CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 1
[ CODE – A ]
JEE Main Exam 2017 (Paper & Solution)
Code – A Date : 02-04-2017
Part A – PHYSICS Q.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-
(1) 9 (2) 91 (3) 81 (4)
811
Ans. [1] Sol. Let volume of man is Lbh As all dimension increases by a factor (K = 9) keeping the density constant
Stress on his legs = area
weight = A
gVρ
Initial stress = Stress1 = A
gVρ
Final stress = Stress2 = AK
gVK2
3 ρ
Stress2 = K Stress1 Where (K = 9) So stress is changed by a factor 9. Q.2 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time ?
(1) v
t
(2) v
t (3) v
t (4) v
t
Ans. [3] Sol. v = u – gt (straight line graph)
t
v
CAREER POINT
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 2
[ CODE – A ]
Q.3 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 20mv
81 , the value of k will be-
(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 Ans. [3]
Sol. a = – m
kv2
dtdv = –
mkv2
∫v
102v
dv = – ∫10
0
dtmk
v
10v1
⎥⎦⎤
⎢⎣⎡− = –
mk × 10
– ⎥⎦⎤
⎢⎣⎡ −
101
v1 =
210k−
× 10
– 101
v1
+ = – k × 1000
According to question
KE = 21 mv2 = 2
0mv81
v = 2v0 =
210
– 101 × 2 +
101 = – k × 1000
101 = k × 1000
k = 10–4
Q.4 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 sec. will be-
(1) 4.5 J (2) 22 J (3) 9 J (4) 18 J Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Work-Power-Energy, Ex.2, Page No.25, Q. No.13]
Ans. [1]
Sol. a = 1t6 = 6 t
dtdv = 6t
v = 1
0
2
2t6
⎥⎦
⎤⎢⎣
⎡ = 3 × 12 = 3
KE = 21 × 1 × 32 = 4.5
W = ΔKE = 4.5 – 0 = 4.5 Joule
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 3
[ CODE – A ]
Q.5 The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum ?
(1) 23 (2)
23 (3) 1 (4)
23
Ans. [1] Sol.
R
l
Moment of inertia of cylinder about perpendicular bisector is I
I = ⎥⎥⎦
⎤
⎢⎢⎣
⎡+
4R
12LM
22
For given mass and density
M = πR2Lρ
R2 = ρπL
M
I = M⎥⎥⎦
⎤
⎢⎢⎣
⎡
ρπ+
L4M
12L2
For maxima or minima of I
dLdI = 0
dLdI = M
⎥⎥⎦
⎤
⎢⎢⎣
⎡
ρπ− 2L4
M12
L2 = 0
6L =
ρπ 2L4M
6L =
ρπ
ρπ2
2
L4LR
2
2
RL =
23
23
RL
=
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 4
[ CODE – A ]
Q.6 A slender uniform rod of mass M and length l is pivoted at one end so that is can rotate in 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 θ with the vertical is-
z
x
θ
(1) θsin2g3l
(2) θsin3g2l
(3) θcos2g3l
(4) θcos3g2l
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Rotational motion, Ex.3, Page No., 36, Q. No.11]
Ans. [1] Sol.
z
x
λ/2
α
θ mg
θsin2l
θ
τ = mg θsin2l
τ = Iα
mg θsin2l = α
3m 2l
α = θsing23
l
Q.7 The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by
(R = Earth's radius)-
(1)
g
O d (2)
g
O d R
(3)
g
O d R
(4)
g
O d R
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 5
[ CODE – A ]
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Gravitation, Ex.4, Page No., 42, Q. No.20]
Ans. [4] Sol.
R
g = 2rGM when r > R
g = 3RrGM when r < R
g
r R
g ∝ r g∝2r
1
Q.8 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 75ºC. T is given by- (Given : room temperature = 30ºC, specific heat of copper = 0.1 cal/gmºC)
(1) 800ºC (2) 885ºC (3) 1250ºC (4) 825ºC Ans. [2] Sol. Total heat gain = Total heat loss 100 × 0.1 (75 – 30) + 170 × 1 × (75 – 30) = 100 × 0.1 (T – 75) 10 × 45 + 170 × 45 = 10T – 750 1200 + 7650 = 10 T T = 885ºC Q.9 An external pressure P is applied on a cube at 0ºC so that it is equally compressed from all sides. K is the bulk
modulus of the material of the cube and α 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-
(1) K3
Pα
(2) KP
α (3)
PK3α (4) 3PKα
Ans. [1]
Sol. B =
VdV
P−
V
dV = BP−
dV = BPV−
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 6
[ CODE – A ]
By heating we have to increase the volume by B
PV
ΔV = VγΔT = V × 3α ΔT
V × 3αΔT = B
PV
ΔT = B3
Pα
Here B = K
∴ ΔT = B3
Pα
Q.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 = b141 (2) a = b (3) a = 14 b (4) a = 28 b
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : KTG, Ex.1, Page No.17, Q. No.35] [JEE Advance, Chapter : KTG, Ex.1, Page No.25,Q. No.22]
Ans. [3] Sol. Cp – Cv = R If Cp and Cv are molar specific heat But if Cp and Cv are specific heat i.e. gram specific heat then C = MSg {Q = 1CΔT ⇒ Q = MSgΔT ⇒ C = MSg}
Sg = MC
MSgp – MSgv = R
Sgp –Sgv = MR
2R = a
28R = b
14 = ba
a = 14 b
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 7
[ CODE – A ]
Q.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. In ni and nf are the number of molecules in the room before and after heating, the nf – ni will be :
(1) – 1.61 × 1023 (2) 1.38 × 1023 (3) 2.5 × 1025 (4) – 2.5 × 1025
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : KTG, Ex.3, Page No.27, Q. No.5]
Ans. [4] Sol. PV = nRT
n = RTPV
Ti = 273 + 17 = 290 K Tf = 273 + 27 = 300 K
nf – ni = 314.8
30105 × ⎥⎦⎤
⎢⎣⎡
2901–
3001 × 6.023 × 1023
= 314.8103 6× ⎥⎦
⎤⎢⎣⎡
× 29030010– × 6.023 × 1023
= –
293314.8023.6103 27
××××
= –
29314.810023.6– 27
××
= – 0.025 × 1027 = – 2.5 ×1025 Q.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 :
(1)
0
KE
T T t 2T
(2) 0
KE
T t
(3)
0
KE
T t 2T
(4) 0
KE
t 4T
2T T
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : SHM, Ex.1, Page No.27, Q. No.33]
Ans. [4] Sol. x = A sin ωt
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 8
[ CODE – A ]
v = dtdx = Aω cos ωt
KE =21 mA2ω2cos2ωt
=21 mA2ω2cos2
T2π t
0
KE
t 4T
2T T
Q.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 Ans. [3] Sol. According to theory of relativity
fapp = f
cv–1
cv1
2
2
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡+
(for approach)
= 2
c2/c–1
c2/c1
⎟⎠⎞
⎜⎝⎛
+ × 10 GHz
=
43
23
× 10 GHz
= 3 × 10 GHz fapp = 17.32 GHz Q.14 An electric dipole has fixed dipole moment p
r, which makes angle θ with respect to x-axis . When subjected
to an electric field 1Er
= E i , it experience a torque 1Tr
= kτ . When subjected to another electric field
2Er
= 3 E1 j it experiences a torque 2Tr
= – 1Tr
. The angle θ is. (1) 30° (2) 45° (3) 60° (4) 90°
Ans. [3] Sol.
