Answers & Solutions

Test Booklet Code

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Answers & Solutions for

M.M. : 360

JEE (MAIN)-2017 (Physics, Chemistry and Mathematics)

Important Instructions : 1.

The test is of 3 hours duration.

2.

The Test Booklet consists of 90 questions. The maximum marks are 360.

3.

There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

4.

Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. ¼ (one-fourth) marks of the total marks allotted to the question (i.e. 1 mark) will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

5.

There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.

6.

For writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet use only Black BallPoint Pen provided in the examination hall.

7.

No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall/room.

1

JEE (MAIN)-2017 (Code-A)

PART–A : PHYSICS 1.

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of

v t

(3) (1) 9

(2)

1 9

(3) 81

(4)

1 81

v

Answer (1) Sol.

t

(4)

vf  93 vi

Answer (3)

∵ Density remains same

Sol. Acceleration is constant and negative

So, mass  Volume

v

mf  93 mi

t

(Area)f  92 (Area)i Stress =

(Mass)  g Area

3.

2 ⎛ mf ⎞ ⎛ Ai ⎞  1 ⎜⎝ mi ⎟⎠ ⎜⎝ Af ⎟⎠ 

2.

93 92

1 mv 02 , the value of k will be 8

9

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

(1) 10–3 kg m–1

(2) 10–3 kg s–1

(3) 10–4 kg m–1

(4) 10–1 kg m–1 s–1

Answer (3) 1 mv 02 kf 1 8 Sol.   1 ki 4 mv 02 2

v

vf 1  vi 2

(1)

vf 

t

v

(2)

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is

v0 2

 kv 2  v0 2

t

dv

mdv dt t0

∫ v2 ∫

v0

2



0

 kdt m

JEE (MAIN)-2017 (Code-A) v0

k ⎡ 1⎤ 2 ⎢  v ⎥  m t0 ⎣ ⎦v 0

Sol. 2R

l

1 2 k    t0 v0 v0 m 

1 k   t0 v0 m

k



m v 0 t0

102 10  10

m ⎡ 2 2 ⎤ ⎢R  ⎥ 4 ⎣⎢ 3 ⎥⎦

m ⎡ v 2 ⎤ ⎢  ⎥ 4 ⎢⎣  3 ⎥⎦

(1) 4.5 J

(2) 22 J

(3) 9 J

(4) 18 J

v 



2

v

Answer (1)

2 3

23 3

R 2 

dv dt

2



v

R2

0

 R

∫ dv  ∫ 6t dt 1

⎡t2 ⎤ v  6⎢ ⎥ ⎣2⎦

6. 0

= 3 ms–1 W = KE  5.

I

dI m ⎡ v 2 ⎤   0 d 4 ⎢⎣ 2 3 ⎥⎦

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be

Sol. 6t  1

mR 2 m2  4 12



= 10–4 kg m–1 4.

I

1  1  9  4.5 J 2

23 3

3 2 3 2

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle  with the vertical is

z

The moment of inertia of a uniform cylinder of length and radius R about its perpendicular bisector is I. What is the ratio such that the moment of inertia R is minimum? (1)

3 2

(3) 1

(2)

(4)



3 2

x

3

(1)

3g sin  2

(2)

2g sin  3

(3)

3g cos  2

(4)

2g cos  3

2

Answer (1) 3


Answer (1)

Answer (4)

Sol. Torque at angle    Mg sin  

g

2

Sol. gs

, (M, l)





O

Mg

Variation of g inside earth surface d Rg 

= I I = Mg sin 

2

 I 

M 2 3

d Rg 

8.

 sin  g 3 2

7.

Gm R2

d  R  gs 

M 2    Mg sin  3 2



d

d=R

3g sin  2

d

Gm R2

Gm d2

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius) :

g

(1) 800°C

(2) 885°C

(3) 1250°C

(4) 825°C

Answer (2) (1)

Sol. 100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

d O

10t – 750 = 450 + 7650 10t = 1200 + 7650

g

10t = 8850 (2)

t = 885°C

d R

O

9.

g (3)

d R

O g (4)

d O

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and  is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by :

R 4

P K

(1)

P 3 K

(2)

(3)

3 PK

(4) 3PK


Answer (1)

Answer (4) Sol. n1 = initial number of moles

P Sol. K = ⎛ V ⎞ ⎜– V ⎟ ⎝ ⎠

5

n1 =

V P = V K

n2 = final number of moles 5

 = 3

 V = V0 (1 + t)

=

V V0 = t



PV 10  30 3 1 1   1.24  10 RT1 8.3  290

P2V2 10  30 3   1.20  10 RT2 8.3  300

Change of number of molecules : nf – ni = (n2 – n1) × 6.023 × 1023

P P P = t  t = = K 3 K K

 – 2.5 × 1025 12. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like :

10. Cp and Cv are specific heats at constant pressure and constant volume respectively. It is observed that Cp – Cv = a for hydrogen gas Cp – Cv = b for nitrogen gas The correct relation between a and b is : (1) a 

1 b 14

(1)

0

(2)

0

(3)

0

T 2

T

(4)

0

T 4

T 2

(2) a = b

(3) a = 14b

T 2

T

T

(4) a = 28b

Answer (3) Sol. Let molar heat capacity at constant pressure = Xp

T

and molar heat capacity at constant volume = Xv Xp – Xv = R MCp – MCv = R Cp – Cv =

R M

For hydrogen; a = For N2; b =

R 2

R 28

a = 14 b

Answer (4)

a = 14b Sol. K.E 

11. The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf – ni will be (1) –1.61 × 1023 (3) 2.5 ×

1025

(2) 1.38 × 1023 (4) –2.5 ×

1 m2 A2 cos2 t 2

T 4

1025 5

T 2

T


T2  ( p cos i  p sin j )  3E1j

13. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms–1) (1) 10.1 GHz

(2) 12.1 GHz

(3) 17.3 GHz

(4) 15.3 GHz

 k  3 pE1 cos  k From (i) and (ii)

pE sin   3 pE cos 

Answer (3)

tan   3

Sol. For relativistic motion

 = 60°

f = f0

c v c v

;

