SECTION I : PHYSICS PART I : Single Correct Answer Type
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1.
A conducting spherical shell of radius R is given a charge Q. The force exerted by one half on the other half is Q2 (A) 16 0 R 2
Q2 (B) 8 0 R 2
Q2 (C) 0 R 2
Q2 (D) 32 0 R 2
2.
Two infinite line charges of linear charge density are parallel to each other and moving in same direction with velocity v each. The value of v such that the force of interaction between them is zero is
(A)
3.
1 0 0
1 (B) 2 0 0
1 (C) 3 0 0
1 (D) 4 0 0
The switch S is kept in contact with 1 for long time and then thrown to 2. The current through the resistor at this instant will be
(A) 1A
(B) 2.5 A
(C) 0.5 A
(D) zero
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4.
In the figure shown the area of each plate is ‘A’ and distance between two adjacent plates is ‘d’. The charge on plate 2 is
(A)
5.
0 AV 3d
0 AV 3d
(C)
2 0 AV 3d
(D)
2 0 AV 3d
An ideal ammeter (of zero resistance ) is connected as shown. The reading of the ammeter is:
(A) 0 6.
(B)
(B)
E 3R
(C)
E 5R
(D)
E 7R
A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity , in gravity free space. The kinetic energy of the particle is (the length of the spring reaches an equilibrium value) (A)
1 m2l 2 2
1 k 2 2 (C) m l 2 2 k m
(B)
1 k m2l 2 2 2 k m
(D)
none of these.
2
7
The velocities of a particle executing S.H.M. are 30 cm/s and 16 cm/s when its displacements are 8 cm and 15 cm from the equilibrium position. Then its amplitude of oscillation in cm is : (A) 25 (B) 21 (C) 17 (D) 13
8
In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper = 4.5eV). The emitter is illuminated by a source of monochromatic light of wavelength 200nm. (A) the minimum kinetic energy of the photoelectrons reaching the collector is 0. (B) the maximum kinetic energy of the photoelectrons reaching the collector is 3.7eV. (C) if the polarity of the battery is reversed then answer to part A will be 0. (D) if the polarity of the battery is reversed then answer to part B will be 1.7eV.
9
Let K1 be the maximum kinetic energy of photoelectrons emitted by a light of wavelength 1 and K2 corresponding to 2. If 1 = 22 , then : (A) 2K1 = K2
(B) K1 = 2K2
(C) K1 <
K2 2
(D) K1 > 2K2
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10
The relation between 1 : wavelength of series limit of lyman 2 : the wavelength of the series limit of Balmer series and 3 : the wavelength of first line of lyman series is (A) 1 = 2 + 3 (B) 3 = 1 + 2 (C) 2 = 3 – 1 (D) none PART II : Multiple Correct Answer(s) Type
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 11.
A magnetic storm from the Sun can disrupt a satellite as well as move it, either toward or away from the Earth radially. Ground-based engineers start it back in a new circular orbit at the new position. Due to the storm (A) The period of a satellite displaced further away is more than the previous period.. (B) The mechanical energy of a satellite displaced towards earth is more than the previous energy.. (C) The speed of a satellite displaced further away is more than the previous speed.. (D) The angular momentum of a satellite displaced towards earth is more than the previous angular momentum.
12.
A series RLC circuit is driven by a generator at frequency 1000 Hz. The inductance is 90.0 mH; capaci tance is 0.500 µF; and the phase constant has a magnitude of 60.0° (Take 2 = 10) (A) Here current leads the voltage in phase (B) Here voltage leads the current in phase (C) Resistance of circuit is (D) At resonance =
80 3
2 104 rad/sec. 3
13.
Each of the following system begins moving upwards with a constant acceleration. Select the cases in which quantity will change due to this upward acceleration : (A) time period of simple pendulum. (B) fraction of floating body submerged in a liquid. (C) time period of a spring block system. (D) pressure on the base of a container containing liquid.
14.
