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ADVANCED PATTERN CUMULATIVE TEST-1 (ACT-1)

TARGET : JEE (MAIN + ADVANCED) 2021

COURSE : VIJETA (JP) | BATCH : JPB-STAR, JPAB

P10-20

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

INSTRUCTIONS TO CANDIDATES

DATE

21-06-2020

PAPER

1

CODE

1

Time : 3 Hours Maximum Marks : 180

Question Paper-1 has three (03) parts: Mathematics, Physics and Chemistry.

Each part has a total eighteen (18) questions divided into three (03) sections (Section-1, Section-2 and

Section-3).

Total number of questions in Question Paper-1 are Fifty Four(54) and Maximum Marks are One Hundred Eighty (180)

TYPE OF QUESTIONS AND MARKING SCHEMES

SECTION-1 (Maximum Marks : 24) This section contains SIX (06) questions.

Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is (are) correct answer(s).

For each question, choose the option(s) corresponding to (all) the correct answer(s).

Answer to each question will be evaluated according to the following marking scheme :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen and both of which

are correct.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct option.

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : –2 In all other cases.

SECTION – 2 : (Maximum Marks : 24) This section contains EIGHT (08) questions

The answer to each question is a DOUBLE DIGIT INTEGER ranging from 00 to 99, both inclusive

Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen

Zero Marks : 0 In all other cases

SECTION-3 : (Maximum Marks : 12) This section contains TWO (02) paragraphs

Based on each paragraph, there will be TWO questions.

Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct

Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered)

Negative Marks : –1 In all other cases

NAME OF THE CANDIDATE : ………………………………………………… ROLL NO. :………………..………………………..

I have read all the instructions I have verified the identity, name and roll number and shall abide by them of the candidate.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Signature of the Candidate Signature of the Invigilator

mailto:[email protected]

http://www.resonance.ac.in/reso/results/jee-main-2014.aspx



INSTRUCTIONS FOR OPTICAL RESPONSE SHEET (ORS)

Darken the appropriate bubbles on the original by applying sufficient pressure.

The original is machine-gradable and will be collected by the invigilator at the end of the examination.

Do not tamper with or mutilate the ORS.

Write your name, roll number and the name of the examination centre and sign with pen in the space provided for this

purpose on the original. Do not write any of these details anywhere else. Darken the appropriate bubble under each

digit of your roll number.

DARKENING THE BUBBLES ON THE ORS :

Use a BLACK BALL POINT to darken the bubbles in the upper sheet.

Darken the bubble COMPLETELY.

Darken the bubble ONLY if you are sure of the answer.

The correct way of darkening a bubble is as shown here :

There is NO way to erase or "un-darkened bubble.

The marking scheme given at the beginning of each section gives details of how darkened and not darkened bubbles

are evaluated.

Zero marks ‘0’ If none of the options is chosen (i.e. the question is unanswered).

FOR INTEGER TYPE QUESTIONS OMR LOOKS LIKE :

For example, if answer ‘SINGLE DIGIT’ integer type below :

0 1 2 3 4 5 6 7 8 9 For example, if answer ‘SINGLE DIGIT’ integer with positive / negative type below :

0– 1

23456789

For example, if answer ‘DOUBLEDIGIT’ integer type below :

0 01 12 23 34 45 56 67 78 89 9

FOR DECIMAL TYPE QUESTIONS OMR LOOKS LIKE :

1 2 . 3 4

0 0 0 01 1 1 12 2 2 23 3 3 34 4 4 45 5 5 56 6 6 67 7 7 78 8 8 89 9 9 9

COLUMN

If answer is 3.7, then fill 3 in either 1st or 2nd column and 7 in 3rd or 4th column.

If answer is 3.07 then fill 3 in 1st or 2nd column ‘0’ in 3rd column and 7 in 4th column.

If answer is, 23 then fill 2 & 3 in 1st and 2nd column respectively, while you can either leave column 3 & 4 or fill ‘0’ in either of them.



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PART : I MATHEMATICS

SECTION – 1 : (Maximum Marks : 24)

This section contains SIX (06) questions.

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is(are) correct For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –2 In all other cases.

For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct answers, then : Choosing ONLY (A),(B) and (D) will get +4 marks Choosing ONLY (A) and (B) will get +2 marks Choosing ONLY (A) and (D) will get +2 marks Choosing ONLY (B) and (D) will get +2 marks Choosing ONLY (A) will get +1 marks Choosing ONLY (B) will get +1 marks Choosing ONLY (D) will get +1 marks Choosing no option (i.e. the question is unanswered) will get 0 marks, and Choosing any other combination of options will get –2 mark

1. If f(x) = sin(tan–1(cos(cot–1x))), then which of the following is/are true.

(A) 2

1–)x(flim

x

(B)

2

1)x(flim

x

(C) Range of f is

2

1,

2

1– (D) f(x) =

2x21

x



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2. Consider the function f(x) = (log3x)4 + 12(log3x)2.

x

27log3 where x [1, 729]. If M be the

maximum and m be the minimum value of f(x), then

(A) M = 81 (B) m = 0 (C) M = 27 (D) m = 3

3. Let f : R [0, 1] be a function defined by f(x) = min(e–|x|, |x2 – 1|). Identify which of the following

statement(s) is/are correct ?

(A) f is surjective but not injective (B) f is neither injective nor surjective

(C) f is discontinuous at exactly three points (D) f is continuous every where

4. Given that 2xtan

)e–1e(nlim

2

xxcosba

0x

c

, where a,b,c R and c > 0, if

(A) a = 6, b = –6, c = 3 (B) a = 6, b = –6, c = 2 (C) a = 4, b = –4, c = 3 (D) a = 4, b = –4, c = 4

5. Let cosf x x and min ,0 , 0 /2

, 32 2

f t t x for x

g xx for x

then

(A) g x is continuous and derivable in 0, 3

(B) g x is continuous but not derivable at 2

x

(C) g x is neither continuous nor derivable at 2

x

(D) Maximum value of g x in 0, 3 is 1



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6. In the expansion of

485

2 x3

, at x = 3

(A) numerically greatest term are 35T and 36T

(B) algebraically greatest term is 35T

(C) algebraically least term is 36T

(D) number of term in expansion is 49


This section contains EIGHT (08) questions


Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen.


7. Let m be the minimum value of the function f(x) = 2x3 – 3x2 – 12x + 6 on the set

A = {x : nx > n(x2 – 3x + 3)}, then find |m|.

8. A couple has one or two or three children with probability 1

4,

1

2 and

1

4 respectively. Probability of

a couple having exactly four grandchildren in such a type of society is p

q then find the value of

(q–4p)

(Where p & q are co-prime natural numbers)



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9. If there are 5 circles and 3 straight lines, in a plane, then their maximum number of point of

intersection is equal to

10.

xx

1

x

1

0xecxcosxcosxsinlim is equal to

11. If f(x) is a monic polynomial of degree 2 such that |x2 – x + 2 + f(x)| – |f(x)| = x2 – x + 2 x R and

f(1) = 0, then find f(3).

12. Let

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

then the value of 2012 a if 'f' is continuous at x = 0,

are

13. Let f : R

2

,0 defined by f(x) = cosec–1(ax2 – 2px + 4) where a,p N is surjective then find

smallest value of 2(a + p).

14. If f : R R, f(x) = x3 – px2 + (q – 10) x, p, q R is bijective odd function and then find minimum

value of (p + q).



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This section contains TWO (02) paragraphs.


Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct Marking scheme :




Paragraph for Question Nos. 15 to 16

If f(x) =

3 3

5

x

3

µ (x – sin x)x 0tan x

x 0

asinx bcosx 1– ce x 0

sin x

, is a continuous function at x = 0, then answer the

following questions.

15. The interval in which g ' (x) 0, where function2–3µ –2(ax bx)g(x) x e , is

(A) 1, 1,2

(B)

1,12

(C) 1, 0,2

(D)

1,22

16. 2 2 2 2n

1 2 3 4cnlim ......

n 1 n 2 n 3 n 4cn

=

(A) – 6 – 3µ (B) – – µ (C) + 2µ (D) – 3 – 2µ



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A multiple choice question has n options, of which only one is correct. If a student does home

work, then it is sure to identify the correct answer; otherwise, answer is choosen at random. Let E

be the event that student does home work with P(E) = p and F be the event that student answers

question correctly.

