JEE MAINS Previous year solwed question paper 2019 - phase 2 -evening shift online -

Physics Chemistry Mathematics - Free pdf download link 2019 JEE Mains 

PART–A : PHYSICS

Physics

1. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when :
   (1) R = 1000r
   (2) R = r
   (3) R = 2r
   (4) R = 0.001r
   Answer (2)
   Sol. For maximum power in external resistance, Internal resistance = External resistance ⇒ R = r

2. A body of mass m1 moving with an unknown velocity of v1i, undergoes a collinear collision with a body of mass m2 moving with a velocity v2i. After collision, m1 and m2 move with velocities of v3i and v4i, respectively.
   If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :
   (1) v4 - v2/2
   (2) v4 - v2
   (3) v4 + v2
   (4) v4 - v2/2
   Answer (2)
   Sol. m1v1 + m2v2 = m1v3 + m2v4
   m1v1 + 0.5m1v2 = 0.5m1v1 + 0.5m1v4
   v1 = v4 - v2

3. In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be :
   (Radius of the Earth = 6.4 × 10^6 m).
   (1) 20 m
   (2) 51 m
   (3) 32 m
   (4) 40 m
   Answer (3)
   Sol. sqrt(2 × 70 × RE) + sqrt(2 × hR × RE) = 50 × 10^3
   Putting RE = 6.4 × 10^6 m and solving we get
   hR = 32 m

4. The electric field in a region is given by E = (Ax + B)i, where E is in NC^-1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = -5 is V2, then V1 - V2 is :
   (1) 180 V
   (2) -520 V
   (3) 320 V
   (4) -48 V
   Answer (1)
   Sol. dV = -E·dr = -(Ax + B)dx
   ∫V2^V1 dV = ∫ - (Ax + B)dx
   V1 - V2 = (-A x^2/2 - Bx)|{-5}^{1}
   = (-A/2 - B) + (A/2 25 + B(-5))
   = 12A - 6B = 240 - 60 = 180 V

5. A parallel plate capacitor has 1 μF capacitance. One of its two plates is given +2 μC charge and the other plate, +4 μC charge. The potential difference developed across the capacitor is :
   (1) 5 V
   (2) 1 V
   (3) 3 V
   (4) 2 V
   Answer (2)
   Sol. V = q/C = 1 μC / 1 μF

6. The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by :
   (1) a d c b
   (2) a d b c
   (3) d a b c
   Answer (3)

7. Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their mid-point P, at angle θ = 45° with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? (d is much larger than the dimensions of the dipole)
   (1) 0
   (2) sqrt(2)(μ0/4π) M/(d/2)^3 × qv
   (3) (μ0/4π) 2M/(d/2)^3 × qv
   (4) (μ0/4π) M/(d/2)^3 × qv
   Answer (1)
   Sol. Fm = q(V × B)
   B = Bx + By
   Since My = 2Mx
   ⇒ |Bx| = |By|
   Bnet is parallel to V
   ⇒ F = 0

8. A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations.
   The time it will take to drop to 1/1000 of the original amplitude is close to:
   (1) 100 s
   (2) 10 s
   (3) 50 s
   (4) 20 s
   Answer (4)
   Sol. Time for 10 oscillations = 10/5 = 2 s
   A = A0 e^{-kt}
   1/2 = e^{-2k} ⇒ ln 2 = 2k
   10^{-3} = e^{-kt} ⇒ 3 ln 10 = kt
   t = 3 ln 10 / k = (3 ln 10 / ln 2) × 2
   = 6 × 2.3 / 0.69 ≈ 20 s

9. If Surface tension (S), Moment of Inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be:
   (1) S^{1/2} I^{3/2} h^{-1}
   (2) S^{3/2} I^{1/2} h^0
   (3) S^{1/2} I^{1/2} h^{-1}
   (4) S^{1/2} I^{1/2} h^0
   Answer (4)
   Sol. [p] = sqrt(IS)

10. A circuit connected to an ac source of emf e = e0 sin(100t) with t in seconds, gives a phase difference of π/4 between the emf e and current i. Which of the following circuits will exhibit this?
    (1) RL circuit with R = 1 kΩ and L = 10 mH
    (2) RL circuit with R = 1 kΩ and L = 1 mH
    (3) RC circuit with R = 1 kΩ and C = 10 μF
    (4) RC circuit with R = 1 kΩ and C = 1 μF
    Answer (3)
    Sol. As φ = π/4, XC = R

11. The magnetic field of an electromagnetic wave is given by:
    B = 1.6 × 10^{-6} cos(2 × 10^7 z + 6 × 10^15 t)(2i + j) Wb/m^2
    The associated electric field will be:
    (1) E = 4.8 × 10^2 cos(2 × 10^7 z + 6 × 10^15 t)(-i + 2j) V/m
    (2) E = 4.8 × 10^2 cos(2 × 10^7 z - 6 × 10^15 t)(2i + j) V/m
    (3) E = 4.8 × 10^2 cos(2 × 10^7 z - 6 × 10^15 t)(-2j + i) V/m
    (4) E = 4.8 × 10^2 cos(2 × 10^7 z + 6 × 10^15 t)(i - 2j) V/m
    Answer (1)
    Sol. Amplitude of electric field, E = B0C
    = 1.6 × 10^{-6} × sqrt(5) × 3 × 10^8
    = 4.8 × 10^2 sqrt(5) V/m
    Also E × B is along -k (the direction of propagation)
    ⇒ E = 4.8 × 10^2 cos(2 × 10^7 z + 6 × 10^15 t)(-i + 2j) V/m

12. In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 Ω/cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:
    (1) 0.75 V
    (2) 0.50 V
    (3) 0.20 V
    (4) 0.25 V
    Answer (4)
    Sol. Resistance of potentiometer wire, Rp = 400 × 0.01 = 4 Ω
    ⇒ I = 3/6 = 0.5 A
    ⇒ Reading of voltmeter = I RAJ
    = 0.5 × 50 × 0.01
    = 0.25 V

13. A nucleus A, with a finite de-Broglie wavelength λA, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelength λB and λC of B and C are respectively:
    (1) 2 λA, λA
    (2) λA, λA/2
    (3) λA/2, λA
    (4) λA, 2 λA
    Answer (3)
    Sol. λA = h / mVA
    Conservation of linear momentum
    ⇒ mVA = m/2 V - m/2 V/2 = mV/4
    ⇒ λA = 4h / mV
    ∴ VA = V/4
    λB = h / (m/2 V) = 2h / mV = λA/2
    λC = h / (m/2 · V/2) = 4h / mV = λA