A source is moving across a circle given by the equation x2 + y2 = R2, with constant speed
330 m/s, in anti6 3 clockwise sense. A detector is at rest at point (2R, 0) w.r.t. the centre of the circle. If the frequency emitted by the source is f and the speed of sound, C = 330 m/s. Then R R 3 (A) the position of the source when the detector records the minimum frequency is 2 , 2 and the
R 3 R position of the source when the detector records the maximum frequency , 2 2 (B) the co-ordinate of the source when the detector records minimum frequency is (0, R)
(C) the minimum frequency recorded by the detector is
6 3 f 6 3
(D) the maximum frequency recorded by the detector is
6 3 f 6 3
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15.
Consider the situation in which ball of mass M is hanging in equilibrium with a string and a spring as shown in figure. If another small ball of mass m collides it then which of the following are correct m FS
FT M FG
k
(A) Tension force due to inextensible string FT is impulsive force (B) Spring force FS is non impulsive force (C) Gravitational force FG is conservative force (D) Normal force due to collision between m and M, is electro magnatic force
PART III : Integer Answer Type
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).
16.
A rod of length R and mass m 2 is free to rotate about a horizontal axis passing through hinge P as shown in the figure. First it is taken aside such that it becomes horizontal and then released. At the lowest point the rod hits the block B of mass m1 and stops. If mass of rod is 60 kg, find mass of the block in Kg if it just completes the circle.
17.
How much water would be filled in a container of height 14 cm, so that it appears half filled to the observer when viewed from the top of the container ? (w = 4/3)
18.
A box and a solid sphere of equal mass are moving with the same velocity across a horizontal floor. The sphere rolls without slipping and the box slides without friction. They encounter an upward slope in the floor and each move up the slope some distance before momentarily stopping and then moving down again. During the motion on the upward slope, sphere rolls without slipping and box slides without friction. The maximum vertical heights reached by the box and the sphere are HB and HS respectively. What is the ratio
5H S ? HB
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19.
Consider the circuit shown in figure. With switch S1 closed and the other two switches open, the circuit has a time constant 0.05 sec. With switch S2 closed and the other two switches open, the circuit has a time constant 2 sec. With switch S3 closed and the other two switches open, the circuit oscillates with a period T. Find T (in sec). (Take 2 = 10) S1 L
20.
S2
S3
C
R
Two identical plates of metal are welded end to end as shown in figure (A), 20 cal of heat flows through it in 4 minutes. If the plates are welded as shown in figure (B) the same amount of heat will flow through the plates in how many minutes ?
A
B
SPACE FOR ROUGH WORK
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SECTIONII : CHEMISTRY
PARTI : Single Correct Answer Type This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
21.
22.
Two students A & B are given (+)-2-butanol (X) for experiment Student A Na C2 H 5 I X Y Student B TsCl C2 H 5 OH X Z Between Y and Z, optically active product will be: (A) Y (B) Z (C) both In PBr3F2, the Br—P—F angle is: (A) 90o (C) 180 o
(D) neither
(B) 120 o (D) 109.5o
O
23.
H / H2O CH2 X
H3C
CH3
The product (X) formed in the above reaction is: OH OH
CH2OH
OH
(A)
CH2 H3C
CH3
(B)
(C)
(D) OH
OH H3C
CH3
CH2OH
H3C
CH3
H3C
CH3
O
24. NH2
NH C
NH2
, which nitrogen attacks on the carbonyl carbon during nucleophillic addition reaction? (A) (B) (C) (D) any
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25.
For a first order reaction, 3R P , conc. vs time graph is given below P
Concentration R
20 t (in days)
Choose correct option regarding this (A) t1 2 20 days
(B) t1 2 10 days
(C) t1 2 5 days
(D) t1 2 15 days
26.
An acid-base indicator has Ka = 3.0 10–5. The acid form of the indicator is red and the basic form is blue. The [H+] required to change the indicator from 75% red to 75% blue is (A) 8 10–5 M (B)9 10 –5 M (C) 1 10–5 M (D)3 10–4 M
27.
Which of the following a compounds, on pinacol-pinacalone rearrangement produces a compound which gives a precrpitate with KOI ? OH
C2H5 CH3
OH
(A)
CH3
OH
C H 2 5
CH3 OH
CH3
OH Ph
CH3
OH
OH
(B)
(C)
OH (D)
3
28.
For a perfectly crystalline solid Cp, m =aT , where a is constant. If Cp, m at 10 K is 0.42 J/K mol, molar entropy at 10 K is: (A) 0.42 J/K mol (B) 0.14 J/K mol (C) 4.2 J/K mol (D) Zero
29.