17. If n = 5, p = 0.75 the value of P(E/F) equals

(A) 8

16 (B)

10

16 (C)

12

16 (D)

15

16

18. Largest set of values of p for which relation P(E/F) P(E) holds is

(A) [0,1] (B) R (C) 1

0,2

(D) 1

, 12




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Website : www.resonance.ac.in | E-mail : [email protected] P1JPBACT1210620-1



PART : II PHYSICS




option(s) is(are) correct

For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option.


Negative Marks : –2 In all other cases.

For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct

answers, then :

Choosing ONLY (A),(B) and (D) will get +4 marks

Choosing ONLY (A) and (B) will get +2 marks

Choosing ONLY (A) and (D) will get +2 marks

Choosing ONLY (B) and (D) will get +2 marks

Choosing ONLY (A) will get +1 marks

Choosing ONLY (B) will get +1 marks

Choosing ONLY (D) will get +1 marks

Choosing no option (i.e. the question is unanswered) will get 0 marks, and

Choosing any other combination of options will get –2 mark

19. A gaseous mixture enclosed in a vessel consist of 1 g mole of gas A with (1 = 5/3) and another

gas B with (2 = 7/5) at a temperature T. The gas A and B do not react with each other and

assume to be ideal. If = 19/13 for the gaseous mixture, then?

(A) No. of gram mole of the gas B is 2 gm mole

(B) The value of (CV) mix is equal to 13

6R

(C) No. of gram moles of the gas B is 1 gm mole

(D) The value of (CV)mix is equal to 19

6R




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20. In the circuit shown in figure, E1 and E2 are two ideal sources of unknown emfs. Some currents are

shown. Potential difference appearing across 6 resistance is VA – VB = 10V.

E1E2

4.00

4.00

3.00 3.00 6.00

3.00A

R2.00A

A

BDC

(A) The current in the 4.00 resistor in between C and D is 5A.

(B) The unknown emf E1 is 36 V.

(C) The unknown emf E2 is 54 V.

(D) The resistance R is equal to 9 .

21. Charge – Q and 2Q are distributed uniformly on surface of two concentric spherical shells of radii

‘R’ and ‘2R’ respectively as shown in the figure. Select correct alternative(s)

–Q +2Q

R2R

(A) the total electrostatic energy stored in the system is 2

0

Q

8 R

(B) electrostatic energy in the space between two shells is

2

0

Q

16 R

(C) electrostatic energy stored outside the system is

2

0

Q

2 R

(D) electrostatic energy in space between two shells is zero.




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22. In displacement method, the distance between object and screen is 96 cm. The ratio of length of

two images formed by a convex lens placed between them is 4.84.

(A) Ratio of the length of object to the length of shorter image is 11/5.

(B) Distance between the two positions of the lens is 36 cm.

(C) Focal length of the lens is 22.5 cm.

(D) Distance of the lens from the shorter image is 30 cm.

23. Two satellites of same mass of a planet in circular orbits have periods of revolution 32 days and

256 days. If the radius of the orbit of the first is R, then the

(A) radius of the orbit of the second is 8R

(B) radius of the orbit of the second is 4R

(C) total mechanical energy of the second is greater than that of the first.

(D) kinetic energy of the second is less than that of the first.

24. In the given circuit potential of the point A is 9V higher than potential of the point B. Choose

correct alternative(s)

(A) Value of resistance R is 1

(B) Value of resistance R is 7

(C) magnitude of potential difference between B and D is 30 V.

(D) magnitude of potential difference between B and C is 15 V.




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Marking scheme :



25. A container (water equivalent of container is negligible) contains 50gm of water at 0ºC. If H amount

of heat is extracted per second from water and keeping the spray closed, then whole water freezes

in 10 sec. If spray is also working and sprays the water with 2 gm/sec at 0ºC, then time delay is

x

3seconds in complete freezing compared to previous case keeping the rate of extraction of heat

from water same. Then calculate 'x'. (Latent heat capacity of ice = 80 cal/gm)

26. A cone made of insulating material has a total charge Q = 1 C spread uniformly over its sloping

surface. Find the work (in Joule) to take a test charge q = 1 mC from infinity to apex of the cone.

Slant length of the cone is = 1m.




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27. Inside water 4

3

there is an air bubble of radius 4 cm as shown in figure. An observer O is

looking into the diametrical axis AB of bubble. Find the distance in mm of a point object from point

A on the axis in water which appears to be at point A as seen by observer.

O

AirBubble

A B

28. A planet is made of two materials of density 1 and 2 as shown in figure.

The acceleration due to gravity at surface of planet is same as a depth ‘R’. The ratio of 1

2

is x

6.

Find the value of x.

29. A monoatomic ideal gas follows the process :

TV3/2 = constant

The molar specific heat for this process is 5R

y

. where R is gas constant. Find the value of '2y'.




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30. A surge suppressor is a device for preventing sudden excessive flows of current in circuit. It is

made of material whose conducting properties are such that the current flowing through it is

directly proportional to third power of the potential difference across it. When the applied potential

difference is 200V, the suppressor dissipates energy at a rate 10 Watt. When the potential

difference rises to 400V, the suppressor dissipates energy at a rate 8X Watt. Calculate X.

31. In the figure, an object is placed at distance 25 cm from the surface of a convex mirror, and a

plane mirror is set so that the image formed by the two mirrors lie adjacent to each other in the

same plane. The plane mirror is placed at 20 cm from the object. What is the radius of curvature of

the convex mirror ?

32. A rectangular sheet of copper is 2.00 mm thick and has surface dimensions 8.0 cm x 24 cm. If the

long edges are joined to form a tube 24 cm in length, find the resistance of the tube in micro-ohms

between the ends of the tube. Resistivity of copper = 1.6 x 10–8 -m.




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Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct

Marking scheme :





A pair of stars rotates about a common center of mass. One of the stars has a mass M and the

other has mass m such that M = 2m. The distance between the centres of the stars is d (d being

large compared to the size of either star).

33. The period of rotation of the stars about their common centre of mass (in terms of d, m, G.) is

(A) 2

34d

Gm

(B)

238

dGm

(C)

232

d3Gm

(D)

234

d3Gm

34. The ratio of kinetic energies of the two stars (Km/KM.) is

(A) 1 (B) 2 (C) 4 (D) 9




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A quantity of an ideal monoatomic gas consists of n moles initially at temperature T1. The pressure

and volume are then slowly doubled in such a manner so as to trace out a straight line on a P-V

diagram.

35. For this process, the ratio 1

W

nRT is equal to (where W is work done by the gas) :

(A) 1.5 (B) 3 (C) 4.5 (D) 6

36. If C is defined as the average molar specific heat for the process then C

Rhas value

(A) 1.5

(B) 2

(C) 3

(D) 6




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PART : III CHEMISTRY

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,

P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,

Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]




option(s) is(are) correct For each question, marks will be awarded in one of the following categories :

Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen

and both of which are correct. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –2 In all other cases.

For example, in a question, if (A),(B) and (D) are the ONLY three options corresponding to correct answers, then : Choosing ONLY (A),(B) and (D) will get +4 marks Choosing ONLY (A) and (B) will get +2 marks Choosing ONLY (A) and (D) will get +2 marks Choosing ONLY (B) and (D) will get +2 marks Choosing ONLY (A) will get +1 marks Choosing ONLY (B) will get +1 marks Choosing ONLY (D) will get +1 marks Choosing no option (i.e. the question is unanswered) will get 0 marks, and Choosing any other combination of options will get –2 mark




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37. Increasing amount of solid Hg2 is added to 1 liter of an aqueous solution containing 0.1 mole K.

Which of the following graph(s) is/are represent the correct variation with the amount of Hg2

added ?

(TB : Boiling point, TF : Freezing point) (Hg2 is insoluble solid which dissolves in excess of –)

(A)

TF

0.05 0.10

Mole of HgI2

(B)

TF

0.05 0.10

Mole of HgI2

(C)

TB

0.05 0.10

Mole of HgI2

(D)

TB

0.05 0.10

Mole of HgI2




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38. Which of the following is/are ore of iron ?

(A) Haematite (B) Magnetite (C) Siderite (D) Calamine

39. Which of the following statement is/are incorrect ?

(A) Free expansion of an ideal gas in an insulated container is a reversible process.

(B) When (Gsystem)T,P < 0, the reaction must be exothermic.