14. A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.
    (1) E/64
    (2) E/4
    (3) E/16
    (4) E/32
    Answer (3)
    Sol. E = GME m / RE
    E' = GMm m / RM
    ρRE^3 = 64 ρRM^3
    ⇒ RE = 4 RM
    E'/E = Mm/ME · RE/RM = 1/64 · 4 = 1/16
    ⇒ E' = E/16

15. In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:
    (1) 6.8%
    (2) 0.2%
    (3) 3.5%
    (4) 0.7%
    Answer (1)
    Sol. T = 2π sqrt(l/g) ⇒ g = 4π^2 l / T^2
    Δg/g = Δl/l + 2ΔT/T = (0.1/55 + 2×1/30) × 100
    ≈ 6.8%

16. An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:
    (1) sqrt(2qE / md)
    (2) 2 sqrt(qE / md)
    (3) sqrt(qE / 2md)
    (4) sqrt(qE / md)
    Answer (1)
    Sol. -Eqdθ = I d^2θ/dt^2 = 2 md^2/4 d^2θ/dt^2
    ⇒ d^2θ/dt^2 = 2Eqθ/md
    ⇒ ω = sqrt(2Eq/md)

17. A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time τ = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:
    (1) 0.3
    (2) 0.02
    (3) 0.28
    (4) 0.5
    Answer (4)
    Sol. Initial angular acceleration before it slips off
    mg l/2 = Iα
    α = 3g / 2l
    Angular speed acquire by the box in time τ = 0.01 s
    ω = αt = 3g/2l × 0.01 = 3×10×0.01 / 2×0.3 = 1/2 rad/sec
    The angle by which it would rotate when hits the ground
    θ = ωt'
    Assuming ω = constant and t' = time of fall = sqrt(2H/g) = 1 sec
    θ = 1/2 radians

18. A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be:
    (1) (2a/3, 2b/3)
    (2) (5a/12, 5b/12)
    (3) (3a/4, 3b/4)
    (4) (5a/3, 5b/3)
    Answer (2)
    Sol. X-coordinate of CM of remaining sheet
    xcm = (Mx - mx)/(M - m)
    = (4m × a/2 - M × 3a/4)/(4m - m) = 5a/12
    Similarly ycm = 5b/12
    ∴ CM (5a/12, 5b/12)

19. In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:
    R1 = 15 Ω, R2 = 10 Ω, R3 = 20 Ω, R4 = 5 Ω, R5 = 25 Ω, R6 = 30 Ω, E = 15 V
    (1) 13/24
    (2) 9/32
    (3) 20/3
    (4) 7/18
    Answer (2)
    Sol. Equivalent resistance of the given circuit
    Req = 45 + (10×50)/(10+50) = (45 + 50/6) Ω = 160/3 Ω
    ∴ I = 15 / (160/3) = 9/32 A

20. Let |A1| = 3, |A2| = 5 and |A1 + A2| = 5. The value of (2A1 + 3A2)·(3A1 - 2A2) is:
    (1) -106.5
    (2) -118.5
    (3) -99.5
    (4) -112.5
    Answer (2)
    Sol. (2A1 + 3A2)·(3A1 - 2A2) = 6|A1|^2 + 5A1·A2 - 6|A2|^2
    = (6×9) + 5A1·A2 - (6×25) ...(i)
    As 25 = 9 + 25 + 2A1·A2
    ⇒ 2A1·A2 = -9 ...(ii)
    From (i) and (ii),
    ⇒ (2A1 + 3A2)·(3A1 - 2A2) = 54 - 22.5 - 150
    = -118.5

21. Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.
    (1) 457.5 × 10^-9 radian
    (2) 305 × 10^-9 radian
    (3) 152.5 × 10^-9 radian
    (4) 610 × 10^-9 radian
    Answer (2)
    Sol. θ = 1.22λ/D
    θ = 1.22×500×10^-9 / (200×10^-2) = 1.22×500×10^-9 / 2
    θ = 305 × 10^-9 radian

22. Two very long, straight, and insulated wires are kept at 90° angle from each other in xy-plane as shown in the figure.
    These wires carry currents of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be
    (1) Zero
    (2) +μ0I/πd (z)
    (3) -μ0I/2πd (x + y)
    (4) μ0I/2πd (x + y)
    Answer (1)
    Sol. B = B1 + B2 = μ0I/2πd [k - k] = 0

23. A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to:
    (1) v ∝ sqrt(ln(r/r0))
    (2) v ∝ e^{+r/r0}
    (3) v ∝ ln(r/r0)
    (4) v ∝ (r/r0)
    Answer (1)
    Sol. E = λ / 2πε0r
    ∫ dV = ∫ -λ/2πε0r dr
    ⇒ Vp - VG = λ/2πε0 ln(r/r0)
    1/2 mv^2 = q(Vp - VG)
    ⇒ v ∝ [ln(r/r0)]^{1/2}

24. A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be:
    (1) 30 cm
    (2) 25 cm
    (3) 20 cm
    (4) 10 cm
    Answer (4)
    Sol. f = 20 cm
    1/v - 1/u = 1/f
    ∴ 1/v + 1/30 = 1/20
    1/v = (30-20)/(20×30) = 10/(20×30)
    v = 60 cm
    So clearly radius of curvature of mirror is 20 cm. Now if the object is placed within focal plane i.e. 10 cm then image formed by mirror is virtual.

25. The ratio of mass densities of nuclei 40Ca and 16O is close to:
    (1) 0.1
    (2) 1
    (3) 2
    (4) 5
    Answer (2)
    Sol. Densities of nucleus happens to be constant, irrespective of mass number.

26. A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, Rc = 1 kΩ and VCC = 10 V. What is the minimum base current for VCE to reach saturation?
    (1) 10 μA
    (2) 100 μA
    (3) 7 μA
    (4) 40 μA
    Answer (4)
    Sol. For saturation, VCC - ic × Rc = 0
    ⇒ ic = VCC/Rc = 10/10^3 = 10^-2 A
    ∴ β = ic/iB = 250
    ∴ iB = ic/250 = 10^-2/250 = 40 μA

27. A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio hsph/hcyl is given by
    (1) 2/sqrt(5)
    (2) 14/15
    Answer (4)
    Sol. mghsph = 1/2 mv^2 + 1/2 (2/5 mv^2)(v/R)^2 ...(i)
    mghcylinder = 1/2 mv^2 + 1/2 (mR^2/2)(v/R)^2 = 3/4 mv^2
    ⇒ hsph/hcylinder = (7×4)/(10×3) = 14/15

28. A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time)
    (1) (A)
    (2) (A), (B), (C)
    (3) (A), (B), (D)
    (4) (B), (C)
    Answer (3)
    Sol. a = Constant
    v = at
    x = 1/2 at^2 [Particle starts from the origin]
    A, B and D are correct graphs.

