Indian Institute of Information Technology 2010 Entrance Exams IIT Entrance Exams Iit-jee part-1 chemistry - Question Paper

well as sin20 = cos40 is Key. (3)

Sol. tan 0 = cot 50

tan 0 = tan I--50

, 2

0 = nn +0 - 50 2

n

60 = nn +

2

_ nn n

0 = +n el 6 12

sin 20 = cos 40

sin 20 = 1 - 2 sin² 20 2sin²20 + sin 20-1 = 0 2sin²20 + 2sin20- sin 20-1 = 0 (2 sin 20 - 1)(sin 20 + 1) = 0

n 5n 12^,12

All three values of 0 which satisfy the eq. (i).

For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [-10, 10] by

Here f(x) is periodic with period 2 and cos nx is also periodic with period 2 f(x) cos nx is periodic with period 2.

"    ²    40

f f(x)cosnx dx = 10f f(x)cosnx dx = -

Hence, f(x)cosnx dx = in J

50. If the distance between the plane Ax - 2y + z = d and the plane containing the lines

x -1 y - 2 z - 3 . x - 2 y - 3 z - 4 r- .

-=-=-and-=-=- is V6 , then |d| is

2 3 4    3 4 5

Key. (6)

Sol. The equation of the plane containing the given lines will be a(x - 1) + b(y - 2) + c(z - 3) = 0 where a, b, c are direction ratios of normal to the plane considering vectors parallel to the two lines 2i + 3j + 4k and 3i + 4j + 5k So 2a₁ + 3b₁ + 4c₁ = 0 3a₁ + 4b₁ + 5c₁ = 0

^a1 ^-b1 =

15 -16 10 -12 8 - 9 So the plane is x - 2y + z = 0 Hence distance between two planes

|d|

V1² + 2² +1

|d|= 6

2 2 x y

51. The line 2x + y = 1 is tangent to the hyperbola ---- = 1 . If this line passes through the point of

a b

intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is Key. (2)

Sol. Since the line 2x + y - 1 = 0 is tangent so, C² = a²m² - b²

1 = 4a² - b²    .....(i)

a

Also line passes through |--,0

e

21 - - | = 1

_4a² _{= e}²

Using (i) and (ii) e = 2

Let S_k, k = 1, 2, ... , 100, denote the sum of the infinite geometric series whose whose first term is

1 100² 100 .

and the common ratio is . Then the value of--h'V fk² - 3k + 1|S,

100! > ^k

k

Key. (4)

k -1 k! ³

53. Let f be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x, y) on the curve y = f(x) is equal to the cube of the abscissa of P, then the value of f(-3) is equal to Key. (9)

Sol. eq. of tangent at P(x, y)

Y - y = (X - x) dx

dy 3

y-integer y - x = x dx

- y = - x dx x

-j -¹dx 1

I.F. = e ^x = x

The solution

1 r ₂ 1 _d

y x = j - x x dx x ^J x

-=-+c

x 2

f(1) = 1 C = ³

2

3x - x

f(x) = y = f (-3) = 9

_T. _ , . . , _ i-2j    , 2^j +j + 31c , , ,

54.    If a and b are vectors in space given by a ==- and b =-==, then the value of

V5    V14

(a + b)(xbjx(-2b) is

Key. (5) _    _

Sol. |a|=| b|= 1 a.b = 0

Let I = (a x b) x (a - 2b) = (a x b) x a - 2(a x b) x b

=| a |² b - (a.b)a - 2(a.b)b + 21 b | a.

= b + 2a

(2a + b)l =| 2a + b |² = 5

55.    The number of all possible values of 0, where 0 < 0 < n, for which the system of equations

(y + z) cos30 = (xyz) sin 30

. 2cos30 2sin30 xsin30 =--1--

y ^z

(xyz) sin30 = (y + 2z) cos30 + ysin 30 have a solution (x₀, y₀, z₀) with y₀ z₀ # 0, is Key. (3)

Sol. (y + z)cos30 = (xyz) sin 30    (A)

. 2cos30 2sin30    _

x sin 30 =-+--(B)

y ^z

(xyz) sin 30 = (y + 2z) cos 30 + y sin 30 (C)

When cos 30 # 0.