(C) All adiabatic processes are isoentropic.

(D) 1 mol N2 at STP has more entropy than 1 mol He gas at 273 K and 1 atm.

40. An amount of 20g potassium sulphate was dissolved in 150 cm3 of water. The solution was then

electrolysed. After electrolysis the content of potassium sulphate in the solution was 15% by mass.

Determine volume of the gases obtained at NTP ? ( OH2d = 1 g cm–3)

(A) 45.6 L H2 gas (B) 22.8L O2 gas (C) 42 L H2 gas (D) None of these

41.

Cl

P3AlCl

QNaH

O O R

C115100

CO)NH( 324

O

SHCl

Among P, Q, R and S, the aromatic compound(s) is/are : (A) P (B) Q (C) R (D) S




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42. Identify correct statements for given reaction sequence.

Major product

(A) R =

(B) R =

(C) In the given three reaction steps, two are electrophilic substitution steps.

(D) Product (P) gives a racemic mixture with (i) MeMgBr (ii) aq.NH4Cl




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Marking scheme : Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 In all other cases

43. The cell notation of the standard galvanic (voltaic) cell containing an unknown metal electrode X is

shown below :

X(s) | X3+ (1 mol dm–3) | | Pb2+ (1 mol dm–3) | Pb (s)

How many of the below statement(s) is/are correct related to the above situation ?

(a) The component of the cell represented by the double vertical lines (II) in the above cell

notation is known as salt bridge.

(b) Pb(s) is the oxidizing agent in the above cell.

(c) The initial reading on a voltmeter connected across the electrodes of the above cell is 1.53 V.

[ o

Pb|Pb2E = – 0.13 V] then the standard reduction potential of the unknown metal X is 1.66 V.

(d) The balanced equation for the net reaction taking place in this cell is :

2X(s) + 3Pb2+ 2X3+ + 3Pb(s)

(e) The initial voltmeter reading will decrease if the concentration of electrolyte in the

X(s) | X3+ (aq) half cell is increased.




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44. 50 mL of 0.1 M CuSO4 solution is electrolyzed with a current of 0.965 A for a period of 200 sec.

The reactions at electrodes are :

Cathode: Cu2+ + 2e– Cu(s)

Anode: 2H2O O2 + 4H+ + 4e–

Assuming no change in volume during electrolysis, calculate the molar concentration of –24SO at

the end of electroysis.

Report your answer after multiplying by 100.

45. How many of the following involves carbon reduction during extraction of metal from ore?

Bauxite Al

Copper pyrites Cu

Haematite Fe

Galena Pb

Carnallite Mg

Sylvine K

Cassiterite Sn

Zinc blende Zn

46. Depression of freezing point of 0.01 molal aqueous CH3COOH is 0.0198ºC. 1 molal urea solution

freezes at –1.8ºC. Assuming molarity is equal to molality, pH of CH3COOH solution is :




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47. In modern periodic table let, a = Group number of a hypothetical element having atomic number 'z'

= 120.

b = Period number of the element 'Pd'.

c = Atomic number of last element of lanthanide series.

Find (a + b + c)

48. Total number of structural isomers of C8H18 is (x) and out of them (y) no. of isomer gives 1 monochloro product. Report (x+y) as your answer.

49. Total number of products formed in the following reaction is :

50. How many isomeric aromatic C6H

2Br

4 are possible which on bromination with Br

2/FeBr

3 gives only

1 type of product.




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Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option is correct Marking scheme :

Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : –1 In all other cases


Copper is the most noble of the first row transition metals and occurs in small deposits in several

countries. Ores of copper include chalcanthite (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite

(Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). However 80% of the world copper

production comes from the ore chalcopyrites (CuFeS2). The extraction of copper from

chalcopyrites involves partial roasting, removal of iron and self reduction.

51. Partial roasting of chalcopyrites with silica added to it, does not produces :

(A) FeSiO3 (B) FeS (C) Cu2S (D) CuO

52. In self reduction the reducing species is ……… . Select the correct option for blank.

(A) Cu+ (B) O–2 (C) S–2 (D) SiO2




CH

EM

IST

RY

CHEMISTRY




Paragraph for Question Nos. 53 to 54 Benzene is an unsaturated hydrocarbon but does not give addition reactions under normal

conditions. It undergoes electrophilic substitution reactions mainly. In substituted derivatives of benzene, orientation of further substitution is decided by resonance effect, steric factors, etc. On the basis of this, answer the following :

53. In which of the following the given product is the major product

(A)

Br

NO2 sulphonation

Br

NO2

SO3H

(B)

CN

NO2

2Br

Fe

CN

NO2

Br

(C)

2 3Cl /FeCl

Cl

(D)

3 2

3

CH CO O

AlCl

COCH3 54. Which of the following reagents used for alkylation of benzene by electrophilic substitution

mechanism will give same product ?

(p)

CH3–CH–CH2–Cl

CH3

in presence of anhy.AlCl3 (q) (CH3)3 COH in presence of H2SO4

(r) (CH3)2 CH-CH2-OH in presence of H2SO4 (s)

CH3–C–CH2–OH

CH3

CH3

in presence of H2SO4.

(A) p, q, r (B) q, r, s (C) p, q, r, s (D) p, q, s



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ADVANCED PATTERN

CUMULATIVE TEST-1(ACT-1) TARGET : JEE (MAIN+ADVANCED) 2021

DATE : 21-06-2020 SET/CODE-1 COURSE : VIJETA (JP) | BATCH : (JPB*)

HINTS & SOLUTIONS

PAPER-1


1. If f(x) = sin(tan–1(cos(cot–1x))), ……………

Sol. 2

1)x(flim

x

;

2

1–)x(flim

x

f(x) = xcotcostansin 1–1–

Range of f(x) is

2

1,

2

1–

f(x) = sin(tan–1(cos(cot–1x)))

=

2

1–

x1

xtansin

Let

2

1–

x1

xtan

tan = 2x1

x

cot–1x = (0, )

x = cot

(i) Let

2

,0

cos = xsin

1

= 2

2

2 x1

x

x

x1

1

(ii) if

,2

x < 0

cos = xsin

1=

2t1

1

= –

2x

11

1

= 2x1

|x|

=

2x1

x

x < 0

f(x) = sin() = ecxcos

1

=

2cot1

1– =

2

2

x

x11

1–

=

1x2

|x|

2

f(x) = 2x21

x

similarly for x > 0 (x > 0)

f(x) = 2x21

x

2. Consider the function ……………

Sol. Let log3x = t x [1, 729]

t [0, 6]

f(x) = t4 + 12t2(3 – t)

= (t(t – 6)2 = = ((t – 3)2 – 9)2 t [0, 6]

M = 81, m = 0

3. Let f : R [0, 1] be…………… Sol.

Look at the graph and we can check options.

4. Given that……………

Sol. 2

x.x

xtan

)e–e1(nlim

2

2

2

xxcosba

0x

c

2x

)e–e1(nlim

2

xxcosba

0x

c

a + b = 0 b = – a

2x

)e–e(

)e–e(

)e–e1(nlim

2

x)xcos–1(a

x)xcos–1(a

x)xcos–1(a

0x

c

c

c

2x

e–elim

2

x)xcos–1(a

0x

c

2

x

2

)xcos–1(a

0x x

1–e–x

1–elim

c

= 2

2x

1–e–x

)xcos–1(alim

2

x

20x

c

2x.x

1–e–x

)xcos–1(alim 2c

c

x

20x

c

(i) for c = 2, 2

a– 1 = 2 a = 6 b = – 6

a + b + c = 2




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(ii) for c > 2, 2

a– 0 = 2 a = 4

a + b + c > 2

(iii) for c < 2 limit does not exist.

5. Let cosf x x ………

Sol.

6. In the expansion of

485

2 x3

, ………

Sol.

4934 r

21

5

35 r 1 36

7. Let m be the minimum………

Sol. A : x > x2 – 3x + 3 x2 – 4x + 3 < 0

x (1, 3)

f (x) = 6x2 – 6x + 12

= 6(x – 2)(x + 1)

f(x) is minimum at x = 2

f(2) = –14

8. A couple has one ……… Sol. A : Exactly one children ; B : Exactly two children; C : Exactly three children

P(A) =1

4 P(B) =

1

2 P(C) =

1

4 ;

E : Couple has exactly 4 grand children

P(E) = P(A).PE

A

+ P(B).PE

B

+ P(C).PE

C

P(E)=1

4.0 +

1

2

21 1 1

. .22 4 4

+ 1

4

1 1 13. . .

