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JEE Main Full Mock Test: Practice Test 03

Question paper with answer key. Selection Pro

Q1. A particle moves in a potential U(x)=ax^4-bx^2, where a,b>0. The angular frequency of small oscillations about either stable equilibrium point is:
  1. sqrt(2b/m)
  2. sqrt(4b/m)
  3. sqrt(b/m)
  4. sqrt(8b/m)

Answer: (B) sqrt(4b/m)

Equilibrium points satisfy dU/dx=4ax^3-2bx=0, giving x=±sqrt(b/2a) as stable points. U''=12ax^2-2b. At either stable point, U''=12a(b/2a)-2b=4b. Hence omega=sqrt(U''/m)=sqrt(4b/m).

Q2. A block of mass m is attached to a spring of force constant k on a smooth horizontal surface. The spring is initially compressed by x0. A constant horizontal force F is suddenly applied in the direction of compression and maintained. The maximum compression from the natural length is:
  1. x0+F/k
  2. F/k+x0/sqrt(2)
  3. F/k+sqrt(x0^2+F^2/k^2)
  4. sqrt(x0^2+2Fx0/k+F^2/k^2)

Answer: (C) F/k+sqrt(x0^2+F^2/k^2)

Taking the natural length as zero, initial energy is (1/2)kx0^2. At maximum compression X, velocity is zero and energy is (1/2)kX^2-FX. Equating gives kX^2-2FX-kx0^2=0. The positive root is X=F/k+sqrt(x0^2+F^2/k^2).

Q3. A satellite is in a circular orbit of radius r around Earth. An impulse changes its speed instantaneously to sqrt(3/2) times the circular speed in the same direction. The resulting orbit has eccentricity:
  1. 1/2
  2. 1/3
  3. sqrt(2)/2
  4. 1

Answer: (A) 1/2

At the impulse point the velocity is tangential, so it becomes an apsis. Using specific energy epsilon=v^2/2-GM/r=(3/2)(GM/r)/2-GM/r=-GM/(4r), hence a=r/(1/2)=2r. At the apoapsis, r=a(1+e), so r=2r(1+e) cannot hold; instead the speed is above circular and the point is periapsis: r=a(1-e)=2r(1-e), giving e=1/2.

Q4. A uniform disc of mass M and radius R rolls without slipping down an incline through a vertical height h. Its speed at the bottom is:
  1. sqrt(2gh)
  2. sqrt(gh)
  3. sqrt(4gh/3)
  4. sqrt(gh/2)

Answer: (C) sqrt(4gh/3)

For a disc I=(1/2)MR^2 and omega=v/R. Conservation of energy gives Mgh=(1/2)Mv^2+(1/2)(1/2 MR^2)(v^2/R^2)=3Mv^2/4. Therefore v=sqrt(4gh/3).

Q5. A conducting sphere of radius R is charged to potential V. It is then connected to an uncharged conducting sphere of radius 2R by a thin wire and separated after equilibrium. The ratio of final electrostatic energies of the spheres, U1:U2, is:
  1. 1:2
  2. 1:4
  3. 1:1
  4. 2:1

Answer: (B) 1:4

At equilibrium both spheres have the same potential. Since Q=CV and capacitance of an isolated sphere is proportional to radius, Q1:Q2=R:2R=1:2. Energy U=Q^2/(2C)=(1/2)CV^2, so U1:U2=C1:C2=1:2, not 1:4.

Q6. Two identical capacitors C are charged to potentials V and 2V separately. They are then connected in parallel with positive plates together. The fraction of initial total electrostatic energy lost is:
  1. 1/10
  2. 1/5
  3. 1/6
  4. 1/3

Answer: (A) 1/10

Initial energy=(1/2)C(V^2+4V^2)=5CV^2/2. Common potential=(V+2V)/2=3V/2. Final energy=(1/2)(2C)(9V^2/4)=9CV^2/4. Energy lost=CV^2/4, so fraction=(CV^2/4)/(5CV^2/2)=1/10.

Q7. A charged particle enters a uniform magnetic field with velocity making an angle theta with the field. The radius of its helical path is R and pitch is p. The ratio p/(2pi R) equals:
  1. tan theta
  2. cot theta
  3. sin theta
  4. cos theta

Answer: (B) cot theta

R=mv sin(theta)/(qB), while pitch=(v cos(theta))(2pi m/qB). Hence p/(2pi R)=cot(theta).

