\sigma bond between B and F in BF3BF_3 as compared to that between C and F in CF4CF_4
  3. C.significant pπ−pπp\pi - p\pi interaction between B and F in BF3BF_3 whereas there is no possibility of such interaction between C and F in CF4CF_4.
  4. D.lower degree of pπ−pπp\pi - p\pi interaction between B and F in BF3BF_3 than that between C and F in CF4CF_4.

Correct answer: C

Explanation: Back-donation from F into the vacant B 2p orbital gives partial double-bond character.

Question 55

Chemistry · 4 marks · −1

In Cannizzaro reaction given below 2PhCHO→:OH−PhCH2OH+PhCOO−2PhCHO \xrightarrow{:OH^-} PhCH_2OH + PhCOO^- the slowest step is :

  1. A.the attack of :OH−:OH^- at the carboxyl group
  2. B.the transfer of hydride to the carbonyl group
  3. C.the abstraction of proton from the carboxylic group
  4. D.the deprotonation of PhCH2OHPhCH_2OH

Correct answer: B

Explanation: Hydride transfer is rate-determining.

Question 56

Chemistry · 4 marks · −1

Which of the following pairs represents linkage isomers ?

  1. A.[Cu(NH3)4][PtCl4][Cu(NH_3)_4][PtCl_4] and [Pt(NH3)4][CuCl4][Pt(NH_3)_4][CuCl_4]
  2. B.[Pd(PPh3)2(NCS)2][Pd(PPh_3)_2(NCS)_2] and [Pd(PPh3)2(SCN)2][Pd(PPh_3)_2(SCN)_2]
  3. C.[Co(NH3)5NO3]SO4[Co(NH_3)_5NO_3]SO_4 and [Co(NH3)5SO4]NO3[Co(NH_3)_5SO_4]NO_3
  4. D.[PtCl2(NH3)4]Br2[PtCl_2(NH_3)_4]Br_2 and [PtBr2(NH3)4]Cl2[PtBr_2(NH_3)_4]Cl_2

Correct answer: B

Explanation: Thiocyanate binds through N or S.

Question 57

Chemistry · 4 marks · −1

Buna-N synthetic rubber is a copolymer of :

  1. A.H2C=CH−C(Cl)=CH2H_2C=CH-C(Cl)=CH_2 and H2C=CH−CH=CH2H_2C=CH-CH=CH_2
  2. B.H2C=CH−CH=CH2H_2C=CH-CH=CH_2 and H5C6−CH=CH2H_5C_6-CH=CH_2
  3. C.H2C=CH−CNH_2C=CH-CN and H2C=CH−CH=CH2H_2C=CH-CH=CH_2
  4. D.H2C=CH−CNH_2C=CH-CN and H2C=CH−C(CH3)=CH2H_2C=CH-C(CH_3)=CH_2

Correct answer: C

Explanation: Acrylonitrile and 1,3-butadiene.

Question 58

Mathematics · 4 marks · −1

Let a, b, c be such that b(a+c)≠0b(a + c) \ne 0. If ∣aa+1a−1−bb+1b−1cc−1c+1∣+∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n+1b(−1)nc∣=0\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix} + \begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n+1}b & (-1)^nc \end{vmatrix} = 0, then the value of 'n' is

  1. A.zero
  2. B.any even integer
  3. C.any odd integer
  4. D.any integer

Correct answer: C

Explanation: The second determinant equals (−1)n(-1)^n times the first (after transposing and row swaps), so the sum vanishes only when n is odd.

Question 59

Mathematics · 4 marks · −1

If the mean deviation of number 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the d is equal to

  1. A.10.0
  2. B.20.0
  3. C.10.1
  4. D.20.2

Correct answer: C

Explanation: Mean 1+50d1+50d; MD =2d(1+⋯+50)101=2550d101=255⇒d=10.1=\frac{2d(1+\dots+50)}{101}=\frac{2550d}{101}=255\Rightarrow d=10.1.

Question 60

Mathematics · 4 marks · −1

If the roots of the equation bx2+cx+a=0bx^2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2+6bcx+2c23b^2x^2 + 6bcx + 2c^2 is

  1. A.greater than 4ab
  2. B.less than 4ab
  3. C.greater than -4ab
  4. D.less than -4ab

Correct answer: C

Explanation: Minimum value −c2-c^2 and c2<4abc^2<4ab, so the expression exceeds −4ab-4ab.

