PHYSICS
1. An equation of real gas
2
a
p V b RT
V
then dimension of
2
a
b
is
P : Pressure
V = Volume
R = Gas constant
T = Temperature
(1) [ML–1T–2] (2) [MLT–2] (3) [ML2T–2] (4) [MLT–1]
Ans. (1)
Sol. Basic theory
2. Assertion: There can be positive zero error in vernier calliper.
Reason: Due to mishandling or rough handling of instrument
(1) Assertion true, reason true and reason is correct explanation of assertion
(2) Assertion true, reason true and reason is not correct explanation of assertion
(3) Assertion true, reason false
(4) Assertion false, reason true
Ans. (1)
3. In a RLC series circuit R = 10, L =
100
mH, C =
3
10
F and frequency is 50 Hz. Find power factor.
Ans. 1
Sol. XL =
3
100 2 50 10
= 10
XC =
3
1
10
2 50
= 10
XL = XC
cos = 1
The Actual Paper will be Updated with Solution After the Official Release
JEE Main 27 Jan 2024
(Shift-2)
(Memory Based)
4. A stone is released and while free-fall stone covers 80 m distance in last 2 sec. Find distance of point A
from top most point.
u = 0
A
B
80m
g
Top most point
Ans. 45
Sol.
A
B
t = 0
80m
t'= t +2
S = ut + gt2
t = t
5 (t + 2)2 – 5t2 = 80 t = 3sec
SA = 0 +
1
2
× 10× 32 = 45m
5. A person is standing on horizontal ground. A rod of mass 12 kg is touching a shoulder of person and other
end is resting on ground. Angle made by rod with horizontal is 60°. Reaction force applied by person on rod
is
(1) 60 N (2) 30 N (3) 90 N (4) 120 N
Ans. (2)
Sol.
N2
N1
mg
60°
Taking torque about N1.
mg.
2
cos 60° = N2.
N2 = 30 N
6. Point 'B' is at highest point of trajectory of object. Magnitude of acceleration at 'A' and 'B' is equal. Find the
angle '' as shown.
B
A
v0
(1) 2 tan–1(1/2) (2) tan–1(1/2) (3) tan–1(1/4) (4) tan–1(2)
Ans. (1)
Sol. Apply work energy theorem
mg(1 – cos) =
1
2
m
2
0
v
2
2
0
v4gsin 2
....(1)
gsin =
2
0
v
....(2)
tan
1
2 2
= 2tan–1
1
2
7. In an adiabatic process, pressure is proportional to cube of temperature. Find the ratio Cp/Cv.
Ans. 3/2
Sol. PT/1– = constant
P
T3
PT–3 = C
3
1
= –3 + 3
2 = 3
= 3/2
8. Assertion: Angular velocity of earth around sun is lesser than the angular velocity of moon about earth.
Reason: Time taken by moon revolve around earth is less than time taken by earth to revolve around sun.
(1) Both Assertion (A) and Reason (R) are true & correct explanation of Assertion 'A'
(2) Both 'A' and 'R' are correct but 'R' is not correct explanation of 'A'
(3) 'A' is correct and 'R' is false
(4) 'A' is false and 'R' is correct
Ans. (1)
Sol.
2
T
Tearth = 365 days
Tmoom = 27 days
9. If wave function of a metal is 6.68eV. Find threshold frequency.
(1) 8 × 1015 Hz (2) 1.6 × 1015 Hz (3) 10 × 1015 Hz (4) 4 × 1015 Hz
Ans. (2)
Sol. 6.68 × 1.6 × 10–19 = 6.626 × 10–34 v0
1.6 × 1015Hz = v0
10. Find magnetic field strength at the centre of loop.
1
R2
2
R4
R2
4A
R1
Ans. 24 × 10–7 Tesla
Sol. Bcentre =
0 0
1 2
(i) i
4R 4R
=
0
1 2
41 1
4 R R
=
0
2 4
=
76
4 10
= 24 × 10–7 Tesla
11. Assertion: If external force is removed, then body will try to regain its actual shape, this is called elasticity.
Reason: Due to intermolecular force, this happens
(1) Assertion True, Reason True & Reason is correct explanation of assertion
(2) Assertion True, Reason True & Reason is not correct explanation of assertion
(3) Assertion True, Reason false
(4) Assertion false, Reason True
Ans. (1)
12. A bullet gets embedded in a fixed target. It is found that bullet losses 1/3rd of its velocity in traveling 4 cm
into target and losses remaining kinetic energy while traveling further d × 10–3 m. Find d.