Q8. A long solenoid carries current I. Its number of turns per unit length is n. If the current is doubled while the solenoid length is halved and total number of turns remains unchanged, the magnetic field inside becomes:
  1. 2B
  2. 4B
  3. B/2
  4. B

Answer: (B) 4B

B=mu0 nI and n=N/L. Halving L doubles n, while doubling I also doubles I. Thus B becomes four times.

Q9. A conducting rod of length l rotates with angular speed omega about one end in a uniform magnetic field B perpendicular to its plane of rotation. The potential difference between its ends is:
  1. Bl^2 omega
  2. (1/2)Bl^2 omega
  3. Bl omega
  4. 2Bl^2 omega

Answer: (B) (1/2)Bl^2 omega

An element dr at distance r has motional emf dE=B(omega r)dr. Integrating from 0 to l gives E=(1/2)B omega l^2.

Q10. In a series LCR circuit at resonance, the voltage across the inductor is 100 V and that across the resistor is 20 V. The voltage across the capacitor is:
  1. 20 V
  2. 80 V
  3. 100 V
  4. 120 V

Answer: (C) 100 V

At resonance XL=XC, so VL and VC are equal in magnitude and opposite in phase. Therefore VC=100 V.

Q11. A particle executes SHM with amplitude A. At what displacement from the mean position are its kinetic and potential energies equal?
  1. A/2
  2. A/sqrt(2)
  3. A/sqrt(3)
  4. A/4

Answer: (B) A/sqrt(2)

Potential energy is (1/2)kx^2 and total energy is (1/2)kA^2. Equality means kx^2/2 = k(A^2-x^2)/2, so 2x^2=A^2 and |x|=A/sqrt(2).

Q12. A gas follows the process PV^gamma=constant. If its volume is doubled, the ratio of final to initial temperature is:
  1. 2^(gamma-1)
  2. 2^(1-gamma)
  3. 2^gamma
  4. 1/2^gamma

Answer: (B) 2^(1-gamma)

Using PV^gamma=constant and PV=nRT, TV^(gamma-1)=constant. Thus T2/T1=(V1/V2)^(gamma-1)=2^(1-gamma).

Q13. One mole of an ideal gas expands reversibly and adiabatically from volume V to 8V. If gamma=5/3, the ratio of final to initial temperature is:
  1. 1/4
  2. 1/2
  3. 1/8
  4. 4

Answer: (A) 1/4

TV^(gamma-1)=constant. gamma-1=2/3, so T2/T1=(1/8)^(2/3)=1/4.

Q14. A monochromatic beam of wavelength lambda is incident on a single slit of width a. If the first minimum occurs at 30 degrees, then a/lambda is:
  1. 1
  2. 2
  3. 1/2
  4. sqrt(3)

Answer: (B) 2

For the first minimum, a sin theta=lambda. At theta=30 degrees, sin theta=1/2, hence a/lambda=2.

Q15. In Young's double-slit experiment, the intensity at a point is maximum when the path difference is delta. If the intensity at that point is one-fourth of maximum for equal-intensity slits, the possible phase difference is:
  1. pi/3
  2. pi/2
  3. 2pi/3
  4. pi

Answer: (C) 2pi/3

For equal slits, I=Imax cos^2(phi/2). I/Imax=1/4 gives |cos(phi/2)|=1/2, so the nonzero principal phase difference is 2pi/3.

Q16. An electron and a proton are accelerated through the same potential difference. The ratio of their de Broglie wavelengths lambda_e/lambda_p is approximately:
  1. 1
  2. sqrt(mp/me)
  3. sqrt(me/mp)
  4. mp/me

Answer: (B) sqrt(mp/me)

For a particle accelerated through V, lambda=h/sqrt(2mqV), so lambda is proportional to 1/sqrt(m). Therefore lambda_e/lambda_p=sqrt(mp/me).

Q17. A radioactive sample has decay constant lambda. The time required for its activity to fall from A0 to A0/16 is:
  1. ln2/lambda
  2. 2ln2/lambda
  3. 4ln2/lambda
  4. 16/lambda

Answer: (C) 4ln2/lambda

A/A0=e^(-lambda t)=1/16=2^-4. Thus lambda t=4ln2 and t=4ln2/lambda.