4 4 2

=27

128

9. If there are 5 circles ………

Sol. 5

2C .2 + 3

2C + 5

1C .3

1C 2 = 20 + 3 + 30 = 53

10.

xx

1

x

1

0xecxcosxcosxsinlim ………

Sol. x1

0xxsinlim

= 0

x1

0xxcoslim

(1)0 = 1ee 0x

1–xcoslim

0x

x0x

ecxcoslim

= 0

=

)ecx(cosnxlim0xe

= x

1

)x(sinnlim–

0x

e

= 2

0x

x

1–

xcotlim–

e

= e0 =1

11. If f(x) is a monic polynomial of degree 2 ………

Sol. |x2 – x + 2 + f(x)| = |x2 – x + 2| + |f(x)| x R

(x2 – x + 2) f(x) 0 x R

f(x) 0 x R but f(1) = 0

f(x) = (x – 1)2 (as f(x) is monic polynomial)

f(3) = 4

12. Let

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

………

Sol. Let

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

For f(x) to be continuous at x 0

limf(x) f(0) limf(x)

x 0 x 0

503a

lim lim |sin x|

x 0 x 0f(x) (1 | sinx |)

limx 0

503a|sinx|

|sinx| 503ae e

Now

tan2xlim lim tan8xx 0 x 0

f(x) e

2

8e

Substituting in equation (i) we get

1503a 4e b e

1503a 4

1 1e e 503a a

4 2012 and

1

4b e

1

a2012

and 1

a2012

13. Let f : R

2

,0 ………

Sol. f(x) is surjective minimum value of ax2 – 2px + 4 is 1

a4

–= 1 4p2 = 12a

p2 = 3a a, p N

for smallest 2(a + p), a = 3, p = 3

2(a + p) = 12

15. The interval in which g ' (x) 0, , …………

Sol. LHL = h 0lim

3 3

5

h sin h

tan h

= h 0lim

3

h sinh

h

2 2

2

h sin h hsinh

h

=.1

6 (3) =

2

RHL =






h 0lim

3 2 2

3

h h h ha h ..... b 1 ...... 1 c 1 ........

3! 2! 1! 2!

h

b + 1 – c = 0 a – c = 0 b + c = 0

b

2

– c

2 = 0 b – c = –1

1 1

b c2 2

1a

2

RHL = a

6 –

c

6 =

1

6

(1) =

1

6

2

=

1

6

= 1

6

,

1

3

g(x) = x.

21 12 x x

2 2e

= x.

2x xe

g'(x) = x. 2x xe

(1 – 2x) +2x xe .1

= 2x xe

(x – 2x2 + 1)

g'(x)= – 2x xe

(2x2 –x – 1)

16. 2 2 2 2n

1 2 3 4cnlim ......

n 1 n 2 n 3 n 4cn

=

Sol. 2c = 1

2

1

n 2n<

2

1

n r<

2

1

n 1

2

r

n 2n<

2

r

n r<

2

r

n 1

2n

2r 1

r

n 2n < 2n

2r 1

r

n r < 2n

2r 1

r

n 1

nlim 2

1

n 2n×

2n(2n 1)

2

<

nlim

2n

2r 1

r

n r <2n

1lim

n 1 ×

2n(2n 1)

2

nlim

2n

2r 1

r

n r = nlim 2

n(2n 1)

(n 2n)

= 2 – 6 – 3µ

17. If n = 5, p = 0.75 ……………… 18. Largest set of values of p ………………

Sol. P(F / E).P(E)P(E / F)

P(F / E).P(E) P(F / E).P(E)

= p

1p (1 p)

n

(17) For p = 0.75 and n = 5

P(E/F) = 15/16

(18) P(E/F) = P(E)

p

p1

p (1 p)n

1

p (1 p) 1n

n 1 1

p 1n n

p 1

p [0,1]

PART : II PHYSICS 19. A gaseous mixture ………………………. Sol. As for ideal gas CP – Cv = R and = (Cp/Cv),

– 1 =

V

R

C or CV =

R

1

(CV)1 = = R

(CV)2 = =

and (CV)mix = = R

Now from the conservation of energy, i.e., U = U1 + U2, we get

(1 + 2) (CV)mixT = [1 (CV)1 + 2 (Cv)2] T

(CV)mix =

R = =

13 + 132= 9 + 152, i.e., 2 = 2 g mole 20. In the circuit shown ……………………….

Sol. after redrawing the circuit

(a) 4 = 5A ,

(b), (c) From loop (1) – 8(3) + E1 – 4(3) = 0

E1 = 36 volt

from loop (2) + 4(5) + 5(2) – E2 + 8(3) = 0

E2 = 54 volt

(d) from loop (3) – 2R – E1 + E2 = 0

R = 2 1E E

2

=54 36

2

= 9

Ans. (a) 5.00 A (b) 36.0 V, (c) 54.0 V (d) 9.00 . 21. Charge – Q and ………………………. Sol.

–Q +2Q

R2R

Total energy of system






=

2 2

0 0 0

Q (2Q) 1 Q2Q

8 R 8 2R 4 2R

=

2

0

Q

8 R

Energy between two shells

= dr.r4r4

Q

2

1R2

R

2

2

20

0

=

2

0

Q

16 R

22. In displacement ………………………. Sol.

Position 1

Principle axis

O

u v

v u

1

2

Position 2

For first & second position v

u=

1

O

, u

v=

2

O

2

2

v

u=

1

2

= 4.84

v

2.2u and v + u = 96

v = 66, u = 30

2

O

= v 11

2.2u 5

A is True Distance between two position of lens = v - u = 36 cm

B is True

Focal length of lens f = uv 66 30

u v 66 30

= 20.63

C is False Distance of lens from shorter image = u = 30 cm

D is True

23. Two satellites ……………………….

Sol.

3 / 2

1 1

2 2

T R

T R

3 / 2

2

32 R

256 R

R2 = 4R

(Total mechanical energy)1 = GM

8R

(Total mechanical energy)2 = GM

2R

(Total mechanical energy)2 > (Total mechanical energy)1

(Kinetic energy)1 = GM

8R

(Kinetic energy)2 = GM

2R

(Kinetic energy)2 < (Kinetic energy)1

24. In the given circuit ……………………….

Sol. 24 15 6

IR 1 2 1

A BV V 6 Ir

33

9 6R 4

R = 7

VBC = 15 – 3(2) = 9V

VBD = 30V 25. A container ………………………. Sol. H × 10 = 50 × 80 H = 400 cal/sec. when spray is working, (50 + 2t)80 = 400 t 4000 + 160 t = 400 t 4000 = 240 t t = 400/24 t = 50/3

t = 50/3 – 10 = 20/3 sec. So x = 20. 26. A cone made ……………………….

Sol. For cone = Q

R

dx x

Area of ring = rdx = R

x dx

Potential of vertex due to ring =

Rk x dx

x

Net potential =

Rk x dx

x

=

2kQ

W = qV = q ·

0

Q

2

=

– 3 – 6 910 10 18 10

1

= 18

27. Inside water ………………………. Sol. For the given case Ray diagram will be as given here

Here say ER will be seen by observer which appear to becoming

from point A. To find x, the distance of object from A we reverse the light rays by considering ER as incident ray & find the position of image after two refractions.

For I refraction we use

431

v 2R =

431

R

v =– 3R

For II refraction we use

43 1

x R

=43 1

R

x = – 2R

Hence OA= 2× 4 = 8 cm.






28. A planet is made ……………………….

Sol. 2

GM

(2R) =

2

GM'

R

M

4 = M

4

3R3 1 +

4

3(8R3 – R3)2 =

31

44 . R .

3

1 + 72 = 41

1

2

= 7

3

29. A monoatomic ………………………. Sol. TV3/2 = constant

3 / 2(PV)V

nR = constant

PV5/2 = constant

PVx = constant x = 5/2

C = Cv + R

1 x

For polytropic process,

C = 3

2R +

R

51

2

=5R

6.