15. A capacitance of 2 F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 F capacitors are available which can withstand a potential difference of not more than 300 V.

v = relative speed of approach

c 2  10 3  17.3 GHz f = 10 c c 2 c

The minimum number of capacitors required to achieve this is

14. An electric dipole has a fixed dipole moment p , which makes angle  with respect to x-axis. When subjected to an electric field E 1  Ei , it experiences a torque T 1   k . When subjected to another electric

(2) 45°

(3) 60°

(4) 90°

(1) 2

(2) 16

(3) 24

(4) 32

Answer (4) Sol. Following arrangement will do the needful : 8 capacitors of 1F in parallel with four such branches in series.

field E 2  3E1j it experiences a torque T 2  T 1 . The angle  is (1) 30°

…(ii)

1 2 3 8

1 2 3 8

1 2 3 8

1 2 3 8

Answer (3) Sol.

y

1000 V 8 F 8 F 8 F 8 F

p



250 V 250 V 250 V 250 V

x

1000 V 16. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

z p  p cos i  p sin j

E

E 1  Ei

r r1

T 1  p  E1

C r2



= ( p cos i  p sin j )  E i

 

 k  pE sin   k

(1) CE

(2) CE

r1 (r2  r )

(4) CE

r1 (r1  r )

…(i) (3) CE

E 2  3E1j 6

r2 (r  r2 )


Answer (3)

Answer (1)

Sol. In steady state, flow of current through capacitor will be zero.

Sol. T = 2π

E

r

i

= 2π

r1 C

 0.01

=

2  1.06 10

= 6.6568 ≈ 6.65 s 19. When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10 V is

Er2C VC = i r2C = r  r 2

17.

6.7  10

–2

t = 10T = 2π × 1.06

E r  r2

VC = CE

7.5  10 –6

For 10 oscillations,

r2 i=

I MB

r2 r  r2

2V

(1) 1.985 × 103 Ω (2) 2.045 × 103 Ω

2V

2V

2V

(4) 4.005 × 103 Ω Answer (1)

1

1

1

(3) 2.535 × 103 Ω

2V

Sol. ig = 5 × 10–3 A G = 15 Ω

2V

Let series resistance be R.

In the above circuit the current in each resistance is

V = ig (R + G)

(1) 1 A

10 = 5 × 10–3 (R + 15)

(2) 0.25 A

R = 2000 – 15 = 1985 = 1.985 × 103 Ω 20. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

(3) 0.5 A (4) 0 A Answer (4) Sol. The potential difference in each loop is zero.

10

∴ No current will flow.

Current (amp.)

18. A magnetic needle of magnetic moment 6.7 × 10–2 Am2 and moment of inertia 7.5 × 10–6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is

Time

(1) 6.65 s

(1) 200 Wb

(2) 8.89 s

(2) 225 Wb

(3) 6.98 s

(3) 250 Wb

(4) 8.76 s

(4) 275 Wb 7

0.5 s


Answer (3)

Answer (1)

d Sol. ε = dt

Sol. In X-ray tube

iR =

min 

d dt

hc eV

⎛ hc ⎞ ln min  ln ⎜ ⎟  lnV ⎝ e ⎠

∫ d   R ∫ idt

Slope is negative

Magnitude of change in flux = R × area under current vs time graph = 100 ×

Intercept on y-axis is positive

log min

1 1 × × 10 2 2

= 250 Wb 21. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in

log V 22. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is

(1) log min

(1) Real and at a distance of 40 cm from convergent lens

log V

(2) Virtual and at a distance of 40 cm from convergent lens (3) Real and at a distance of 40 cm from the divergent lens

(2) log min

(4) Real and at a distance of 6 cm from the convergent lens

log V

Answer (1) f1 = 25 cm

f2 = 20 cm

Sol. (3) log min I1

log V

25 cm

15 cm

For converging lens

(4) log min

u = –40 cm which is equal to 2f  Image will be real and at a distance of 40 cm from convergent lens.

log V 8


Answer (2)

23. In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is



⎡v ⎤ p1  m. ⎢ ⎥ ⎣3 ⎦

v2 

2m1v 0 m1  m2

(1) 1.56 mm

(3) 9.75 mm =

(4) 15.6 mm Answer (2) Sol. For 1 y

m1D d



m 2 4   n 1 5

p2 

For 2 y

n 2 D d

m ⎡ 4v ⎤ 2mv  2 ⎢⎣ 3 ⎥⎦ 3

 de-Broglie wavelength

–E 4  E 3

m1D , 1 = 650 nm d

A 1  B 2

(1) r 

4 3

(2) r 

2 3

(3) r 

3 4

(4) r 

1 3

Answer (4) Sol. From energy level diagram

A (2) 2 B

(4)

1

–3 E

A 1  B 3

A 2  B 3

2

–2 E

24. A particle A of mass m and initial velocity v collides m with a particle B of mass which is at rest. The 2 collision is head on, and elastic. The ratio of the de-Broglie wavelengths A to B after the collision is

(3)

 A p2   2 :1 B p1

25. Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = 1/2, is given by

= 7.8 mm

(1)

m 2

4v 3

For 1 y

m2 

v 3

=

(2) 7.8 mm

m1  m

(m1  m2 )v 0 m1  m2

Sol. v1 

1  2 

 9

hc E hc ⎛E ⎞ ⎜3⎟ ⎝ ⎠

1 1  2 3


28. In amplitude modulation, sinusoidal carrier frequency

26. A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by

used is denoted by c and the signal frequency is denoted by m . The bandwidth ( m ) of the signal is such that m  c . Which of the following frequencies is not contained in the modulated wave?

T log2 (1) t  2 log1.3 log1.3 (2) t  T log2

Answer (2)

(4) c – m

(2) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed

N0  N0 e t  0.3 N0 e t

(3) A rheostat can be used as a potential divider (4) Kirchhoff’s second law represents energy conservation Answer (2) Sol. In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged. 30. The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m. Using g = 9.80 m/s 2 and the simplified relation

 et = 1.3  t = ln 1.3

⎛ ln 2 ⎞ ⎜ T ⎟ t  ln 1.3 ⎝ ⎠

t T

(3) m  c

29. Which of the following statements is false? (1) Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude

T (4) t  log 1.3 

t  T.