Question 61

Mathematics · 4 marks · −1

Let A and B denote the statements A: cos⁡α+cos⁡β+cos⁡γ=0\cos\alpha + \cos\beta + \cos\gamma = 0, B: sin⁡α+sin⁡β+sin⁡γ=0\sin\alpha + \sin\beta + \sin\gamma = 0. If cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}, then

  1. A.A is true and B is false
  2. B.A is false and B is true
  3. C.both A and B are true
  4. D.both A and B are false

Correct answer: C

Explanation: The condition is (∑cos⁡)2+(∑sin⁡)2=0(\sum\cos)^2+(\sum\sin)^2=0.

Question 62

Mathematics · 4 marks · −1

The lines p(p2+1)x−y+q=0p(p^2 + 1)x - y + q = 0 and (p2+1)2x+(p2+1)y+2q=0(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0 are perpendicular to a common line for

  1. A.no value of p
  2. B.exactly one value of p
  3. C.exactly two values of p
  4. D.more than two values of p

Correct answer: B

Explanation: They must be parallel: p(p2+1)=−(p2+1)⇒p=−1p(p^2+1)=-(p^2+1)\Rightarrow p=-1.

Question 63

Mathematics · 4 marks · −1

If A, B and C are three sets such that A∩B=A∩CA\cap B = A\cap C and A∪B=A∪CA\cup B = A\cup C, then

  1. A.A = B
  2. B.A = C
  3. C.B = C
  4. D.A∩B=ϕA\cap B = \phi

Correct answer: C

Explanation: Standard set identity: B = C.

Question 64

Mathematics · 4 marks · −1

If u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} are non-coplanar vectors and p, q are real numbers, then the equality [3u⃗ pv⃗ pw⃗]−[pv⃗ w⃗ qu⃗]−[2w⃗ qv⃗ qu⃗]=0[3\vec{u}\ p\vec{v}\ p\vec{w}] - [p\vec{v}\ \vec{w}\ q\vec{u}] - [2\vec{w}\ q\vec{v}\ q\vec{u}] = 0 holds for

  1. A.exactly one value of (p, q)
  2. B.exactly two values of (p, q)
  3. C.more than two but not all values of (p, q)
  4. D.all values of (p, q)

Correct answer: A

Explanation: (3p2−pq+2q2)[u⃗ v⃗ w⃗]=0(3p^2-pq+2q^2)[\vec u\,\vec v\,\vec w]=0 and 3p2−pq+2q2=03p^2-pq+2q^2=0 only at p=q=0p=q=0.

Question 65

Mathematics · 4 marks · −1

Let the line x−23=y−1−5=z+22\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2} lie in the plane x+3y−αz+β=0x + 3y - \alpha z + \beta = 0. Then (α,β)(\alpha, \beta) equals

  1. A.(6, -17)
  2. B.(-6, 7)
  3. C.(5, -15)
  4. D.(-5, 15)

Correct answer: B

Explanation: 3−15−2α=0⇒α=−63-15-2\alpha=0\Rightarrow\alpha=-6; point (2,1,−2)(2,1,-2): 2+3−12+β=0⇒β=72+3-12+\beta=0\Rightarrow\beta=7.

Question 66

Mathematics · 4 marks · −1

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is

  1. A.less than 500
  2. B.at least 500 but less than 750
  3. C.at least 750 but less than 1000
  4. D.at least 1000

Correct answer: D

Explanation: (64)⋅3⋅4!=1080\binom64\cdot3\cdot4!=1080.

Question 67

Mathematics · 4 marks · −1

∫0π[cot⁡x] dx\int_0^\pi[\cot x]\,dx, [⋅][\cdot] denotes the greatest integer function, is equal to

  1. A.π2\frac{\pi}{2}
  2. B.1
  3. C.−1-1
  4. D.−π2-\frac{\pi}{2}

Correct answer: D

Explanation: [cot⁡x]+[−cot⁡x]=−1[\cot x]+[-\cot x]=-1 a.e., so 2I=−π2I=-\pi.

Question 68

Mathematics · 4 marks · −1

For real x, let f(x)=x3+5x+1f(x) = x^3 + 5x + 1, then

  1. A.f is one-one but not onto R
  2. B.f is onto R but not one-one
  3. C.f is one-one and onto R
  4. D.f is neither one-one nor onto R

Correct answer: C

Explanation: Strictly increasing odd-degree polynomial.