30. A surge suppressor ………………………. Sol. P = V = KV3 P = KV4

2

1

P

P =

4

2

1

V

V

2P

10 =

4

400

200

P2 = 160 W

X = 20 31. In the figure ………………………. Sol. Image due to plane mirror will form at a distance of 20 cm

left of the mirror. Since image formed by two mirrrors lie adjacent to each

other. For convex mirror, image position is 15 cm towards left. u = – 25 cm v = + 15 cm

Using 1

v + 1

u = 1

f = 2

R

1

15 – 1

25 = 2

R

R = 75 cm. Ans. R = 75 cm. 32. A rectangular ……………………….

Sol. Using R = A

A = 8 cm x 2 mm = 16 x 10–5 m2

R = 5

28

10x16

10x24x10x6.1

= 24 x 10–6 = 24 µ Ans. 24

34. The ratio of ………………………. Sol. Let the angular speed of revolution of both stars be about the

common centre, that is, centre of mass of system.

The centripetal force on star of mass m is

m22d

3 =

2

Gm(2m)

d

Solving we get T =

234

d3Gm

The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass of system.

+

2

2m

m

22M

M

2dm

K 32

K d2m

3

1I

2

1I

2

35. For this process ……………………….

Sol. W = Area under the curve = 3

2P1V1

2 1

2 1

V 2V

P 2P

and P1V1 = nRT1

Therefore

1

w

nRT.

1 1

1 1

3. P V

2

P V

36. If C is defined ……………………….

Sol. nC T = Q nCT = 6n RT1

dT = 4T1 – T1 = 3T1

n . C . 3T1 = 6nRT1

C

R = 2







37. Increasing amount of solid Hg2 is added ………………. Sol. HgI2 + 2KI(aq) K2[HgI4] 2K+ + [HgI4]

2–

Due to complex formation number of particle is decreasing. 38. Which of the following is/are ore ………………. Sol. Haematite – Fe2O3 Magnetite – Fe3O4 Siderite – FeCO3 Calamine – ZnCO3 39. Which of the following statement ………………. Sol. (B) Spontaneous reaction need not to be exothermic. (C) reversible adiabatic is isoentropic. (D) for same amount of gas in lesser volume has lower

entropy. 40. An amount of 20g potassium sulphate ………………. Sol. Before electrolysis mass (H2O) = 150 g After electrolysis m (H2O) = 150 – 113.3 = 36.7g

2H O

n = 2.04 mol

Since 2H2O 2H2 + O2

2H

n = 2.04 mol V = 45.6L

2O

n = 1.02 mol. V = 22.8L

41.

Cl

P3AlCl ……………….

Sol. ;

42. Identify correct statements for given ……………….

Sol.

43. The cell notation of the standard galvanic ………………. Sol. Eºcell = Eºred(c) – Eºred(a) Eºred(a) = – 1.66 V 45. How many of the following involves ……………….

Sol. Haematite Fe

Galena Pb

Cassiterite Sn

Zinc blende Zn

46. Depression of freezing point of ……………….

Sol. For urea, Tf = Kf × m Kf = m

Tf =

1.8

1 = 1.8

Now, for CH3COOH

Tf = i × Kf × m i = 1.1

Thus, i = 1 + so = 0.1

Now, [H+] = C = 0.01 × 0.1 = 10–3 M

pH = 3

47. In modern periodic table let, a = Group ……………….

Sol. a = 2, b = 5, c = 71

a + b + c = 2 + 5 + 71 = 78.






49. Total number of products formed ……………….

Sol.

50. How many isomeric aromatic ……………….

Sol. 32 FeBr/Br

32 FeBr/Br

32 FeBr/Br

51. Partial roasting of chalcopyrites ………………. 52. In self reduction the reducing s………………. Sol. (Q.51 & Q.52)

2CuFeS2 + 4O2 Cu2S + 2FeO + 3SO2

Cu2S + FeO + SiO2 FeSiO3 (fusible slag) + Cu2S

(matte) 53. In which of the following the given ………………. 54. Which of the following reagents used ………………. Sol. (Q.53 & Q.54)

3AlCl

6 5 3 318 C to 80 Ctert butylbenzene

(only product)

C H C(CH )

3BF

6 5 3 2 2 360 C(only product)

tert pentylbenzene

C H C(CH ) CH CH

2 4H SO

6 6 3 6 5 3 33tert butylbenzene

C H CH COH C H C(CH )

2 4H SO

6 6 3 2 6 5 3 32tert butylbenzene

C H CH C CH C H C(CH )

PAPER-2

PART : I MATHEMATICS 1. Match the number ……….

Sol. f(x) = 2

1

x 7x 12 the points of discontinuous are x = 3, 4

(A) Now g(x) = 4x – 3 – x2 = 3 iff x2 – 4x + 6 = 0 has no solution

and g(x) = 4x – 3 – x2 = 4 iff x2 – 4x + 7 = 0 has no solution

fog has no point of discontinuity

(B) f(x) = x2 is continuous every where

g(x) = x 1 x 0

x 1 x 0

g(0) = – 1 x 0Lim

g(x) = 1 and

x 0Lim

g(x)

= – 1

g(x) is not continuous at x = 0

but x 0Lim

fog(x) =

x 1Lim

x2 = 1

x 0Lim

fog (x) =

x 1Lim

x2 = 1, fog (0) = f(–1) = 1

fog is continuous at x = 0

Number of discontinuous is 0.

(C) f(x) = 2

1

15x 8x 1 is discontinuous at x = – 1

3, – 1

5

g(x) = 2

1

x 3x 2 is discontinuous at x = 1, 2

1 and 2 are points of discontinuity

Now g(x) = 2

1

x 3x 2 = –

1

3 iff x2 – 3x + 5 = 0 (No

solution)

g(x) = 2

1

x 3x 2 =

1

5 iff x2 – 3x + 7 = 0 (no solution)

thus there are 2 points of discontinuity

(D) f(x) =2

1

6x 25x 14 is discontinuous of x = –

2

3, –

7

2

g(x) = 2

1

x x 2 is discontinuous of x = –1, 2

Now g(x) = 2

1

x x 2 = –

2

3 iff 2x2 – 2x – 4 + 3 = 0

iff 2x2 – 2x – 1 = 0

x = 2 4 8

4

=

1 3

2

g(x) = 2

1

x x 2 = –

7

2 iff 7x2 – 7x – 12 = 0

iff x = 7 385

14

Thus there are 6 points of discontinuity






2. Match List I with List II ……….

Sol. (P) 0 < 16

x ; 2

6

x1 ; 3

6

x2 ;

46

x3 ; 5

6

x4 ; 5

6

x

f(x) = 0 f(x) = –1 f(x) = –2 f(x) = –3 f(x) = – 4 f(x) = –5

6 elements

(Q) sin3 = 3sin – 4sin3

4sin3 = 3sin – sin3

E=3sin

1

)34sin(–3

4sin3)32sin(–

3

2sin33sin–sin3

=)3sin(

1

3sin–3

4sin33sin–

3

2sin3)3sin–sin3(

=

3sin3–3

4sin

3

2sin(sin3

)3sin(

1

=

3sin3–3

cossin2(sin3)3sin(

1

=

3sin3–)sin–(sin3)3sin(

1 = –3

(R) sinx cosx – 3cosx + 4sinx – 12 – 1 > 0

sinx(cosx + 4) – 3(cosx + 4) – 1 > 0 (cosx + 4)(sinx – 3) > 1 always negative

no solution

(S) sin2x + sinx cosx = n 2sin2x + sin2x = 2n 1 – cos2x + sin2x = 2n sin2x – cos2x = 2n – 1

for solution to exist

21–n22–

12n212–

2

1

2

1n

2

1

2

1–

number of integral values of n are 2

3. Consider f(x) =

2x

2x

,x2–3

,3x4–x2

, ……….

Sol. y = x2 – 4x + 3; n 2 ; y = 3 – 2x ; x > 2

y = x2 – 4x + 3; (–, 2] [–1, )

at x = 2; y = –1 f is one-one and onto

f(x)

2

3

–1

x2 – 4x + 3 – y = 0

x = 2

)y–3(14–164

x = 2

y444

x = y12 at x = 2, y = –1

at x = 0, y = 3 we consider negative sign

f–1 = 2 – x1 ; x –1 y = 3 – 2x, x > 2

x = 2

y–3, x > 2

f–1 =2

x–3, x < – 1

f(f(1)) = 3 and f(f(2)) = 8

sin–1(sin3) + cos–1(cos8) = ( – 3) + (8 – 2) = 5 –

4. Let f(x) =

4x

4x2

2x

,

,

,

3bx6

x–a

1bx2

……….