(2) c

Answer (1) Sol. Modulated wave has frequency range. c ± m  Since c >> m  m is excluded.

(3) t  T log 1.3 

Sol.

(1) m

ln(1.3) ln 2

log(1.3) log2

rhg  103 N / m , the possible error in surface 2 tension is closest to T 

27. In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be

(1) 0.15% (3) 2.4% Answer (2)

(1) 45°

Sol.

(2) 90°

T D h  100   100   100 T D h

=

0.01 0.01  100   100 1.25 1.45

Answer (4)

=

100 100  125 145

Sol. In common emitter configuration for n-p-n transistor, phase difference between output and input voltage is 180°.

= 0.8 + 0.689 = 1.489

(3) 135° (4) 180°

1.5% 10

(2) 1.5% (4) 10%


PART–B : CHEMISTRY 33. U is equal to

31. Given C(graphite) + O2(g)  CO2(g);

(1) Adiabatic work

rHº = –393.5 kJ

(2) Isothermal work

H2(g) +

mol–1

1 O2 (g)  H2O(l); 2

rHº = –285.8 kJ

(3) Isochoric work (4) Isobaric work

mol–1

Answer (1)

CO2(g) + 2H2O(l)  CH4(g) + 2O2(g);

Sol. For adiabatic process, q = 0

rHº = +890.3 kJ mol–1

 As per 1st law of thermodynamics,

Based on the above thermochemical equations, the value of rHº at 298 K for the reaction

U = W 34. The Tyndall effect is observed only when following conditions are satisfied

C(graphite) + 2H2(g)  CH4(g) will be (1) –74.8 kJ mol–1

(2) –144.0 kJ mol–1

(3) +74.8 kJ mol–1

(4) +144.0 kJ mol–1

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.

Answer (1)

(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used

 CO2(g);

Sol. C(graphite) + O2(g)

rH° = –393.5 kJ mol–1 ...(i) H2 (g) 

1 O2 (g) 2

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude

 H2O(l); rH° = –285.8 kJ mol–1 ...(ii)

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude

 CH4(g) + 2O2(g);

CO2(g) + 2H2O(l)

rH° = 890.3 kJ mol–1

...(iii)

(1) (a) and (c)

By applying the operation

(2) (b) and (c)

(i) + 2 × (ii) + (iii), we get  CH4(g);

C(graphite) + 2H2(g)

(3) (a) and (d)

rH° = –393.5 –285.8 × 2 + 890.3 = –74.8 kJ

(4) (b) and (d)

mol–1

Answer (4)

32. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is (1) 118.6

(2) 11.86

(3) 1186

(4) 84.3

Sol. For Tyndall effect refractive index of dispersion phase and dispersion medium must differ significantly. Secondly, size of dispersed phase should not differ much from wavelength used. 35. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be

Answer (4) Sol. M2CO3 + 2HCl  2MCl + H2O + CO2

nM2CO3  nCO2 1 MM2CO3

(1)

 0.01186

MM2CO3 =

(2)

1 0.01186

2a

a 2

(3) 2a (4) 2 2a

= 84.3 g/mol 11


Answer (2)

Answer (2) Sol. In FCC, one of the face is like

Sol. 0.45 = i(5.12)

A

 i = 0.527 2CH3 COOH  CH3COOH2

a C

0.2 / 60  1000 20

1– 

B  i  1–

By ABC,

 2

 2

2a2 = 16r2  0.527 = 1 –

1  r 2  a2 8

 r

1 2 2



a

Distance of closest approach = 2r =

2 /Cl

–

 = 0.473 2

  = 0.946

a

 % association = 94.6%

2

36. Given º ECl

 2

38. The radius of the second Bohr orbit for hydrogen atom is

º  1.36 V, ECr  –0.74 V 3 /Cr

(Planck's Const. h = 6.6262 × 10–34 Js;

º º ECr  1.33 V, EMnO  1.51 V – O2– /Cr 3  /Mn2 

mass of electron = 9.1091 × 10–31 kg;

Among the following, the strongest reducing agent is

permittivity of vacuum

(1) Cr3+

(2) Cl–

0 = 8.854185 × 10–12 kg–1 m–3 A2)

(3) Cr

(4) Mn2+

(1) 0.529 Å

2

7

4

charge of electron e = 1.60210 × 10–19 C;

Answer (3) Sol. For Cr3+ ,

(2) 2.12 Å (3) 1.65 Å

E°Cr 3+ /Cr O2– = – 1.33 V 2 7

(4) 4.76 Å

For Cl–,

E°Cl– /Cl = – 1.36 V 2

Answer (2)

For Cr,

E°Cr/Cr 3 = 0.74 V

Sol. r = a0

For Mn2 , E°Mn2 /MnO – = – 1.51 V

n2 = 0.529 ×4 Z

= 2.12 Å

4

39. Two reactions R 1 and R 2 have identical preexponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k 1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to

Positive E° is for Cr, hence it is strongest reducing agent. 37. The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be

(R = 8.314 J mole–1 K–1) (1) 6

(Kf for benzene = 5.12 K kg mol–1)

(2) 4

(1) 74.6%

(2) 94.6%

(3) 8

(3) 64.6%

(4) 80.4%

(4) 12 12


Answer (2) Sol. k1 = Ae k2 = Ae

42. Which of the following species is not paramagnetic? (1) O2

–Ea /RT 1

(2) B2 –Ea /RT

(3) NO

2

(4) CO

1 k2 Ea – Ea2  e RT 1 = k1

Answer (4) Sol. CO has 14 electrons (even)  it is diamagnetic

Ea – Ea2 k ln 2 = 1 k1 RT

NO has 15e–(odd)  it is paramagnetic and has 1 unpaired electron in 2p molecular orbital. B2 has 10e– (even) but still paramagnetic and has two unpaired electrons in 2p x and 2p y (s-p mixing).