29. Young's moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to
    (1) 1.3 mm
    (2) 1.9 mm
    (3) 1.5 mm
    (4) 1.7 mm
    Answer (4)
    Sol. ΔL = FL/YA
    ⇒ LA/(YA rA^2) = LB/(YB rB^2)
    ⇒ rA^2 = sqrt(LA/LB · YB/YA · rB)
    = sqrt(2×2×4/(3×7)) × 2 mm
    = (4/4.58) × 2 = 1.7 mm

30. The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to
    [Boltzmann constant kB = 1.38 × 10^-23 J/K
    Avogadro Number NA = 6.02 × 10^26 / kg
    Radius of Earth : 6.4 × 10^6 m
    Gravitational acceleration on Earth = 10 ms^-2]
    (1) 10^4 K
    (2) 650 K
    (3) 800 K
    (4) 3 × 10^5 K
    Answer (1)
    Sol. Vrms = sqrt(3RT/M) = 11.2 × 10^3 m/s
    ⇒ T = M/3R × (11.2 × 10^3)^2
    = (2 × 10^-3 / 3 × 8.3) × 125.44 × 10^6 = 10^4 K

Chemistry

1. 0.27 g of a long chain fatty acid was dissolved in 100 cm^3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?
   [Density of fatty acid = 0.9 g cm^-3; π = 3]
   (1) 10^-8 m
   (2) 10^-4 m
   (3) 10^-2 m
   (4) 10^-6 m
   Answer (4)
   Sol. 0.27 gm in 100 ml of hexane
   in 10 ml of aqueous solution only 0.027 gm acid is present
   volume of 0.027 g acid = 0.027/0.9 ml
   ∴ πr^2h = 0.027/0.9 (given r = 10 cm, π = 3)
   ∴ h = 10^-4 cm = 10^-6 m

2. The Mond process is used for the :
   (1) purification of Zr and Ti
   (2) extraction of Mo
   (3) purification of Ni
   (4) extraction of Zn
   Answer (3)
   Sol. Nickel is purified by Mond's process
   Ni + 4CO → 330-350 K → Ni(CO)4
   Ni(CO)4 → 450-470 K → Ni + 4CO

3. The correct statement about ICl5 and ICl4^- is :
   (1) ICl5 is square pyramidal and ICl4^- is tetrahedral.
   (2) both are isostructural.
   (3) ICl5 is square pyramidal and ICl4^- is square planar.
   (4) ICl5 is trigonal bipyramidal and ICl4^- is tetrahedral.
   Answer (3)
   Sol. ICl5 is sp^3d^6 hybridised (5 bond pairs, 1 lone pair) square pyramidal
   ICl4^- is sp^3d^6 hybridised (4 bond pairs, 2 lone pairs) square planar

4. For the following reactions, equilibrium constants are given :
   S(s) + O2(g) ? SO2(g); K1 = 10^52
   2S(s) + 3O2(g) ? 2SO3(g); K2 = 10^129
   The equilibrium constant for the reaction,
   2SO2(g) + O2(g) ? 2SO3(g) is :
   (1) 10^154
   (2) 10^25
   (3) 10^77
   (4) 10^181
   Answer (2)
   Sol. eq1: S + O2 ? SO2 K1 = 10^52
   eq2: 2S + 3O2 ? 2SO3 K2 = 10^129
   eq3: 2SO2 + O2 ? 2SO3
   eq3 = eq2 - 2(eq1)
   = 10^129/(10^52)^2 = 10^25

5. Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?
   (1) 45%
   (2) 65%
   (3) 75%
   (4) 90%
   Answer (4)
   Sol. Volume occupied by atoms in solid 2
   = 4/3 πr^3 + 4/3 π(2r)^3 = 12πr^3
   relationship between edge length (a) and radius of atom (r)
   = 6r = sqrt(3)a ⇒ a = 6r/sqrt(3)
   packing efficiency = 12πr^3 / (6r/sqrt(3))^3 × 100 = 90%

6. The compound that inhibits the growth of tumors is
   (1) cis-[Pt(Cl)2(NH3)2]
   (2) trans-[Pt(Cl)2(NH3)2]
   (3) cis-[Pd(Cl)2(NH3)2]
   (4) trans-[Pd(Cl)2(NH3)2]
   Answer (1)
   Sol. Cis-platin is used as an anti-cancer drug.
   Cis [PtCl2(NH3)2]

7. The structure of Nylon-6 is
   (1) [-(CH2)4-C(=O)-NH-]n
   (2) [-C(=O)-(CH2)6-NH-]n
   (3) [-(CH2)6-C(=O)-NH-]n
   (4) [-C(=O)-(CH2)5-NH-]n
   Answer (4)
   Sol. Nylon-6 is produced from caprolactam

8. Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?
   (1) F3C-CH=CH2
   (2) CH3O-CH=CH2
   (3) H2N-CH=CH2
   (4) Cl-CH=CH2
   Answer (1)
   Sol. CF3-CH=CH2
   CF3-CH=CH2 + H+ → CF3-CH2-CH2+ (more stable)
   anti-Markovnikov

9. The strength of 11.2 volume solution of H2O2 is
   [Given that molar mass of H = 1 g mol^-1 and O = 16 g mol^-1]
   (1) 13.6%
   (2) 1.7%
   (3) 3.4%
   (4) 34%
   Answer (3)
   Sol. 11.2 V of H2O2
   H2O2 → H2O + 1/2 O2
   11.2 L of O2 at STP = 0.5 mol
   It means 1 L of given H2O2 solution consist 1 mole of H2O2 (i.e., 34 g)
   strength = 34/1000 × 100 = 3.4%

10. The covalent alkaline earth metal halide (X = Cl, Br, I) is
    (1) BeX2
    (2) SrX2
    (3) CaX2
    (4) MgX2
    Answer (1)
    Sol. According to Fajan's rule, greater the polarising power of cation greater would be the covalent character.
    Since Be^2+ has maximum polarising power among given cation, therefore, BeX2 would be most covalent among given alkaline with metal halides.