Ans. 32
Sol. v2 = u2 + 2ax
2
2
2u u 2(–a)(4cm)
3
….(1)
for next
O =
2
2u 2(–a)(x)]
3
….(2)
using equation (i) &(ii)
x = 32× 10–3 m
So d = 32
13. 1 mole of an ideal O2 gas is at 27ºC. Find its total kinetic energy?
(1) 1250 J (2) 6250 J (3) 645 J (4) 1025 J
Ans. (2)
Sol. Kinetic Energy =
nfRT
2
KE =
1 25
5 300
2 3
= 6250 J
14. Light of intensity I = 6 × 108
2
W
m
is incident on an object kept in medium of refractive index, = 3
assuming 100% absorption. Find radiation pressure (N/m2)?
Ans. 6
Sol. Radiation =
IA h 1
hv A
=
I I
v C
=
8
8
6 10 3
3 10
= 6
15. A ring and a solid sphere of same mass and radius are released from same point of inclined plane. Find the
ratio of their KE when they reach to bottom without slipping
(1) 1 : 7 (2) 1 : 3 (3) 1 : 5 (4) 1 : 1
Ans. (4)
Sol. EC
mgh = kf – ki
kf = mgh
so KERing = K.E.solid sphere
16. Three voltmeters of different internal resistances are connected as shown in figure and a certain voltage is
applied across AB. State which is true?
V3
V1
V2
A
B
(1) V1 + V2 > V3 (2) V1 + V2 V3 (3) V1 + V2 = V3 (4) V1 + V2 < V3
Ans. (3)
Sol. By series and parallel combination
V1 + V2 = V3
17. Specific resistance S is given as S =
RA
. If length is doubled, find corresponding change in S.
(1) S is halved
(2) S is doubled
(3) S is quadrupled
(4) No change in S
Ans. (4)
18. Assertion : Static friction depends on area of contact but independent of material.
Reason : Kinetic friction is independent of area of contact but depends on material.
(1) Assertion true, reason true and reason is correct explanation of assertion.
(2) Assertion true, reason true and reason is not correct explanation of assertion.
(3) Assertion true, reason false.
(4) Assertion false, reason true.
Ans. (4)
19. Assertion : Work done by electrostatics force on an object when moved on equipotential surface is always
zero.
Reason : Electric field lines falls perpendicular to the equipotential surface
(1) Assertion true, reason true and reason is correct explanation of assertion.
(2) Assertion true, reason true and reason is not correct explanation of assertion.
(3) Assertion true, reason false.
(4) Assertion false, reason true.
Ans. (1)
Sol. Assertion is true and reason is true and correct explanation.
20. A nucleus of C13 breaks into C12 and neutron. Find energy released.
Atomic mass of C12 = 12.000 u
C13 = 13.013975 u
n = 1.008665 u
(1) 3.04 MeV (2) 4.1 MeV (3) 4.94 MeV (4) 6 MeV
Ans. (3)
Sol. Mass defect = 13.013975 – (12 + 1.008665) = 0.00531 U
Energy released = 0.00531 × 931 = 4.94 MeV
21. For given logic circuit. The truth table will be
A
B
Y
(1)
A B Y
000
011
101
111
(2)
A B Y
001
011
101
110
(3)
A B Y
000
010
100
111
(4)
A B Y
000
011
101
110
Ans. (4)
Sol. Using Boolean algebra
Y AB AB
A B Y
000
011
101
110
22. In a transformer, ratio of turns in primary to secondary coil is 10 : 1. If primary side voltage is 230 volt and
frequency is 50 Hz and resistance of secondary side is 46 then find power output.