10  103 4 = 8.314  300

O2 has 16 e– (even) but still paramagnetic and has two unpaired electrons in *2px and *2py molecular orbitals.

40. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively, The pH of their salt (AB) solution is

43. Which of the following reactions is an example of a redox reaction?

(1) 7.0 (2) 1.0 (3) 7.2

(1) XeF6 + H2O  XeOF4 + 2HF

(4) 6.9

(2) XeF6 + 2H2O  XeO2F2 + 4HF (3) XeF4 + O2F2  XeF6 + O2

Answer (4)

(4) XeF2 + PF5  [XeF]+ PF6–

1 Sol. pH = 7 +  pK a – pK b  2

Answer (3)

1 = 7 +  3.2 – 3.4 2

Sol. Xe is oxidised from +4(in XeF4) to +6(in XeF6)

= 6.9

44. A water sample has ppm level concentration of following anions

Oxygen is reduced from +1 (in O2F2) to zero (in O2)

41. Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the one which is incorrect, is

F– = 10; SO42– = 100; NO3– = 50

(1) Both form nitrides

The anion/anions that make/makes the water sample unsuitable for drinking is/are

(2) Nitrates of both Li and Mg yield NO2 and O2 on heating

(1) Only F– (2) Only SO42–

(3) Both form basic carbonates

(3) Only NO3–

(4) Both form soluble bicarbonates

(4) Both SO42– and NO3–

Answer (3)

Answer (1)

Sol. Mg forms basic carbonate

–

3MgCO3 · Mg  OH2 ·3H2O

Sol. Permissible limit of F in drinking water is upto – 1 ppm. Excess concentration of F  10 ppm causes decay of bones.

but no such basic

carbonate is formed by Li. 13


45. The group having isoelectronic species is

Answer (2)

(1) O2–, F–, Na, Mg2+

Sol. Na2C2O4  H2SO4   Na2SO4  H2 C2 O4 ( X)

(2) O–, F–, Na+, Mg2+ (3) O2–, F–, Na+, Mg2+

Conc.

oxalic acid

Conc. H SO

2 4 H2 C2 O 4   CO  CO2  

–H2O

(4) O–, F–, Na, Mg+ Answer (3)

(effervescence)

Na2 C2 O 4  CaCl2  CaC2 O 4   2NaCl ( X)

Sol. Mg2+, Na+, O2– and F– all have 10 electrons each.

 2 2MnO4–  5C2O2–  10CO2  8H2O 4  16H  2Mn

46. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are (1) Cl– and ClO–

(2) Cl– and ClO2–

(3) ClO– and ClO3–

(4) ClO2– and ClO3–

white ppt.

49. The most abundant elements by mass in the body of a healthy human adult are :

Answer (1)

Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%) and Nitrogen (2.6%).

Sol. Cl2  2NaOH  NaCl  NaOCl  H2 O

The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

Cold & dilute

Sodium hypochlorite

(1) 7.5 kg (2) 10 kg

47. In the following reactions, ZnO is respectively acting as a/an

(3) 15 kg

(a) ZnO + Na2O  Na2ZnO2

(4) 37.5 kg

(b) ZnO + CO2  ZnCO3

Answer (1)

(1) Acid and acid Sol. Mass of hydrogen =

(2) Acid and base

10  75 = 7.5 kg 100

(3) Base and acid

Replacing 1H by 2H would replace 7.5 kg with 15 kg

(4) Base and base

 Net gain = 7.5 kg 50. On treatment of 100 mL of 0.1 M solution of CoCl3  6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is

Answer (2) Sol. In (a), ZnO acts as acidic oxide as Na2O is basic oxide.

(1) [Co(H2O)6]Cl3

In (b), ZnO acts as basic oxide as CO2 is acidic oxide.

(2) [Co(H2O)5Cl]Cl2  H2O (3) [Co(H2O)4Cl2]Cl  2H2O

48. Sodium salt of an organic acid 'X' produces effervescence with conc. H2SO4. 'X' reacts with the acidified aqueous CaCl2 solution to give a white precipitate which decolourises acidic solution of KMnO4. 'X' is

(4) [Co(H2O)3Cl3]  3H2O Answer (2) Sol. Millimoles of AgNO3 =

(1) CH3COONa

1.2  1022 6  1023

 1000 = 20

Millimoles of CoCl3·6H2O = 0.1 × 100 = 10

(2) Na2C2O4 (3) C6H5COONa

 Each mole of CoCl3·6H2O gives two chloride ions.

(4) HCOONa

 [Co(H2O)5Cl]Cl2·H2O 14


51. Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

Answer (3)

CH3

NH2

+

–

Na O

C

O

(1)

Sol.

CH3

CH3

Br

NHCOCH3 CH3

(2)

O

O

C

CH3

CH3

OH (Product)

(3) The above product does not have any C = C or C  C bond, so, it will not give Br2-water test.

OCOCH3

53. The formation of which of the following polymers involves hydrolysis reaction?

(4)

(1) Nylon 6, 6 (2) Terylene

Answer (1)

(3) Nylon 6

NH2

NH3

(4) Bakelite

H+

Sol.

Answer (3)

NO2

Sol. Caprolactam is hydrolysed to produce caproic acid which undergoes condensation to produce Nylon-6.

NH3

(51%)

NH3

NO2

+

NH3

NO2

O

NO2

+

NH (Caprolactam)

52. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine?

(1)

N

Br O (2)

(2)

O

Br

O (3)

Br

C6H5 (4)

(CH2)5 – NH2 (Caproic acid)

54. Which of the following molecules is least resonance stabilized?

O

(3)

C

H3O+

HO

(2%)

(47%)

(1)

O

(4)

Br

O 15


Answer (2)

Answer (4)

Sol. However, all molecules given in options are stabilised by resonance but compound given in option (2) is least resonance stabilised (other three are aromatic)

Br Sol.