56. The maximum value of the expression Key. (2)

1

Sol. Let y =

sin² 0 + 3sin0cos0 + 5cos² 0 , , ~ ~_n , 3 .

3 + 2 cos 20 +sin 20

2

5    3    5

< 2cos20 + sin20 <  2 2 2

max. value of y =

3-

2

PART III: PHYSICS

_Single Correct Choice Type_

This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

An AC voltage source of variable angular frequency ro and fixed amplitude V₀ is connected in series with a

capacitance C and an electric bulb of resistance R (inductance zero). When ro is increased

(A) the bulb glows dimmer    (B) the bulb glows brighter

(C) total impedance of the circuit is unchanged (D) total impedance of the circuit increases.

(C)

2T - iB(2R)

iBL

2n

"ence correct option is (C).

59. A block of mass m is on an inclined plane of angle 0. The coefficient of friction between the block and the plane is |i and tan 0 > |i. The block is held stationary by applying a force P parallel to the plane. The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg (sin

0 - |i cos 0) to P₂ mg (sin 0 + |i cos 0), the frictional force f versus P graph

will look like (A)

f

(C)
	-P

Sol.

	x
V = -2tcGct J rdr

3R Vr² + x²

2r dr = dz r rdr dz Vr² + x² 2\/z

= "T7 = ²>2

V = -2tcGct! Vr²+ x

= -2nGr [4R>/2 - 5]

W = (1) [0 + 2tcGct(4rV2 - 5R)]

= 2nG--^M-- (4R>/2 - 5R)

n(16 - 9)R 2nGM

-(W2 - 5).

7R

Hence correct option is (A).

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40, respectively, the relation between these resistances is

(A) = - + -

^{(B) R}100 = ^R40 + ^R6'

^R100 ^R40 ^R60

As temperature increase, resistance increases

^So, ^R40 > ^R60 > ^R100 .

Hence correct option is (D).

To verify Ohm's law, a student is provided with a test resistor R_T, a high resistance R₁, a small resistance R₂, two identical galvanometers G₁ and G₂, and a variable voltage source V. The correct circuit to carry out the experiment is

(B)

(A)

R

-to*-

V    ' V

Key. (C)

Sol. An ideal voltmeter should have large resistance and an ideal ammeter should have low resistance.

Hence correct option is (C).

SECTION - II

_Multiple Correct Choice Type_

This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

65. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms^-1. Which of the following statement(s) is (are) correct for the system of these two masses ?

(A)    total momentum of the system is 3 kg ms^-1

(B)    momentum of 5 kg mass after collision is 4 kg ms^-1

(C)    kinetic energy of the center of mass is 0.75 J

(D)    total kinetic energy of the system is 4 J.

Key. (A), (C)

Sol.    (1) (V) + (5) (0) = (1) (-2) + 5 V'

V    = 5V' - 2    (!)

V'+ 2 = ₁

V    - 0 =

V' = V - 2    (ii)

V    = 5 (V - 2) - 2 From equation (i) and (ii)

V    = 5V - 10 - 2 4V = 12

V    = 3 m/s.

P_i = (1) (3) = 3 kg - m/s.

V    = (¹)(³) + (⁵)() = 1m/s

CM    ~^m/ *

6 2

1 1 3

^kcm = 2(⁶)4 = 4 = ⁰.^75J

Ktotal = 2(1)(3)² = 4.5 J Hence correct options are (A), (C).

66. A few electric field lines for a system of two charges Q₁ and Q₂ fixed at two different points on the x-axis are shown in the figure. These lines suggest that

(A)    |Q1 > |Q2|

(B)    |Q1| < |Q2|

(C)    at a finite distance to the left of Q1 the electric field is zero

(D)    at a finite distance to the right of Q₂ the electric field is zero.

Key. (A), (D)

Sol. Density of field lines is more are Q1

|Q1 > |Q2|

Q₁ and Q₂ are of opposite signs

So, null point will be closer to charge of smaller magnitude i.e., Q2 Hence correct options are (A), (D).

67. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60 (see figure). If the refractive index of the material of the prism is y/3 , which of the following is (are) correct ?

O

D

(A)    the ray gets totally internally reflected at face CD

(B)    the ray comes out through face AD

(C)    the angle between the incident ray and the emergent ray is 90

(D)    the angle between the incident ray and the emergent ray is 120.