E1 = E p
x θ
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 9
[ CODE – A ]
τ = 1Tr
= PE sinθ k
p
x θ
E2 = jE3 1
τ = 2T
r = )k(–cosE3P θ
PE sin θ ( k ) = )k(–cosE3P– θ
PE sin θ = θcosE3
tan θ = 3 θ = 60° Q.15 A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large
number of 1μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :
(1) 2 (2) 16 (3) 24 (4) 32 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Ex.3, Page No.47, Q. No.12]
Ans. [4] Sol.
n capacitor in a row
m rows
Potential on each capacitor V = n
1000
n
1000 = 300
n = 3
10 ≈ 4
Ceq = mnC
×
mn1
× = 2
m = n × 2 = 4 × 2 = 8 Minimum number of capacitor = 8 × 4 = 32
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 10
[ CODE – A ]
Q.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:
E r
C
r1
r2
(1) CE (2) CE )rr(
r
2
1
+ (3) CE
)rr(r
2
2
+ (4) CE
)rr(r
1
1
+
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Ex.2, Page No.44, Q. No.25]
Ans. [3] Sol.
E r
C
r1
r2
i
At steady state current through capacitor branch become zero.
i = 2rr
E+
Potential difference across capacitor ΔV ΔV = ir2
ΔV = ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 2rrE r2
charge on capacitor = C ΔV
= CE ⎟⎟⎠
⎞⎜⎜⎝
⎛+ 2
2
rrr
Q.17 In the above circuit the current in each resistance is :
2V
2V
1Ω 1Ω
2V
2V
2V
2V
1Ω
(1) 1A (2) 0.25 A (3) 0.5 A (4) 0 A
Ans. [4]
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 11
[ CODE – A ]
Sol. 2V
2V
1Ω 1Ω
2V
2V
2V
2V
1Ω
6 4 2 0
024
Potential difference across each resistor is zero so current in each resistor also zero. Q.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 : (1) 6.65 s (2) 8.89 s (3) 6.98 s (4) 8.76 s
Ans. [1]
Sol. M = 6.7 × 10–2 Am2
I = 7.5 × 10–6 kg m2, B = 0.01 T τ = – MB sin θ
Ια = – MB θ (for small oscillations )
α = ⎟⎠⎞
⎜⎝⎛
IMB θ ⇒ ω =
IMB ⇒ T = 2π
MBI
T = 2π 01.0107.6
105.72–
6–
×××
⇒ T = 0.6644 sec Time for 10 oscillation Δt = 10 T ⇒ Δt = 6.65 sec Q.19 When a current of 5 mA is passed through a galvanometer having a coil of resistance 15Ω, 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
(1) 1.985 × 103 Ω (2) 2.045 × 103 Ω (3) 2.535 × 103 Ω (4) 4.005 × 103 Ω Ans. [1] Sol. Ig max = 5 mA , Rg = 15 Ω Range of voltmeter = 10 volt
RH Rg
ΔV
Ig
ΔV = Ig (Rg + RH) Range ΔVmax = Ig max (Rg + RH) 10 = 5 × 10–3 (15 + RH) RH = 1985 Ω
RH = 1.985 × 103 Ω
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 12
[ CODE – A ]
Q.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 :
Time 0.5 sec
Current(amp.)
10
(1) 200 Wb (2) 225 Wb (3) 250 Wb (4) 275 Wb
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : EMI, Ex.1, Page No.31, Q. No.1]
Ans. [3] Sol.
Time 0.5 sec
Current(amp.)
10
Emf = iR
– dtdφ = iR
( )∫ φd– = ( )∫ idt R
(– Δφ) = ∫5.0
0
idtR
( )φΔ– = R (area of i – t curve)
Δφ = R (21 × 0.5 × 10)
Δφ = 100 (2.5) Δφ = 250 Wb Q.21 An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-ray. 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 -
(1)
log λmin
log V (2)
log λmin
log V
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 13
[ CODE – A ]
(3)
log λmin
log V (4)
log λmin
log V
Ans. [1] Sol.
log λmin
log V
K.E. = eV (K.E. = kinetic energy of electron) EPmax = K.E.
min
hcλ
= eV ⇒ λmin V = ⎟⎠⎞
⎜⎝⎛
ehc = constant
ln λmin + ln V = ln constant ln λmin = – ln V + ln (constant) Straight line of – ve slope
Q.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 as -
(1) real and at a distance of 40 cm from convergent lens (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 (4) real and at a distance of 6 cm from the convergent lens
Ans. [1] Sol.
15 cm
ƒ = 25 cm ƒ = 20 cm Image form by diverging is at the focus of diverging lens. Now image form by diverging act as a source for converging lens. For converging lens object real at a distance 40 cm from it which is at (2ƒ).
u = – 2 ƒ, ƒ = + ƒ, v = + 2ƒ ⎟⎟⎠
⎞⎜⎜⎝
⎛=−
ƒ1
u1
v1
Final image real at a distance 2ƒ = 40 cm from converging lens.
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 14
[ CODE – A ]
Q.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 -
(1) 1.56 mm (2) 7.8 mm (3) 9.75 mm (4) 15.6 mm
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Wave nature of light : Interference, Ex.6, Page No.43, Q. No.23]
Ans. [2] Sol. d = 0.5 mm, D = 1.5 m, λ1 = 650 nm, λ2 = 520 nm
β1 = dD1λ = 3–
9–
105.05.110650
×
×× = 1.95 mm
β2 = dD2λ = 3–
9–
105.05.110520
×
×× = 1.56 mm
Least distance where their maxima again coincides from central maxima is = LCM of β1 & β2 = 7.8 mm
Q.24 A particle A of mass m and initial velocity v collides with a particle B of mass 2m which is at rest. The
collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is -
(1) 31
B
A =λλ (2)
B
A
λλ = 2 (3)
32
B
A =λλ (4)
21
B
A =λλ
Ans. [2] Sol.
m v
m/2 mv1
m/2v2⇒
v1 = 2m/m2m/m
+− v + 0
v1 = 2m/32m/ v
v1 = 3v
Similarly
v2 = 0 + 2m/m
m2+
× v
v2 = 2m/3
m2 v
v2 = 34 v
Q λ1 = 11vm
h , λ2 = 22vm
h
2
1
λλ =
11
22
vmvm =
m2/m .
3/v3/v4 = 2
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 15
[ CODE – A ]
Q.25 Some energy levels of a molecule are shown in the figure. The ratio of the wavelength r = 2
1
λλ , is given by -
– E
– 34 E
– 2 E
– 3 E
λ2
λ1
(1) r = 34 (2) r =
32 (3) r =
43 (4) r =
31
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Ex.5, Q. No.27]
Ans. [4]
Sol. 1
hcλ
= (– E) – (– 2E)
1
hcλ
= E ….(i)
2
hcλ
= (– E) – ⎟⎠⎞
⎜⎝⎛−
3E4 = – E +
3E4 =
3E4E3 +− =
3E ….(ii)
By (i) & (ii)
r = 2
1
λλ =
3/EhcEhc
= 31
Q.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 -
(1) t = 2T
3.1log2log (2) t = T
2log3.1log (3) t = T log (1.3) (4) t =
)3.1log(T
Ans. [2] Sol. A ⎯⎯→ B
A = A0e–λt ; B = A0 (1 – e–λt) ; AB = t
0
t0
eA)e1(A
λ−
λ−−
0.3 = eλt – 1 eλt = 1.3
λt = ln (1.3)
T
)2ln( t = ln (1.3)
t = T )2ln()3.1ln(
⇒ t = )2log(
)3.1log(T
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 16
[ CODE – A ]
Q.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) 45º (2) 90º (3) 135º (4) 180º Ans. [4] Sol. In C-E amplifier phase difference between input-output voltage is 180º. Q.28 In amplitude modulation, sinusoidal carrier frequency used is denoted by ωc and the signal frequency is
denoted ωm. The bandwidth (Δωm) of the signal is such that Δωm << ωc. Which of the following frequencies is not contained in the modulated wave?