How many maximum number of electrons of an atom will have the following set and combination of quantum numbers? n+ l =5 m = 0, −1 s
(A) 8
(B) 12
1 2
(C) 10
(D) 6
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CH3
30.
CH2
BH3 H / OH A B
CH2
Product (B) in the above reaction is: CH3
CH3
(A)
CH3
CH3
CH3
(B)
B
CH3
CH3
CH3
B
CH3
CH3
CH3
CH3
CH3
(C)
CH3
CH3
CH3
CH3
(D)
B
CH3
CH3 B H3C
H3C
CH3
CH3
CH3
PARTII: Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) or (D) out of whichONE or MORE are correct.
31.
O2 Na
A
H2O
B
C
White residue of a mixture of compounds
D Na CO H O 2 3 2
HCl H2O D E
Choose correct statement(s) regarding the properties of the products of the above reaction sequence. (A) Gas (D) is a linear molecule. (B) Gas (C) is paramagnetic in nature (C) Aqueous solution of (B) cannot be stored in aluminium vessels. (D) Compound(E) is hygroscopic in nature. 32.
Which of the following equations is/are correctly formulated : (A) 3Cu + 8HNO3(dil.) 3Cu(NO3)2 + 2NO + 4H2O (B) 3Zn + 8HNO3 (very dil.) 3Zn(NO3)2 + 2NO + 4H2O (C) 4Sn + 10HNO3 (very dil.) 4Sn(NO3)2 + NH4NO3 + 3H2O (D) As + 3HNO3 (dil.) H3AsO3 + 3NO2
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33.
Which of the following form(s) more than one monosubstituted product with NBS? CH 3
(A)
34.
(B)
(C)
(D)
Find out the correct properties for [Co(ox)3 ]3 (A) sp 3d 2 hybridization
(B) d 2 sp 3 hybridization (D) Diamagnetic
(C) Chelating reagent
35.
CH 3
Choose correct statement(s) out of the following. (A) CO2 shows only negative deviation from ideal behavior at any P and T. (B) The r.m.s speed of CO2 is higher than that of N2O at constant temperature. (C) KE of 4g of N2 at NTP is lower than that of 6g of CO under same conditions. (D) The rate of diffusion of PH3 is same as that of H2 S under identical conditions.
PART III: Integer Answer Type This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive)
36.
In fuel cell H2 gas is consumed at anode while O2 at cathode. If volume of H2 consumed is same as the volume of He produced at STP from the 0.5 gm atom of U235 which is disintegrated into Pb207 after 3×108 yrs. ( t1 2 of U235 =108 yrs. ). Approximately How many faradays have left at the anode of fuel cell?
37.
38.
Calculate the molality of NaCl solution whose elevation in boiling point is numerically equal to the depression in freezing point of 0.12m Al2(SO4)3 solution in water. Assume complete dissociation of salts. (Kf and Kb of water are 1.86 and 0.558 Kkg/mole) The number of acetal (or ketal) groups in the following saccharide is: OH
OH
O HO HO
O OH
OH O HO
OH
39.
How many alcohols with formula C5H12O will give an anesthetic substance on treatment with Cl2/Ca(OH)2?
40.
What is the number of unpaired electrons in the complex which is formed when aqueous solution of NaNO2 is treated with saturated solution of FeSO4 followed by addition of small amount of conc. H2SO4
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SECTION III : MATHEMATICS PART I : Single Correct Answer Type
This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
41.
Let L be the line of intersection of the planes r · iˆ 2ˆj 3kˆ 0 and r · 3ˆi 2ˆj kˆ 0 . The
acute angle which line L makes with ˆj , is (A) cos–1
42.
3 4
(B) cos–1
2 3
(C) cos–1
1 2
(D) cos–1
1 3
Let f (x) be a function defined for 0 < x < such that f = 0 and f '(x) tan x = 2 6
x
2 cos t dt , sin t
6
then f (x) is equal to (A) ln2(2 sin x)
(B) ln2(sin 3x)
(C) ln2
3 tan x
(D) ln2(2 cos 2x)
43.
In triangle ABC, let b = 6, c = 10 and r1 = r2 + r3 + r then the area of triangle ABC is equal to [Note: All symbols used have usual meaning in a triangle.] (A) 15 (B) 18 (C) 24 (D) 30
44.