(1) 11.5 W (2) 12 W (3) 12.5 W (4) 23 W
Ans. (1)
Sol.
1 1
2 2
N V
N V
2
10 230
1 V
V2 = 23 V
P2 =
2
2
2
V23 23 23 11.5W
R 46 2
23. Between two polaroid placed in crossed position, a third polaroid is introduced. By what angle (in degree)
the introduced polaroid placed should be rotated to get maximum intensity of the out coming light.
Ans. 45
Sol.
I = I0 cos2 sin2
Imax at = 45°
24. If their fundamental frequencies are sounded together, beat frequency is 7 Hz. Find velocity (in m/s) of
sound in air?
150cm
350cm
Ans. 24
Sol. f1 =
1
v
4
f2 =
2
v
2
f1 =
v 100
4 150
f2 =
v 100
2 350
f1 =
v
6
f2 =
v
7
v v
6 7
=7
v7
42
, v = 42 × 7
v = 294 m/sec
25. For 200 A current galvanometer deflects by /3 radians. For what value of current, it will deflect by /10
radians ?
Ans. 60
Sol. i
(angle of deflection)
1 1
2 2
i
i
2
200 A / 3 10
i /10 3
60 A = i2
26. Two charges of magnitude –4 C kept at (1, 0, 4) and another charge of +4 C kept at (2, –1, 5) in the
presence of external electric field E = 0.2
ˆ
i
V/cm. The torque on the system of charges is
5
8 10 N m
. Find .
Ans. 2
Sol.
P E
r r
r
6ˆ ˆ ˆ
(i j k)
ˆ
P P r 4 10 3 3
r
6ˆ ˆ ˆ
P 4 10 (i j k)
r
2ˆˆ
E 0.2 10 i 20i
r
V/m
6ˆ ˆ ˆ ˆ
4 10 20[(i j k) i]
r
5ˆˆ
8 10 (k j)
r
5
8 2 10
r
Nm
= 2
27. A nucleus with atomic number '50' and having radius of nucleus is 9 × 10–13 cm. Calculate the potential
(in MV) at the surface of the nucleus.
Ans. 8
Sol.
r
Q = 50e
Vsurface =
9 19
15
kQ 9 10 50 1.6 10
r 9 10
= 80 × 105 volt
Vsurface = 8 MV
28. A pressure inside wall pipe before hole is 4.5 × 104 N/m2. When a small hole is made in pipe, pressure is
changed to 2.0 × 104 N/m2. If speed of water flux after hole is
v
m/s. Find out v :
Ans. 50
Sol. P =
2
1v
2
2.5 × 104 =
3 2
0
110 v
2
v0 =
50
m/s
v = 50
CHEMISTRY
1. For Ist order reaction, time required for 99.9% completion is :
(1) 2t1/2 (2) 4t1/2 (3) 5t1/2 (4) 10t1/2
Ans. (4)
Sol.
99.9%
1/2
1 100
ln
tk 100 99.9
1
tln 2
k
=
3
ln(10 ) 3 10
ln 2 0.3
t99.9% = 10t1/2
2. Number of non polar molecules among following are :
HF, H2O, CO2, NH3, SO2, H2, CH4, BF3
Ans. (4)
Sol. CO2, H2, CH4, BF3
3. 3M NaOH solution is to be prepared using 84 g NaOH, then the volume of solution in litre is
____ × 10–1
Ans. (7)
Sol.
sol( L)
84 / 40
3V
Vsolution = 0.7 L
4. Select incorrect match :
(1) Haber process : Fe
(2) Polythene : Ziegler-Natta catalyst [Al2(CH3)6 + TiCl4]
(3) Wacker's process : PtCl2
(4) Photography : AgBr
Ans. (3)
Sol. Wacker's process : PdCl2
5. 1 mole PbS is oxidised by x mole O3 liberating y mole O2.
Determine (x + y).
Ans. (8)
Sol. PbS + 4O3 PbSO4 + 4O2
x = 4 ; y = 4
6. Spin only magnetic moment of [Pt(NH3)2Cl(CH3NH2)]Cl is :
Ans. (0)
Sol. Pt+2 : 5d8 dsp2 & unpaired e– = 0 Magnetic moment = 0
7. S-1: Formation of Ce4+ is favoured by inert gas configuration.
S-2: Ce4+ acts as strong oxidising agent & converts to Ce3+.
