FIITJEE INTERNAL TEST

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website: www.fiitjee.com.

PHYSICS, CHEMISTRY & MATHEMATICS

Time Allotted: 3 Hours Set-A Maximum Marks: 201

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS

Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results.

A. General Instructions 1. Attempt ALL the questions. Answers have to be marked on the OMR sheets. 2. This question paper contains Three Parts. 3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics. 4. Each part is further divided into two sections: Section A & B. 5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided

for rough work. 6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices,

in any form, are not allowed.

B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR

sheet.

2. On the OMR sheet, darken the appropriate bubble with Blue/Black Ball Point Pen for each

character of your Enrolment No. and write in ink your Name, Test Centre and other details at the designated places.

3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

(i) Section-A (01 – 06) contains 6 multiple choice questions which have only one correct answer. Each question carries +3 marks for correct answer and – 1 mark for wrong answer.

Section-A (07 – 09) contains 3 Assertion-Reasoning multiple choice questions which have only one correct

answer. Each question carries +3 marks for correct answer and – 1 mark for wrong answer. Section-A (10 – 15) contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have

to be answered. Each question has only one correct answer and carries +4 marks for the correct answer and – 1 mark for wrong answer.

(ii) Section-B (01 – 02) contains 2 Matrix Match Type Questions which have statements given in 2 columns.

Statements in the first column have to be matched with statements in the second column. There may be One or More Than One Correct choices. Each question carries +8 marks for all correct answer however for each correct row +2 marks will be awarded and –1 mark for each row matched incorrectly.

Name of the Candidate :______________________________________________

Batch :____________________ Date of Examination :_____________________

Enrolment Number :_________________________________________________

BA

TC

HE

S-

Tw

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FIITJEE INTERNAL TEST

CPT-1

CODE: 110674

PAPER - 2



IT2020Two Year CTY (2018-20)-A LOT-PH-I (PAPER-2)-SET A-PCM-JEE ADVANCED-(0000.1)-2

PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSIIICCCSSS

SECTION – A : (Single Correct Answer 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. A circular disc of radius R is rotating about its axis through O with a

uniform angular velocity rad/s as shown in the figure. P and Q are two points on the disc. At any instant of time the magnitude of the relative velocity of P with respect to Q is

(A) 0 (B) R

2

(C) R

32

(D)

5R

2

O Q

R/2

P

2. A car accelerates from rest at a constant rate of 2 m/s2 for some time. Then, its retards at a constant

rate of 4 m/s2 and comes to rest. What is the maximum speed attained by the car if it remains in motion for 3 seconds?

(A) 2 m/s (B) 3 m/s (C) 4 m/s (D) 6 m/s 3. For the system shown in the figure, the pulleys are light and

frictionless. Assume wedge to be fixed and smooth. The tension in the string will be

(A) 2

3mgsin (B)

3

4mgsin

(C) 1

2mgsin (D) mgsin

m

m

Space for rough work




4. A stone is projected at time t = 0 with a speed v0 at an angle with the horizontal in a uniform gravitational field. The rate of work done (P) by the gravitational force plotted against time (t) will be as

(A)

P

t O

(B)

P

t O

(C)

P

t O

(D)

t O

P

5. A projectile is projected at an angle (>45°) with an initial velocity u. The time t, at which its horizontal velocity will equal the vertical velocity, is

(A) t = u

g (cos – sin ) (B) t =

u

g (cos + sin )

(C) t = g

u (sin cos ) (D) t =

u

g (sin2 – cos2 )

6. Figure shows regular hexagon PQRSTU. Find the value

ofPQ PR PS PT PU .

(A) PO (B) 2PO

(C) 4PO (D) 6PO

S T

P Q

O U R





(Assertion – Reason Type)

This section contains 3 reasoning type questions. Each question has 4 choices (A), (B), (C) and (D), out of

which ONLY ONE is correct.