Number of positive integers which have no two digits having the same value with sum of their digits being 45, is (A) 10 ! (B) 9 ! (C) 9 · 9 ! (D) 17 · 8 !
45.
The equation of the line passing through M(1, 1, 1) and intersects at right angle to the line of intersection of the y 1 z 1 x 1 = = , then a : b : c equals b c a (C) 5 : – 1 : – 2(D) 5 : 1 : 2
planes x + 2y – 4z = 0 and 2x – y + 2z = 0 is (A) 5 : – 1 : 2
46.
(B) – 5 : 1 : 2
n 3 195 P3 Let xn be the sequence of numbers denoted by xn = – (n N) where Pn denotes the number 4 Pn Pn 1
of ways in which n distinct things can be arranged on n different places. The sum of all possible values of n N for which xn > 0, is (A) 10 (B) 9 (C) 8 (D) 6
47.
1 3 The range of values of a for which the function f(x) = x cos a, 0 x 1 x, 1 x 3 has the smallest value at x = 1, is (A) [cos 2, 1] (B) [–1, cos 2] (C) [0, 1] (D) [–1, 1]
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48.
Suppose f(x) = x2 – 2x. The value of f f (2i 1) is equal to (where i = 1 ) (A) 15 (B) 5i + 5 (C) 35 (D) 15i – 10
49.
The locus of the midpoints of the chords drawn from the point M (1, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0, is equal to (A) x2 + y2 – 4x + 10y – 19 = 0 (B) x2 + y2 + 4x + 10y – 19 = 0 2 2 (C) x + y + 4x – 10y – 19 = 0 (D) x2 + y2 – 4x – 10y + 19 = 0
50.
Given a triangle ABC with AB fixed in length and position. As the vertex C moves on a fixed straight line, the intersection point of the three medians of the ABC moves on (A) a straight line (B) a circle (C) a parabola (D) an ellipse
PART II : Multiple Correct Answer(s) Type This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
51.
3x x 2 , x2 2 2x 3 If f (x) = [ x 1], 2 x 3 x 8x 17,
then which of the following hold(s) good ? [Note : [x] denotes largest integer less than or equal to x. ] (A) Lim f ( x ) 1 .
(B) f(x) is differentiable at x = 2.
(C) f (x) is continuous at x = 2.
(D) f(x) is discontinuous at x = 3.
x 2
52.
If the position vector of a point P is r x ˆi yˆj zkˆ , where x, y, z N and projection of r on a ˆi ˆj kˆ
10 then number of possible position of P is also equal to 3 (A) number of natural solution of the equation x + y + z = 7. (B) number of ways of selecting two objects out of 10 distinct objects arranged in a row so that no two of them are next to each other. (C) the total number of outcomes when a pair of dice are rolled once. (D) number of positive divisors of 1800.
is
53.
For x, y, z 0, , let x, y, z be first three consecutive terms of an arithmetic progression such that cos 2
x + cos y + cos z = 1 and sin x + sin y + sin z =
(A) cot y = (C) tan 2 y
1 , then which of the following is/are correct? 2
(B) cos (x – y) =
2 2 2 3
3 1 2 2
(D) sin (x – y) + sin (z – y) = 0
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54.
Which of the following statement(s) is(are) correct? (A) If P, Q, R are 3 different points in space and they do not lie on same line, then PQ RQ is a vector orthogonal to the plane containing P, Q and R. (B) If x, y, z are positive real numbers satisfying x + y + z = 10, then the minimum value of
xy yz zx z x y
is equal to 10. n
(C) The value of Lim
n
i 1
e ln 2i ln n is equal to ln n 2
(D) Number of ordered pairs (x, y) of real numbers satisfying 4x – 2x + 2 + 5 = | sin y | are infinite. 55.
In which of the following cases unique equation of the plane can be established? (A) Given a normal vector to the plane and a perpendicular distance from origin to the plane is known. (B) Given a point on the plane and a line which intersects the plane at right angles. (C) A plane which consists of all points that are equidistant from two given points. (D) A plane which is parallel to a given line L1 and as well as line L2. PART III : Integer Answer Type
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 56.