Ans. Both S-1 & S-2 are correct.
8. Which of the following can't act as oxidising agent ?
(1) MnO4– (2) N3– (3) BrO3– (4) SO42–
Ans. (2)
Sol. In N–3, nitrogen is present in minimum O.N. & hence it cannot act as oxidising agent.
9. The quantity which changes with temperature is:
(1) Molarity (2) Molality (3) Mole fraction (4) Mass %
Ans. (1)
Sol. Quantities involving volume are temperature dependent.
10. Reduction potential of hydrogen electrode at pH = 3 is.........
2.303RT 0.059
F
Ans. (–0.177 volt)
Sol. H+ (aq) + e–
1
2
H2(g)
R.P. = –
0.059
1
log
1
H
= –0.059 log(10+3)
= –0.059 × 3 = –0.177 volt
11. Identify the species in which central atom is in d2sp3 hybridisation :
(1) SF6 (2) BrF5 (3) [PtCl4]2– (4) [Co(NH3)6]3+
Ans. (4)
Sol. SF6 : sp3d2
BrF5 : sp3d2
[PtCl4]2– : dsp2
[Co(NH3)6]3+ : d2sp3
12. H° = +77.2 kJ, S° = 122 J/mol-K, T = 300 K, logK = ?
Ans. (–7.07)
Sol. G° = –2.303RTlogk
77.2 –
300 122
1000
=
–2.303 8.314 300 logK
1000
logK = –7.07
13. In group 16
Statement-I : Oxygen shows only –2 oxidation state.
Statement-II : On moving top to bottom, stability of +4 oxidation state decreases, whereas that of
+6 oxidation state increases.
(1) Both Statement I and Statement II are correct.
(2) Both Statement I and Statement II are incorrect.
(3) Statement I is correct but Statement II is incorrect.
(4) Statement I is incorrect but Statement II is correct.
Ans. (2)
Sol. Statement-I : Since electronegativity of oxygen is very high, it shows only negative oxidation
state as –2 except in the case of OF2 where its oxidation state is + 2.
Statement-II : The stability of + 6 oxidation state decreases down the group and stability of + 4
oxidation state increases (inert pair effect).
14. How many of following has/have noble gas configuration ?
Sr2+, Cs+, Yb+2, La2+
Ans. (2)
Sol. (Sr2+, Cs+)
15. Which of the following has d10 configuration ?
(1) Cr, Cd, Cu, Ag (2) Cd, Cr, Ag, Zn
(3) Ag, Cr, Cu, Zn (4) Cu, Cd, Zn, Ag
Ans. (4)
Sol. Cr : [Ar] 3d5 4s1
Cu : [Ar] 3d10 4s1
Ag : [Kr] 4d10 5s1
Zn : [Ar] 3d10 4s2
Cd : [Kr] 4d10 5s2
16. Which of the following is used to identify the phenolic group test?
(1) Carbylamine test (2) Lucas test
(3) Tollen’s test (4) Phthalein dye test
Ans. (4)
17.
O
HI
Product is :
(1)
+
I
(2)
OH
+
I
(3)
OH
+
H–O
(4)
I
+
H–O
Ans. (2)
18. Match the column
(P)
OH
(i) CHCl3/NaOH
(ii) H+
(1)
O
O
(Q)
OH
Na2Cr2O7
(2)
OH
CHO
(R)
OH
(i) NaOH (1eq.)
(ii) CH3–Cl
(3)
OH
COOH
(S)
OH
(i) CO2/NaOH
(ii) H+
(4)
OCH3
Ans. (P) – (2) ; (Q) – (1) ; (R) – (4) ; (S) – (3)
19. When egg is boiled then which of the following structure of protein remains intact?
(1) Quaternary structure (2) Primary structure
(3) Secondary structure (4) Tertiary structure
Ans. (2)
20. Which of the following compound will not give SN1 reaction?
(1) CH2=CH–CH2Cl (2) Ph–CH2–Cl
(3)
CH–Cl
H3C
H3C
(4) CH3–CH=CH–Cl
Ans. (4)
21. The second homologue of monocarboxylic acid is
(1) HCOOH (2) CH3COOH (3) CH3CH2COOH (4) CH3CH2CH2–COOH
Ans. (2)
22.