Q18. A semiconductor has an energy gap Eg. The wavelength corresponding to a photon just capable of exciting an electron across the gap is:
  1. hc/Eg
  2. Eg/hc
  3. hEg/c
  4. cEg/h

Answer: (A) hc/Eg

Threshold photon energy equals Eg. Since E=hc/lambda, lambda=hc/Eg.

Q19. A wire of resistance R is stretched uniformly to twice its original length without change of volume. Its new resistance is:
  1. 2R
  2. 4R
  3. R/2
  4. R/4

Answer: (B) 4R

Volume constant means area becomes A/2. Since R=rho L/A, new resistance=rho(2L)/(A/2)=4R.

Q20. A dipole of moment p is rotated quasistatically in a uniform electric field E from a direction perpendicular to the field to a direction making 60 degrees with the field. The work done by the external agent is:
  1. -pE/2
  2. pE/2
  3. pE
  4. -pE

Answer: (A) -pE/2

U=-pE cos theta. Initially theta=90 degrees, Ui=0. Finally theta=60 degrees, Uf=-pE/2. For quasistatic rotation, external work equals Delta U=-pE/2.

Q21. A particle is projected vertically upward with speed u from Earth's surface. Neglecting air resistance and assuming g constant, its maximum height is h. If the initial speed is doubled, the maximum height becomes:
  1. 2h
  2. 3h
  3. 4h
  4. h/2

Answer: (C) 4h

Maximum height is u^2/(2g), so it is proportional to u^2. Doubling u makes the height four times.

Q22. A body of mass m moving with speed v collides elastically head-on with an identical body initially at rest. The speed of the first body after collision is:
  1. v
  2. v/2
  3. 0
  4. 2v

Answer: (C) 0

For a one-dimensional perfectly elastic collision between equal masses, the moving body transfers its velocity completely to the stationary body. Hence its final speed is zero.

Q23. A liquid rises to height h in a capillary of radius r. If the same capillary is taken to a place where g becomes g/4, neglecting changes in surface tension, the new capillary rise is:
  1. h/4
  2. h/2
  3. 2h
  4. 4h

Answer: (D) 4h

Capillary rise h=2T cos(theta)/(rho g r), hence h is inversely proportional to g. Reducing g to g/4 makes h four times.

Q24. A convex lens of focal length 20 cm forms a real image twice the size of the object. The object distance is:
  1. 10 cm
  2. 20 cm
  3. 30 cm
  4. 40 cm

Answer: (C) 30 cm

Magnification magnitude 2 gives v=2u. Lens formula 1/f=1/u+1/v=1/u+1/(2u)=3/(2u). Thus u=3f/2=30 cm.

Q25. A satellite's orbital period around Earth is T at radius R. Its period at radius 4R is:
  1. 2T
  2. 4T
  3. 8T
  4. T/2

Answer: (C) 8T

Kepler's third law gives T proportional to r^(3/2). Hence T2/T1=4^(3/2)=8.

Q26. For a weak monoprotic acid HA of concentration C, if its degree of ionization is alpha, the exact expression for Ka is:
  1. C alpha
  2. C alpha^2/(1-alpha)
  3. C alpha^2
  4. C(1-alpha)/alpha^2

Answer: (B) C alpha^2/(1-alpha)

At equilibrium [H+]=C alpha, [A-]=C alpha and [HA]=C(1-alpha). Therefore Ka=(C alpha)^2/[C(1-alpha)]=C alpha^2/(1-alpha).

Q27. A buffer contains equal concentrations of a weak acid HA and its salt NaA. If 0.01 mol HCl is added to 1 L of this buffer initially containing 0.10 mol each of HA and A-, the pH becomes (pKa=4.76):
  1. 4.76
  2. 4.67
  3. 4.85
  4. 5.00

Answer: (B) 4.67

H+ converts A- to HA. New amounts are A-=0.09 and HA=0.11. pH=pKa+log(0.09/0.11)=4.76+log(0.818)=approximately 4.67.

Q28. For a reaction with rate law rate=k[A]^2[B], the unit of k when concentration is in mol L^-1 and time in seconds is:
  1. L mol^-1 s^-1
  2. L^2 mol^-2 s^-1
  3. mol L^-1 s^-1
  4. s^-1

Answer: (B) L^2 mol^-2 s^-1

Overall order is 3. k=rate/[A]^2[B], giving units (mol L^-1 s^-1)/(mol L^-1)^3=L^2 mol^-2 s^-1.