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is not the correct explanation of A

(C) A is true , R is false

(D) A is false , R is true

7. ASSERTION: The safe velocity limit for taking a turn on an unbanked road is v rg (symbols

have usual meanings). REASON: Banking of roads will increase the value of limiting velocity. 8. ASSERTION: When a particle moves on a circular path with constant speed, it is called uniform circular

motion. REASON: For a uniform circular motion, it’s necessary that a force is always acting parallel to the

velocity direction.

9. ASSERTION: Work done by a conservative force is always zero in a round trip.

REASON: Work done by a non conservative force is not necessarily zero in a round trip.





(Paragraph Type)

This section contains 2 paragraphs. Based upon the paragraphs 3 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question Nos. 10 to 12

In the adjacent figure, two identical beads of mass m free to slide on a fixed

smooth vertical ring are attached to the free ends of two identical light springs

of spring constant k= (2 3)mg

3R

.

O

The other ends of the springs are fixed to the highest point O on the ring. Initially beads are at rest and angle between the springs is 60º at the fixed point O. Normal length of each spring is 2R, where R is the radius of the ring. Beads are initially symmetric w.r.t. the vertical diameter. 10. Normal reaction on one of the bead at initial moment due to ring is

(A) mg

2 (B)

3mg

2

(C) mg (D) zero 11. Relative acceleration between two beads at the initial moment

(A) g

2vertically away from each other (B)

g

2horizontally towards each other

(C) 2g

3vertically away from each other (D)

2g

3horizontally towards each other

12. The magnitude of spring force on one of the beads at the initial moment is

(A) mg (B) mg

2

(C) mg

3 (D)

mg 3

2





Paragraph for Question Nos. 13 to 15

The potential energy function for the force between two atoms in a diatomic molecule can approximately be

expressed as U(x) = 12 6

a b

x x , where a and b are positive constants, and x is the distance between the

atoms. Answer the following questions by selecting most appropriate alternative. 13. The graph between potential energy vs x will be

(A)

U (x)

x

(B)

U (x)

x

(C)

U (x)

x

(D)

U (x)

x

14. The work done by an external agent to slowly separate atoms of a diatomic molecules to infinity is

(initially atoms are at rest at equilibrium position)

(A) 2b

4a (B)

2b

2a

(C) +2b

4a (D) +

2b

2a

15 The intermolecular separation for equilibrium is :

(A)

1/6a

b (B)

1/62a

b

(C)

1/6a

2b (D)

1/6b

a





SECTION – B : (MatrixMatch Type)

This section contains two questions, each having two matching Columns. Choices for the correct combination of elements from Column-I and Column-II are given as option (A), (B), (C) and (D) out of which one or more than one are correct.

1. Match the column:

Column - I Column - II

(A) A particle is moving in circle (p) The acceleration may be opposite to its velocity

(B) A particle is moving in a straight line (q) The acceleration may be at some angle

02

with the velocity

(C) A particle is undergoing ground to ground projectile motion with angle of projection

,o

2

(r) The acceleration may be in the direction of velocity

(D) A particle is moving in space (s) The acceleration may be perpendicular

to its velocity

(t) none of these

2. The potential energy for a conservative system is given by 2U ax bx (a, b are positive constants)

then

Column - I Column - II

(A) The net force acting on the system (p) stable

(B) The equilibrium position (q) 2b / 4a

(C) The potential energy at the equilibrium position (r) b – 2ax

(D) The equilibrium (s) b/2a

(t) unstable





PART – II :CHEMISTRY

SECTION – A : (Single Correct Answer 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.

Compare bond length from the above given molecules: (A) X > Y (B) X < Y (C) X = Y (D) None of these

2. Elemental analysis of a sample of ionic compound gave the following results : 2.82 g sodium , 4.35 g chlorine , and 7.83 g oxygen. What is the name of empirical compound (A) Soduim hypochlorous (B) sodium chlorous (C) sodium chlorate (D) sodium pelchlorate