CH=CH2
(1) B2H6/H2O2,OH
(2) HBr
(3) Mg/dry ether
(4) H–C–H
O
(5) H3O+
–
Product is
(1)
CH2–OH
Ph–CH–CH3
(2) Ph–CH2–CH2–CH2–OH
(3) Ph–CH2–CH2–O–CH3 (4)
Ph–CH–CH2–CH3
OH
Ans. (1)
23. When 9.3 gm of aniline in reacted with acetic anhydride then mass of acetanilide formed is [X]
gm. Report your answer as 10X.
Sol.
NH2
(CH3CO)2O
NH–C–CH3
O
9.3 gm
Mole of Aniline =
9.3 0.1
93
Mole of acetanilide = 0.1
Mass of acetanilide = 0.1 × 135 = 13.5 gm
10x = 13.5 × 10 = 135 gm
24. The correct stability order of following resonating structures is
(I) CH2=CH–CH=O (II)
CH2–CH=C–H
O
(III)
CH2–CH=C–H
O
(1) II > III > I (2) I > II > III (3) I > III > II (4) III > II > I
Ans. (2)
25. Steam volatile and water immiscible substances are separated by
(1) Steam distillation (2) Fractional distillation under reduced pressure
(3) Fractional distillation (4) Distillation.
Ans. (1)
26. How many of the following compounds contain chiral centre ?
(I)
O
(II)
O
O
(III)
CH3–CH2–CH–COOH
OH
(IV)
NO2
COOH
(V)
I
I
(VI)
CH3–CH2–CH–C2H5
OH
Ans. 4 (I, III, IV, V)
27. The bond line representation of following compound is
CH(OH)(CN)2
(1)
CN
CN
(2)
CN
OH
CN
(3)
OH
CN
NC
(4)
CN
CN
HO
Ans. (3)
MATHEMATICS
1. An urn contains 6 black and 9 red balls. Four balls are drawn from the urn twice without
replacement. The probability that first four balls are black & 2nd four balls are red in colour is:
(1)
3
765
(2)
6
715
(3)
3
715
(4)
6
615
Ans. (3)
Sol.
6
4
15
4
C
C
×
9
4
11
4
C
C
=
3
715
2. A line x + y = 0 touches the circle (x – )2 + (y – )2 = 50, , , > 0. The distance of origin from
its points of contact is 4
2
. Find 2 + 2.
Ans. 82
Sol. Point of contact is (0 + 4
2
cos 135º, 0 + 4
2
sin 135º) = (–4, 4)
2
= 5
2
+ = 10 ….(i)
(–4 – )2 + (4 – )2 = 50
( + )2 + (4 – 10 + )2 = 50
( + )2 + ( – 6)2 = 50
= 1, = 9
2 + 2 = 82
Also point of contact is (4, –4)
Satisfying this point of contact in the equation of circle we get
(4 – )2 + (–4 – )2 = 50
(4 – )2 + ( + )2 = 50
(4 – )2 + ( – )2 = 50
= 9, = 1
2 + 2 = 82
3. Let 2tan2x – 5secx – 1 = 0 has 7 solutions in x
n
0, 2
, then the minimum value of n is N find
N
k
k 1
k
2
(1)
13
13 13
2 –1 13
2. –
2 2
(2)
13
13 13
2 –1 13
–
2 2
(3)
13
13 14
2 –1 13
2. 2 2
(4)
13
13 14
2 –1 13
2. –
2 2
Ans. (1)
Sol.