Key. (A), (B), (C)

Sol.    1 sin 60 = V3 sinr

r = 30

sin 0_C =

^C V3

0C = 35

D

At CD angle of incidence is greater than 0_C .

At AD angle of incidence is less than critical angle So ray will come out of AD.

Angle of deviation

- 30 + 90 + 30 = 90

Hence correct options are (A), (B), (C)

(D) temperature at C is

Key. (A), (B)

Sol. Internal energy of an ideal gas depends on temperature

W_BC = nRT fn^V

^BC    V₁

P V 4V = (1)(R)-^0VL in-R V₀

= Wn4 Hence (A), (B) options are correct.

69. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 second fore this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true ?

(A)    error AT in measuring T, the time period, is 0.05 seconds

(B)    error AT in measuring T, the time period, is 1 second

(C)    percentage error in the determination of g is 5%

(D)    percentage error in the determination of g is 2.5%.

Key. (A), (C)

Sol. Error in measurement of T = s = 0.05 s

20

dg = 2dl g T

g- = 2 x-L

g 40

% error in calculation of g = 5%.

SECTION - III

_Linked Comprehension Type_

This section contains 2 paragraphs. Based upon the first paragraph, 3 multiple choice questions and based upon the second paragraph 2 Multiple choice questions have to be answered. Each of these questions have four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question Nos. 70 to 72

When a particle of mass m moves on the x-axis in a potential of the form V (x) = kx², it performs simple

harmonic motion. The corresponding time period is proportional to , , as can be seen easily using

k

dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of x = 0 in a way different from kx² and its total energy is such that the particle does not escape to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = ax⁴ (a > 0) for |x| near the origin and becomes a constant equal to V₀ for |x| > X₀ (see figure).

If the total energy of the particle is E, it will perform periodic motion only if (A) E < 0    (B) E > 0

(C) V₀ > E > 0    (D) E > V₀.

(C)

For periodic motion

Total energy should be less than V₀ but greater than zero.

Hence (C) is correct.

Paragraph for Question Nos. 73 to 74

Electrical resistance of certain materials, known as superconductors, changes abruptly from a nonzero value to zero as their temperature is lowered below a critical temperature T_C(0). An interesting property of superconductors is that their critical temperature becomes smaller than T_C(0) if they are placed in magnetic field, i.e., the critical temperature T_C(B) is a function of the magnetic field strength B. The dependence of T_C(B) on B is shown in the figure.

73. In the graphs below, the resistance R of a superconductor is shown as a function of its temperature T for two different magnetic field B₁ (solid line) and B₂ (dashed line). If B₂ is larger than B₁, which of the following graphs shows the correct variation of R with T in these fields ?

T T

(A)

T

Key. (A)

Sol. As B increases, critical temperature decreases.

74. A superconductors has TC(0) = 100 K. When a magnetic field of 7.5 Tesla is applied, its TC decreases to 75 K. For this material one can definitely say that when

(A) B = 5 Tesla, T_C(B) = 80 K    (B) B = 5 Tesla, 75 K < T_C(B) < 100 K

(C) B = 10 Tesla, 75 K < T_C(B) < 100 K    (D) B = 10 Tesla, T_C(B) = 70 K.

Key. (B)

SECTION - IV

Integer Answer Type

This Section contains TEN questions. The answer to each question is a Single Digit Integer ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

lm (1X0.1) 140 x140 = y(49) x10^-7 y = 4 x 10⁹ n = 4.

78. When two progressive waves y. 4 sin (2x - 6t) and y₂ = 3 sin 2x - 6t - | are superimposed, the

amplitude of the resultant wave is Key. 5.

Amplitude - V4² + 3²

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of total energy radiated by A to that of B ?

9.

(T₁) (500 nm) = T₂ (1500 nm)

T₁ = 3T₂

E_a -ct-4n(6cm)²(T₁ )⁴ E_b - ct-4n(18cm)² (T₁)⁴

^: (3)⁴ - 9.

V6

Gravitational acceleration on the surface of a planet is ng, where g is the gravitational acceleration on

2

the surface of the earth. The average mass density of the planet is 3 times that of the earth. If the escape

speed on the surface of the earth is taken to be 11 kms^-1, the escape speed on the surface of the planet in kms^-1 will be

3.