Question 69

Mathematics · 4 marks · −1

In a binomial distribution B(n,p=14)B\left(n, p = \frac{1}{4}\right), if the probability of at least one success is greater than or equal to 910\frac{9}{10}, then n is greater than

  1. A.1log⁡104−log⁡103\frac{1}{\log_{10}4 - \log_{10}3}
  2. B.1log⁡104+log⁡103\frac{1}{\log_{10}4 + \log_{10}3}
  3. C.9log⁡104−log⁡103\frac{9}{\log_{10}4 - \log_{10}3}
  4. D.4log⁡104−log⁡103\frac{4}{\log_{10}4 - \log_{10}3}

Correct answer: A

Explanation: (34)n≤110⇒n≥1log⁡4−log⁡3\left(\frac34\right)^n\le\frac1{10}\Rightarrow n\ge\frac{1}{\log4-\log3}.

Question 70

Mathematics · 4 marks · −1

The projections of a vector on the three coordinate axis are 6, -3, 2 respectively. The direction cosines of the vector are

  1. A.6, -3, 2
  2. B.65,−35,25\frac{6}{5}, -\frac{3}{5}, \frac{2}{5}
  3. C.67,−37,27\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}
  4. D.−67,−37,27-\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}

Correct answer: C

Explanation: Magnitude 7.

Question 71

Mathematics · 4 marks · −1

If ∣Z−4Z∣=2\left|Z - \frac{4}{Z}\right| = 2, then the maximum value of |Z| is equal to

  1. A.3+1\sqrt{3} + 1
  2. B.5+1\sqrt{5} + 1
  3. C.2
  4. D.2+22 + \sqrt{2}

Correct answer: B

Explanation: ∣Z∣≤2+4∣Z∣⇒∣Z∣2−2∣Z∣−4≤0|Z|\le2+\frac{4}{|Z|}\Rightarrow|Z|^2-2|Z|-4\le0.

Question 72

Mathematics · 4 marks · −1

Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (-1, 0) is equal to 13\frac{1}{3}. Then the circumcentre of the triangle ABC is at the point

  1. A.(0, 0)
  2. B.(54,0)\left(\frac{5}{4}, 0\right)
  3. C.(52,0)\left(\frac{5}{2}, 0\right)
  4. D.(53,0)\left(\frac{5}{3}, 0\right)

Correct answer: B

Explanation: Apollonius circle x2+y2−52x+1=0x^2+y^2-\frac52x+1=0, centre (54,0)\left(\frac54,0\right).

Question 73

Mathematics · 4 marks · −1

The remainder left out when 82n−(62)2n+18^{2n} - (62)^{2n+1} is divided by 9 is

  1. A.0
  2. B.2
  3. C.7
  4. D.8

Correct answer: B

Explanation: 82n≡18^{2n}\equiv1, 62≡−162\equiv-1: 1−(−1)=21-(-1)=2.

Question 74

Mathematics · 4 marks · −1

The ellipse x2+4y2=4x^2 + 4y^2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn in inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is

  1. A.x2+16y2=16x^2 + 16y^2 = 16
  2. B.x2+12y2=16x^2 + 12y^2 = 16
  3. C.4x2+48y2=484x^2 + 48y^2 = 48
  4. D.4x2+64y2=484x^2 + 64y^2 = 48

Correct answer: B

Explanation: Rectangle corner (2, 1) on x216+y2b2=1\frac{x^2}{16}+\frac{y^2}{b^2}=1: b2=43b^2=\frac43.

Question 75

Mathematics · 4 marks · −1

The sum to the infinity of the series 1+23+632+1033+1434+…1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots is

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

Correct answer: B

Explanation: Arithmetico-geometric series: S=3S=3.

Question 76

Mathematics · 4 marks · −1

The differential equation which represents the family of curves y=c1ec2xy = c_1e^{c_2x}, where c1c_1 and c2c_2 are arbitrary constants is

  1. A.y′=y2y' = y^2
  2. B.y′′=y′yy'' = y'y
  3. C.yy′′=y′yy'' = y'
  4. D.yy′′=(y′)2yy'' = (y')^2

Correct answer: D

Explanation: y′=c2yy'=c_2y, y′′=c2y′y''=c_2y'.

Question 77

Mathematics · 4 marks · −1

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ...., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

  1. A.114\frac{1}{14}
  2. B.17\frac{1}{7}
  3. C.514\frac{5}{14}
  4. D.150\frac{1}{50}

Correct answer: A

Explanation: Product zero: 14 tickets; among them only 08 has digit sum 8.