13 2
2tan2x – 5sec x – 1 = 0
2sec2x – 5sec x – 3 = 0
2sec2x – 6sec x + sec x – 3 = 0
(2sec x + 1)(sec x – 3) = 0
sec x = 3, –
1
2
sec x = 3
cos x =
1
3
For 7 solutions, n = 13 = N
so
13
k
k 1
k
2
let S =
1
2
+
2
2
2
+
3
3
2
+ …… +
13
13
2
1
2
S =
2
1
2
+
3
2
2
+ …….. +
14
13
2
1
2
S =
1
2
+
2
1
2
+
3
1
2
+ …….
13
1
2
–
14
13
2
1S
2
=
1
2
.
13
14
1
1– 13
2–
12
1– 2
S =
13
13 13
2 –1 13
2. –
2 2
4. The vertices of a triangle are A(1, 2, 2), B(2, 1, 2) & C(2, 2, 1). The perpendicular distance of its
orthocentre from the given sides are 1, 2 & 3. Find the value of
2 2 2
1 2 3
+ +
.
(1) 1 (2)
1
2
(3)
1
3
(4)
1
4
Ans. (2)
orthocentre & centroid will be same
555
, ,
333
midpoint of AB is
3 3
, ,2
2 2
1
1 1 1
36 36 9
=
1
6
= 2 =3
5. Let two sets A and B having 'm' & 'n' elements respectively such that difference of the number of
subsets of A and that of B is 56, then (m, n) is
(1) (8, 3) (2) (8, 5) (3) (6, 3) (4) (7, 4)
Ans. (3)
Sol. 2m – 2n = 56; m > n
2n(2m–n – 1) = 8 (23 – 1)
m = 6, n = 3
6. If 'A' is a square matrix of order '2' such that roots of the equation det(A – I) = 0 are 1 and –3,
then sum of diagonal elements of matrix 'A2' is
(1) 2 (2) –3 (3) 9 (4) 10
Ans. (4)
Sol. Let A =
a b
c d
|A – I| =
a b
c d
= 0
2 – (a + d) + ad – bc = 0
Sum of the roots = a + d = 2
Product of roots = ad – bc = –3
Now A2 =
a b
c d
a b
c d
=
2
2
a bc b(a d)
c(a d) d bc
tr(A2) = a2 + d2 + 2bc = (a + d)2 – 2(ad – bc) = 4 + 6 = 10
7. Let tan–1x + tan–12x =
4
; x > 0, then number of positive values of x is/are
(1) 0 (2) 1 (3) 2 (4) 3
Ans. (2)
Sol. ABC is equilateral
Sol. tan–1
2
x 2x
1 2x
=
4
2
3x
1 2x
= 1 x =
3 17
4
x =
17 3
4
8. Find the coefficient of x2012 in (1 – x)2008.(1 + x + x2)2007
Ans. (0)
Sol. Coefficient of x2012 in (1 – x3)2007.(1 – x)
Coefficient of x2012 in (1 – x3)2007 – x(1 – x3)2007
Coefficient of x2012 in
122
1 2
rr3r 1
2007 3 2007
r r
C x C 1 x
3r1 = 2012 Which is not possible for any r1 w
and 3r2 + 1 = 2012 also not possible for any r2 w
no term containing x2012
Coefficient of x2012 is 0
9. An ellipse is passing through focii of hyperbola
2 2
x y 1
16 9
and product of their eccentricities is
1, then the length of chord of ellipse passing through (0, 2) and parallel to x-axis is
(1)
5 5
3
(2)
3
5 5
(3)
10 5
3
(4)
20 5
3
Ans. (3)
Sol.