GM_P

(i)

RP 11 11 R

-y/2g_eR_e - 11 km / s

V²2p^Rp ^{- x} (11)²

ge^Re

^gP^RP

GM

M_{P -} 2 M,

Rf ^- 3^- r|

x = 3.

81. A stationary source is emitting sound at a fixed frequency f₀, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f₀. What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms^-1.

Key. 7.

V + V_C    V + V_C

Sol.    f1 -- f; f     f

V - V_C

C₁

V + V_C V + V_C

= 2AV_f = 1f

V ⁰ 100 AV_C = 7 km/hr.

82. When two identical batteries of internal resistance 1Q each are connected in series across a resistor R, the rate of heat produced in R is J₁. When the same batteries are connected in parallel across R, the rate is J₂. If J₁ = 2.25 J₂ then the value of R in Q is Key. 4.

^So1' ^J1=( i!R I^R

J₂ = ⁽R

² l 0.5 + R )

2.25 = -^4(a5 + ^R)2

(2 + R)²

9 = 4(R + 0.5)

4 = 2 + R

3 2R + 2

2 2 + R

6 + 3R = 4R + 2 R = 4Q.

83.    A piece of ice (heat capacity = 2100 J kg^-1 C^-1 and latent heat = 3.36 x 10⁵ J kg^-1) of mass m grams is at -5C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is

Key. 8.

Sol. [ m(2100)(5) +1(3.36 x 10⁵)] x 10^-3 = 420

11m + 336 = 420 11m = 420 - 336 = 84 m = 8 gm .

84.    An a-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de

X_p

Broglie wavelengths are X_a and X_p respectively. The ratio -, to the nearest integer, is

^Xa

Key. 3.

Sol.    X ^h

V2mk

k = qV X ^h

V^2(m)q⁴

^Xp = |m7q X = |^(4m)(2q⁾ = ₂2 = ₃

^Xa v ^mpqp v ^(m)q

IIT - JEE (2010) PAPER II QUESTION & SOLUTIONS CODE 0

PART I : CHEMISTRY

PAPER - II SECTION - I

_Single Correct Choice Type_

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. The compounds P, Q and S

NO

Key: (C)

COOH
	no2
OH

COOH
P
OH

Sol.:

Q S

= nL² = 16

<100

Area =2 n

Packing efficiency =

3.14

<100 = 78.5%.

The complex showing a spin-only magnetic moment of 2.82 B.M. is (A) Ni (CO)₄    (B) [NiCl4 ]

(C) Ni (PPh3 )4    (D) [Ni (CN) ]²

Key: (B)

Ni₂₈ 4S²3d⁸

Ni+²4S⁰3d⁸

4p

t t

2 unpaired electron (n = 2) |i = yjn (n + 2) = V8 BM = 2.82 BM.

5.

H,C-

NH /?' O

Key:

O

Sol.:

SECTION - II

_Integer Type_

This section contains a group of 5 questions. The answer to each of the questions is a single-digit integer. ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled.

7.    Silver (atomic weight = 108 g mol^-1) has a density of 10.5 g cm^-3. The number of silver atoms on a surface of area 10^-12 m² can be expressed in scientific notation as y x 10^x . The value of x is

Key: (7)

Sol.: Consider a single layer square shaped arrangement of n x n silver atoms. Also assume the radius of each silver atom r.

Area of the layer = (2rn)²= 10^-12m² = 10^-8cm² Mass of the layer

Surface area x thickness x density = number of atoms x mass of single atom.

10^-8 x 2r x10.5 = n² x 108x 1.66 x10^-24    [put 2rn = 10^-4]

n³ = 5.855 x10¹⁰ n = 3.82 x10³ .. number of silver atoms on the surface

= n² =(3.82 x10³ )²

= 1.4592 x 10⁷ .. x = 7.

8.    Among the following, the number of elements showing only one non-zero oxidation state is

O, Cl, F, N, P, Sn, Tl, Na, Ti

Key: (2)

Sol.: Fluorine and sodium shown only one non zero oxidation state, fluorine show - 1 and sodium shown + 1.

9.    One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is w_s and that along the dotted line path is w_d, then the integer closest to the ratio w_d/w_s is

V(lit.)