Sol.

4

x

lim f(g(x)) = a – 4

and –

4x

lim

f(g(x)) = 4b + 1

f(g(x)) to be continuous at x = 4

a – 4 = 4b + 1 a – 4b – 5 = 0 …….(1) ax + by + 1 = 0 is passing through (–1, 3)

–a + 3b + 1 = 0 ………(2) Solving (1) and (2), we get a = –11, b = – 4

5. If = 2

2

0x x

x4x–2–x2cosbxsinalim

……….

Sol. = 2

223

0x x

x4x–2–.....!2

)x2(–1b.....!3

x–xa

lim

= 2

32

0x x

.......6

a–x)b2–4(x)1–a(x)2–b(

lim

for limit to exist b = 2, a = 1 and = 0






6. If 2)y(sin2–)y(sin2 21–41–)x(cos 21– = 1……….

Sol. 1)1–)y((sin2 221–)x(cos 21– = 1 ……….(1)

equation (1) is possible if

21– )x(cos2 = 1 and 01–)y(sin 21–

cos–1x = 0 and sin–1y = ±1

x = 1 and y = ±sin1

x – y = 1 + sin1 or x – y = 1 – sin1 7. Find the value of sin……….

Sol. sin(4) = 2sin2 cos2 ; =3

1tan 1–

2 = 2tan–1

3

1= tan–1

4

3

= 2

4

3tancos

4

3tansin 1–1–

= 2 × 25

24

5

4

5

3

24

7tancos

7

1tan2cos 1–1–

= 25

24

8. There are 15 seats in ………. Sol. Case-I : 2 left to 6 and 1 right to 6 two can sit in the ways (1,

3), (1, 4), (2, 4) and one who sit right to 6 can sit in 8 ways

total number of ways = 24

Case-II : 1 left to 6 and 2 right to 6 the person left to 6 in 4 ways

Two right to 6 can be done in following way x1 + x2 + x3 = 7

x1 1, x2 1, x3 0

Number of integral solution = 21 total ways 21 × 4 = 84 ways Case-III : All 3 right to 6

x1 + x2 + x3 + x4 = 6(x1 1, x2 1, x3 1, x4 1)

coefficient t3 in (1 – t)–4 6C3 = 20

Case-I+ Case-II + Case-III = 128

Total arrangements = 128 × 4! = 384Hence = 8

9. Three balls marked ………. Sol. He picked 3 times, the following outcome add up to 6

1, 2, 3 1, 3, 2 2, 3, 1 2, 1, 3 2, 2, 2 3, 1, 2 3, 2, 1

all these events are equally likely, ME and exhaustive. Out of 7 cases only one favours

Therefore, there is 1/7 chance. ] 10. Find the number of values ……….

Sol. {sinx} + {–sinx} =

xsinif1

xsinif0

{sinx} + {–sinx} =

xsin1

xsin0

fn;

fn;

sin x x = 0, 2

, ,

2

3, 2,

2

5,3

11. Find the number of ……….

Sol. –1 [2 – 4x2] 1

–1 2 – 4x2 <2

–3 – 4x2 <0

0 < x2 4

3

x

2

3,

2

3– – {0}

12. Find the number of ………. Sol. x = 0 and x = 1

13. Let f(x) =

x2cos–x2sin1

x2cosx2sin1cot 1–

……….

Sol. f(x) = cot–1(cotx)

5

1x

)x(f = cot–1 = (cot1) + cot–1(cot2) + cot–1(cot3) + cot–1(cot4) +

cot–1(cot5)

= 1 + 2 + 3 + 4 – + 5 – = 15 – 2

14. If

0x|2–x|

0xx–)x(f , ……….

Sol.

f(x) = –x 2

–1

1

1 2 3 x

y

The solutions of the equation f(x) = 1 are x = –1, 1, 3

to solve f(f(x)) = 1 Let f(x) = p

f(p) = 1

f(p) = –1,1,3

(i) p = – 1 ; f(x) = –1; No solutions

(ii) p = 1 ; f(x) = 1 ; 3 solutions (iii) p = 3 ; f(x) = 3 ; 2 solutions

Total 5 solutions

15. Let f(x) be a monic polynomial ……….

Sol. f(x) = (x2 – 1)(x2 – 4)(x2 – 9) + x2 + 4

2f = 20 and 3f = 19

2f – 3f = 1






16.

xtan2–x1

x2sin

)1–x(4lim

1–2

1–1x

.

Sol. xtan2–xtan2–

)1–x(4lim

1–1–1x

=

xtan–4

4

)1–x(4lim

1–1x

apply LHR

=

2

1x

x1

1–4

4lim = –2

17. Let f(x) =

3x,}x{–

}x{–sinb

3x,a

3x

23

4–e

3–x

1

3–x

1

……….

Sol.

3–x

1

3–x

1

3–x

1

3x3x

3

21

3

4–3

e

lim)x(flim

= 01

0–0

= 0

)3x(–

)3xsin(–blim)x(flim

–– 3x3x

= b

for function to be continuous at x = 3

f(3+) = f(3) = f(3–) = 0 = a = b a = b = 0 18. If the second, third ………. Sol. t

2 = nC

1 an – 1b = 135

t3 = nC

2 an – 2b2 = 30

t4 = nC

3 an – 3 b3 = 3

10

, then

nC1 . nC

3 a2n – 4 b4 = 135 × 3

10

2

2n

3n

1n

)C(

C.C

= 2

1

n 3

)2n)(1n(n

= 4

)1n(n 22

3

)2n(

= 4

1n

n = 5

3

2

t

t

= 25

15

C

C

b

a

b

a

= 2

9

× 2.1

45

. 5

1

a = 9b

5a4b = 135 9

a5 5

= 135 a5 = 27 × 9

a = 3 b = 3

1

PART : II PHYSICS 19. Two dipoles ………………………

Sol. Torque = P E

= 0 b/c = 0 [In figure (i)]

In figure (ii)

= 2

13

2KPP

d =

1 2

3

2KPP

d

force on 1P in figure (i)

1

dP

dr

=

1 2

4

6KPP

d

force on 1P in figure II by action –reaction pair

F =1 2

4

3KPP

d

20. Consider an ………………………

Sol. (A) Time period T is :

22 3

earth

4T R

GM

(B) Orbital speed v is 2 earthGM

vR

(C) Total energy = earth sGM m

2R

(D) Magnitude of Gravitational field at centre of satellite =

earth

2

GM

R

21. Two spherical ………………………

Sol. Ves = 2GM

R =

342.G . R

3

R

=

4G

3

R

Ves R

Sarface area of P = A = 4RP2

Surface area of Q = 4A = 4 RQ2

RQ = 2Rp mass R is MR = MP + MQ

3

R

4R

3 =

3

P

4R

3 +

3

Q

4R

3

RR3 = RP

3 + RQ3

= 9RP3

RR = 91/3 RP RR > RQ > RP Therefore VR > VQ > VP

R

P

V

V = 91/3 and

P

Q

V

V =

1

2

22. In the shown ……………………… Sol.

130 cm 23. Two particles ……………………… Sol. Due to gravitational force, the two particles will fall same height

in same time H. hence the electrostatic force will not have






component in vertical direction. hence the time to hit the

ground wil be 2h

g, independent of v0.