(1) ωm (2) ωc (3) ωm + ωc (4) ωc – ωm Ans. [1] Sol.
ωc – ωm ωc ωc + ωm
LSB USB
because Δωm << ωc
∴ ωm is not present in modulated wave. Q.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 (2) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is
disturbed (3) A rheostat can be used as a potential divider (4) Kirchhoff’s second law represents energy conservation
Ans. [2] Sol.
G A B
P Q
S R
(I)
P
G
Q
SR
(II)
In Ist case for balance
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 17
[ CODE – A ]
RP =
SQ
QP =
SR ….(i)
In IInd case for balance
QP =
SR ….(ii)
In both case Null point is same. Q.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/s2 and the simplified relation T = 2
rhg × 103 N/m, the possible error in surface tension is
closest to - (1) 0.15% (2) 1.5% (3) 2.4% (4) 10%
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Practical Physics, Ex.2, Page No.70, Q. No.7]
Ans. [2] Sol. D = 1.25 × 10–2 m, ΔD = 0.01 × 10–2 m, h = 1.45 × 10–2 m, Δh = 0.01 ×10–2 m g = 9.80 m/s2
T = 2
rhg × 103 N/m
Tr–1h–1 = 2g ×103
Tr–1h–1 = constant
hh
rr
TT Δ
−Δ
−Δ = 0
⎟⎠⎞
⎜⎝⎛ Δ
+Δ
=Δ
hh
rr
TT
TTΔ = ⎟
⎟⎠
⎞⎜⎜⎝
⎛
×
×2–
2–
1025.11001.0 + ⎟
⎟⎠
⎞⎜⎜⎝
⎛
×
×2–
2–
1045.11001.0
⎪⎪
⎭
⎪⎪
⎬
⎫
⎪⎪
⎩
⎪⎪
⎨
⎧
Δ=
ΔΔ=Δ
=
rr
DD
r2Dr2D
% error = TTΔ × 100
≈ 1.5 %
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 18
[ CODE – A ]
Part B – CHEMISTRY Q.31 Given
)g(CO)g(OC 22)graphite( ⎯→⎯+ ;
ΔrH° = – 393.5 kJ mol–1
H2(g) + 21 O2(g) ⎯→ H2O(1);
ΔrH° = – 285.8 kJ mol–1
CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); ΔrH° = + 890.3 kJ mol–1
Based on the above thermochemical equations, the value of ΔrH° at 298 K for the reaction 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
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Energetic ,Exercise-1 Q. No.12] [JEE Advance, Chapter : Chemical Energetic ,Exercise-4 Q. No.11]
Ans. [1]
Sol. )g(CO)g(OC 22)graphite( ⎯→⎯+ ; ΔrH° = – 393.5 kJ .... (i)
H2(g) + 21 O2(g) ⎯→ H2O(1); ΔrH° = – 285.8 kJ .... (ii)
CO2(g) + 2H2O(1) ⎯→ CH4(g) + 2O2(g); ΔrH° = + 890.3 ..... (iii)
)g(CH)g(H2C 42)graphite( ⎯→⎯+ ; ΔrH° = ? .... (iv)
eq. (iv) = eq. (i) + 2 × eq (ii) + eq. (iii) = –393. 5 + 2 (–285.8) + 890.3 = –74.8 kJ/mol Q.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
Ans. [4]
Sol. 01186.
21gm
32 CO HCl COM ⎯→⎯+
POAC on carbon
1
01186.x1
= ⇒ x = 84.3
Q.33 ΔU is equal to -
(1) Adiabatic work (2) Isothermal work (3) Isochoric work (4) Isobaric work
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 19
[ CODE – A ]
Ans. [1]
Sol. FLOT (According to first law of thermodynamics)
ΔE = q + w
If q = 0, ΔE = w ∴ adiabatic process
Q.34 The Tyndall effect is observed only when following conditions are satisfied -
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
(1) (a) and (c) (2) (b) and (c) (3) (a) and (d) (4) (b) and (d)
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter : Surface Chemistry, Key Concept, Optical Properties]
Ans. [4]
Sol. Facts
Q.35 A metal crystallises in the 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 :
(1) a2 (2) 2
a (3) 2a (4) a22
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Solid State, Solved Example, Q. No.21]
[JEE Advance, Chapter : Solid State, Solved Exercise-4, Q. No.4]
Ans. [2]
Sol. ∴ nearest distance = 2
a
2
a
fcc
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 20
[ CODE – A ]
Q.36 Given
V36.1E Cl/Cl2=°
− , V74.0E Cr/Cr3 −=°+
V33.1E 3272 Cr/OCr =°
+− , V51.1E 24 Mn/MnO =°
+− ,
Among the following, the strongest reducing agent is - (1) Cr3+ (2) Cl– (3) Cr (4) Mn2+
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Electro Chemistry, Exercise-4, Q. No.26] [JEE Advance, Chapter : Electro Chemistry, Exercise-5, Section [A], Q. No.11]
Ans. [3] Sol. Less is the SRP, more is the reducing power & strongest is the reducing agent. Q.37 The freezing point of benzene decreases by 0.45°C 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 (Kf for benzene = 5.12 K kg mol–1)
(1) 74.6% (2) 94.6% (3) 64.6% (4) 80.4%
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Solution, Solved Example Q. No.40] [JEE Advance, Chapter : Solution, Exercise-5, Section [B], Q. No.2]
Ans. [2] Sol. ΔTf = i × Kf × m
0.45 = 2060
10002.0 5.12 i×
×××
i = .527
n/11
i−
−1=β = 946.
2/11527.
=−−1 or 94.6 %
Q.38 The radius of the second Bohr orbit for hydrogen atom is – (Planck's Const. h = 6.6262 × 10–34 Js; mass of electron = 9.1091 × 10–31 kg; charge of electron
e = 1.60210 ×10–19 C; permittivity of vaccum ∈0 = 8.854185 ×10–12 kg–1m–3A2) (1) 0.529 Å (2) 2.12 Å (3) 1.65 Å (4) 7.76 Å
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Exercise-1, Q. No.14] [JEE Advance, Chapter : Atomic Structure, Exercise-1, Q. No.21]
Ans. [2]
Sol. r = Åz
n529.02
×
= 1)2(529.0
2× = 2.116 Å ≅ 2.12 Å
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 21
[ CODE – A ]
Q.39 Two reactions, R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1 . If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to. (R = 8.314 J mol–1 K–1).