Let two circles of radii r1 and r2 (r1 > r2 ) in the first quadrant are tangent to co-ordinate axes. If the length of common chord of circles is maximum, then find the value of
57.
Find the value of
.
Tr A10 671
6 7 8 9 10 11 .................... so on. 12 13 14
.
Find the number of points of intersection of the curves y = cos x, y = sin 3x, if
ln 2
59.
r2
The set of natural numbers is divided into arrays of rows and columns in the form of matrices as 2 3 A1 = (1), A2 = 4 5 , A3 =
58.
r1
If the value of the definite integral
x ln 2 e 0
x
eln 2 e ln (1 e
x 2
)
x . 2 2
dx = ln a where a is rational in the b b
lowest form, then find (b – a). (a,b > 0) 60.
Let the variable line ax + by + c = 0, where a, b, c are in arithmetic progression be normal to a family of circles. If r be the radius of the circle of the family which intersects the circle x2 + y2 – 4x – 4y – 1 = 0 orthogonally, then find the value of r2.
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Paper - I (Solution) 1.
(d)
P
2 2
f P.R 2 2
1 Q . . R 2 2 2 4R
2.
Q2 320 R 2
(a)
E
20 d
FE x l repulsion 20 d
FE 2 l 20 d
B
... (1)
0 2d
FB 2B 0 l 2d
2
force of interaction is 0
2 2 2 0 2 0 d 2d
1 0 0
... (2)
3.
(a)
L1 i1 L1 L2 i 2 2n 5v 5 2 i 2 1A
4.
(b)
There is charge on both sides on one side it is 5.
i1 i1 i2
E 6E i3 i4 Reff 7 R
2R & R in parallel
i1 2 i1 i2 1
2 6E 3 7R
R & R in parallel
i3 1 1 6E i3 i4 1 2 7R i A i1 i3
E 7R
2 0 AV 0 AV and other is 3d 3d
6.
(c)
K x mw 2 l x
K
x
mw 2 l K mw 2
1 1 2 mw 2 m l x w 2 2 2
1 kx mw 2 2 mw 2
1 k 2x 2 2 mw 2
k
2
1 k 2 mw 2l 2 mw 2 k mw 2
1 k mw 2 .l2 2 2 k mw 7.
(c)
A2 x2 30 A 2 82 16 A 2 152
8.
A 2 82
30 16
A 2 152
A = 17
(b)
E photon
1242 eV nm 6.2eV 200nm
K max 6.2 W 6.2 4.5
= 1.7 eV () at the collector = (1.7 + 2) eV = 3.7 eV
9.
(c)
hc w k1 1 he w k2 2 2 w k1 1 1 w k 2 2
2w 2k1 w k 2
k 2 2k1 w or
k 2 2k1 k1
10
k2 2
(d)
h1 h2 h3 hc hc hc 1 2 3 1 1 1 1 2 3
11.
(a)
(A) T2 r3
;
(B) E =
GMm 2r
GM ; (D) L = mvr = m GMr ] r
(C) v =
12.
(b, c, d) L = 90 × 10–3 × 2 × 1000 = 180 1 1 2 1000 = 1000 = 6 C 0.5 10 circuit is inductive VL > VR voltage leads the current
tan =
R=
1 80 C = R R
L
80 3
at resonance, =
1
=
13.
45 19 9
=
1 = LC
g eff
Pb (B) fraction = P same L
(C) T = 2
90 10
105 2 10 4 = 15 2 3
(a, d)
(A) T = 2
1 3
m k
(D) P = P0 + hPgeff
0.5 106
14.
(a, b, c, d) dia. 330 f max f 330 330 6 3
6 3.f 6 3
f min
6 3.f 6 3
here cos
1 3 and sin 2 2
max. frequency position = R cos , R sin R R 3 , 2 2
min. frequency position R cos , R sin R R 3 , 2 2
15.
(a, b, c, d)
16.
(2 kg) Let the angular velocity of rod at the time of collision be According to the law of conservation of energy For the rod at the horizontal and vertical positions, we get m2gR =
1 2 R I m2 g 2 2
m 2 gR 3g 1 m 2R 2 2 · = = = 2 R 2 3 Applying the law of conservation of angular momentum about P Let the angular speed of block about P after the collision be 0. I = m1R20
m 2 3g m 2R 2 = m1R20 0 = 3m1 R 3 Linear velocity of ball is
m2 v0 = 0R v0 = 3m 3gR 1
For ball to complete the circle m2 m2 5gR = 3m 3gR m = 15 1 1
v0 = 17.