Q29. For a first-order reaction, the time required for 90% completion is related to half-life t1/2 by:
  1. t1/2
  2. 2t1/2
  3. 3.322t1/2
  4. 10t1/2

Answer: (C) 3.322t1/2

For 90% completion, 10% remains. t=(2.303/k) and t1/2=0.693/k, so t/t1/2=2.303/0.693=3.322.

Q30. The van't Hoff factor for complete dissociation of Al2(SO4)3 in dilute solution is:
  1. 2
  2. 3
  3. 4
  4. 5

Answer: (D) 5

Al2(SO4)3 dissociates into 2 Al3+ and 3 SO4^2-, giving 5 ions per formula unit. Hence i=5 for complete dissociation.

Q31. For an ideal solution, which thermodynamic quantity is necessarily zero on mixing?
  1. Delta Gmix
  2. Delta Smix
  3. Delta Hmix
  4. Delta Vmix

Answer: (C) Delta Hmix

For an ideal solution, intermolecular interactions are effectively unchanged on mixing, so Delta Hmix=0. Entropy and Gibbs free energy of mixing are not zero.

Q32. Which complex has the greatest number of unpaired electrons among the following?
  1. [Fe(CN)6]^4-
  2. [FeF6]^3-
  3. [Co(NH3)6]^3+
  4. [Ni(CN)4]^2-

Answer: (B) [FeF6]^3-

[FeF6]^3- contains Fe3+, d5, and F- is weak field, giving high-spin t2g^3 eg^2 with five unpaired electrons. The others have fewer.

Q33. The CFSE of a low-spin d5 octahedral complex, excluding pairing energy, is:
  1. -1.0 Delta_o
  2. -2.0 Delta_o
  3. -1.5 Delta_o
  4. 0

Answer: (B) -2.0 Delta_o

Low-spin d5 has t2g^5 eg^0. CFSE=5(-0.4 Delta_o)=-2.0 Delta_o, excluding pairing energy.

Q34. Which species has the highest bond order according to molecular orbital theory?
  1. O2
  2. O2+
  3. O2-
  4. O2^2-

Answer: (B) O2+

O2 has bond order 2. Removing one electron from antibonding pi* in O2+ increases bond order to 2.5. Adding electrons decreases it.

Q35. The shape of XeF4 is best described as:
  1. Tetrahedral
  2. Square planar
  3. See-saw
  4. Trigonal bipyramidal

Answer: (B) Square planar

XeF4 has six electron pairs around Xe: four bonding and two lone pairs. The lone pairs occupy opposite axial sites of an octahedral electron arrangement, leaving a square-planar molecular geometry.

Q36. Which compound is expected to have the highest boiling point among the following?
  1. n-pentane
  2. isopentane
  3. neopentane
  4. n-hexane

Answer: (D) n-hexane

For comparable nonpolar hydrocarbons, boiling point generally increases with molecular mass and surface area. n-Hexane has greater molar mass and dispersion forces than the pentane isomers.

Q37. The major product when 2-bromobutane is heated with alcoholic KOH is predominantly:
  1. 1-butene
  2. cis-2-butene
  3. trans-2-butene
  4. 2-butanol

Answer: (C) trans-2-butene

Alcoholic KOH favors elimination. Zaitsev product is 2-butene, and among its stereoisomers the more stable trans-2-butene is predominant under thermodynamic control.

Q38. Which reagent can distinguish between ethanal and propanone most directly?
  1. 2,4-DNP
  2. Iodoform test
  3. Tollens' reagent
  4. NaHSO3

Answer: (C) Tollens' reagent

Ethanal is an aldehyde and gives a positive Tollens' test, while propanone is a ketone and does not. Iodoform is positive for propanone too, so it does not distinguish these two as directly.

Q39. The product of Reimer-Tiemann reaction of phenol is predominantly:
  1. Benzoic acid
  2. o-Hydroxybenzaldehyde
  3. p-Hydroxybenzoic acid
  4. Anisole

Answer: (B) o-Hydroxybenzaldehyde

Phenoxide reacts with dichlorocarbene generated from CHCl3/NaOH, followed by hydrolysis, giving predominantly o-hydroxybenzaldehyde (salicylaldehyde).