(1) 6 (2) 4 (3) 8 (4) 12 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Kinetics, Exercise-5, Section [B], Q. No.27] [JEE Advance, Chapter : Chemical Kinetics, Exercise-4, Q. No.25]
Ans. [2] Sol. R1 R2
A A
Ea +10 Ea k1 k2
k1 = A e–(Ea + 10)/RT k2 = A e–Ea /RT
RT/)10EaEa(
1
2 ekk ++−=
RT/10
1
2 ekk
= = 300 8.314/ 1010 3e ××
= 4 2494.2 / 10000 e e =
4kkln
1
2 =
Q.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- (1) 7.0 (2) 1.0 (3) 7.2 (4) 6.9
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ionic Equilibrium, Exercise-4, Q. No.8] [JEE Advance, Chapter : Ionic Equilibrium, Exercise-1, Q. No.28]
Ans. [4]
Sol. pH = baw pK21pK
21pK
21
−+
= 4.3212.3
2114
21
×−×+×
= 7 + 1.6 – 1.7 = 6.9 Q.41 Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one
which is incorrect, is : (1) Both form nitrides (2) Nitrates of both Li and Mg yield NO2 and O2 on heating (3) Both form basic carbonates (4) Both form soluble bicarbonates
Ans. [3] Sol. It is the best possible option but, it should be bonus because Li2CO3 is less basic and MgCO3 is basic
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 22
[ CODE – A ]
Q.42 Which of the following species is not paramagnetic ? (1) O2 (2) B2 (3) NO (4) CO
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical bonding, Exercise # 1, Q. No.92] [JEE Advance, Chapter : Chemical bonding, Exercise # 2, Q. No.43]
Ans. [4] Sol. CO is diamagnetic because all electrons are paired in it. Q.43 Which of the following reactions is an example of a redox reaction ?
(1) XeF6 + H2O XeOF4 + 2HF (2) XeF6 + 2H2O XeO2F2 + 4HF (3) XeF4 + O2F2 XeF6 + O2 (4) XeF2 + PF5 [XeF]+ PF6
– Ans. [3]
Sol. +4 XeF4 + O2F2 → XeF6 + O2
+1 +6 0
This reaction is a redox reaction Q.44 A water sample has ppm level concentration of following anions
F– = 10 ; −24SO = 100 ; −
3NO = 50
The anion / anions that make / makes the water sample unsuitable for drinking is / are
(1) Only F– (2) Only −24SO
(3) Only −3NO (4) Both −2
4SO and −3NO
Ans. [1] Sol. F– ion concentration above 2 ppm causes brown mottling of teeth Q.45 The group having isoelectronic species is
(1) O2–, F–, Na, Mg2+ (2) O–, F–, Na+, Mg2+ (3) O2–, F–, Na+, Mg2+ (4) O–, F–, Na, Mg+
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Periodic Table, Exercise # 4, Q. No.5] [JEE Advance, Chapter : Periodic Table, Exercise # 2, Q. No.3]
Ans. [3] Sol. O2–, F–, Na+, Mg2+ are isoelectronic species. They all have 10e– Q.46 The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are
(1) Cl– and ClO– (2) Cl– and −2ClO (3) ClO– and −
3ClO (4) −3ClO and −
3ClO
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : p-Block elements, Exercise # 2, Q. No. 39] [JEE Advance, Chapter : p-Block elements (Halogens), Exercise # 2, Q. No. 9]
Ans. [1] Sol. Cl2 + NaOH(cold/dilute) → NaCl + NaOCl + H2O
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 23
[ CODE – A ]
Q.47 In the following reactions, ZnO is respectively acting as a / an
(a) Zn + Na2O Na2ZnO2
(b) Zn + CO2 ZnCO3
(1) acid and acid (2) acid and base (3) base and acid (4) base and base
Ans. [2]
Sol. ZnO + Na2O Na2ZnO2
acid
ZnO + CO2 ZnCO3
Base
ZnO is an amphoteric oxide
Q.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 -
(1) CH3COONa (2) Na2C2O4 (3) C6H5COONa (4) HCOONa
Students may find similar question in CP exercise sheet :
[JEE Advance, Chapter : Salt analysis, Exercise # 1, Q. No. 19]
Ans. [2]
[X] Na2C2O4
(Oxalate-ion)
CaCl2
CaC2O4 ↓ + 2NaClWhite ppt
Conc. H2SO4 Na2CO3 + H2C2O4
CO2 ↑Effervesce
Sol.
KMnO4 + 242OC − H+ CO2↑ + Mn+2
Pink Colour
Oxalate ion
Colourless
Q.49 The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%), Carbon (22.9 %),
Hydrogen (10.0%) and Nitrogen (2.6%). The weight which a 75kg person would gain if all 1H atoms are replaced by 2H atoms is
(1) 7.5 kg (2) 10 kg (3) 15 kg (4) 37.5 kg
Ans. [1]
Sol. Weight of H11 in person = 75 ×
10010 = 7.5 kg
Now if 11H is replaced by 2
1H
Than person will gain weight by = 7.5 kg
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 24
[ CODE – A ]
Q.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 :
(1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O
(3) [Co(H2O)4Cl2]Cl.2H2O (4) [Co(H2O)3Cl3].3H2O Ans. [2]
CoCl3.6H2O + AgNO3 (excess) 100 ml, 0.1 M No. of moles = 0.01 mol
AgClSol.1.2×1022 ion1.2×1022
6×1023 = 0.02 mol
Q 0.01 mol of CoCl3.6H2O produce 0.02 mol of AgCl ∴ 1 mol of CoCl3.6H2O produce 2 mol of AgCl ∴ Correct complex is [Co(H2O)5Cl]Cl2.H2O Q.51 Which of the following compounds will form significant amount of meta product during mono-nitration reaction?
(1)
NH2
(2)
NHCOCH3
(3)
OH
(4)
OCOCH3
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Amines and Nitrogen Compounds, Example-4]
Ans. [1]
Sol.
NH2
HNO3
NH2
NO2
+
[47%]
NH2
NO2
[2%]
+
NH2
[51%]
NO2
In aniline protonation of NH2 Group make it deactivating and meta directing. So meta product is significant. Q.52 Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to
decolourize the colour of bromine?
(1)
Br
O
(2) Br
O
(3)
Br
O
(4) Br
C6H5
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 25
[ CODE – A ]
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Halogen Derivatives, Exercise # 2, Q. No.6]
Ans. [3]
Sol.
Br
O
Elimination reaction not possible
CH3 – C – OΘ
CH3
CH3
Non Nucleophillic base
CH3 – C – CH3
O
It do not give unsaturation test
CH2
CH3
O
Q.53 The formation of which of the following polymers involves hydrolysis reaction?
(1) Nylon 6, 6 (2) Terylene (3) Nylon 6 (4) Bakelite
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Polymer, Exercise # 5, Q. No.1] [JEE Advance, Chapter : Carbohydrate, Amino acid, Protein & Polymer, Example-21]
Ans. [3] Sol. Nylon-6 is formed by monomer which is obtain by hydrolysis of CAPROLACTAM
NH C
O
H2O/Δ
Caprolactam
NH2–(CH2)5–COOHAmino caproic acid
Polymerisation ––NH–(CH2)5–C––––– O n
Nylon-6 Q.54 Which of the following molecules is least resonance stabilized?
(1)
N (2)
O
(3) (4)
O
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : GOC, Page 51, Q. No.32] [JEE Advance, Chapter :GOC, Exercise # 1, Q. No.54]
Ans. [2]
Sol.
N
Aromatic
O Non Aromatic Aromatic
O
Aromatic
..
..