11. Polysubstitution is a major drawback in :
    (1) Reimer Tiemann reaction
    (2) Acetylation of aniline
    (3) Friedel Craft's acylation
    (4) Friedel Craft's alkylation
    Answer (4)
    Sol. Polysubstitution is a major drawback in Friedel Craft's alkylation

12. The IUPAC symbol for the element with atomic number 119 would be :
    (1) une
    (2) uun
    (3) uue
    (4) unh
    Answer (3)
    Sol. Symbol for 1 is u and for 9 is e
    ∴ IUPAC symbol for 119 is uue

13. 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 J K^-1 mol^-1, calculate ΔU and Δ(pV) for this process. (R = 8.0 J K^-1 mol^-1)
    (1) ΔU = 14 kJ, Δ(pV) = 18 kJ
    (2) ΔU = 2.8 kJ, Δ(pV) = 0.8 kJ
    (3) ΔU = 14 J, Δ(pV) = 0.8 J
    (4) ΔU = 14 kJ, Δ(pV) = 4 kJ
    Answer (4)
    Sol. ΔU = nCv,mΔT = 5 × 28 × 100 = 14 kJ
    Δ(PV) = nR(T2 - T1)
    = 5 × 8 × 100
    = 4 kJ

14. The calculated spin-only magnetic moments (BM) of the anionic and cationic species of [Fe(H2O)6]2 and [Fe(CN)6], respectively, are:
    (1) 2.84 and 5.92
    (2) 4.9 and 0
    (3) 0 and 5.92
    (4) 0 and 4.9
    Answer (4)
    Sol. [Fe(H2O)6]2+ [Fe(CN)6]4-
    4 unpaired electrons no unpaired electron
    μ = 4.9 μ = 0

15. The major product of the following reaction is:
    CH3-C6H4-Cl + (1) Cl2/hv (2) H2O, Δ →
    (1) CHCl2-C6H4-Cl
    (2) CO2H-C6H4-Cl
    (3) CHO-C6H4-Cl
    (4) CH2OH-C6H4-Cl
    Answer (1)

16. The major product in the following reaction is:
    (adenine derivative) + CH3I/base →
    (1) N-methylated product at 9 position
    (2) NHCH3 product
    (3) N-methylated product at 3 position
    (4) N-methylated product at 7 position
    Answer (4)
    Sol. In the given compound, H-atom attached to secondary N-atom is more acidic. The base removes the more acidic H-atom and the conjugate base of the given compound attacks at CH3 group to give the final product.

17. The major product of the following reaction is:
    (aryl ketone with ClCH2CH2Cl side chain) (1) tBuOK (2) Conc. H2SO4/Δ →
    (1) cyclopentanone fused product
    (2) cyclohexanone fused product
    (3) cyclopentanone fused product with isopropyl
    (4) cyclohexanone fused product with isopropyl
    Answer (2)

18. The ion that has sp^3d^2 hybridization for the central atom, is:
    (1) [ICl2]^-
    (2) [IF6]^-
    (3) [BrF2]^-
    (4) [ICl4]^-
    Answer (4)
    Sol. Species Hybridisation
    ICl2 sp3d
    ICl4 sp3d2
    BrF2 sp3d
    IF6 sp3d2

19. Calculate the standard cell potential (in V) of the cell in which following reaction takes place:
    Fe^2+(aq) + Ag+(aq) → Fe^3+(aq) + Ag(s)
    Given that
    E°(Ag+/Ag) = x V
    E°(Fe2+/Fe) = y V
    E°(Fe3+/Fe) = z V
    (1) x - y
    (2) x + y - z
    (3) x + 2y - 3z
    (4) x - z
    Answer (3)
    Sol. Ag+ + Fe2+ → Fe3+ + Ag
    E°cell = E°(Ag+/Ag) - E°(Fe3+/Fe2+)
    To calculate E°(Fe3+/Fe2+)
    Fe3+ → Fe2+ → Fe E° = z
    E°(Fe3+/Fe2+) = 3z - 2y
    E°(Ag+/Ag) = x
    E°cell = x - 3z + 2y

20. Which of the following compounds will show the maximum 'enol' content?
    (1) CH3COCH2COOC2H5
    (2) CH3COCH3
    (3) CH3COCH2COCH3
    (4) CH3COCH2CONH2
    Answer (3)
    Sol. enol content ∝ acidity of active methene hydrogens.
    Maximum enol content

21. The major product obtained in the following reaction is:
    (2-amino-3-cyanophenyl ketone) (i) CHCl3/KOH (ii) Pd/C/H2 →
    (1) 2-(methylamino)-3-(aminomethyl)phenyl methanol
    (2) 2-(methylamino)-3-cyanophenyl methanol
    (3) 2-(methylamino)-3-cyanophenyl ketone
    (4) 2-(N-methylformamide)-3-cyanophenyl methanol
    Answer (1)

22. For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is :
    (1) graph with z highest slope, y, x, w
    (2) graph with z highest, y, x, w
    (3) graph with z highest, y, x, w
    (4) graph with z highest, y, x, w
    Answer (1)
    Sol. According to Henry's law
    P = KH · Xgas
    ∴ Xgas = 1 - XH2O
    ∴ P = KH - KH · XH2O
    y = C + mx
    gas KH
    w 0.5
    x 2
    y 35
    z 50

23. If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be :
    (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function):
    (1) 3/4 λ
    (2) 4/9 λ
    (3) 2/3 λ
    (4) 1/2 λ
    Answer (2)
    Sol. In photoelectric effect,
    hc/λ = w + KE of electron
    It is given that KE of ejected electron is very high in comparison to w.
    hc/λ = KE ⇒ hc/λ = p^2/2m
    New wavelength
    hc/λ1 = (1.5P)^2/2m ⇒ λ' = 4/9 λ

24. Fructose and glucose can be distinguished by:
    (1) Fehling's test
    (2) Seliwanoff's test
    (3) Barfoed's test
    (4) Benedict's test
    Answer (2)
    Sol. Selivanoff's test is used to distinguish aldose and ketose