Key: (2)

Sol.: w_d = 4x 1.5 + 1 x 15 + 0.80 x 2.5 = 9.375

w_s = 2.33 pV log

^vi

= 2.33 x 4 x 0.5 log ⁵-⁵

0.5

4.606

H3C    OH
	O
O

O
(A)
O

K2CO3

"3CC .oC C    _C    C"n 3 I I 2 CO2" C"3

Paragraph for Questions 15 to 17

The hydrogen-like species Li²+ is in a spherically symmetric state S₁ with one radial node. Upon absorbing light the ion undergoes transition to a state S₂. The state S₂ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.

The state S1 is

(A)    1s    (B) 2s

(C) 2p (D) 3 s

(B)

No. of radial node = n - I - 1

Since state S₁ has 1 radial node it must be 2s orbital with n = 2 and I - 0 .

(B)

Energy of the state S₁ in units of the hydrogen atom ground state energy is

(A) 0.75

(C) 2.25

(C)

Energy of state S1 =

13.6

3 eV/atom

= -13.6 x 2.25 eV / atom

= 2.25 times energy of ground state of hydrogen atom.

(C)

__,O

OH / -CH

CH3

(C)

OAlH3 ,/ VC-

HO I C-1 Cch3

19. All the compounds listed in Column I react with water. Match the result of the respective reactions with the appropriate options listed in Column II.

PART - II: MATHEMATICS

SECTION - I

Single Correct Choice Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

If the distance of the point P(1, -2, 1) from the plane x + 2y - 2z = a, where a > 0, is 5, then the foot of the perpendicular from P to the plane is

x.-1 = y. + 2 = z. -1 = -(-15)

If (x., y., z. ) is foot of perpendicular (x., y., z.) =(8/3, 4/3, -7/3)

4 1

A signal which can be green or red with probability 5 and 5 respectively, is received by station A and then

3

transmitted to station B. The probability of each station receiving the signal correctly is . If the signal

4

received at station B is green, then the probability that the original signal was green is

(A)

22. Two adjacent sides of a parallelogram ABCD are given by AB = 2i + 10j+11k and AD = -i + 2j + 2k . The side AD is rotated by an acute angle a in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle a is given by

(B)

AB.AD > 0,

So 0 is acute.

AB.AD

8

cos0 =

AB||AD| 9 cos(a + 0) = 0 cosa.cos0 - sina. sin0 = 0 8cosa - -s/T7 .sina = 0

Vn

9

64cos²a = 17 sin²a cosa =

For r = 0, 1, ... , 10, let A_r, B_r and C_r denote, respectively, the coefficient of x^r in the expansions of (1+x)¹⁰, (1+x)²⁰ and (1 + x)³⁰ . Then

A_r (B₁₀B_r - C₁₀A_r) is equal to

A_r = ¹⁰C_r B_r = ²⁰C_r

C_r = 30C_r 10

X A_r(B1₀B_r - CA)

= B10 A_rB_r - C_w A_r²

=1    r=1

10 10 ₂ ¹⁰c_r.²⁰c_r - C1₀ (¹⁰c_r)

r=1    r=1

10 20 10 20 10 20 10 ² = ^B10 L ^C1 ^C1 + ^C2 ^C2 .... + ^C10 ^C10 J ^{- C}10 ^Cr J

= B10 .[³⁰C10 - 1] - C10 [²⁰C10 - 1J ^{= C}10 ^{- B}10

x

Let f be a real-valued function defined on the interval (-1, 1) such that e^-x f(x) = 2 + jyl t⁴ +1 dt, for all xe

(-1, 1), and let f ^-1 be the inverse function of f. Then (f¹)' (2) is equal to

(A) 1

(D) I

e

Key (B)

Sol. v f (f ^-1(x)) = x

f ' (f⁵ (x)) . (f⁶(x))' = 1

(f¹(x))' =-¹-

f '(f^-1(x))

We have to find (f ¹(2))'

(f ¹(2))' = -¹-

f '(f(2))

When f^-1 (x) = 2 f(x) = 2 x = 0

x

given e"^x f(x) = 2 + JV t⁷ +1 dt

0

e"^x (-f(x) + f '(x)) = x⁸ + 1 put x = 0 f '(0) = 3 (f¹ (2))' = 1/3.