2 is least stable as other are aromatic and 2 is non aromatic
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 26
[ CODE – A ]
Q.55 The increasing order of the reactivity of the following halides for the SN1 reaction is :
CH3CHCH2CH3
Cl
(I)
CH3CH2CH2Cl
(II)
p – H3CO – C6H4 – CH2Cl
(III)
(1) (I) < (III) < (II) (2) (II) < (III) < (I) (3) (III) < (II) < (I) (4) (II) < (I) < (III) Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Haloalkane, page 28, Q. No.17] [JEE Advance, Chapter : Halogen Derivatives, Exercise # 1, Q. No.14]
Ans. [4] Sol. Rate of SN1 reaction ∝ stability of carbocation
CH3CHCH2CH3
Cl
(I)
CH3 – CH – CH2 – CH3 + Cl
2º Carbocation
⊕ Θ
CH3 – CH2 – CH2 – Cl CH3 – CH2 – CH2
1º Carbocation
⊕
CH3O
Resonance Stabilize
⊕CH2 – Cl CH3O CH2
Order of SN1 = II < I < III Q.56 The major product obtained in the following reaction is :
C6H5
Br
H
C6H5
( + )
Δ⎯⎯⎯ →⎯ BuOKT
(1) ( + )C6H5CH(OtBu)CH2C6H5
(2) ( – )C6H5CH(OtBu)CH2C6H5 (3) ( ± )C6H5CH(OtBu)CH2C6H5
(4) C6H5CH = CHC6H5 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Hydrocarbon, page 37, Q. No. 31] [JEE Advance, Chapter : Hydrocarbon, Exercise # 2, Q. No. 19]
Ans. [4]
Sol.
C6H5
Br
H
C6H5 tBuOK
Δ C6H5 – CH = CH – C6H5
It is example of E2 elimination as t butoxide is stronger base and heating is also used.
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 27
[ CODE – A ]
Q.57 Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
(1)
HOH2C
OH
HO
O CH2OH
OCH3
(2)
HOH2C
OH
O CH2OCH3
OH
OH
(3)
HOH2C
OH
HO
O CH2OH
OCOCH3 (4)
HOH2C
OH
HO
O CH2OH
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Carbohydrate, page 74, Q. No. 9]
[JEE Advance, Chapter : Carbohydrate, Protein & Polymer, Exercise # 5, Q. No. 2]
Ans. [3]
Sol.
HOCH2
OH
HO
O CH2OH
O – C – CH3 O
KOHAq.
Hydrolysis
HOCH2
OH
HO
O CH2OH
OH
It is hemiacetal that is why it can behave as reducing sugar.
Q.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 :
(1) Two (2) Four (3) Six (4) Zero
Students may find similar question in CP exercise sheet :
[JEE Main, Chapter : Hydrocarbon, page 30, Q. No.50]
[JEE Advance, Chapter : Hydrocarbon, Example-14, Q. No.50]
Ans. [2]
Sol.
H3C – CH2 – C = CH – CH3
CH3
Peroxide
HBr⎯⎯ →⎯ CH3 – CH2 – CH – CH – CH3
CH3
Br
Major Product
*
*
Two new chiral center are formed so total 4 stereoisomer
This HBr/Peroxide gives antimarkownikov product as major product
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 28
[ CODE – A ]
Q.59 The correct sequence of reagents for the following conversion will be : O
CHO
HO CH3
HO CH3
CH3 (1) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH (2) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH (3) [Ag(NH3)2]+OH–, H+/CH3OH, CH3MgBr (4) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–
Ans. [3]
Sol.
O
C – H
HO CH3
HO CH3CH3
O
[Ag(NH3)2]+ OHΘ
Tollen’s reagent oxidise Aldehyde in carboxylic acid
O
C – OΘ
O
H+/CH3OH CH3MgBr
O
C – OCH3
O
Q.60 The major product obtained in the following reaction is :
COOH
OO
⎯⎯⎯ →⎯ H–DIBAL
(1) CHO
COOH
(2) CHO
CHO
(3) CHO
COOH
OH
(4) CHO
CHO
OH
Ans. [4]
Sol.
COOH
O O
DiBAL–H
CHO
HO
CHO
DiBAL-H is selective reducing agent which reduce carboxylic acid and it’s derivative up to aldehyde only
[further reduction not possible]
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 29
[ CODE – A ]
Part C – MATHS
Q.61 The function f : R → ⎥⎦⎤
⎢⎣⎡−
21,
21 defined as f(x) = 2x1
x+
, is
(1) injective but not surjective (2) surjective but not injective (3) neither injective nor surjective (4) invertible
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Function, Page 55, Ex. 5A, Q. No. 23] [JEE Advance, Chapter : Function, Page 20, Ex. 1, ]
Ans. [2] Sol.
–1
min. –1/2
1
max. 1/2
O
y-axis
x-axis
f(x) = 2x1x
+ (odd)
(symmetry about origin)
dxdy = 22
2
)x1()x20(x1·)x1(
++−+
= 22
2
)x1(x1
+− = 0 ⇒ x = 1, –1
+ – –
–1 1
max. min.
Line parallel to x-axis cuts the graph more than one points hence function is many one.
Range = ⎥⎦⎤
⎢⎣⎡−
21,
21 = codomain hence function is onto
Q.62 If, for a positive integer n, the quadratic equation, x (x + 1) + (x + 1) (x + 2) + .... + (x + 1n − ) (x + n) = 10n has two consecutive integral solutions, then n is equal to
(1) 9 (2) 10 (3) 11 (4) 12 Ans. [3]
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 30
[ CODE – A ]
Sol. x (x + 1) + (x + 1) (x + 2) + .... + (x + (n – 1)) (x + n) = 10n After simplify nx2 + (1 + 3 + 5 + 7 + ...... + (2n –1)x + (0·1 + 1·2 + 2·3 + ... + (n –1)n) = 10n
nx2 + n2x + 3
)1n(n 2 − – 10n = 0
x2 + nx + 3
301n2 −− = 0
x2 + nx + 3
31n2 − = 0
Put n = 11 (where n ∈ I+)
x2 + 11x + 3
31121− = 0
x2 + 11x + 30 = 0 (x + 6) (x + 5) = 0 i.e. x = –5, –6 (Two consecutive integral solutions) So, n = 11
Q.63 Let ω be a complex number such that 2ω + 1 = z where z = 3− . If 72
22
111
111
ωωω−ω− = 3k, then k is equal to
(1) z (2) –1 (3) 1 (4) –z
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Complex Number, Page 29, Ex. 3 ]
Ans. [4] Sol. Apply operation C1 = C1 + C2 + C3
ωω
ωω+−2
22
0)1(0
113 = 3k
ωω
ωω2
2
00
113 = 3k (Because 1 + ω + ω2 = 0)
open by C1 3(ω2 – ω4) = 3k 3(ω2 – ω) = 3k 3(–1 – ω – ω) = 3k –3(1 + 2ω) = 3k Given that 2ω + 1 = z –3z = 3k k = – z
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 31
[ CODE – A ]
Q.64 If A = ⎥⎦
⎤⎢⎣
⎡−
−1432
, then adj (3A2 + 12A) is equal to
(1) ⎥⎦
⎤⎢⎣
⎡72846351
(2) ⎥⎦
⎤⎢⎣
⎡72638451
(3) ⎥⎦
⎤⎢⎣
⎡−
−51846372
(4) ⎥⎦
⎤⎢⎣
⎡−
−51638472
Ans. [1]
Sol. A = ⎥⎦
⎤⎢⎣
⎡−
−1432
A2 = ⎥⎦
⎤⎢⎣
⎡−
−1432
⎥⎦
⎤⎢⎣
⎡−
−1432
= ⎥⎦
⎤⎢⎣
⎡−
−1312
916
3A2 + 12A = ⎥⎦
⎤⎢⎣
⎡−
−39362748
+ ⎥⎦
⎤⎢⎣
⎡−
−12483624
= ⎥⎦
⎤⎢⎣
⎡−
−51846372
adj (3A2 + 12A) = ⎥⎦
⎤⎢⎣
⎡72846351
Q.65 If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is
(1) an infinite set (2) a finite set containing two or more elements (3) a singleton (4) an empty set
Ans. [3] Sol. Δ = 0 and at the one of Δ1 or Δ2 or Δ3 ≠ 0
Δ = 1ba1a1111
= 0
1[a – b] –1[1 – a] + 1[b – a2] = 0 2a – b – 1 + b – a2 = 0 a2 – 2a + 1 = 0 a = 1 x + y + z = 1 x + y + z = 1 x + by + z = 0 only one value of b, S is singleton set
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 32
[ CODE – A ]
Q.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
(1) 468 (2) 469 (3) 484 (4) 485
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :P & C, Page 16, Ex. 1, Q. No. 29]
Ans. [4] Sol.