2 2
x y 1
16 9
eH =
5
4
eE =
4
5
Ellipse is passing through (±5, 0)
ellipse :
2 2
x y 1
25 9
and points of chord :
5 5 ,2
3
Length of chord =
10 5
3
10. If and
1
are two complex numbers which satisfy the equations |z – z0|2 = 4 and |z – z0|2 = 16
respectively, where z0 = 1 + i, then the value of 5||2 is
Ans. (1)
Sol. | – z0|2 = 4
( – z0)
0
( z )
= 4
–
0
z
–
0
z
+ |z0|2 = 4
–
0
z
–
0
z
= 2 …(i)
(ii)
2
0
1z
= 16 (1 –
z0) (1 –
0
z
) = 16||2
1 –
0
z
–
0
z
+ ||2 . 2 = 16||2 ….(ii)
from (i) & (ii) – 1 – ||2 = 2 – 16 ||2 15||2 = 3 5||2 = 1
11. Let x2 – x – 1 = 0 has roots and such that Sn = 2023 n + 2024 n, then
(1) S12 = S11 – S10 (2) S12 = S10 – S11
(3) S12 = S10 + S11 (4) S12 = –S10 – S11
Ans. (3)
Sol. Sn = 2023 n + 2024 n
Sn – Sn–1 – Sn–2 = 0
S12 = S11 + S10
12. For the series 20, 19
1
4
, 18
1
2
, ……., –129
1
4
, the 20th term from end is
(1) –115 (2) – 119 (3) –117 (4) –120
Ans. (1)
Sol. T20 for a = – 129
1
4
= –
517
4
, d =
3
4
T20 = –
517
4
+ 19 .
3
4
= –
460
4
= –115
13.
8 2
12 6 1 3
3
x x dx
1
x 3x 1 tan x x
(1)
1
3
n tan–1
3
3
1
x C
x
(2) n tan–1
3
3
1
x C
x
(3) tan–1
3
3
1
x C
x
(4) None of these
Ans. (1)
Sol.
8 2
12 6 1 3
3
x x dx
1
x 3x 1 tan x x
Let
1 3
3
1
tan x t
x
2
24
3
3
1 3
3x x
1
1 x x
dx = dt
6 6
12 6 6 4
x 3x 3
x 2x 1 x x
dx = dt
1 1 1
dt n t C
3 t 3
=
1 3
3
1 1
n tan x C
3 x
14. If
2
x 0
sin x cos x ln(1 – x) 3 1
lim 3
3tan x
find 2 –
(1) 10 (2) 7 (3) 8 (4) 5
Ans. (4)
Sol.
2
x 0
2
sin x cosx ln(1– x) 3 1
lim 3
3tan x
x
+ 3 = 0 = – 3
x 0
1
cosx – sin x – 1 – x
lim 2x
– 1 = 0 = 1
so 2 – = 5
15. Let a1, a2, a3 ……. a15 are 15 observations having mean and variance as 12 & 9 respectively. One
of the observation which was 12, misread as 10. The correct mean and variance are and
2 respectively, then 15( + 2 + 2)
(1) 2521 (2) 2522 (3) 2518 (4) 2621
Ans. (1)
Sol. old mean 12 =
i
x
n
1 2 14
a a .....a 10
12 15
14
i
i 1
a 170
old variance = 9 9 + (12)2 =
2 2 2 2
1 2 14
a a ...... a 10
15
14
2
i
i 1
a 2195
new mean () =
14
i
i 1
a 12
15
=
170 12 182
15 15
new variance (2)
2 + 2 =
14
2 2
i
i 1
a 12 2339
15 15
2 + 2 + =
2339 182 2521
15 15 15
15(2 + 2 + ) = 2521
16. Values of for which
3 3
12 2
1 1
13 3
2 3 3 1 0
= 0 lies in the interval
(1) (0, 3) (2) (–3, 0) (3) (–2, 1) (4) (–2, 0)
Ans. (2)
Sol. C3 C3 – (C1 + C2)
2
3
1 0
2
1
1 0
3
2 3 3 1 (2 6 1)
= 0
7
6
(22 + 6 + 1) = 0
=
3 7
2
,
3 7
2
17. Let g(x) = 3f
x
3
+ f(3 – x), where f(x) > 0 and x (0, 3), g(x) is decreasing in x (0, ) and
increasing in (, 3), then 8 is
Ans. (18)
Sol. g(x) = 3 .
1
3
f
x
3
– f(3 – x) = f
x
3
– f(3 – x)
g(x) is decreasing g(x) < 0
f
x
3
< f(3 – x)
f(x) > 0 f(x) is increasing
x
3
< 3 – x
4x
3
< 3 x <
9
4
=
9
4
8 = 18
18. Let ABC have vertices A(2, 2, 3), B(2, –3, 3) C(–1, –2, 3) and length of internal angle bisector
of angle A is , then the value of 22 is
Ans. (45)
Sol.