24. If E denotes ……………………… Sol. (A, B, D)

i = ne A Vd i Vd (A)

P =

2V

R=

22E

R P E2 (B)

P = i2R P i2 (D) 25. An ideal diatomic ………………………

Sol. Equation of straight line AB P = 0

0

0

2PV 5P

V

Slope of the straight line AB = 0

0

2P

V

When process changes from endo to exo. slope should be

equal to adiabatic P-V curve slope = P 7 P

V 5 V

0

0

2P 7P

V 5V

P = 0

0

P10V

7 V

10

7

0

0

PV

V =

00

0

2PV 5P

V

0

35V V

24

0

0

P10P

7 V

35

24 V0 = 0

25P

12

3P0V0 = nRT0

0

25P

12

0

35V

24

= 825

288 P0 V0 = nRT

0

T 875

T 288(3) =

875

864

T = 875

864 T0

26. A liquid of density ………………………

Sol. 0.8 × 5 × S × 15 = 210

4.2

S = 5

6cal/gm-°C

27. Electric field ………………………

Sol. = jdA = EdA =

R

0

kr . 2r dr =

3k2 R

3

= 3

3

2 kR

=

6

38

3 16 10

22 10

100

= 3 × 10–8

28. Consider a ……………………… Sol. (Moderate) Flux through ABCD.

A B

C D z

G

F E x

y

H

1 =EA

= (2ˆ ˆx i y j ). (

2 â i )

= 0 as x = 0 Flux through EFGH

2 = (2ˆ ˆx i y j ). (

2 â i )

= x2.a2 = a4 = 1.0 × 10–4 Nm2/C Flux through BCGF

3 = (2ˆ ˆx i y j ). (

2â j )

= a3 = 1.0 × 10–3 Nm2/C Flux through EADH

4 = (2ˆ ˆx i y j ). (

2â j ) = 0 as y = 0

Flux through ABFE

5 = (2ˆ ˆx i y j ). (

2 â k ) = 0

Flux through CDHG

6 = 0

Net flux = (1.0 × 10–4 + 1.0 × 10–3) N-m2/C = 11 × 10–4 N-

m2/C 29. Two thin symmetrical ………………………

Sol. 1

43

11

f

2

R

+ 2

43

1

2

R

=1

24

f = 24 30. There is a long ………………………

Sol. | J | = 1

|E | sin30°

| J | =106 × 10–4 × 1

2= 100

2A/m2 = 50 A/m2

31. One mole of an ……………………… Sol. Using 1st law of TD Q = W + U

0 = fan

f( W ) P v n R T

2

Wfan = n RT + nf

2RT

Wfan = n Cp T

Wfan = (1) f

R R2

(500 k) (as the gas is expending

slowly so p = constant, so T v) Wfan = 14 kJ

32. About 0.014 kg ……………………… Sol. As gas is enclosed in the cylinder, V = constant






(Q)v = nCvT

Here, n = (0.014 103)/28 = (1/2) mol And as nitrogen is diatomic, Cv = (5/2) R, Further, as according to the given problem,

rms 2 2

rms 11

V T

V T = 2, i.e., T2 = 4T1

T = 4T1 – T1 = 3T1 = 3 300 = 900 K

(Q)v = 1 5

2 2 2 × 900 = 2250 cal

33. A conducting ………………………

Sol. R = A

10 = A

103 3

A

=

3

104

Electric flux = E.A

AV

= x × 10–3

310

204 = x × 10–3

x = 6 34. Figure shows two ……………………… Sol. Suppose after earthing charge on inner shell becomes q’

then for its potential to become zero.

3

q'q0

R3

Kq

r

'Kq

So charge flow from shell to earth is 3

q .

35. The apparatus shown ……………………… Sol. Applying pascals law we have

P0 = P0 + AghA – BghB + CghC – DghD

95°C g(52.8 cm) – 5°g (49 cm) + 95°Cg(49 cm) –

5°Cg(51) = 0

5 C

95 C

52.8 49

49 51

= 1.018 ......(i)

Coefficient of volume expansion

y = 95 C 5 C

5 C

v v

V (95 C 5 C)

=5 C 95 C

95 C(90 C)

y = 5 C

95 C

11

90 C

from equation

y = (1.018 – 1) 1

90 =

90

018.0=

90000

18 = 2 × 10–4

Coefficient of linear expansion =

4y 2 10

3 3

36. An infinitely long ………………………

Sol. Flux from total cylindrical surface (angle = 2)

30° 30°

aA B

O

= in

0

Q

Flux from cylindrical surface AB = flux from the given surface

= in

0

Q

6=

06

n = 6


40. Titanium metal is extensively used ……………….

Sol. ºrH = 140.5 kJ

ºrS = – 0.058 kJ

ºrG 0

41. In the following reactions ……………….

Sol.

42. The compounds P, Q and S were ……………….

Sol. 3 2 4

2

HNO /H SO

(NO )






(–OH is o/p director)

3 2 4

2

HNO /H SO

(NO )

(–OCH3 is stronger activator)

3 2 4

2

HNO /H SO

(NO )

(Substitution takes place in activated ring at least crowded

p-position)

43. It is observed that the voltage of a galvanic ……………….

Sol. Ecell = XM|ME +

2H |HE

= X

0

M|ME –

0.059

x log (MX+) +

2H |HE –

0.059

2

log 2H

2

P

[H ]

= XM|ME –

0.059

1log (M+x)1/x –

0.059

1log

2

1/2

HP

[H ]

= X

0

M|ME – 0.0591 log

2

1/2x 1/x

H(M ) . P

[H ]

x = 3

44. 10–6 mole of AgNO3 are added in 1 Litre ……………….

Sol. AgX Ag+ + X–

10–6 + S S

27 × 10–6 = 6 (10–6 + S) + 8S + 7 × 10–6

S = 10–6

Ksp = (10–6 + S) S = 2 × 10–12 M2

45. A sample of an ideal nonlinear ……………….

Sol. W = –Pext (V2 – V1) = –1 (8 – 2) = – 6 L atm

as q = 0 E = W

3(8Pf – 12) = – 6

Pf = 4

5 atm

i

f

T

T =

58

46 2

=

12

10

S = 3 × 300

12 ln

12

10 +

300

12ln

8

2

= 3.34 J/K

46. MX3 dissociates in M3+ and X– ions in an ……………….

Sol. MX3 M3+ + 3X–

1 – 3

i = 1 + 3

= 2

47. How many of the following are ……………….

Sol. Ores of A :

(i) Bauxite A2O3.2H2O

(ii) Cryolite Na3 AF6

48. If in long form of ……………….

Sol. (i) 9F (ii) 17Cl (iii) 2He (iv) 3Li

49. Total number of compound(s) or ……………….

Sol. H2SO3 Sulfurous acid ; HClO Hypochlorous acid






HClO2 Chlorous acid ;

Sn2+ Stannous ion ; Fe+2 Ferrous ion

50. In how many reactions major products ……………….

Sol. (6), (7)

51. The number of products formed ……………….

Sol. HCN

52. Number of carbocations which are ……………….

Sol. (i), (iii), (iv), (v), (vi), (vii), (viii)

54. Give your final answer by adding ……………….

Sol. (4 + 3) = 7






ADVANCED PATTERN



ANSWER KEY

PAPER-1


1. (CD) 2. (AB) 3. (AD) 4. (BCD) 5. (AD) 6. (ABCD)

7. 14 8. 20 9. 53 10. 02 11. 04 12. 01

13. 12 14. 10 15. (B) 16. (A) 17. (D) 18. (A)

PART : II PHYSICS

19. (AB) 20. (ABCD) 21. (AB) 22. (ABD) 23. (BCD) 24. (BC) 25. (20)

26. (18) 27. (80) 28. (14) 29. (12) 30. (20) 31. (75) 32. (24)

33. (D) 34. (B) 35. (A) 36. (B)


37. (AD)

38. (ABC) 39. (ABC) 40. (AB) 41. (ABCD) 42. (BCD) 43. (03)

44. (10) 45. (04) 46. (03) 47. (78) 48. (19) 49. (02) 50. (03)

51. (D) 52. (C) 53. (A)

54. (A)

PAPER-2


1. (B) 2. (D) 3. (AD) 4. (BC) 5. (BD) 6. (CD) 7. 0

8. 8 9. 8 10. 7 11. 0 12. 2 13. 5 14. 5

15. 1 16. 2 17. 6 18. 1

PART : II PHYSICS

19. (A) 20. (C) 21. (BD) 22. (ABD) 23. (AB) 24. (ABD) 25. (5)

26. (5) 27. (3) 28. (1) 29. (8) 30. (5) 31. (7) 32. (5)

33. (6) 34. (3) 35. (5) 36. (6)


37. (B) 38. (D) 39. (AC) 40. (ACD) 41. (BCD) 42. (ABD) 43. (3)

44. (2) 45. (3) 46. (2) 47. (2) 48. (9) 49. (5) 50. (2)

51. (2) 52. (7) 53. (5) 54. (7)





®

ADVANCED PATTERN CUMULATIVE TEST-1 (ACT-1)