X
7 friends
3M 4L
Y
7 friends
4M 3L Case I : 3L from X side and 3M from Y side 4C3 × 4C3 = 4 × 4 = 16 Case II : 3M from X side and 3L from Y side 3C3 × 3C3 = 1 × 1 = 1 Case III : 2L and 1M from X side and 2M and 1L from Y side (4C2 × 3C1) × (4C2 × 3C1) = (6 × 3) × (6 × 3) = 18 × 18 = 324 Case IV : 2M and 1L from X side and 1M and 2L from Y side (3C2 × 4C1) × (4C1 × 3C2) = (3 × 4) × (4 × 3) = 12 × 12 = 144 Total number of ways = Case I + Case II + Case III + Case IV = 16 + 1 + 324 + 144 = 485 Q.67 The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + ... + (21C10 – 10C10) is
(1) 221 – 210 (2) 220 – 29 (3) 220– 210 (4) 221 – 211
Ans. [3]
Sol. (21C1 + 21C2 + .... + 21C10) – (10C1 + 10C2 + ..... + 10C10)
= [ ]1021
221
121 C2...C2C2
21
×++×+× – (210 – 1)
= [ ])CC(–CC...CC...CCC21
2121
021
2121
2021
1121
1021
221
121
021 +++++++++ – (210 – 1)
= 21 (221 – 2) – (210 – 1)
= 220 – 1 – 210 + 1 = 220 – 210
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 33
[ CODE – A ]
Q.68 For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then (1) b, c and a are in A.P. (2) a, b and c are in A.P. (3) a, b and c are in G.P. (4) b, c and a are in G.P.
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Progression, Page 27, Ex. 4, Q. No. 19] [JEE Advance, Chapter : Progression, Page 20, Ex. 2, Q. No. 26 ]
Ans. [1] Sol. 225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0 450a2 + 18b2 + 50c2 – 150ac – 10ab – 30bc = 0 (15a – 3b)2 + (3b – 5c)2 + (15a – 5c)2 = 0 (15a – 3b)2 = 0, (3b – 5c)2 = 0, (15a – 5c)2 = 0 15a = 3b, 3b = 5c 15a = 3b = 5c
3c
5b
1a
== = k (let)
a = k, b = 5k, c = 3k Then a, c and b are in A.P. Q.69 Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀ x, y ∈ R, then
∑=
10
1n
)n(f is equal to
(1) 165 (2) 190 (3) 255 (4) 330 Ans. [4] Sol. As a + b + c = 3 So, f(1) = 3 f(x + y) = f(x) + f(y) + xy Put x = 1, y = 1 f(2) = 2f(1) + 1 = 7 Put x = 2, y = 1 f(3) = f(2) + f(1) + 2 = 12 Put x = 2, y = 2 f(4) = 2f(2) + 4 = 18
So, ∑=
10
1n
)n(f = f(1) + f(2) + f(3) + .... + f(10)
Let S = 3 + 7 + 12 + 18 + .... + f(n) S = 3 + 7 + 12 + ........ + f(n) 0 = 3 + 4 + 5 + 6 + ....... – f(n)
f(n) = 3 + 4 + 5 + 6 + ....... = 2n [6 + (n – 1)]
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 34
[ CODE – A ]
f(n) = 2
)5n(n +
∴ ∑=
10
1n
)n(f = ∑=
10
1n
2n21 + ∑
=
10
1n
n25
= 21 ×
6211110 ×× +
25 ×
21110×
= 330
Q.70 3
2x )x2(
xcosxcotlim−π−
π→
equals :
(1) 161 (2)
81 (3)
41 (4)
241
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Limit, Page 20, Ex. 2, Q. No. 30] [JEE Advance, Chapter :Limit, Page 16, Ex. 1, Q. No. 18]
Ans. [1]
Sol. Put x = 2π + h
30hh
22
)h2
cosh2
cotlim
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛ +
π−π
⎟⎠⎞
⎜⎝⎛ +
π−⎟
⎠⎞
⎜⎝⎛ +
π
→
30h h8hsinhtanlim
−+−
→
30h h8hsinhtanlim −
→
by expansion method
3
535
3
0h h8
...!5
h!3
hh....h152
3hh
lim⎟⎟⎠
⎞⎜⎜⎝
⎛+−−⎟⎟
⎠
⎞⎜⎜⎝
⎛++
→
3
53
0h h8
...!5
1152h
!31
31h
lim+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛+
→
= 8
061
31
++ =
161
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 35
[ CODE – A ]
Q.71 If for ⎟⎠⎞
⎜⎝⎛∈
41,0x , the derivative of ⎟
⎟⎠
⎞⎜⎜⎝
⎛
−−
31
x91xx6tan is )x(gx ⋅ , then g(x) equals :
(1) 3x91xx3
− (2) 3x91
x3−
(3) 3x913
+ (4) 3x91
9+
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Differentiation, Page 25, Ex. 2, Q. No. 26]
Ans. [4]
Sol. ⎟⎟⎠
⎞⎜⎜⎝
⎛
−= −
31
x91xx6tany
⎟⎟⎠
⎞⎜⎜⎝
⎛
−⋅
= −2
1
)xx3(1xx32tan
)xx3(tan2 1−=
)1xx2
1x(3)xx3(1
2dxdy
2 ⋅+⋅⋅+
= (differentiating w.r.t. x)
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+= x
2x
x916
3
3x91x9
+=
3x919x
+⋅=
Q.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 :
(1) ⎟⎠⎞
⎜⎝⎛
21,
21 (2) ⎟
⎠⎞
⎜⎝⎛ −
31,
21 (3) ⎟
⎠⎞
⎜⎝⎛
31,
21 (4) ⎟
⎠⎞
⎜⎝⎛ −−
21,
21