(8 cm)
18.
(7) mgHB =
1 mv2 2
mgHS =
1 1 1 1 2 1 7 mv2 + I2 = mv2 + × mr22 = × mv2 2 2 2 2 5 2 5
HS 7 H B = 5 = 1.4
19.
(0002) 1 = RC 2 =
L R
LC = 1 2 = 0.1 sec
T = 2 20.
LC = 2
1 = 2 sec. 10
(1) 20 dQ kA = = (T) 4 dt 2L 20 k 2A T = t L
reqd. time = 1min
PAPER-1 21. 22.
(A) (A)
More electronegative element always occupy axial position in trigonal bipyramidal geometry F
90 Br
Br
o
P Br F
23.
(C) +
O CH2 H3C
CH3
O H H
+
OH
CH2OH
CH2 H3C
CH3
H 2O
H3C
CH3
25.
(A)
H
+ H3C
24.
CH2OH
CH3
OH H3C
CH3
As corresponding lone pair does not participate in resonance and thus better nucleophillic site.
(B) t 0, t 20 d ,
3R P a 0 ax x/3
For intersection point, [R] = [P], a x x 3,
x 3a 4,
a x a 4
2t1 2 20 days
26.
27.
(C)
H In 75 a 100 [H ][In ] Ka [HIn] [H ] 75 or 3 10 5 25 + –5 [H ] = 1 10 . HIn 25 a 100
(A)
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OH
OH CH3
CH3 anti-group migration CH3
H
CH3 OH
OH C
+OH2
+
CH3
CH3
H O C
CH3
CH3
28.
(B)
According to 3rd law of thermodynamics, T
S
C p ,m
0
29.
(C)
31.
(A, B, C)
T
dT
30.
(C)
H2O O2 Na Na2O Na2O2 NaOH O2
CO2 Na2CO3 H2O
HCl H2O CO2 NaCl
32.
(A, C, D)
4Zn 10HNO 3 4 Zn ( NO 3 ) 2 NH 4 NO 3 3H 2 O ( V. dil.)
33.
(B, D)
35.
(C, D)
36.
6
34.
(B, C, D)
U 235 Pb 207 7 42 He2
3×108 yrs = 3 t1 2 Thus, total moles of He produced = 7 0.5 2 0.5 4 0.5 8 3.5 8 3.06 Total moles of H2 produced for same volume =3.06 Total Faraday = 2 3.08 6 IIT – ian’s PACE Education Pvt. Ltd. : Andheri / Dadar / Chembur / Thane / Churchgate / Nerul / Powai
37.
1
Tb NaCl( aq ) T f Al2 SO4 3(aq ) ikb m ik f m 2 0.558 m 5 1.86 0.12, 38.
2
39.
2
40.
m 1
3
IIT – ian’s PACE Education Pvt. Ltd. : Andheri / Dadar / Chembur / Thane / Churchgate / Nerul / Powai
PAPER - 1 (SOLUTION) 41.
B The line of intersection of the two planes will be perpendicular to the normal's to the planes. Hence it is parallel
ˆi ˆj kˆ to the vector = ˆi 2ˆj 3kˆ × 3ˆi 2ˆj kˆ = 1 2 3 = 4ˆi 8ˆj 4kˆ 3 2 1
Also,
4ˆi 8ˆj 4kˆ · ˆj 16 64 16
=
8 96
If is the required angle, then cos = 42.
2 3
2 = cos–1 3
2 . 3
A x
We have f '(x) tan x =
2 cos t dt f '(x) tan x = 2ln (sin x) + 2 ln 2 sin t 6
f '(x) = 2 cot x ln (sin x) + 2 cot x ln 2 f '(x) =
d dx
(ln2(sin x) + 2 ln2 ln(sin x))
f (x) = ln2(sin x) + 2 ln 2 ln (sin x) + C ; 0 = ln2 2 – 2 ln2 2 + C C = ln2 2 6 So, f(x) = ln2(sin x) + 2 ln 2 ln(sin x) + ln2 2 f(x) = [ln 2 + ln(sin x)]2 = ln2 (2sin x)
For x =
43.