3. Calculate n-factor of HCl:

4 2 2 2KMnO HCl MnCl Cl KCl H O

(A) 3/7 (B) 7/3 (C) 5/8 (D) 8/5

4. Two gases A and B have critical temperature 35°C and 25°C respectively . Which of the following statements are correct? (i) B has higher boiling point than A (ii) A is more easily liquefiable than B (iii)at 30°C and under very high pressure A exists as a liquid and B as a gas (iv) at 35°C and under very high pressure both A a& B exists as gases (A) i, ii (B) ii,iii (C) ii, iv (D) iii, iv

5. The radial wave equation for hydrogen atom is

3/2

2 r/2

0

1 1x 1 x 8x 12 e

a16 4

where

0

2rx

a ; 0a radius of first Bohr orbit . The minimum and maximum position of radial nodes

from nucleus are:

(A) 0 0a ,3a (B)

00

a,3a

2 (C)

00

a,a

2 (D)

00

a,2a

2

6. In the following reaction (x) and (y) are respectively

2 2

2 2

BaC N x

CaC N y

(A) 2 2

Ba CN and Ca CN (B) 22Ba CN and CaCN

(C) 2 2BaCN and Ca CN (D) 2 2BaCN and CaCN




Reasoning Type This section contains 3 reasoning type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Choose the correct option from the following (A) Statement – 1 is true Statement – 2 is true ; statement-2 is a correct explanation for statement -1 (B) Statement – 1 is true Statement – 2 is true ; statement-2 is a NOT correct explanation for statement -1 (C)Statement – 1 is true, statement – 2 is false (D) Statement – 1 is false, statement 2 is true

7. Statement 1 : In the roasting of FeS2 , ore is converted into ferric oxide and SO2 gas. The equivalent mass of FeS2 is equal to molecular mass/11 Statement – 2 The n-factor for reducing agent is Total net change in oxidation number per formula unit .

8. Statement – 1 All low pressure Vander Waal’s equation is reduced to

2

aP V RT

V

Statement – 2 the compressibility factor corresponding to low pressure is given by RTV

1a

9. Statement -1 :First ionization energy of “P” is greater than “S” Statement - 2 Effective nuclear charge of “S” is greater than “P”.

Paragraph type (One Option Correct)

This section contains TWO paragraphs. Based on each paragraph, there will be THREE questions. Each question has FOUR options (A), (B), (C) and (D). ONE of these four option(s) is correct.

Paragraph for question Nos. 10 to 12

Heisenberg gave the principle there is uncertainty in simultaneous measurement of position and momentum of small particles. If any one of these two quantities is measured with higher accuracy, the

measurement of the other becomes less accurate. The product of the uncertainty in position ( x)

and uncertainty in momentum p is always constant and is equal to or greater than h/4 , where h

is Planck’s constant.

10. The uncertainty in position of an electron ( m=9.1×10––28g) moving with a velocity 3×104 cm s–1 accurate upto 0.001% will be (A) 3.84cm (B) 1.93 cm (C) 7.68cm (D) 5.76cm

11. Uncertainty in position is twice the uncertainty in momentum, uncertainty in velocity is

(A)

h (B)

h

m2

1 (C)

1h

2m (D)

1 h

2 2m

12. If the uncertainty in velocity and position is same, then the uncertainty in momentum will be

(A)4

hm (B)

4

hm (C)

m

h

4 (D)

4

1 h

m




Paragraph for question Nos. 13 to 15

Bent’s rule is consistent with VSEPR theory, According to bent’s rule, the more electronegative substituent attach itself of hybrid orbital that contains more p character or less s – characters

13. The molecule/ion which bond angle is less than 109°:

(A) 4FN

(B) 4CCl (C)

4ClO (D) None of these

14. Select pair of compounds in which both have the same molecular geometry:

(A) BF3BrF3 (B) 2 2ICl ,BeCl

(C) 3 3BCl ,PCl (D) 3 3PCl ,BrF

15. Find the species which is not perfectly planar:

(A) 3HC

(B) 2HFC

(C) 2HC (D) 2ON

SECTION – B: Match the Column Type (One or More than one option Correct) This section contains 2 questions. Each question contains statements given in two columns which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example: If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s and t; then the correct darkening of bubbles will look like the following.

p q r s

p q r s

p q r s

p q r s

p q r s

D

C

B

A t

t

t

t

t

1. Match the following Column

Column – I Column – II

(A) 9.8 % H2SO4 by weight (density = 1.8 g ml–1)

(p) 3.6 N

(B) 9.8 % H3PO4 by weight (density = 1.2 g ml–1)

(q) 1.2 M

(C) 1.8 NA molecules of HCl is 500 mL

(r)1.8 equivalent

(D) 250 mL of 4N NaOH +250 mL of 1.6 M Ca(OH)2

(s) 1.10 m

.