Ans. [1] Sol. y(x – 2) (x – 3) = x + 6 at y axis, x = 0 y(–2) (–3) = 0 + 6 y = 1 Now y(x2 – 5x + 6) = x + 6
6x5x
6xy 2 +−+
=
22
2
)6x5x()5x2()6x(1)6x5x(
dxdy
+−−+−⋅+−
=
at x = 0, y = 1 16
)5()6(62 =
−−=
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 36
[ CODE – A ]
equation of normal y – 1 = – 1 (x – 0) x + y = 1
passes ⎟⎠⎞
⎜⎝⎛
21,
21 (by option)
Q.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
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Maxima & Minima, Page 96, Ex. 2, Q. No. 13]
Ans. [2] Sol. Given r + r + rθ = 20
r
r220 −=θ
θ
r r
rθ
Area θ= 2r21
⎟⎠⎞
⎜⎝⎛ −
⋅=r
r220r21 2
)r2r20(21z 2−=
0)r420(21
drdz
=−= ⇒ r = 5
at r = 5, θ = 2, 0dr
zd2
2< (hence maxima)
maximum area
θ= 2r21z
22 m252521
=××=
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 37
[ CODE – A ]
Q.74 Let ∫ >= ).1n(,dxxtanI nn If I4 + I6 = a tan5 x + bx5 + C, where C is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) ⎟⎠⎞
⎜⎝⎛ 0,
51 (2) ⎟
⎠⎞
⎜⎝⎛ −1,
51 (3) ⎟
⎠⎞
⎜⎝⎛− 0,
51 (4) ⎟
⎠⎞
⎜⎝⎛− 1,
51
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Indefinite Integration, Page 33, Ex. 3, Q. No. 2] [JEE Advance, Chapter : Indefinite Integration, Page 35, Ex. 4, Q. No. 1]
Ans. [1]
Sol. In ∫= dxxtann
I4 + I6 ∫ += dx)xtanx(tan 64
∫ += dx)xtan1(xtan 24
∫ ⋅= dxxsecxtan 24 put t = tan x
∫ ⋅= dtt4
C5t5
+=
C5
xtan5+=
On comparison, we get
0b,51a ==
Q.75 The integral ∫π
π +
43
4
xcos1dx is equal to :
(1) 2 (2) 4 (3) – 1 (4) – 2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter :Definite Integration, Page 42, Ex. 5A, Q. No. 17]
Ans. [1]
Sol. ∫π
π+
=4/3
4/xcos1
dxI
∫π
π−π+
=4/3
4/)x(cos1
dxI by using ∫ ∫ −+=b
a
b
a
dx)xba(fdx)x(f
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 38
[ CODE – A ]
∫π
π
⎟⎠⎞
⎜⎝⎛
−+
+=
4/3
4/
dxxcos1
1xcos1
1I2 ∫π
π
⎟⎠⎞
⎜⎝⎛
−=
4/3
4/2 dx
xcos12
∫π
π
=4/3
4/
2 dxxeccos2I2
4/34/)x(cotI π
π−=
– (– 1 – 1) = 2
Q.76 The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4y and y ≤ 1 + x } is :
(1) 23 (2)
37 (3)
25 (4)
1259
Ans. [3] Sol.
(0, 3)
(1, 2)
(2, 1)
(3, 0)
(0, 1)
y
xO
2
1
Required Area = ∫ ∫ ⎟⎟⎠
⎞⎜⎜⎝
⎛−−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+
1
0
2
1
22dx
4xx3dx
4xx1
2
1
321
0
32/3
12x
2xx3
12x
2/3xx ⎥
⎦
⎤⎢⎣
⎡−−+⎥
⎦
⎤⎢⎣
⎡−+=
25
1211
1219
=+=
Q.77 If (2 + sin x) )1y(dxdy
++ cos x = 0 and y(0) = 1, then y ⎟⎠⎞
⎜⎝⎛ π
2 is equal to :
(1) 32
− (2) 31
− (3) 34 (4)
31
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Differential Equation, Page 70, Ex. 5A, Q. No. 6]
Ans. [4]
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 39
[ CODE – A ]
Sol. 0xcos)1y(dxdy)xsin2( =+++
xsin2
xcos)1y(dxdy
++−
=
⇒ dxxsin2
xcos1y
dy∫ ∫ ⎟
⎠⎞
⎜⎝⎛
+−=
+
⇒ log(y + 1) = – log(2 + sin x) + log c
⇒ y + 1 = xsin2
c+
…(1)
Given that y(0) = 1
∴ 1 + 1 = 2c ⇒ c = 4
∴ Equation of curve
xsin2
41y+
=+
at 2
x π= ⇒
1241y+
=+
⇒ 134y −=
31y =
Q.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 :
(1) ⎟⎠⎞
⎜⎝⎛
43,1 (2) ⎟
⎠⎞
⎜⎝⎛ −
43,1 (3) ⎟
⎠⎞
⎜⎝⎛
21,2 (4) ⎟
⎠⎞
⎜⎝⎛ −
21,2
Ans. [3] Sol. Area
2812k1k51k3k
21
±=−
−
k(k – 2) + 3k(5 + k) + 1(10 + k2) = ± 56 k2 – 2k + 15k + 3k2 + 10 + k2 = ± 56 5k2 + 13k + 10 = ± 56 5k2 + 13k – 46 = 0 5k2 + 13k + 66 = 0 5k2 + 23k – 10k – 46 = 0 D = 169 – 4 × 5 × 66 < 0 k(5k + 23) – 2(5k + 23) = 0 No solution (5k + 23) (k – 2) = 0 k = 2 (k is integer) Hence co-ordinate (2, –6) (5, 2) (–2, 2)
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 40
[ CODE – A ]
E(2, β) AD ⊥ BC
138
222
−=×+−β
83
42 −
=−β
232 −=−β
232 −=β
21
=
⎟⎠⎞
⎜⎝⎛
21,2
(–2, 2) A
C (5, 2)
(2, β)
D
B (2, –6)
x = 2
x
E
y
O
Q.79 The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :
(1) )12(2 − (2) )12(4 − (3) )12(4 + (4) )12(2 +
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Area under the curve, Ex. 3]
Ans. [2]
Sol.