25. Among the following molecules/ions,
    C2^2-, N2^2-, O2^2-, O2
    Which one is diamagnetic and has the shortest bond length?
    (1) O2
    (2) O2^2-
    (3) N2^2-
    (4) C2^2-
    Answer (4)
    Sol. Bond length ∝ 1/bond order
    and diamagnetic species has no unpaired electron in their molecular orbitals.
    Bond order Magnetic character
    C2^2- 3 diamagnetic
    N2^2- 2 paramagnetic
    O2^2- 1 diamagnetic
    O2 2 paramagnetic
    C2^2- has least bond length and is diamagnetic

26. For a reaction scheme A →(k1) B →(k2) C, if the rate of formation of B is set to be zero then the concentration of B is given by:
    (1) k1k2[A]
    (2) (k1 - k2)[A]
    (3) [k1/k2][A]
    (4) (k1 + k2)[A]
    Answer (3)
    Sol. A →(k1) B →(k2) C
    d[B]/dt = k1[A] - k2[B] = 0
    [B] = k1[A]/k2

27. The maximum prescribed concentration of copper in drinking water is:
    (1) 3 ppm
    (2) 0.05 ppm
    (3) 0.5 ppm
    (4) 5 ppm
    Answer (1)
    Sol. Maximum prescribed concentration of Cu in drinking water is 3 ppm.

28. The major product obtained in the following reaction is
    (aldehyde with ketone side chain) NaOH/Δ →
    (1) cyclopentane carbaldehyde with methylene
    (2) cyclopentene carbaldehyde with methyl
    (3) cyclopentene with acetyl methyl
    (4) cyclopentene with acetyl
    Answer (3)
    Sol. Intramolecular aldol condensation

29. The percentage composition of carbon by mole in methane is
    (1) 80%
    (2) 75%
    (3) 20%
    (4) 25%
    Answer (3)
    Sol. In CH4
    one atom of carbon among 5 atoms (1C + 4H atoms)
    ∴ Mole % of C = 1/5 × 100 = 20%

30. The statement that is INCORRECT about the interstitial compounds is
    (1) They are chemically reactive.
    (2) They are very hard.
    (3) They have high melting points.
    (4) They have metallic conductivity.
    Answer (1)
    Sol. Interstitial compounds are inert.

Mathematics

1. The vector equation of plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0 is:
   (1) r·(i - k) + 2 = 0
   (2) r·(i - k) - 2 = 0
   (3) r × (i - k) + 2 = 0
   (4) r × (i + k) + 2 = 0
   Answer (1)
   Sol. Equation of the plane passing through the line of intersection of x + y + z = 1 and 2x + 3y + 4z = 5 is
   (2x + 3y + 4z - 5) + λ(x + y + z - 1) = 0
   (2 + λ)x + (3 + λ)y + (4 + λ)z + (-5 - λ) = 0 ...(i)
   (i) is perpendicular to x - y + z = 0
   ⇒ (2 + λ)(1) + (3 + λ)(-1) + (4 + λ)(1) = 0
   2 + λ - 3 - λ + 4 + λ = 0
   λ = -3
   ⇒ Equation of required plane is
   -x + z - 2 = 0
   ⇒ x - z + 2 = 0
   ⇒ r·(i - k) + 2 = 0

2. Which one of the following statements is not a tautology?
   (1) (p ∧ q) → (¬p) ∨ q
   (2) (p ∧ q) → p
   (3) (p ∨ q) → (p ∨ (¬q))
   (4) p → (p ∨ q)
   Answer (3)
   Sol. By help of truth table.

3. [Not available in provided text extraction]

4. [Not available in provided text extraction]

5. A student scores the following marks in five tests: 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is:
   (1) 100/√3
   (2) 10/√3
   (3) 100/3
   (4) 10/3
   Answer (2)
   Sol. x? = (41 + 45 + 54 + 57 + 43 + x)/6 = 48
   x + 240 = 288
   x = 48
   σ^2 = 1/6 [(48 - 41)^2 + (48 - 45)^2 + (48 - 54)^2 + (48 - 57)^2 + (48 - 43)^2 + (48 - 48)^2]
   = 1/6 (49 + 9 + 36 + 81 + 25)
   = 200/6 = 100/3
   σ = 10/√3

6. The tangent and the normal lines at the point (√3, 1) to the circle x^2 + y^2 = 4 and the x-axis form a triangle. The area of this triangle (in square units) is:
   (1) 2/√3
   (2) 4/√3
   (3) 1/3
   (4) 1/√3
   Answer (1)
   Sol. Equation of tangent to circle at point (√3, 1) is √3x + y = 4
   ∴ Coordinate of A = (4/√3, 0)
   Area = 1/2 × OA × PM
   = 1/2 × 4/√3 × 1 = 2/√3 square units

7. If the fourth term in the binomial expansion of (sqrt(1/x^{1+log10 x}) + x^{1/12})^6 is equal to 200, and x > 1, then the value of x is:
   (1) 10
   (2) 10^3
   (3) 100
   (4) 10^4
   Answer (1)
   Sol. T4 = 6C3 (sqrt(1/x^{1+log10 x}))^3 (x^{1/12})^3 = 200
   ⇒ 3/(20 x^{2(1+log10 x)/4} · x^{1/4}) = 200
   x^{1/4 - 3/(2(1+log10 x))} = 10
   Taking log10 on both sides and put log10 x = t
   (1/4 + 3/(2(1+t)))t = 1
   ((1+t)+6)/(4(1+t)) t = 1 ⇒ t^2 + 7t = 4 + 4t
   t^2 + 3t - 4 = 0 ⇒ t^2 + 4t - t - 4 = 0
   ⇒ t(t+4) - 1(t+4) = 0
   ⇒ t = 1 or t = -4
   log10 x = 1 ⇒ x = 10 or if log10 x = -4 ⇒ x = 10^-4
   Note: There seems a printing error in this question in the original question paper.