B(2, –3, 3)
D
C(–1, –2, 3)
A(2, 2, 3)
AB 5j
AC 3i 4j
AB AC
1 2 2 3 3 3
D , ,
2 2 2
D
1 5
, ,3
2 2
AD =
2 2
2
1 5
2 2 3 3
2 2
=
9 81
4 4
22 = 45
=
90 45
4 2
19. Let S1 =
3!
4!
(4!)
and S2 =
4!
5!
(5!)
, then
(1) S1 N and S2 N (2) S1 N and S2 N
(3) S1 N and S2 N (2) S1 N and S2 N
Ans. (2)
Sol. Make 6 groups of 4 each
24 (4, 4, 4, 4, 4, 4, 4)
Number of ways of making groups =
6
24!
(4!) .6!
= I1
6
(24)!
(4!)
=
3!
4!
(4!)
= (6!I1)
S1 N
(5!) (5, 5, 5, 5 …….., 5(24 times))
24
5!
(5!) . 24!
= I2 S2 = (24!) I2
Hence S2 N
20. Let area bounded by y = min.(3x, 6x – x2); y 0 is A, then 2A is
Ans. (63)
Sol.
3
6
2x = 6x – x2
A =
6
2
3
13 9 6x – x dx
2
A =
27
2
+
6
2
3
9 – (x – 3) dx
A =
27
2
+
6
2 –1
4
x – 3 9 x – 3
9 – (x – 3) sin
2 2 3
A =
27
2
+
9
4
12A = 162 + 27
21. Let (x2 – 4)dy = y(y – 3)dx satisfying y(4) =
3
2
then y(10) is equal to
(1)
1
4
3
1 8
(2)
1
4
3
1 8
(3)
1
4
3
1 2
(4)
1
4
3
1 2
Ans. (2)
Sol. =
y y 3 x 2 x 2
1 1
dy dx
3 y y 3 4 x 2 x 2
1 1
n | y 3 | n | y | n | x 2 | n | x 2 | C
3 4
1 y 3 1 x 2
n n C
3 y 4 x 2
33
1 1 4 2
2
n n C
3
3 4 4 2
2
C =
1n 3
4
1 y 3 1 x 2 1
n n n 3
3 y 4 x 2 4
x = 10
1 y 3 1 2 1
n n n 3
3 y 4 3 4
1 y 3 1
n n 2
3 y 4
3
4
y 3
n n 2
y
3
4
y 3 2
y
– y + 3 =
1
4
8 y
y =
1
4
3
1 8
22. Three lines 2x – y – 3 = 0, 6x + 3y + 4 = 0, x + 2y + 4 = 0 does not form triangle then find [2]
(where [.] denotes the greatest integer function)
Ans. (32)
Sol. If two lines are parallel
2 –1
2
= –4
6 3
2
= 4
If lines are concurrent
2 –1 –3
6 3 4
2 4
= 0
2(12 – 8) + 1(24 – 4) – 3(12 – 3) = 0
8 + 24 – 4 – 36 + 9 = 0
5 = 4 =
4
5
2 = 16 + 16 +
16
25
[2] = 32
23. Let f(x) =
x
0
g(t)
log
1 t
1 t
dt, (where g(x) is cont. odd function).
If
/2 2
x
2
x cos x
f (x) dx
1 e
=
2
, then find
Ans. = 2
Sol. I =
/2 2 2
x x
0
x cos x x cos x
f (x) f ( x) 1 e 1 e
dx =
/2
0
(f (x)
+ f(–x) + x2cos x) dx ….(i)
Now f(–x) =
x
0
g(t)
log
1 t
1 t
dt
t = –p
=
x
0
g( p)
log
1 p
1 p
dp = –f(x)
(i) becomes I =
/2
2
0
x
cos x dx = x2 sin x – 2
x
sin x dx = (x2 sin x – 2) (–x cos x + sin x)
/2
0
)
=
2
4
– 2
2
4
– 2