Question 78

Mathematics · 4 marks · −1

Let y be an implicit function of x defined by x2x−2xxcot⁡y−1=0x^{2x} - 2x^x\cot y - 1 = 0. Then y'(1) equals

  1. A.−1-1
  2. B.1
  3. C.log 2
  4. D.−log⁡2-\log2

Correct answer: A

Explanation: At x = 1, cot⁡y=0\cot y=0; differentiating gives 2+2y′=02+2y'=0.

Question 79

Mathematics · 8 marks · −2

The area of the region bounded by the parabola (y−2)2=x−1(y - 2)^2 = x - 1, the tangent to the parabola at the point (2, 3) and the x-axis is

  1. A.3
  2. B.6
  3. C.9
  4. D.12

Correct answer: C

Explanation: Tangent x=2y−4x=2y-4; area =∫03[(y−2)2+1−(2y−4)]dy=∫03(y−3)2dy=9=\int_0^3[(y-2)^2+1-(2y-4)]dy=\int_0^3(y-3)^2dy=9.

Question 80

Mathematics · 8 marks · −2

Given P(x)=x4+ax3+bx2+cx+dP(x) = x^4 + ax^3 + bx^2 + cx + d such that x = 0 is the only real root of P'(x) = 0. If P(-1) < P(1), then in the interval [-1, 1]

  1. A.P(-1) is the minimum and P(1) is the maximum of P
  2. B.P(-1) is not minimum but P(1) is the maximum of P
  3. C.P(-1) is the minimum and P(1) is not the maximum of P
  4. D.neither P(-1) is the minimum nor P(1) is the maximum of P

Correct answer: B

Explanation: c=0c=0, P decreases on [-1, 0] and increases on [0, 1]; the minimum is P(0), the maximum P(1).

Question 81

Mathematics · 8 marks · −2

The shortest distance between the line y−x=1y - x = 1 and the curve x=y2x = y^2 is

  1. A.328\frac{3\sqrt{2}}{8}
  2. B.238\frac{2\sqrt{3}}{8}
  3. C.325\frac{3\sqrt{2}}{5}
  4. D.34\frac{\sqrt{3}}{4}

Correct answer: A

Explanation: Tangent slope 1 at (14,12)(\frac14,\frac12); distance 3/42\frac{3/4}{\sqrt2}.

Question 82

Mathematics · 8 marks · −2

Let f(x)=(x+1)2−1f(x) = (x + 1)^2 - 1, x≥−1x \ge -1. Statement-1 : The set {x:f(x)=f−1(x)}={0,−1}\{x : f(x) = f^{-1}(x)\} = \{0, -1\}. Statement-2 : f is a bijection.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. C.Statement-1 is true, Statement-2 is false
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: C

Explanation: f(x)=xf(x)=x gives x=0,−1x=0,-1; f is not onto R (co-domain unspecified), so Statement-2 is false.

Question 83

Mathematics · 8 marks · −2

Let f(x)=x∣x∣f(x) = x|x| and g(x)=sin⁡xg(x) = \sin x. Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. C.Statement-1 is true, Statement-2 is false
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: C

Explanation: (gof)′(x)=2∣x∣cos⁡x2(gof)'(x)=2|x|\cos x^2 is continuous, but the second derivative jumps from -2 to 2.

Question 84

Mathematics · 8 marks · −2

Statement-1 : The variance of first n even natural numbers is n2−14\frac{n^2 - 1}{4}. Statement-2 : The sum of first n natural numbers is n(n+1)2\frac{n(n+1)}{2} and the sum of squares of first n natural numbers is n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. C.Statement-1 is true, Statement-2 is false
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: D

Explanation: The variance is n2−13\frac{n^2-1}{3}, so Statement-1 is false.

Question 85

Mathematics · 4 marks · −1

Statement-1 : ∼(p↔∼q)\sim(p \leftrightarrow \sim q) is equivalent to p↔qp \leftrightarrow q. Statement-2 : ∼(p↔∼q)\sim(p \leftrightarrow \sim q) is a tautology.

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. C.Statement-1 is true, Statement-2 is false
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: C

Explanation: Truth table: equivalent to p↔qp\leftrightarrow q, not a tautology.

Question 86

Mathematics · 4 marks · −1

Let A be a 2 x 2 matrix. Statement-1 : adj(adj A) = A. Statement-2 : |adj A| = |A|

  1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. C.Statement-1 is true, Statement-2 is false
  4. D.Statement-1 is false, Statement-2 is true

Correct answer: B

Explanation: For n = 2: adj(adjA)=∣A∣0A=A\text{adj}(\text{adj}A)=|A|^0A=A and ∣adjA∣=∣A∣|\text{adj}A|=|A|; both true but independent.

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