8. The point of intersection of the tangent to y^2 = 4x at the point where it intersects the circle x^2 + y^2 = 5 in the first quadrant, passes through the point:
   (1) (3/4, 7/4)
   (2) (-1/3, 4/3)
   (3) (1/4, 3/4)
   (4) (-1/4, 1/2)
   Answer (1)
   Sol. Intersection point of x^2 + y^2 = 5, y^2 = 4x
   ⇒ x^2 + 4x - 5 = 0
   ⇒ x^2 + 5x - x - 5 = 0
   ⇒ x(x+5) - 1(x+5) = 0
   ∴ x = 1, -5
   Intersection point in 1st quadrant be (1, 2)
   equation of tangent to y^2 = 4x at (1, 2) is
   y × 2 = 2(x + 1)
   ⇒ y = x + 1
   ⇒ x - y + 1 = 0 ...(i)
   (3/4, 7/4) lies on (i)

9. If three distinct numbers a, b, c are in G.P. and the equations ax^2 + 2bx + c = 0 and dx^2 + 2ex + f = 0 have a common root, then which one of the following statements is correct?
   (1) d, e, f are in A.P.
   (2) d/a, e/b, f/c are in G.P.
   (3) d/a, e/b, f/c are in A.P.
   (4) d, e, f are in G.P.
   Answer (3)
   Sol. Since a, b, c are in G.P.
   ⇒ b^2 = ac
   Given, ax^2 + 2bx + c = 0
   ⇒ ax^2 + 2√ac x + c = 0
   ⇒ (√a x + √c)^2 = 0
   ⇒ x = -√(c/a)
   ∴ ax^2 + 2bx + c = 0 and dx^2 + 2ex + f = 0 have common root
   ⇒ x = -√(c/a) must satisfy dx^2 + 2ex + f = 0
   ⇒ d c/a + 2e(-√(c/a)) + f = 0
   ⇒ d/a - 2e/√(ac) + f/c = 0
   ⇒ d/a - 2e/b + f/c = 0
   ⇒ 2e/b = d/a + f/c
   ⇒ d/a, e/b, f/c are in A.P.

10. If z = √3/2 + i/2 (i = √-1), then (1 + iz + z^5 + iz^8)^9 is equal to:
    (1) 0
    (2) (-1 + 2i)^9
    (3) -1
    (4) 1
    Answer (3)
    Sol. z = √3/2 + i/2 = -i(-1/2 + √3/2 i) = -iω
    where ω is not real cube root of unity
    ⇒ (1 + iz + z^5 + iz^8)^9 = (1 + ω - iω^2 + iω^2)^9
    = (1 + ω)^9
    = (-ω^2)^9
    = -ω^18
    = -1

11. The number of integral values of m for which the equation (1 + m^2)x^2 - 2(1 + 3m)x + (1 + 8m) = 0 has no real root is:
    (1) Infinitely many
    (2) 3
    (3) 2
    (4) 1
    Answer (1)
    Sol. (1 + m^2)x^2 - 2(1 + 3m)x + (1 + 8m) = 0
    equation has no real solution
    ⇒ D < 0
    4(1 + 3m)^2 < 4(1 + m^2)(1 + 8m)
    1 + 9m^2 + 6m < 1 + 8m + m^2 + 8m^3
    8m^3 - 8m^2 + 2m > 0
    2m(4m^2 - 4m + 1) > 0
    2m(2m - 1)^2 > 0
    m > 0, m ≠ 1/2
    ⇒ number of integral values of m are infinitely many.

12. Let the numbers 2, b, c be in an A.P. and A = [[1, 1, 1], [2, b, c], [4, b^2, c^2]]. If det(A) ∈ [2, 16], then c lies in the interval:
    (1) [2, 3)
    (2) (2 + 2^{3/4}, 4)
    (3) [3, 2 + 2^{3/4}]
    (4) [4, 6]
    Answer (4)
    Sol. c2 → c2 - c1, c3 → c3 - c1
    = (b - 2)(c - 2)(c - b)
    2, b, c are in A.P. ⇒ (b - 2) = (c - b) = d, c - 2 = 2d
    ⇒ |A| = d · 2d · d = 2d^3
    ∴ |A| ∈ [2, 16] ⇒ 1 ≤ d^3 ≤ 8 ⇒ 1 ≤ d ≤ 2
    4 ≤ 2d + 2 ≤ 6 ⇒ 4 ≤ c ≤ 6

13. The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:
    (1) 5
    (2) 2
    (3) 4
    (4) 3
    Answer (3)
    Sol. p = P(H) = 1/2, q = 1 - p = 1/2
    P(x ≥ 1) ≥ 9/10
    1 - P(x = 0) ≥ 9/10
    1 - nC0 (1/2)^n ≥ 9/10
    1/2^n ≤ 1 - 9/10 ⇒ 1/2^n ≤ 1/10
    2^n ≥ 10 ⇒ n ≥ 4
    ⇒ nmin = 4

14. Let a = 3i + 2j + xk and b = i - j + k, for some real x. Then |a × b| = r is possible if:
    (1) 3√(3/2) < r < 5√(3/2)
    (2) 2√(3/2) < r ≤ 3√(3/2)
    (3) 0 < r ≤ √(3/2)
    (4) r ≥ 5√(3/2)
    Answer (4)
    Sol. a × b = |i j k; 3 2 x; 1 -1 1|
    = (2 + x)i + (x - 3)j - 5k
    |a × b| = r = sqrt((2 + x)^2 + (x - 3)^2 + (-5)^2)
    ⇒ r = sqrt(4 + x^2 + 4x + x^2 + 9 - 6x + 25)
    = sqrt(2x^2 - 2x + 38) = sqrt(2(x^2 - x + 1/4) + 38 - 1/2)
    = sqrt(2(x - 1/2)^2 + 75/2)
    ⇒ r ≥ sqrt(75/2) ⇒ r ≥ 5√(3/2)

15. The sum ∑{k=1}^{20} k · 1/2^k is equal to:
    (1) 2 - 3/2^{17}
    (2) 1 - 11/2^{20}
    (3) 2 - 21/2^{20}
    (4) 2 - 11/2^{19}
    Answer (4)
    Sol. S = ∑{k=1}^{20} k · 1/2^k
    S = 1/2 + 2·1/2^2 + 3·1/2^3 + ... + 20·1/2^{20}
    1/2 S = 1/2^2 + 2·1/2^3 + ... + 19·1/2^{20} + 20·1/2^{21}
    On subtracting
    S/2 = (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{20}) - 20·1/2^{21}
    = 1/2(1 - 1/2^{20}) - 20·1/2^{21} = 1 - 1/2^{20} - 10·1/2^{20}
    S/2 = 1 - 11·1/2^{20} ⇒ S = 2 - 11·1/2^{19} = 2 - 11/2^{19}

16. Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is:
    (1) 16
    (2) 18
    (3) 15
    (4) 12
    Answer (1)
    Sol. equation of line OB and AC are respectively
    y = 80/x1 x ...(i)
    x/x1 + y/20 = 1 ...(ii)
    For intersection point, from equations (i) and (ii)
    y/80 + y/20 = 1
    ⇒ y + 4y = 80
    ⇒ y = 16 m
    ⇒ Height of intersection point is 16 m