2. For a fixed amount of the gas match the two column :

Column – I Column – II

(A)

(p) 1 2 3T T T

(B)

(q) 1 2 3P P P

(C)

(r) 1 2 3V V V

(D)

(s) 1 2 3d d d





PPPAAARRRTTT ––– IIIIIIIII ::: MMMAAATTTHHHEEEMMMAAATTTIIICCCSSS

SECTION – A : (Single Correct Answer 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. Value of x satisfying inequation 2xlog x 1 0

(A) (0, 2) (B) (1, 2)

(C) (0, 7) (D) (1, 1)

2. Number of integral coordinate inside the triangle made by the line x + y = 15 and coordinate axes is (A) 91 (B) 90 (C) 78 (D) none of these

3. If a b c

2c bbc

, where a, b, c > 0, then family of lines ax by c 0 passes through the

fixed point, given by

(A) (1, 2) (B) (1, 2)

(C) (1, 2) (D) (1, 1)

4. The maximum value of cos2x sin2x27 81 is

(A) 23 (B) 53

(C) 73 (D) 3

5. The value of 5 7

sin sin sin18 18 18

is

(A) 1

8 (B)

1

2

(C) 1

4 (D)

1

16

6. If the line l x + my = 1 be a tangent to the circle 2 2

2

1x y

a , then ( l, m) lies on

(A) 2 2

2

1x y

a (B) 2 2 2x y a

(C) 2 2 2x y a (D) 2 2

2

1x y

a





(Assertion – Reason Type)

This section contains 3 questions numbered 7 to 9. Each question contains Statement – 1 (Assertion) and Statement – 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

(A) Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True

7. Statement -1 : The equation of the chord of the circle x2 + y2 6x + 10y 9 = 0, which is bisected at

the point (2, 4) must be x + y 2 = 0. Statement -2 : In notations the equation of the chord of the circle S = 0 bisected at x1, y1 must be

T = S1.

8. Statement -1 : If tan sin m and tan sin n then 2 2m n 4 mn

Statement -2 : 2 2cosec 1 cot , n ,n I

9. Statement -1 : If the Point (2a – 5, a2) is on the same side of the line x + y – 3 = 0 as that of the

origin, then the set of values of a (2, 4) Statement -2 : The points (x1, y1) and (x2, y2) lies on the same or opposite side of the line ax+by+c=0,

as ax1 + by1 + c and ax2 + by2 + c have the same or opposite signs.





(Paragraph Type)

This section contains 2 paragraphs. Based upon the paragraphs 2 multiple choice questions and 3 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question No. 10 to 12

A triangle ABC is given, where vertex A is (1, 1) and the orthocentre is (2, 4). Also sides AB and BC are member of the family of lines ax + by + c = 0, where a, b, c are in A.P. 10. The vertex B is

(A) (2, 1) (B) (1, 2)

(C) (1, 2) (D) (1, 2) 11. The vertex C is

(A) (4, 16) (B) (17, 4)

(C) (4, 17) (D) (17, 4) 12. Triangle ABC is a, (A) obtuse angled triangle (B) right angled triangle (C) acute angled triangle (D) equilateral triangle





Paragraph for Question No. 13 to 15

If a line 3x 4y 20 0 intersect the circle 2 2x y 25 at the point A, B then

13. Distance between A, B is (A) 4 (B) 6

(C) 8 (D) 4 2

14. The radius of the circle which is orthogonal to the given circle and which pass through points A and B

is (A) 3 (B) 4

(C) 15

4 (D)

15

2

15. If tangents drawn at A, B intersect at P then area of triangle PAB is

(A) 25

2 (B) 9

(C) 27

2 (D)

27

4





SECTION – B: Match Column Type (One or More than one option Correct)

This section contains two questions, each having two matching Columns. Choices for the correct combination of elements from Column-I and Column-II are given as option (A), (B), (C) and (D) out of which one or more than one are correct.