D
MB
We have r1 = r2 + r3 + r (Given) (r1 – r) = (r2 + r3)
2s ( b c) s (s a ) = (s b ) (s c ) s (s a )
A (s b)(s c) A 45 1 tan2 1 2 2 s (s a ) Hence, A = 90°
Now, area of ABC =
1 1 bc sin A = (6) (10) sin 90° = 30 (square units). 2 2
44.
A 1 + 2 + 3 + ............. + 9 = 45 = 0 + 1 + 2 + 3 + .................. + 9 All 9 digit such numbers = 9 ! All 10 digit such numbers when '0' included = 10 ! – 9 ! So, total = 9 ! + (10 ! – 9 !) = (10) !
45.
A Solving the equation of planes, we get equation of line containing planes x y z ...........(1) 0 10 5 Any point P on (1) is (0, – 10, – 5).
Now, direction ratios of the line joining P and M is 1, 1 10, 1 5
As line MP is perpendicular to line (1), so 0 (1) – 10 (1 + 10) – 5 (1 + 5) = 0 = So, d.r's of MP are 1,
3 6 3 P 0, , 25 5 5 0, – (0, – 1
1 2 , 5 5
x + 2y – 4z
5 )
P (0,0,0) 2x
M(1,1,1)
–
y
x 1 y 1 z 1 So, equation of required line is = = . 1 2 5
46.
=0
+
2z
=
0
A We must have, 19 9 < n < n {1, 2, 3, 4} 2 2 Hence, sum = 1 + 2 + 3 + 4 = 10
(2n + 19) (2n – 9) < 0
47.
B Clearly f(x) is decreasing in [0, 1) and increasing in [1, 3] So, for f(x) to be minimum at x = 1, we must have
Lim f(x) f(1) Lim x 3 cos 1 a 1 – 1 + cos–1 a 1 cos–1a 2
x 1
x 1
a [–1, cos 2].
48.
C We have f (2i + 1) = (2i + 1)2 – 2(2i + 1) = – 4 + 4i + 1 – 4i – 2 = – 5, and f(– 5) = (– 5)2 – 2(– 5) = 25 + 10 = 35.
49.
D
82MB
Clearly, mCP × mAB = – 1 k 2 k 8 =–1 h 3 h 1 Locus of (h, k) is (x – 1) (x – 3) + (y – 2) (y – 8) = 0 i.e., x2 + y2 – 4x – 10y + 19 = 0
50.
A line lx + my = 1 where l, m are constants
C (3, 2)
A
y
1 lb Let C b, m Hence 3h = b ....(1)
and
1 lb 3k = ....(2) m 3km = 1 – l (3h)
P (h, k)
C
lx + my = 1 fixed line
G(h,k) A(–a, 0) O
Locus of (h, k), is (3l)x + (3m)y = 1
lx + my =
B(a, 0)
1 3
x
B
M (1, 8)
51.
A, C, D y
(4,1)
y=1 O (0,0)
x
x=2 x=3 x=4
Interpret from graph.
52.
(A) (B)
(C) (D)
53.
B, C, D
r ·a Projection of r on a is x + y + z = 10 (x, y, z N) |a| 10–1 number of solution C3–1 = 9C2 = 36 x + y + z = 7, x, y , z 1 x + y + z = 4 6C2 = 15 number of non negative integral solution = 7 + 3 – 1C3 – 1 = 9C2 = 36. 10 objects 2 to be selected O O ||||||||| 6C = 36 2 9C = 36 2 1800 = 23 32 52 Number of divisors = (3 + 1) (2 + 1) (2 + 1) = 36.
A, D We have y d , y , y z in A.P.. x
z
Now,
cosx 1
Also,
sin x
Equation (1) Equation (2) cot y =
cos (y – d) + cos y + cos(y + d) = 1 cos y (2 cos d + 1) = 1 .....(1)
1 1 1 sin(y – d) + sin y + sin (y + d) = sin y (2 cos d + 1) = .....(2) 2 2 2
Now, putting cos y =
2 cos d + 1 =
2
2 in (1), we get 3
2 cos d = 3
3 2 = cos (y – x) = cos (x – y) 2 2
1 2 2 tan y 2 Also, tan 2y = = = 1 1 tan 2 y 1 2
2 2 2 1 2
Clearly, sin (x – y) + sin (z – y) = sin (– d) + sin d = 0.