C6H5

C6H5

(+)

–

O

H

O

t-BuOK  (E-2)

C6H5 C6H5

57. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

O

HOH2C 55. The increasing order of the reactivity of the following halides for the SN1 reaction is I.

HO OCH 3

(1)

OH

CH 3CHCH 2CH 3 Cl

II.

CH2OH

HOH2C

O CH OCH 2 3

CH3CH2CH2Cl

III. p–H3CO – C6H4 – CH2Cl (1) (I) < (III) < (II)

(2) (II) < (III) < (I)

(3) (III) < (II) < (I)

(4) (II) < (I) < (III)

(2)

OH

OH OH

Answer (4)

I.

CH3 – CH – CH2 – CH3

(3)

CH3 – CH – CH2 – CH3

CH2OH

HO OCOCH 3

Cl

II.

O

HOH2C

Sol. Rate of SN1 reaction  stability of carbocation

OH

CH3 – CH2 – CH2 – Cl CH2 – Cl

CH3 – CH2 – CH2

O

HOH2C (4)

CH2

CH2OH

HO

III.

OH OCH3

OCH3

Answer (3) Sol. Sugars in which there is free anomeric –OH group are reducing sugars

So, II < I < III Increase stability of carbocation and hence increase reactivity of halides.

OH CH2 O

56. The major product obtained in the following reaction is

Br C6H5

H

C6H5

CH2 – OH O

KOH(aq.)

HO O – C– CH 3 OH

t-BuOK 

OH

(+)

CH2 O

(1) (+)C6H5CH(OtBu)CH2C6H5

HO

(2) (–)C6H5CH(OtBu)CH2C6H5 (3) (±)C6H5CH(OtBu)CH2C6H5

OH

(4) C6H5CH = CHC6H5 16

CH2 – OH OH

+ CH3COOK

Free anomeric group


58. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

O

(i) CH3MgBr (3 moles) (ii) H2O

HO

CH3

(1) Two (2) Four

O

(3) Six (4) Zero

HO – C

C – OCH3

CH3 CH3

60. The major product obtained in the following reaction is

Answer (2)

CH3 Sol. CH3 – CH = C – CH2 – CH3 3-methyl pent-2-ene

O

O

HBr R2O2

DIBAL-H

COOH

CH3 CH3 – *CH –*C – CH2 – CH3 Br

H

CHO

(1)

Product (X)

COOH

Since product (X) contains two chiral centres and it is unsymmetrical.

CHO (2)

So, its total stereoisomers = 22 = 4.

CHO

59. The correct sequence of reagents for the following conversion will be

OH

O

HO

CH3

CHO

(3)

COOH

HO

CHO

CH3 CH3

OH

(1) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH

CHO

(4)

CHO

(2) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (3) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr

Answer (4)

(4) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–

Sol. DIBAL — H reduces esters and carboxylic acids into aldehydes

Answer (3)

O

O [Ag(NH3)2]+OH–

O

O

CH3 – OH/H+

Sol.

DIBAL-H

Esterification

H

C=O

HO

OH H

C=O

COOH 17

CHO

O


PART–A : MATHEMATICS

61. The

⎡ 1 1⎤ f : R  ⎢ , ⎥ ⎣ 2 2⎦

function

f (x) 

x 1 x

Given difference of roots = 1 defined

as

 || = 1  D1

, is

2

4 2 (n  31)  1 3

 n2 

(1) Injective but not surjective (2) Surjective but not injective

So, n = 11

(3) Neither injective nor surjective

63. Let  be a complex number such that 2 + 1 = z

(4) Invertible

where z  3 . If

Answer (2) Sol. f ( x ) 

f ( x ) 

x

1

1 x2 2

(1  x )  1  x  2 x 2 2

(1  x )



1 x

1

1

1 2  1 2  3k , then k is equal to

2

1

2 2

(1  x )

2

7

f(x) changes sign in different intervals.

(1) z

(2) –1

 Not injective.

(3) 1

(4) –z

y

x

Answer (4)

1 x2

Sol. 2 + 1 = z , z  3 i

yx 2  x  y  0 For y  0



⎡ 1 1⎤ D  1  4 y 2  0 ⇒ y  ⎢  , ⎥  {0} ⎣ 2 2⎦

1  3i Cube root of unity. 2 C1  C1 + C2 + C3

For, y = 0 x = 0

1

 Part of range

1

1 2

1 1  

⎡ 1 1⎤  Range : ⎢  , ⎥ ⎣ 2 2⎦  Surjective but not injective.

1

2

1 1

2

1

3

1

1

  1    0  2 7

2

1 2



0 2



= 3 (2 – 4)

62. If, for a positive integer n, the quadratic equation,

⎡⎛ 1  3i ⎞ ⎛ 1  3i ⎞ ⎤ ⎟⎜ ⎟⎥ = 3 ⎢⎜ 2 ⎠ ⎝ 2 ⎠⎦ ⎣⎝

x( x  1)  ( x  1)( x  2)  ...  ( x  n  1)( x  n )  10n has two consecutive integral solutions, then n is equal to (1) 9

(2) 10

= 3 3i

(3) 11

(4) 12

= –3z  k = –z

Answer (3) Sol. Rearranging equation, we get

⎡ 2 3 ⎤ 2 64. If A  ⎢ ⎥ , then adj (3A + 12A) is equal to ⎣ 4 1 ⎦

nx 2  1  3  5  ....  (2n  1) x  1 2  2  3  ...  ( n  1)n  10n

(n  1)n(n  1) ⇒ nx 2  n 2 x   10n 3

⎛ n 2  31 ⎞ ⇒ x 2  nx  ⎜ ⎟0 ⎝ 3 ⎠

18

⎡ 51 63 ⎤ (1) ⎢ ⎥ ⎣84 72 ⎦

⎡ 51 84 ⎤ (2) ⎢ ⎥ ⎣ 63 72 ⎦

⎡ 72 63 ⎤ (3) ⎢ ⎥ ⎣ 84 51 ⎦

⎡ 72 84 ⎤ (4) ⎢ ⎥ ⎣ 63 51 ⎦


Answer (1)

Eq. (1) & (2) are identical i.e.,x + y + z = 1

⎡ 2 3 ⎤ Sol. A  ⎢ ⎥ ⎣ 4 1 ⎦

To have no solution with x + by + z = 0.