(0, 4) y = x
r (0, k)
y = –x
y
xO
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 41
[ CODE – A ]
Circle touches the line
By graph radius = 4 – k
Perpendicular distance from centre = radius
⇒ 2k0k4 −
=−
⇒ 16 + k2 – 8k = 2k2
⇒ k2 – 16k + 32 = 0
⇒ 2
)32(425616k
−±=
⇒ 2
12816k ±=
⇒ 2
2816k ±=
⇒ 248k ±=
⇒ 248k −= (k should be 0 < k < 4) Radius = 4 – k
= )248(4 −−
= )12(4 −
Q.80 The eccentricity of an ellipse whose centre is at the origin is .21 If one of its directrices is x = – 4, then the
equation of the normal to it at ⎟⎠⎞
⎜⎝⎛
23,1 is :
(1) 4x – 2y = 1 (2) 4x + 2y = 7 (3) x + 2y = 4 (4) 2y – x = 2
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ellipse, Page 58, Ex. 3, Q. No. 3] [JEE Advance, Chapter : Ellipse, Page 24, Ex. 3, Q. No. 13]
Ans. [1]
Sol. Let the equation of ellipse 2
2
2
2
by
ax
+ = 1
given that e = 21
and directrix ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
−=
−=
4xor
e/ax
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 42
[ CODE – A ]
⇒ ea = 4
⇒ a = 2 Now, b2 = a2 (1 – e2)
b2 = 4 ⎟⎠⎞
⎜⎝⎛ −
411 = 3
Equation of ellipse
3y
4x 22
+ = 1
diff. w.r.t. x
dxdy
3y2
4x2
+ = 0
21
dxdy
)2/3,1(−=⎟
⎠⎞
⎜⎝⎛
∴ Equation of normal at ⎟⎠⎞
⎜⎝⎛
23,1 is
y – y1 = )xx(
dxdy1
1−⎟⎠⎞
⎜⎝⎛
−
⇒ y – 23 = 2(x – 1)
⇒ 2y – 3 = 4x – 4 ⇒ 4x – 2y = 1
Q.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 :
(1) (2 2 , 3 3 ) (2) ( 3 , 2 ) (3) (– 2 , – 3 ) (4) ( 23 , 32 )
Ans. [1] Sol. Let the equation of hyperbola
2
2
ax – 2
2
by = 1
given that foci (±ae, 0) = (±2, 0) ⇒ ae = 2
Hyperbola passes through P( 2 , 3 )
∴ 2a2 – 2b
3 = 1
⇒ 2a2 –
)1–e(a322 = 1
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 43
[ CODE – A ]
⇒ 2a2 – 2a–4
3 = 1
⇒ 8 – 2a2 – 3a2 = 4a2 – a4
⇒ a4 – 9a2 + 8 = 0
⇒ (a2 – 8) (a2 – 1) = 0 ⇒ a2 = 8 and a2 = 1 b2 = a2(e2 – 1) b2 = a2(e2 – 1) b2 = 4 – 8 b2 = 4 – 1 = 3 b2 = – 4 (not possible) ∴ Equation of hyperbola
1
x2–
3y2
= 1
tangent at P( 2 , 3 )
T = 0
x2 – 3
y = 1
By option it passes through ( 22 , 33 )
Q.82 The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal
perpendicular to both the lines 1
1–x =2–2y + =
34–z and
22–x =
1–1y + =
1–7z + , is :
(1) 83
10 (2) 835 (3)
7410 (4)
7420
Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : 3D, Page 110, Ex. 2, Q. No. 17]
Ans. [1]
Sol. n = 1–1–2
32–1kji
= k3)7(–j–i5 +
Equation of plane 5(x – 1) + 7(y + 1) + 3(z + 1) = 0
5x + 7y + 3z + 5 = 0
Perpendicular distance of the plane from (1, 3, –7) is = 94925
|521–215|++++ =
8310
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 44
[ CODE – A ]
Q.83 If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,
1x =
4y =
5z is Q, then PQ is equal to :
(1) 422 (2) 42 (3) 56 (4) 53
Ans. [1] Sol.
R (2, 2, 8)
P(1, –2, 3)given line
Q The equation of line PR
1
1–x = 4
2y + = 5
3–z = k
let R(k + 1, 4k – 2, 5k + 3) it lies on the plane 2x + 3y – 4z + 22 = 0 ∴ 2(k + 1) + 3(4k – 2) – 4(5k + 3) + 22 = 0 ⇒ – 6k + 6 = 0 ⇒ k = 1 ∴ R(2, 2, 8) Image of P in the plane is (R is the mid-point of PQ) ∴ Q(3, 6, 13)
PQ = 100644 ++ = 168 = 422
Q.84 Let →a = k2–ji2 + and
→b = ji + . Let
→c be a vector such that |
→c –
→a | = 3, |(
→a ×
→b ) ×
→c | = 3 and the angle
between →c and
→a ×
→b be 30º. Then
→a ·
→c is equal to :
(1) 2 (2) 5 (3) 81 (4)
825
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Vector, Page 55, Ex. 5A, Q. No. 10] [JEE Advance, Chapter : Vector, Page 38, Ex. 4, Q. No. 35]
Ans. [1]
Sol. |→a ×
→b | =
0112–12
kji
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 45
[ CODE – A ]
|→a ×
→b | = kj2–i2 +
|→a ×
→b | = 3
given | (→a ×
→b ) ×
→c | = 3
|→a ×
→b | |
→c | sin 30º = 3
⇒ 3|→c | ·
21 = 3
⇒ |→c | = 2
Now |→c –
→a | = 3
|→c |2 + |
→a |2 – 2
→a .
→c = 9
⇒ 4 + 9 – 2→a .
→c = 9
⇒ →a .
→c = 2
Q.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 :
(1) 6 (2) 4 (3) 256 (4)
512
Ans. [4]
Sol. Total no. of balls = 25
(15 green and 10 yellow balls)
(Varience) σ2 = npq
where n → No. of Trial
p → Probability of happening of that event
q → Probability of not happening of that event
n = 10, p =2515 =
53 , q =
2510 =
52
So, σ2 = 10 × 53 ×
52
σ2 = 2560 =
512
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 46
[ CODE – A ]
Q.86 For three events A, B and C,
P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 41
and P(All the three events occur simultaneously) = 161 .
Then the probability that at least one of the events occurs, is :
(1) 167 (2)
647 (3)
163 (4)
327
Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Probability, Page 35, Ex. 3, Q. No. 6]
Ans. [1]
Sol. P(Exactly one of A or B occurs) = P(A) + P(B) – 2P(A ∩ B) = 41 ... (1)
P(Exactly one of B or C occurs) = P(B) + P(C) – 2P(B ∩ C) = 41 ... (2)
P(Exactly one of C or A occurs) = P(C) + P(A) – 2P(C ∩ A) = 41 ... (3)
Adding (1), (2) and (3)
2[P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A)] = 43
P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) = 83
P(All the three events occurs simultaneously) = P(A ∩ B ∩ C) =161
P(Atleast one of the events occurs) = P(A ∪ B ∪ C) = 83 +
161 =
167
Q.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) 5512 (2)
4514 (3)
557 (4)
556
Ans. [4] Sol. n(S) = 11C2 = 55
(0, 4) (0, 8) (2, 6), (2, 10)(4, 8), (6, 10)
Favorable events
So, required probability = EventsTotalEvents.Fav =
556
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 47
[ CODE – A ]
Q.88 If 5(tan2x – cos2x) = 2cos2x + 9, then the value of cos4x is :
(1) 31 (2)
92 (3) –
97 (4) –
53
Ans. [3] Sol. 5(tan2x – cos2x) = 2cos2x + 9
⇒ 5 ⎥⎦
⎤⎢⎣
⎡xcos–
xcosxsin 2
2
2= 2(2cos2x – 1) + 9
⇒ 5[(1 – cos2x) – cos4x] = 4cos4x – 2cos2x + 9cos2x ⇒ 9cos4x + 12cos2x – 5 = 0 ⇒ 9cos4x + 15cos2x – 3cos2x – 5 = 0 ⇒ 3cos2x (3cos2x + 5) – (3cos2x + 5) = 0
⇒ cos2x = 31
⇒ cos2x = 2cos2x – 1
⇒ cos2x =32 – 1 =
31–
Now cos4x = 2cos22x – 1
cos4x = 2 ⎟⎠⎞
⎜⎝⎛
91 – 1
cos4x = 97–
Q.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 = β, then tan β is equal to :
(1) 41 (2)
92 (3)
94 (4)
76
Ans. [2] Sol.
B
x
C
A P4x
x βγ α
β = α – γ
tanβ = γα+
γαtan.tan1
tan–tan =
811
41–
21
+=
92
CAREER POINT
CAREER POINT Ltd., CP Tower, IPIA, Road No.1, Kota (Raj.), Ph: 0744-5151200 www.careerpoint.ac.in 48
[ CODE – A ]
Q.90 The following statement (p → q) → [(~p → q) → q] is :
(1) equivalent to ~p → q (2) equivalent to p → ~q (3) a fallacy (4) a tautology Ans. [4] Sol.
q p ~p (p → q) (~p → q) (~p → q) → q (p → q) → [(~p → q) → q] T F T F
T T F F
F F T T
T F T T
T T T F
T F T T
T T T T
It is a tautology.