Q40. Aniline is less basic than cyclohexylamine primarily because in aniline the nitrogen lone pair is:
  1. localized on nitrogen
  2. involved in resonance with the benzene ring
  3. used in sigma bonding to hydrogen
  4. absent

Answer: (B) involved in resonance with the benzene ring

The lone pair on aniline nitrogen is delocalized into the benzene ring through resonance, reducing its availability for protonation. Cyclohexylamine has a localized lone pair.

Q41. Which statement about benzene is correct?
  1. All C-C bonds alternate between single and double
  2. All six C-C bonds have equal length
  3. It readily undergoes addition rather than substitution
  4. It contains only localized pi electrons

Answer: (B) All six C-C bonds have equal length

Benzene has six delocalized pi electrons distributed over the ring, making all C-C bonds equivalent with intermediate bond length.

Q42. The oxidation state of chromium in K2Cr2O7 is:
  1. +3
  2. +4
  3. +6
  4. +7

Answer: (C) +6

2(+1)+2x+7(-2)=0 gives 2x=12, so x=+6.

Q43. Which oxide is amphoteric?
  1. MgO
  2. Al2O3
  3. SO2
  4. K2O

Answer: (B) Al2O3

Al2O3 reacts with both acids and strong bases, so it is amphoteric. MgO and K2O are basic while SO2 is acidic.

Q44. For the cell Zn|Zn2+(1 M)||Cu2+(0.01 M)|Cu, the cell potential at 298 K is closest to (E°cell=1.10 V):
  1. 1.04 V
  2. 1.16 V
  3. 1.10 V
  4. 0.98 V

Answer: (A) 1.04 V

For Zn oxidation and Cu2+ reduction, Q=[Zn2+]/[Cu2+]=100. E=E°-(0.0591/2)log100=1.10-0.0591=1.0409 V.

Q45. Which colligative property is most suitable for determining the molar mass of a protein in aqueous solution?
  1. Elevation in boiling point
  2. Depression in freezing point
  3. Osmotic pressure
  4. Relative lowering of vapor pressure

Answer: (C) Osmotic pressure

Proteins have very large molar masses and can decompose on heating or freezing. Osmotic pressure can be measured at relatively low temperatures and is highly sensitive to molar concentration.

Q46. The major product of hydration of propene in acidic medium is:
  1. 1-propanol
  2. 2-propanol
  3. propanal
  4. propane

Answer: (B) 2-propanol

Acid-catalyzed hydration proceeds by Markovnikov addition through the more stable secondary carbocation, producing 2-propanol.

Q47. Which of the following has the highest acidic strength?
  1. Ethanol
  2. Phenol
  3. p-Nitrophenol
  4. p-Cresol

Answer: (C) p-Nitrophenol

The nitro group strongly withdraws electron density and stabilizes the phenoxide ion by -I and -M effects, making p-nitrophenol more acidic than phenol and p-cresol.

Q48. The number of sigma and pi bonds in benzene (C6H6) is respectively:
  1. 6 and 6
  2. 12 and 3
  3. 9 and 3
  4. 12 and 6

Answer: (B) 12 and 3

There are six C-C sigma bonds and six C-H sigma bonds, giving 12 sigma bonds. The aromatic ring contains three pi bonds in the conventional bond-counting representation.

Q49. A compound gives a positive iodoform test and also reacts with sodium metal with liberation of hydrogen. Which compound can satisfy both tests?
  1. Methanol
  2. Ethanol
  3. Propanal
  4. Benzaldehyde

Answer: (B) Ethanol

Ethanol gives the iodoform test because it is oxidized to ethanal under the test conditions, and its O-H group reacts with sodium to release hydrogen.

Q50. For a reaction with Delta H=+40 kJ mol^-1 and Delta S=+100 J K^-1 mol^-1, the reaction becomes spontaneous when temperature is:
  1. below 200 K
  2. above 400 K
  3. exactly 300 K only
  4. at all temperatures

Answer: (B) above 400 K

Spontaneity requires Delta G=Delta H-TDelta S<0. Thus T>Delta H/Delta S=40000/100=400 K.

Q51. If f(x)=x^x for x>0, then f''(1) is:
  1. 0
  2. 1
  3. 2
  4. -1

Answer: (B) 1

ln f=x ln x. Thus f'/f=ln x+1. Differentiating, f''/f=(ln x+1)^2+1/x. At x=1, f=1, giving f''(1)=2.