25.    Let S = {1, 2, 3, 4). The total number of unordered pairs of disjoint subsets of S is equal to

(A) 25    (B) 34

(C) 42    (D) 41

Key (D)

Sol. S = {1, 2, 3,4}, then No. of unordered pairs of disjoint subsets of S is

= 41

2

SECTION - II

_Integer Type_

This section contains a group of 5 questions. The answer to each of the questions is a single-digit integer. ranging from 0 to 9. The correct digit below the question no. in the ORS is to be bubbled.

2009    2010    2011    2012

The local maximum of f(x) occurs at x = 2009

Hence, local maximum of g(x) also occurs at x = 2009. Hence the number of point of local maximum = 1.

det (adj A) + det (adj B) = 10 , then [k] is equal to

{Note: adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

(4)

|A| = (2k - 1) (-1 + 4k⁹) + 2>/k (2 + 4Wk ) + 2>/k(4kVk + 2>/k)

(2k - 1) (4k² - 1) + 4k + 8k² + 8k² + 4k = (2k - 1) (4k² - 1) + 8k + 16k²

= 8k¹⁰ - 4k² - 2k + 1 + 8k + 16k² = 8k³ + 12k² + 6k +1

|B| = 0 as B is skew symmetric matrix of odd order.

(8k³ + 12k² + 6k + 1)² = (10³)² (2k + 1)³ = 10³ 2k + 1 = 10 k = 4.5 [k] = 4.

29. Two parallel chords of a circle of radius 2 are at a distance V3 +1 apart. If the chords subtend at the center, n 2n

angles of and , where k > 0, then the value of [k] is k k

[Note : [k] denotes the largest integer less than or equal to k]

Key (3)

Sol. 2cos +2cos= -v/3 +1 k 2k

3 + V3

2cos-- cos-

2k 2k

-1 (2/3 + 1)² = yf3 ->/3 - 1

30. Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15>/3 . If ZACB is obtuse and if r denotes the radius of the incircle of the triangle, then r² is equal to ^{Key (3)} _B Sol. The area = 153

C b = 10

10² + 6² - c²

2 x 10 x 6

-60 = 136 - c² c² = 196 c = 14.

A 1573 15>/3 x 2 Since r = -- =-

s (10 + 6 +14)/ 2 30

SECTION - III

Paragraph Type

This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions have to be answered. Each of these question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions Nos. 31 to 33

Consider the polynomial f(x) = 1 + 2x + 3x² + 4x³

Let s be the sum of all distinct real roots of f(x) and let t = |s|.

The area bounded by the curve y = f(x) and the lines x = 0, y = 0 and x = t, lies in the interval

(C) (9, 10)

(A)

Required area t

A = g(t) = J f(x)dx

1 -1 ;'(t) -f(x) > 0 Vx > 0

Paragraph for Questions Nos. 34 to 36

2 2 x y

Tangents are drawn from the point P(3, 4) to the ellipse + = 1 touching the ellipse at points A and B.

.(i)

The equation of the locus of the point whose distances from the point P and the line AB are equal, is

(A) 9x² + y² - 6xy - 54x - 62y + 241 = 0

(C) 9x² + 9y² - 6xy - 54x - 62y - 241 = 0 (A)

V(a-3)² +(P-4)² = ^(a+P_~³ 10(a² -6a + 9 + p² -8p +16) = a² + 9p² + 9 + 6ap-6a- 18p

Required locus is 9x + y - 6xy - 54x - 62y + 241 = 0

SECTION - IV

_Matrix Type_

This section contains 2 questions. Each question four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for that particular question, against statement B, darken the bubbles corresponding to q and r in the ORS._

Key. (A-q, r), (B-p), (C-p, s, t), (D-q, r, s, t)

^{|z - i|z||}=^|z+^i|z|| (A)

Putting z = x + iy

We get y -x² + y² = 0

i.e., Im (z) = 0.

2ae = 8, 2a = 10

(B)

(0, 3)

3    5

x = cos0, y = sin0

2 2

f) +(f

2x1 zl

-9. + 25 = 1

9 = 25(1 - e2)

4 4

e2 = 1 -A = 16

25 25

4

5

(D) Let w = cos 0 + isin0

1

z = x + iy = w +--

w

x + iy = 2cos 0 x = 2cos0, y = 0

(q), (s)

38. Match the statements in Column - l with the values in Column-ll