1. Match the values of ‘a’ from Column II.

Column I Column II

(A)

4(cos310° + sin320°) = a(cos10° + sin20°) (p) 2

(B)

1 1cot(a )

tan3 tan cot3 cot

(q) 1

(C) 1 cos2 sin2tana

1 cos2 sin2

(r) 4

(D) cot 1 1 sina

cot 1 cosa

(s) 3

2. Match the column

Column I Column II

(A)

If a, b, c are in H.P., then the straight line

x y 10

a b c always passes through a point

(p) (1, 1)

(B)

All chords of the curve 3x2 y2 2x + 4y = 0 which subtend a right angle at the origin pass through the point

(q) (1, 1)

(C) The circumcentre of the triangle formed by the lines xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is

(r) (0, 0)

(D) The circle passing through three distinct points (1, t) , (t, 1) and (t, t) passes through the point

(s) (1, 2)





FIITJEE INTERNAL TEST

BATCHES : Two Year CTY (2018-20) A LOT

PHASE -I: PAPER-2

JEE ADVANCED LEVEL

ANSWKER KEY-SET-A

PHYSICS (PART-I)

SECTION – A

1. D 2. C 3. C 4. D

5. C 6. D 7 B 8 C

9 B 10. C 11. D 12. C

13. B 14 C 15. B

SECTION – B

1. A (s, q); B (p, r); C (q, s); D (p, q, r, s)

2. A (r); B s, C q, D p

Part – II

(CHEMISTRY)

1. B

2. D

3. C

4. B

5. B

6. B

7. A

8. C

9. B

10. B

11. D

12. A

13. D

14. B

15. B




1.

A→PS B→P,Q,S C→P,R D→R

2.

A→S B→Q,S C→R D→P

MATHEMATICS (PART-III)

SECTION – A

1. B 2. A 3. D 4. B

5. A 6. C 7. D 8. A

9. D 10. B 11. D 12. A

13. A 14. C 15. D

SECTION – B

1. (A) (s) ; (B) (r) ; (C) (q) ; (D) (p)

2. (A) (s) ; (B) (s) ; (C) (p) ; (D) (q)




HINTS AND SOLUTIONS

PHYSICS

SECTION – A

1. 2

2 R Rx R 5

4 2

P,Q

5 RV x

2

2. max

TV

Where a

, magnitudes of acceleration and deceleration. T is the total time taken.

3. mgsin gsin

a2m 2

mgsin

T ma2

4. P F.V

ˆ ˆ ˆmgj . ucos i usin gt j

or P mg gt usin

5. x yV V

u cos = u sin - gt

t = u sin cos

g

6. PS 2PO

PR PU PS

and PT PQ PS

10. FBD of one bead

x 2R 3R is the compression in the spring.

N = kx cos 30 + mg cos 60

N

30

kx mg

30 30

12. Fspring = kx = mg

3

14. For equilibrium

6dU 2a0 x

dx b

w U

= –Uequilibrium = 2b

4a




MATHEMATICS

SECTION A

1. 2xlog x 1 0

Case I : if x2 > 1

x (, 1) (1, ) …(i)

Now 2xlog x 1 0 ||x| 1| < 1

1 < |x| 1 < 1 0 < |x| < 2

x (2, 2) {0} …(ii) equ. (i) and (ii)

x (2, 1) (1, 2)