A  I 

b=1

2

3

4

1 

66. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is

= (2 – 2– + 2) – 12

f ( )   2  3  10 ∵ A satisfies f ( )  A2 – 3A –10I = 0

(1) 468

(2) 469

(3) 484

(4) 485

Answer (4)

A2 – 3A = 10I

Sol. X(4 L 3 G)

Y(3 L 4 G)

3A2 – 9A = 30I

3L0G

0L3G

3A2 + 12A = 30I + 21A

2L1G

1L2G

1L2G

2L1G

0L3G

3L0G

⎡30 0 ⎤ ⎡ 42 63 ⎤ ⎢ ⎥⎢ ⎥ ⎢⎣ 0 30 ⎥⎦ ⎢⎣ 84 21 ⎥⎦

Required number of ways

⎡ 72 63 ⎤ ⎢ ⎥ ⎣⎢ 84 51 ⎦⎥

= 4C3  4C3 



4

C2  3C1

  2

4

C1  3C2

  C  2

3

2

3

= 16 + 324 + 144 + 1

⎡ 51 63 ⎤ adj(3 A2  12 A)  ⎢ ⎥ ⎢⎣84 72 ⎥⎦

= 485 67. The value of

( 21C1  10C1 )  ( 21C2  10C2 )  ( 21C3  10C3 ) 

65. If S is the set of distinct values of b for which the following system of linear equations

( 21C4  10C4 )  ...  ( 21C10  10C10 ) is

x y z 1

(1) 221 – 210

x  ay  z  1

(2) 220 – 29 (3) 220 – 210

ax  by  z  0

(4) 221 – 211 Answer (3)

has no solution, then S is (1) An infinite set

Sol.

C1  21C2  ...  21C10 

(2) A finite set containing two or more elements (3) A singleton



(4) An empty set





1 21 C0  21C1  ...  21C21  1 2 = 220 – 1

21

10



C1  10C2  ...  10C10  210  1

 Required sum = (220 – 1) – (210 – 1)

Answer (3)

= 220 – 210

Sol.

68. For any three positive real numbers a, b and c,

1 1 1

9(25a2  b2 )  25(c 2  3ac )  15b(3a  c ).

1 a 10

Then

a b 1

(1) b, c and a are in A.P.

 –(1 – a)2 = 0

(2) a, b and c are in A.P.

 a=1

(3) a, b and c are in G.P.

For a = 1

(4) b, c and a are in G.P. 19


Answer (1) 70.

Sol. 9(25a2  b2 )  25(c 2  3ac )  15b (3a  c )  (15a )2  (3b)2  (5c )2  45ab  15bc  75ac  0

lim

1 4 Answer (1)

15a  3b  0 and 3b  5c  0 and 15a  5c  0

Sol. lim

1 8

(4)

1 24

(   2 x )3

 x t 2 tan t  sin t lim t 0 8t 3

a b c   1 5 3

Put,

 b, c, a are in A.P. 69. Let a, b, c  R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and

= lim =

10

∑ f (n )

n 1

(2) 190

(3) 255

(4) 330

Answer (4)

(1)

Sol. As, f ( x  y )  f ( x )  f ( y )  xy

f ( x ) 

f (1)  f (2)  f (3)  ...  f (10)

g( x ) 

= 3 + 7 + 12 + 18 + ... = S (let)

Again, Sn  3  7  12  ...  t n 1  t n

n 1

(4)

3x 1 9x3

9 1  9x 3

1⎞ ⎛ For x  ⎜ 0, ⎟ 4⎠ ⎝

9 x 1 9x3

9 1 9x3

⎛ 1 1⎞ (3) ⎜ , ⎟ ⎝2 3⎠ Answer (1)

n (n  5) = 2 1

3

3

⎛ 1 1⎞ (1) ⎜ , ⎟ ⎝2 2⎠

We get, t n  3  4  5  ... n terms

∑ tn = 2 ∑ n 2  5 ∑ n =

(2)

72. The normal to the curve y ( x  2)( x  3)  x  6 at the point where the curve intersects the y-axis passes through the point

Now, Sn  3  7  12  18  ...  t n

n

1 9x

3

Sol. f ( x )  2 tan1(3 x x )

Similarly, x  1, y  2  f (3)  f (1)  f (2)  2  12

n 1

3x x

1  9x Answer (4)

Putting, x  y  1  f (2)  2f (1)  1  7

∑ f (n ) =

1 . 16

(3)

Given, f (1)  3

10

8t 3

t 2

1 ⎛ 6 x x ⎞ 1⎞ ⎛ ⎟ is 71. If for x  ⎜ 0, ⎟ , the derivative of tan ⎜ ⎝ 1  9x 3 ⎠ 4⎠ ⎝ x  g ( x ) , then g(x) equals

is equal to

(1) 165

sin t  2 sin2

t 0

f ( x  y )  f ( x )  f ( y )  xy ,  x, y  R,

So, S10 =

(2)

cot x  cos x

 x 2

15a  3b  5c

i.e., Sn =

equals

(3)

It is possible when

Now,

(   2)3 1 16

(1)

 (15a  3b)2  (3b  5c )2  (15a  5c )2  0

then

cot x  cos x

 x 2

⎛ 1 1⎞ (2) ⎜ ,  ⎟ ⎝2 3⎠ ⎛ 1 1⎞ (4) ⎜  ,  ⎟ ⎝ 2 2⎠

Sol. y ( x  2)( x  3)  x  6

n (n  1)( n  8) 6

At y-axis, x = 0, y = 1 Now, on differentiation.