17. Let f:[-1,3] → R be defined as
    f(x) = [x] for -1 ≤ x < 0
    f(x) = |x| for 0 ≤ x < 1
    f(x) = 2x^2 - 5x + 7 for 1 ≤ x ≤ 3
    where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at:
    (1) Only one point
    (2) Only two points
    (3) Only three points
    (4) Four or more points
    Answer (3)
    Sol. ⇒ f(-1) = 0, f(-1+) = 0
    f(0-) = -1, f(0) = 0, f(0+) = 0
    f(1-) = 1, f(1) = 2, f(1+) = 2
    f(2-) = 4, f(2) = 4, f(2+) = 4
    f(3-) = 5, f(3) = 6
    f(x) is discontinuous at x = {0, 1, 3}

18. Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is 2y/x^2. If the curve passes through the centre of the circle x^2 + y^2 - 2x - 2y = 0, then its equation is
    (1) x log_e |y| = -2(x - 1)
    (2) x log_e |y| = x - 1
    (3) x log_e |y| = 2(x - 1)
    (4) x^2 log_e |y| = -2(x - 1)
    Answer (3)
    Sol. dy/dx = 2y/x^2 ⇒ ∫ dy/y = 2∫ dx/x^2
    ⇒ ln |y| = -2/x + C ...(i)
    (i) passes through (1, 1)
    ⇒ C = 2
    ⇒ ln |y| = -2/x + 2
    x ln |y| = -2 + 2x
    x ln |y| = -2(1 - x) = 2(x - 1)

19. Suppose that the points (h, k), (1, 2) and (-3, 4) lie on the line L. If a line L1 passing through the points (h, k) and (4, 3) is perpendicular to L1, then k/h equals
    (1) 3
    (2) 1/3
    (3) 0
    (4) -1/7
    Answer (4)
    Sol. (h, k), (1, 2) and (-3, 4) are collinear
    2b - 2a = 10 ⇒ b - a = 5 (ii)
    From (i) and (ii)
    b + a = 15 (iii)
    ⇒ b = 10, a = 5
    Length of L·R·R = 2a^2/b = 50/10 = 5

20. Let S(α) = {(x, y) : y^2 ≤ x, 0 ≤ x ≤ α} and A(α) is area of the region S(α). If for a λ, 0 < λ < 4, A(λ) : A(4) = 2 : 5 then λ equals
    (1) 2(2/5)^{1/3}
    (2) 2(4/25)^{1/3}
    (3) 4(2/5)^{1/3}
    (4) 4(4/25)^{1/3}
    Answer (4)
    Sol. A(λ) = 2 × 2/3 (λ × √λ) = 4/3 λ^{3/2}
    ⇒ A(λ)/A(4) = 2/5 ⇒ λ^{3/2}/8 = 2/5
    λ = (16/5)^{2/3} = 4(4/25)^{1/3}

21. If a point R(4, y, z) lies on the line segment joining the points P(2, -3, 4) and Q(8, 0, 10) then the distance of R from the origin is
    (1) √53
    (2) 2√21
    (3) 6
    (4) 2√14
    Answer (4)
    Sol. P, Q, R are collinear
    PR = λ PQ
    2i + (y + 3)j + (z - 4)k = λ[6i + 3j + 6k]
    6λ = 2, y + 3 = 3λ, z - 4 = 6λ
    λ = 1/3, y = -2, z = 6
    point R(4, -2, 6)
    OR = sqrt(4^2 + (-2)^2 + 6^2) = sqrt(16 + 4 + 36)
    = √56 = 2√14

22. If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is
    (1) 4 : 5 : 6
    (2) 3 : 4 : 5
    (3) 5 : 9 : 13
    (4) 5 : 6 : 7
    Answer (1)
    Sol. Let a > b > c
    ∴ A = 2C
    ⇒ A + B + C = π
    ⇒ B = π - 3C
    ∴ a + c = 2b
    ⇒ sin A + sin C = 2 sin B
    ⇒ sin A = sin(2C), sin B = sin 3C
    ⇒ From (i), sin 2C + sin C = 2 sin 3C
    (2 cos C + 1) sin C = 2 sin C(3 - 4 sin^2 C)
    ⇒ 2 cos C + 1 = 6 - 8(1 - cos^2 C)
    ⇒ 8 cos^2 C - 2 cos C - 3 = 0
    ⇒ cos C = 3/4 or cos C = -1/2
    ∴ C is acute angle
    ⇒ cos C = 3/4, sin A = 2 sin C cos C = 2 × √7/4 × 3/4
    ⇒ sin C = √7/4, sin B = 3√7/4 - 4√7/4 × 7/16 = 5√7/16
    ⇒ sin A : sin B : sin C : a : b : c is 6 : 5 : 4

23. The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is
    (1) 2/3 √3
    (2) 2√3
    (3) √3
    (4) √6
    Answer (2)
    Sol. Let radius of base and height of cylinder be r and h respectively.
    ∴ r^2 + h^2/4 = 9 (i)
    Volume of cylinder
    V = πr^2h
    V = πh(9 - h^2/4)
    V = 9πh - π/4 h^3
    ∴ dV/dh = 9π - 3/4 πh^2
    For maxima/minima
    dV/dh = 0
    ⇒ h = √12
    and d^2V/dh^2 = -3/2 πh
    ∴ (d^2V/dh^2)_{h=√12} < 0
    ⇒ Volume is maximum when h = 2√3

24. If the system of linear equations
    x - 2y + kz = 1
    2x + y + z = 2
    3x - y - kz = 3
    has a solution (x, y, z), z ≠ 0 then (x, y) lies on the straight line whose equation is
    (1) 3x - 4y - 4 = 0
    (2) 3x - 4y - 1 = 0
    (3) 4x - 3y - 1 = 0
    (4) 4x - 3y - 4 = 0
    Answer (4)
    Sol. x - 2y + kz = 1, 2x + y + z = 2, 3x - y - kz = 3
    Δ = |1 -2 k; 2 1 1; 3 -1 -k| = 1(-k + 1) + 2(-2k - 3) + k(-2 - 3)
    = -k + 1 - 4k - 6 - 5k = -10k - 5 = -5(2k + 1)
    Δ1 = |1 -2 k; 2 1 1; 3 -1 -k| = -5(2k + 1)
    Δ2 = |1 1 k; 2 2 1; 3 3 -k| = 0, Δ3 = |1 -2 1; 2 1 2; 3 -1 3| = 0
    ∴ z ≠ 0
    ⇒ Δ = 0
    ⇒ k = -1/2
    ∴ System of equation has infinite many solutions.
    Let z = λ ≠ 0 then x = (10 - 3λ)/10 and y = -2λ/5
    ∴ (x, y) must lie on line 4x - 3y - 4 = 0