Case II : 0 < x2 < 1 x (1, 1) …(iii)

now 2xlog x 1 0 ||x| 1| > 1

x (, 2) (2, ) …(iv)

equ. (iii) and (iv) x

final solution case I case II

x (2, 1) (1, 2) 2. Total integral coordinate inside the triangle

= 13 + 12 + 11 + ….. + 1 = 13 14

912

4. cos2x sin2x 3cos2x 4sin2x27 81 3

Maximum value of 2 23cos2x 4sin2x 3 4 5

maximum value of cos2x sin2x 527 81 3

5. We have 5 7

sin sin sin18 18 18

5 7

cos cos cos2 18 2 18 2 18

4 2

cos cos cos9 9 9

1 2 2 4

sin cos cos9 9 9

2sin9

2

1 4 4sin cos

9 92 sin

9

3

1sin

9 82 sin

9

6. Hint : Use condition of tangency.

Solution : Since | x + y = 1 is a tangent to 2 2

2

1x y

a

2 2

l(0) m(0) 1 1

al m

2 2 2

2 2 2

1 1l m a

l m a




(l, m) lies on 2 2 2x y a

8. 2tan m n and 2sin m n

So, 2 2

cot , cosecm n m n

Since, 2 2cosec cot 1

2 2

1 14 1

m n m n

2 2m n 4 mn

Both I and II are correct and II is correct explanation of I

9. Since (2a – 5, a2) and O(0, 0) lie on the same side x + y – 3 = 0

-3 < 0 and 2a – 5 + a2 – 3 < 0 a2 + 2a – 8 < 0 -4 < a < 2

a (-4, 2) 10. Given that AB and BC are the members of

the family of line ax + by + c = 0, where a, b, c are in A.P.

b = a c

2

a 2b + c = 0

Hence ax + by + c = 0 always passes

through (1, 2).

A(1, 1)

B C

(2, 4)

11. Now consider coordinate c is (a, b) then we have

b 1 4 2

1a 1 2 1

and

b 1 4 21

a 1 2 1

On solving these two equation we get c(17, 4).

12. Slope of AC(m1) = 1 4 3 1

1 17 18 6

Slope of BC (m2) = 2 4 6 1

1 17 18 3

.

tan A = 2 1

1 2

m m

1 m m

= 1/ 3 1/ 6

01 1

16 3

ACB is obtuse.

13. Perpendicular distance from centre to the chord 20

45

OA2 = OM2 + AM2 => AM = 3 => AB = 6




14. let equation of required circle is 2 2x y 25 3x 4y 20 0 it is orthorgonal to given circle x2 +

y2 – 25 = 0 1 1 1 52gg 2ff c c 0 25 25 20

2

equation of circle

2 2 15x y x 10y 25 0

2 . Radius =

2

215 155 25

4 4

15. For point P (x1, y1), 3x + 4y – 20 = 0 is a chord of contact xx1+yy1 = 25, 3x + 4y = 20

1 11 1

x y 25 15x ,y ,5

3 4 20 4

Length of perpendicular drawn from P on AB.

153. 4.5 20

94

5 4

Area of 1 1 9

PAB AB PN 62 2 4

SECTION – B

1. (A) L.H.S. = (cos30° + 3cos10°) + (3sin20° sin60°)

= 3 3

3cos10°+3sin20°-2 2

= 3(cos10° + sin20°)

(B) L.H.S. = cos cos3 sin3 sin cos4

cot 4sin4 sin4 sin4

(C) 2

2

1 cos2 sin2 2sin 2sin costan

1 cos2 sin2 2cos 2sin cos

(D) cot 1 1 sin2

cot 1 cos2

2. (A) 2 1 1

b a c if a, b, c in H.P. point (1, 2)

(B) Let chord be lx + my = 1, then 3x2 y2 2x(lx + my) + 4y(lx + my) = 0

3 2l 1 + 4m = 0 l 2m = 1 point (1, 2) (C) x + 2 = 0, y + 2 = 0, x + y + 2 = 0

A(2, 0), B(0, 2), C(2, 2)

In ABC is right angled at C

circumcentre is (1, 1). (D) From figure required point be (1, 1).

(1, t) (t, t)

(t, 1) (1, 1)

O x

y

FIITJEE INTERNAL TEST

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