54.
A, B D
(A)
Obviously true.
(B)
Using A.M. G.M. in
xy yz yz zx zx xy , ; , ; , and add, we get z x x y y z
xy yz zx (x + y + z) z x y xy yz zx 10 z x y
(C)
Put
2i ln n = ln 2x dx i 1 0
2x = t
1 S= 2 (D)
1
n
1 Sn = n 2
1 1 2 t ln t t 20 = 2 ln 2 2 0 ln 2 2 e 0 x 2 (2 – 2) + 1 = | sin y | L.H.S. 1 and R.H.S. 1
ln t dt
=
x = 1 and y = 2n ±
, nI] 2
55.
B, C
(A) (B)
There can be two parallel planes equidistant from the origin. Obviously the equation of the plane is A(x – x1) + B(y – y1) + c(z – z1) = 0 where A, B, C are proportional to the dr's of the line.
(C)
P (x1, y1, z1) and Q(x2, y2, z2) Apply condition of locus on R(x, y,. z) to get a unique plane, 2nd degree term cancels. Plane can be parallel.
(D)
56.
3 Let equation of circles be S (x – r1)2 + (y – r1)2 = r12 ... (1) and S' (x – r2)2 + (y – r2)2 = r22 ... (2) where r1 > r2 Equation of common chord is S – S' = 0 given by 2(x + y) = r1 + r2 ... (3) For maximum length of common chord, above equation (3) must pass through the centre (r2, r2) of the smaller circle.
4r2 = r1 + r2 3r2 = r1
r1 r2
= 3
y
A B O (0,0)
x
57.
5 First term of An is the nth term of series S = 1 + 2 + 6 + 15 + .......... + Tn (Using method of difference) S = 0 + 1 + 2 + 6 + .............+ Tn – 1 + Tn (Subtracting) ————————————————— 0 = 1 + 12 + 22 + 33 + .......... + Tn – Tn – 1 – Tn n 1
Tn = 1 +
n2 n 1
Tn = 1 +
(n 1) n (2 n 1) 2n 3 3n 2 n 6 = 6 6
2000 300 10 6 = 286 6 Now common difference of the diagonal elements of A10 is 11 and number of diagonal elements is 10.
So,
T10 =
Hence Tr(A10) = Sum of its diagonal elements = 58.
3 For point of intersection, we have cos x = sin 3x cos x = cos 3x 2
(+ ve) sign,
x=
x=
x=
x = 2n ± 3x 2
n (n I) Possible values of n are – 1 and 0. 2 8
3 and x = 8 8
x = – n
(– ve) sign,
59.
10 2 · 286 9 ·11 = 3355 2
(n I) Possible values of n is 0. 4
4
3 3 Hence, points of intersection are , cos , , cos , , cos . 8 4 8 4 8 8 1 ln 2
I=
x 2
x ln 2 ln (1 e ) x ln 2e · e · e
dx
0 ln 2
I=
1 x x ln 2e · 2 · (1 e x ) 2 dx 0
ln 2
I=
0
ln 2 ln 2 dx 1 2 x l n 2 2 x ln 2 dx = e x 1 (1 e x ) 2 ex 1 0 0 I
ex II
ln 2 x = 2 0 ln 1 e 2
ln 2 3 1 9 = 2 2 ln 2 ln 2 = 2 2 2 ln 2 ln 3 = ln 2 ln 4
ln 2 0
2 4 8 a = ln = ln (b – a) = 9 – 8 = 1. = ln 9 9 b
60.
8 As a, b, c are in A.P., so ax + by + c = 0 represents a family of lines passing through the fixed point (1 ,–2). Since each member of the family is normal to a circle hence its centre must be (1, – 2). So, the family of circles with centre (1, – 2) will be given by (x – 1)2 + (y + 2)2 = r2 x2 + y2 – 2x + 4y + (5 – r2) = 0 And using condition of orthogonality, we get 2 [(–1) (– 2) + (2) (– 2)] = – 1 + 5 – r2 r2 = 8