10  11 18  330 6

dy ( x  2)( x  3)  y (2 x  5)  1 dx

20


Answer (1)

dy (6)  1( 5)  1 dx

n Sol. In  ∫ tan xdx, n  1

dy 6  1 dx 6 Now slope of normal = –1 Equation of normal y – 1 = –1(x – 0) y+x–1=0 ... (i)

I4  I6  ∫ (tan4 x  tan6 x )dx  ∫ tan4 x sec 2 xdx

Let tanx = t

⎛ 1 1⎞ Line (i) passes through ⎜ , ⎟ ⎝2 2⎠ 73. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is

(1) 10

(2) 25

(3) 30

(4) 12.5

sec2x dx = dt  ∫ t 4 dt 

t5 C 5



1 tan5 x  C 5

Answer (2)

r

Sol.

r



a

1 ,b 0 5 3 4

r

2r  r  20

A = area =

... (ii)

r 2 ⎛ 20  2r ⎞ A ⎜ ⎟ r ⎠ 2⎝

(2) 4

(3) –1

(4) –2

3 4

Sol.

A to be maximum

∫  4

dx x 2cos 2 2

dx 

1 2

3 4

∫ sec  4

2

x dx 2

3

dA  10  2r  0 ⇒ r  5 dr

dr 2

(1) 2 Answer (1)

⎛ 20r  2r 2 ⎞ 2 A⎜ ⎟  10r  r ⎝ 2 ⎠

d2A

is equal to

 4

... (i)

 r 2  r 2  2 2

dx

∫ 1  cos x

75. The integral

x⎤4 ⎡ tan ⎥ 1⎢ 2  ⎢ ⎥ 2⎢ 1 ⎥ ⎣ 2 ⎦

 2  0

4

 tan

Hence for r = 5, A is maximum Now, 10 + ·5 = 20  = 2 (radian)

⎡  1  cos ⎢  4  ⎢ tan   8 ⎢ 1  cos ⎢⎣ 4

2   5 2  25 sq m Area = 2 n

74. Let In  ∫ tan xdx,(n  1) . If

3 3 4  tan  3 8 1  cos 4

I4  I6  a tan5 x  bx 5  C, where C is a constant of integration, then the ordered pair (a, b) is equal to ⎛1 ⎞ (1) ⎜ ,0 ⎟ ⎝5 ⎠

⎛1 ⎞ (2) ⎜ , 1⎟ ⎝5 ⎠

⎛ 1 ⎞ (3) ⎜  ,0 ⎟ ⎝ 5 ⎠

⎛ 1 ⎞ (4) ⎜  ,1⎟ ⎝ 5 ⎠

3   tan 8 8

1  cos

 ( 2  1)  ( 2  1) 2 21

2 1 2 1



2 1 1

⎤ ⎥ 2 1  2  1⎥ 2 1 ⎥ ⎥⎦


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

(1  1)  2  C  C = 4

{(x, y) : x  0, x + y  3, x2  4y and y  1  x }

Now, ( y  1)(2  sin x )  4

is

For, x 

3 (1) 2

7 (2) 3

5 (3) 2

59 (4) 12

( y  1)(2  1)  4 y 1

Answer (3) y

Sol.

y ) ,2 (1

x

O

x=1 x=2

x+

x y=

3

x=0 y

Area of shaded region 2 ⎛ ⎛ x2 ⎞ x2 ⎞  ∫ ⎜ x  1 ⎟ dx  ∫ ⎜ (3  x )  ⎟ dx ⎝ 4 ⎠ ⎝ 4 ⎠ 0 1

(3)

2 3

1⎞ ⎛ (4) ⎜ 2,  ⎟ ⎝ 2⎠

(4)

5k 2  13k  10   56

1 3

5k 2  13k  46  0

1 3

5K 2  13K  66  0

5k 2  13k  46  0

Answer (4) Sol. (2  sin x )

1⎞ ⎛ (3) ⎜ 2, ⎟ 2⎠ ⎝

(k 2  7k  10)  4k 2  20k   56 (2) 

4 3

3⎞ ⎛ (2) ⎜ 1,  ⎟ ⎝ 4⎠

k  5 4k 0 5  k k  2 0   56 k 2 1

dy   y  1 cos x  0 and y(0) = 1, then dx

⎛ ⎞ y ⎜ ⎟ is equal to ⎝ 2⎠ (1) 

⎛ 3⎞ (1) ⎜⎝ 1, ⎟⎠ 4

k 3 k 1 1 k 1  28 5 Sol. Area = 2 k 2 1

5 sq. unit 2

77. If  2  sin x 

4 1 1 3 3

Answer (3)

1



4 3

78. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point

(2, 1)

(0, 1)

 2

k = dy  ( y  1)cos x  0 dx

13  169  920 10

= 2, –4.6 reject

⎛⎞ y (0)  1, y ⎜ ⎟  ? ⎝2⎠

For k = 2

A (2, –6) =

8

1 cos x dy  dx  0 y 1 2  sin x

m

E

ln| y  1|  ln(2  sin x )  ln C

(5, 2) B

( y  1)(2  sin x )  C

D m=0

Put x = 0, y = 1 22

m= –2 C (–2, 2)


Equation of AD, x=2

 r 

...(i)

k 2

Also equation of BE, Equation of circle is x 2  ( y  k )2 

1 y  2  ( x  5) 2

It passes through point P

2y  4  x  5

2

x  2y  1  0

⎞ t2 ⎛ t2 k2  ⎜⎜ 4   k ⎟⎟  4 ⎝ 4 2 ⎠

...(ii)

Solving (i) & (ii), 2y = 1

t 4  t 2 (8k  28)  8k 2  128k  256  0

1 y 2

For t = 0 

k 84 2



79. The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is (1) 2  2  1

(2) 4  2  1

(3) 4  2  1

(4) 2  2  1

2

 4( 2  1) (discarding 4( 2  1) ) ...(iii)

2k 2  4k  15  0

k

2  34 2



r

P x

k 2



17  2 2

(Ignoring negative

...(iv)

value of r)

From (iii) & (iv),

rmin 

x 2  ( y  4) ⎛t t2 Let a point on the parabola P ⎜⎜ , 4  4 ⎝2 Equation of normal at P is



k

(14  4k )2  (14  4k )(8k  28)  8k 2  128k  256  0

y

y

r 

For t   14  4k

Answer (2)