25. If f(1) = 1, f'(1) = 3, then the derivative of f(f(f(x))) + (f(x))^2 at x = 1 is:
    (1) 33
    (2) 12
    (3) 9
    (4) 15
    Answer (1)
    Sol. Let g(x) = f(f(f(x))) + (f(x))^2
    On differentiating both sides w.r.t. x we get
    g'(x) = f'(f(f(x))) f'(f(x)) f'(x) + 2f(x) f'(x)
    g'(1) = f'(f(f(1))) f'(f(1)) f'(1) + 2f(1) f'(1)
    = f'(f(1)) f'(1) f'(1) + 2f(1) f'(1)
    = 3 × 3 × 3 + 2 × 1 × 3 = 27 + 6 = 33

26. Let f(x) = ∫_0^x g(t)dt, where g is a non-zero even function. If f(x + 5) = g(x), then ∫_0^x f(t)dt, equals:
    (1) ∫_5^{x+5} g(t)dt
    (2) 2∫5^{x+5} g(t)dt
    (3) ∫{x+5}^5 g(t)dt
    (4) 5∫_5^{x+5} g(t)dt
    Answer (1)
    Sol. f(x) = ∫_0^x g(t)dt, ...(i)
    g(-x) = g(x), ...(ii)
    f(x + 5) = g(x) ...(iii)
    From (i)
    f'(x) = g(x)
    Let I = ∫_0^x f(t)dt,
    Put t = λ - 5
    ⇒ I = ∫_5^{x+5} f(λ - 5)dλ
    ? f(x + 5) = g(x)
    ⇒ f(-x + 5) = g(-x) = g(x) ...(iv)
    I = ∫_5^{x+5} f(λ - 5)dλ
    I = ∫_5^{x+5} -f(5 - λ)dλ
    (? f(0) = 0, g(x) is even ⇒ f(x) is odd)
    I = -∫5^{x+5} g(λ)dλ = ∫{x+5}^5 g(t)dt (from (iv))
    Wait the correct option is (1) ∫_5^{x+5} g(t)dt.

27. Let f(x) = a^x (a > 0) be written as f(x) = f1(x) + f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) + f1(x - y) equals:
    (1) 2f1(x)f1(y)
    (2) 2f1(x + y)f1(x - y)
    (3) 2f1(x + y)f2(x - y)
    (4) 2f1(x)f2(y)
    Answer (1)
    Sol. f(x) = a^x = (a^x + a^{-x})/2 + (a^x - a^{-x})/2
    where f1(x) = (a^x + a^{-x})/2 is even function
    f2(x) = (a^x - a^{-x})/2 is odd function
    ⇒ f1(x + y) + f1(x - y)
    = (a^{x+y} + a^{-x-y})/2 + (a^{x-y} + a^{-x+y})/2
    = 1/2 [a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})]
    = (a^x + a^{-x})(a^y + a^{-y})/2
    = 2f1(x) · f1(y)

28. If ∫ dx/(x^3(1 + x^6)^{2/3}) = x f(x)(1 + x^6)^{1/3} + C where C is a constant of integration, then the function f(x) is equal to:
    (1) -1/(2x^3)
    (2) 3/x^2
    (3) -1/(6x^3)
    (4) -1/(2x^2)
    Answer (1)
    Sol. I = ∫ dx/(x^3(1 + x^6)^{2/3}) = ∫ dx/(x^7(1 + x^{-6})^{2/3})
    Put 1 + x^{-6} = t^3
    ⇒ -6x^{-7}dx = 3t^2 dt
    ⇒ dx/x^7 = -1/2 t^2 dt
    ⇒ I = ∫ -1/2 t^2 dt / t^2
    = -1/2 t + C
    = -1/2 (1 + x^{-6})^{1/3} + C
    = -1/2 ((1 + x^6)/x^6)^{1/3} + C
    = -1/(2x^2) (1 + x^6)^{1/3} + C
    ⇒ f(x) = -1/(2x^3)

29. If the eccentricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is:
    (1) 2x - 3y + 10 = 0
    (2) x - 2y + 8 = 0
    (3) 3x - 2y = 0
    (4) 2x - y - 2 = 0
    Answer (4)
    Sol. Let equation of hyperbola be
    x^2/a^2 - y^2/b^2 = 1 ...(i)
    ⇒ b^2 = a^2(e^2 - 1)
    ? e = 2 ⇒ b^2 = 3a^2 ...(ii)
    (i) passes through (4, 6)
    ⇒ 16/a^2 - 36/b^2 = 1 ...(iii)
    From (ii) and (iii)
    a^2 = 4, b^2 = 12
    ⇒ Equation of hyperbola is x^2/4 - y^2/12 = 1
    Equation of tangent to the hyperbola at (4, 6) is
    4x/4 - 6y/12 = 1
    ⇒ x - y/2 = 1
    ⇒ 2x - y = 2

30. Let f : R → R be a differentiable function satisfying f(3) + f(2) = 0.
    Then lim_{x→0} ( (1 + f(3 + x) - f(3)) / (1 + f(2 - x) - f(2)) )^{1/x} is equal to:
    (1) e
    (2) 1
    (3) e^2
    (4) e^{-1}
    Answer (2)
    Sol. I = lim_{x→0} ( (1 + f(3 + x) - f(3)) / (1 + f(2 - x) - f(2)) )^{1/x}
    form : 1^∞
    ⇒ I = e^{I1}, where
    I1 = lim_{x→0} ( (1 + f(3 + x) - f(3)) / (1 + f(2 - x) - f(2)) - 1 ) 1/x
    = lim_{x→0} ( (f(3 + x) - f(3) - f(2 - x) + f(2)) / (1 + f(2 - x) - f(2)) ) 1/x
    form : 0/0
    Using L.H. Rule
    I1 = lim_{x→0} ( f'(3 + x) + f'(2 - x) ) / 1 · lim_{x→0} 1/(1 + f(2 - x) - f(2))
    = f'(3) + f'(2) = 0
    ⇒ I = e^{I1} = 1

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