Q52. The maximum value of x(1-x)^3 for 0
  1. 27/256
  2. 1/4
  3. 3/16
  4. 1/8

Answer: (A) 27/256

Let f=x(1-x)^3. f'=(1-x)^2(1-4x), so the interior maximum occurs at x=1/4. Hence fmax=(1/4)(3/4)^3=27/256.

Q53. Evaluate integral from 0 to pi/2 of ln(sin x) dx.
  1. 0
  2. -(pi/2)ln2
  3. -pi ln2
  4. (pi/2)ln2

Answer: (B) -(pi/2)ln2

The standard integral is ∫0^(pi/2) ln(sin x) dx=-(pi/2)ln2. It follows by using the substitution x->pi/2-x and adding the resulting integral.

Q54. If z satisfies |z-1|=|z+1| and |z-2i|=2, then the number of possible z is:
  1. 0
  2. 1
  3. 2
  4. 4

Answer: (C) 2

The first condition gives Re(z)=0, the imaginary axis. Intersecting it with the circle centered at 2i with radius 2 gives y=0 or 4, hence two points z=0 and z=4i.

Q55. If the roots of x^2-2x+lambda=0 are both positive and distinct, then lambda must satisfy:
  1. 0
  2. 1
  3. 0
  4. lambda>4

Answer: (A) 0

For roots to be real and distinct, discriminant 4-4lambda>0, so lambda<1. For both roots positive, product lambda>0 and sum is 2>0. Hence 0

Q56. The determinant of the matrix with entries a_ij=i+j for i,j=1,2,3 is:
  1. 0
  2. 1
  3. 2
  4. -1

Answer: (A) 0

The matrix is [[2,3,4],[3,4,5],[4,5,6]]. Its rows are linearly dependent because row 2-row 1=row 3-row 2, so the determinant is zero.

Q57. If A is a 3x3 matrix satisfying A^2=A and rank(A)=2, then det(I+A) is:
  1. 1
  2. 2
  3. 4
  4. 8

Answer: (C) 4

An idempotent matrix has eigenvalues only 0 or 1. Rank 2 means two eigenvalues are 1 and one is 0. Therefore det(I+A)=2*2*1=4.

Q58. The number of solutions of sin x = x/3 in the interval [-pi,pi] is:
  1. 1
  2. 2
  3. 3
  4. 4

Answer: (C) 3

Both sides are odd, so roots occur symmetrically. x=0 is one root. On (0,pi), sin x starts above x/3 and ends below it, giving one additional positive root; symmetry gives one negative root. Total=3.

Q59. If a,b,c are positive and abc=1, the minimum value of (1+a)(1+b)(1+c) is:
  1. 4
  2. 6
  3. 8
  4. 9

Answer: (C) 8

Expanding gives 2+(a+b+c)+(ab+bc+ca). By AM-GM, a+b+c>=3 and ab+bc+ca>=3 because abc=1. Hence expression>=8, with equality at a=b=c=1.

Q60. The area enclosed between y=x^2 and y=2x+3 is:
  1. 32/3
  2. 16/3
  3. 20/3
  4. 8

Answer: (A) 32/3

Intersections satisfy x^2=2x+3, giving x=-1,3. Area=∫[-1,3](2x+3-x^2)dx=32/3.

Q61. The solution of the differential equation dy/dx + y tan x = sin x is:
  1. y sec x=-ln|cos x|+C
  2. y cos x=sin x+C
  3. y sec x=ln|cos x|+C
  4. y sin x=cos x+C

Answer: (A) y sec x=-ln|cos x|+C

The integrating factor is sec x. Hence d(y sec x)/dx=sin x sec x=tan x. Integrating gives y sec x=-ln|cos x|+C.

Q62. If the vectors a=(1,2,3), b=(2,-1,1), c=(lambda,1,2) are coplanar, then lambda is:
  1. -3/5
  2. -2
  3. 2
  4. 3

Answer: (A) -3/5

Coplanarity requires the scalar triple product to vanish. det[[1,2,3],[2,-1,1],[lambda,1,2]]=0 gives -5lambda-3=0, hence lambda=-3/5.