O

...(ii)

k 2  16k  32  0

⎛ 1⎞ Orthocentre is ⎜ 2, ⎟ ⎝ 2⎠

Sol.

k2 2

⎞ ⎟⎟ ⎠

17  2 2

But from options, r  4( 2  1) y (0, 4)

2

t 1⎛ t⎞ 4  ⎜x  ⎟ 4 t⎝ 2⎠ (0, 0)

t3 7 x  ty   t  0 4 2

80. The eccentricity of an ellipse whose centre is at the

It passes through centre of circle, say (0, k)

t3 7 tk   t  0 4 2

origin is

...(i)

0k 2

1 . If one of its directrices is x = – 4, then 2

⎛ 3⎞ the equation of the normal to it at ⎜ 1, ⎟ is ⎝ 2⎠

t = 0, t 2  14  4k Radius = r 

x

(Length of perpendicular

(1) 4x – 2y = 1

(2) 4x + 2y = 7

from (0, k) to y = x)

(3) x + 2y = 4

(4) 2y – x = 2

23


Answer (1)

 x2 

Sol.

y2 1 3

 Tangent at P ( 2, 3) is

2x 

y 3

1

Clearly it passes through (2 2, 3 3)

x = –4

82. The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal x 1 y  2 z  4   perpendicular to both the lines 1 2 3 x  2 y 1 z  7   and , is 2 1 1

1 e 2 a  4 e a  4  e

(1)

a2 Now, b2  a2 (1  e2 )  3

(2)

Equation to ellipse

x2 y 2  1 4 3

(3)

Equation of normal is

(4)

3 y x 1 2  4 x  2y  1  0  1 3 4 23

10 83 5 83 10 74 20 74

Answer (1) Sol. Let the plane be a( x  1)  b( y  1)  c( z  1)  0

81. A hyperbola passes through the point P ( 2, 3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point

It is perpendicular to the given lines a – 2b + 3c = 0

(1) (2 2, 3 3)

2a – b – c = 0

(2) ( 3, 2)

Solving, a : b : c = 5 : 7 : 3 (3) (  2,  3)

 The plane is 5x + 7y + 3z + 5 = 0

(4) (3 2, 2 3) Distance of (1, 3, –7) from this plane =

Answer (1) Sol.

a2  b2  4 2 3  1 a2 b2

2 4b

2



3 b

2

83

83. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x y z   is Q, then PQ is equal to 1 4 5

x2 y 2  1 a2 b2

and

10

(1) 2 42 (2)

1

42

 b 3

(3) 6 5

 a2  1

(4) 3 5

2

24


Answer (1)

(1) 6

Sol. Equation of PQ,

(2) 4

x 1 y  2 z  3   1 4 5

Let M be (  1, 4  2, 5  3)

(3)

6 25

(4)

12 5

P

M

Answer (4) Sol. n = 10 p(Probability of drawing a green ball) =

Q

3 2 , q 5 5 var(X) = n.p.q

 p

As it lies on 2x + 3y – 4z + 22 = 0 =1 For Q,  = 2

6 12  25 5 86. For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = 10 

Distance PQ  2 12  42  52  2 42 84. Let a  2iˆ  ˆj  2kˆ and b  iˆ  ˆj . Let c be a vector

= P (Exactly one of C or A occurs) =

such that | c  a | 3, (a  b )  c  3 and the angle between c and a  b be 30°. Then a  c is equal to

1 . 16 Then the probability that at least one of the events occurs, is

(2) 5

(4)

1 8

25 8

(1)

7 16

(2)

7 64

(3)

3 16

(4)

7 32

Answer (1)

Answer (1) Sol. | (a  b )  c |  3

1 and 4

P(All the three events occur simultaneously) =

(1) 2

(3)

15 25

a  b  2iˆ  2 ˆj  kˆ

⇒ | a  b | | c | sin 30  3

|a |  3  ab

⇒ |c | 2 |c a | 3

Sol. P ( A)  P (B )  P ( A  B ) 

1 4

P (B )  P (C )  P (B  C ) 

1 4

P (C )  P ( A)  P ( A  C ) 

1 4

P ( A)  P (B )  P (C )  P ( A  B )  P (B  C )

⇒ | c |2  | a |2 2(a  c )  9

 P( A  C ) 

932 a c  2 2

∵ P( A  B  C ) 

85. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is



25

P( A  B  C ) 

1 16 3 1 7   8 16 16

3 8


87. If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is

1 3 cos4x = 2 cos2 2x – 1 = 

12 (1) 55

=

= 

14 (2) 45

(3)

7 55

(4)

6 55

Sol. Total number of ways =

(1)

1 4

11C 2

(2)

2 9

(3)

4 9

(4)

6 7

= 55 Favourable ways are (0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10) Probability =

6 55

Answer (2)

88. If 5 (tan2x – cos2x) = 2cos 2x + 9, then the value of cos 4x is

Sol. tan  

1 (1) 3 2 9 7 9

(4) 

3 5



1 2

x C x



1  tan  1 4  1 2 1  tan  4

P



A

4x

2 9 90. The following statement (p q) [(~ p q) q] is

Sol. 5 tan2x = 9 cos2x + 7

(1) Equivalent to ~ p q (2) Equivalent to p ~ q

5 sec2x – 5 = 9 cos2x + 7 cos2x

(3) A fallacy

=t

(4) A tautology

5  9t  12 t

Answer (4)

9t2 + 12t – 5 = 0

1 t 3

B

Solving tan  

Answer (3)

Let

1 4

tan     

(3) 

7 9

89. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If BPC =  then tan  is

Answer (4)

(2)

2 1 9

as

cos2 x 

Sol.

5 t 3

1 , cos 2x = 2cos2x – 1 3

p

q

T T F

T F T

F

F

p  q (~p  q) (~p  q)  q (p  q)  [(~p  q)  q] T F T

T T T

T F T

T T T

T

F

T

T

(a tautology)

26