Q63. The distance of point (1,2,3) from the plane 2x-y+2z-9=0 is:
  1. 1/3
  2. 2/3
  3. 1
  4. 2

Answer: (C) 1

Distance=|2(1)-2+2(3)-9|/sqrt(4+1+4)=3/3=1.

Q64. A circle touches both coordinate axes in the first quadrant and passes through (4,2). Its radius is:
  1. 1
  2. 2
  3. 3
  4. 4

Answer: (B) 2

A circle tangent to both positive axes has center (r,r). Passing through (4,2): (4-r)^2+(2-r)^2=r^2. This gives r^2-12r+20=0, so r=2 or 10. The first-quadrant tangent circle through the point has r=2.

Q65. The eccentricity of the ellipse x^2/25 + y^2/9 =1 is:
  1. 3/5
  2. 4/5
  3. 5/4
  4. 2/5

Answer: (B) 4/5

a=5,b=3, so e=sqrt(1-b^2/a^2)=sqrt(1-9/25)=4/5.

Q66. The number of integers n for which n^2+5n+6 is divisible by n+2 is:
  1. 0
  2. 1
  3. 2
  4. infinitely many

Answer: (D) infinitely many

n^2+5n+6=(n+2)(n+3). For every integer n except n=-2, the expression is divisible by n+2. At n=-2 the divisor is zero and divisibility is undefined. Hence infinitely many integers.

Q67. If a,b are roots of x^2-5x+3=0, then a^3+b^3 is:
  1. 35
  2. 80
  3. 110
  4. 125

Answer: (B) 80

a+b=5 and ab=3. Therefore a^3+b^3=(a+b)^3-3ab(a+b)=125-45=80.

Q68. The coefficient of x^5 in (1+x)^10(1-x)^6 is:
  1. -56
  2. -48
  3. 48
  4. 56

Answer: (A) -56

Write the product as (1-x^2)^6(1+x)^4. The x^5 coefficient comes from x^1 in (1+x)^4 times x^4 in (1-x^2)^6, plus x^5 in (1+x)^4: 4(-15)+4=-56.

Q69. Two fair dice are thrown. The probability that the sum is prime and the product is even is:
  1. 5/18
  2. 1/3
  3. 7/18
  4. 1/2

Answer: (C) 7/18

Prime sums are 2,3,5,7,11. Counting outcomes with an even product: sum 3 gives 2, sum 5 gives 4, sum 7 gives 6, sum 11 gives 2; sum 2 gives none. Favorable=14 of 36=7/18.

Q70. If X is a binomial random variable with n=10 and p=1/2, its variance is:
  1. 2.5
  2. 5
  3. 10
  4. 1.25

Answer: (A) 2.5

Variance of Binomial(n,p) is npq=10(1/2)(1/2)=2.5.

Q71. The minimum value of x^2+4/x for x>0 is:
  1. 3*2^(2/3)
  2. 4
  3. 5
  4. 6

Answer: (A) 3*2^(2/3)

Differentiating gives 2x-4/x^2=0, so x^3=2. At x=2^(1/3), the value is 2^(2/3)+4/2^(1/3)=3*2^(2/3).

Q72. If log_2(x)+log_2(x-2)=3, then x is:
  1. 2
  2. 4
  3. -2
  4. 1

Answer: (B) 4

Domain requires x>2. Combining logs gives log2[x(x-2)]=3, so x^2-2x=8. Thus x=4 or -2; only x=4 satisfies the domain.

Q73. The number of distinct real roots of |x^2-5x+6|=2 is:
  1. 2
  2. 3
  3. 4
  4. 1

Answer: (A) 2

For x^2-5x+6=2, we get x^2-5x+4=0 with roots 1 and 4. For x^2-5x+6=-2, the discriminant is 25-32<0. Hence there are exactly 2 real roots.

Q74. The number of onto functions from a 3-element set to a 2-element set is:
  1. 4
  2. 6
  3. 8
  4. 2

Answer: (B) 6

Total functions are 2^3=8. Two constant functions are not onto. Therefore onto functions=8-2=6.

Q75. If the sum of first n terms of an AP is S_n=3n^2+2n, then the 10th term is:
  1. 57
  2. 59
  3. 61
  4. 63

Answer: (B) 59

The nth term is a_n=S_n-S_{n-1}. S_10=320 and S_9=261, so a_10=59.