TARGET : JEE (MAIN + ADVANCED) 2021

COURSE : VIJETA (JP) | BATCH : JPB-STAR, JPAB

P10-20

Ñi;k bu funsZ'kksa dks /;ku ls i<+saA vkidks 5 feuV fo'ks"k :i ls bl dke ds fy, fn;s x;s gSaA

vH;fFkZ;ksa ds fy, vuqns'k

DATE

21-06-2020

PAPER

1

CODE

1

Time (le;) : 3 Hours (?k.Vs) Maximum Marks (vf/kdre vad) : 180

ç'u&i=k 1 eas rhu (03) Hkkx gS % (xf.kr ] HkkSfrdh ,oa jlk;u foKku)

çR;sd Hkkx esa dqy vBkjg (18) iz'u gS tks rhu (03) [kaMks esa foHkkftr gS ¼[kaM 1, [kaM 2 vkSj [kaM 3)

iz'u&i=k 1 esa ç'uksa dh dqy la[;k % pkSou (54) ,oa vf/kdre vad % ,d lkS vLlh (180) gSaA

ç'uksa ds çdkj vkSj ewY;kadu ;kstuk,¡ [kaM–1 ¼vf/kdre vad : 24½

bl [kaM esa N% (06) iz'u gSA izR;sd iz'u ds fy, pkj fodYi fn, x, gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gS¼gSa½A izR;sd iz'u ds fy,] fn, gq, fodYiksa esa ls lgh mÙkj ¼mÙkjksa½ ls lacf/kr fodYi ¼fodYiksa½ dks pqfu,A izR;sd iz'u ds mÙkj dk ewY;kadu fuEu ;kstuk ds vuqlkj gksxk % iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA vkaf'kd vad % +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gS ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj nksauks pqus

gq, fodYi lgh fodYi gSaA vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj pquk gqvk fodYi lgh fodYi gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –2 vU; lHkh ifjfLFkfr;ksa esaA

[kaM–2 ¼vf/kdre vad : 24½ bl [kaM esa vkB (08) iz'u gSaA izR;sd iz'u dk mÙkj 00 ls 99 rd, nksuksa 'kkfey] ds chp dk ,d f} vadh; iw.kk±d gSA vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA a

[kaM–3 ¼vf/kdre vad : 12½ bl [kaM esa nks (02) vuqPNsn gSaA izR;sd vuqPNsn ij nks iz'u gaSA izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSA vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

ijh{kkFkhZ dk uke : ………………………………………………………………jksy uEcj : ………………………..…………………

eSaus lHkh funsZ'kksa dks i<+ fy;k gS vkSj eSa mudk eSusa ijh{kkFkhZ dk ifjp;] uke vkSj jksy uEcj vo'; ikyu d:¡xk@d:¡xhA dks iwjh rjg tk¡p fy;k gSA

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ijh{kkFkhZ ds gLrk{kj ijh{kd ds gLrk{kj





vkWfIVdy fjLikal 'khV (ORS) Hkjus ds funsZ’'k Åijh ewy i`"B ds vuq:i cqycqyksa (BUBBLES) dks i;kZIr ncko Mkydj dkyk djsaA

ewy i`"B e'khu&tk¡p gS rFkk ;g ijh{kk ds lekiu ij fujh{kd ds }kjk ,d=k dj fy;k tk;sxkA

vks-vkj-,l- dks gsj&Qsj@fod`fr u djsaA

viuk uke] jksy ua- vkSj ijh{kk dsanz dk uke ewy i`"B esa fn, x, [kkuksa esa dye ls Hkjsa vkSj vius gLRkk{kj djsaA buesa ls dksbZ Hkh

tkudkjh dgha vkSj u fy[ksaA jksy uEcj ds gj vad ds uhps vuq:i cqycqys dks dkyk djsaA

ORS ij cqycqyksa dks dkyk djus dh fof/k %

Åijh ewy i`"B ds cqycqyksa dks dkys ckWy ikbUV dye ls dkyk djsaA

cqycqys dks iw.kZ :i ls dkyk djsaA

cqycqyksa dks rHkh dkyk djsa tc vkidk mÙkj fuf'pr gksA

cqycqyksa dks dkyk djus dk mi;qDr rjhdk ;gk¡ n'kkZ;k x;k gS %

dkys fd;s gq;s cqycqys dks feVkus dk dksbZ rjhdk ugha gSA

gj [k.M ds izkjEHk esa nh x;h vadu ;kstuk esa dkys fd;s x;s rFkk dkys u fd;s x;s cqycqyksa dks ewY;kafdr djus dk rjhdk fn;k x;k gSA.

iw.kkZad@la[;kRed vad ds fy, ORS fuEu çdkj gS :

mnkgj.k ds fy, , ;fn mÙkj ‘,dy vadh;’ iw.kkZad gS rc :

0 1 2 3 4 5 6 7 8 9 mnkgj.k ds fy, , ;fn mÙkj ‘,dy vadh;’ iw.kkZad /kukRed @ _.kkRed gS rc :

0– 1

23456789

mnkgj.k ds fy, , ;fn mÙkj ‘f)&vadh;’ iw.kkZad gS rc :

0 01 12 23 34 45 56 67 78 89 9

n'keyo iw.kkZad@la[;kRed vadksa ds fy, ORS fuEu çdkj gS :

1 2 . 3 4

0 0 0 01 1 1 12 2 2 23 3 3 34 4 4 45 5 5 56 6 6 67 7 7 78 8 8 89 9 9 9

COLUMN

;fn mÙkj 3.7 gS rc 3 dks 1st ;k 2nd dkWye esa Hkjsa rFkk 7 dks 3rd ;k 4th dkWye esa HkjsaA

;fn mÙkj 3.07 gS rks 3 dks 1st dkWye ;k 2nd dkWye esa Hkjsa rFkk ‘0’ dks 3rd dkWye esa rFkk 7 dks 4th dkWye esa HkjsaA

;fn mÙkj 23 gS rc 2 dks 1st dkWye esa ] 3 dks 2nd dkWye esa tcfd 3rd vkSj 4th dkWye dks [kkyh NksM+ nsa ;k ‘'kwU; Hkj nsaA



dPps dk;Z ds fy, LFkku





Hkkx % I xf.kr

[kaM 1 : (vf/kdre vad : 24)

bl [kaM esa N% (06) iz'u gSaA

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls ,d ;k ,d ls vf/kd fodYi lgh gSaA izR;sd iz'u ds fy, vad fuEufyf[kr ifjfLFkfr;ksa esa ls fdlh ,d ds vuqlkj fn;s tk;saxs %

iw.kZ vad % +4 ;fn dsoy ¼lkjs½ lgh fodYi ¼fodYiksa½ dks pquk x;k gSA

vkaf'kd vad % +3 ;fn pkjksa fodYi lgh gSa ijUrq dsoy rhu fodYiksa dks pquk x;k gSA

vkaf'kd vad % +2 ;fn rhu ;k rhu ls vf/kd fodYi lgh gS ijUrq dsoy nks fodYiksa dks pquk x;k gS vkSj nksauks pqus gq, fodYi lgh fodYi gSaA vkaf'kd vad % +1 ;fn nks ;k nks ls vf/kd fodYi lgh gSa ijUrq dsoy ,d fodYi dks pquk x;k gS vkSj pquk gqvk fodYi lgh fodYi gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –2 vU; lHkh ifjfLFkfr;ksa esaA

mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc %

dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –2 vad feysaxs

1. ;fn f(x) = sin(tan–1(cos(cot–1x))), gks rc fuEu esa ls dkSuls lR; gS -

(A) 2

1–)x(flim

x

(B)

2

1)x(flim

x

(C) f dk ifjlj

2

1,

2

1– (D) f(x) =

2x21

x







2. ekukfd Qyu f(x) = (log3x)4 + 12(log3x)2.

x

27log3 tgk¡ x [1, 729] ;fn f(x) dk vf/kdre eku M

rFkk U;wure eku m gS rc -

(A) M = 81 (B) m = 0 (C) M = 27 (D) m = 3

3. ekukfd f : R [0, 1] ,d Qyu gS tks f(x) = min(e–|x|, |x2 – 1|) ls ifjHkkf"kr gSA rc fuEu esa ls dkSuls dFku lgh

gS ?

(A) f vkPNknd gS ijUrq ,dSdh ugh (B) f u rks ,dSdh vkSj u gha vkPNknd gSA

(C) f Bhd rhu fcUnqvksa ij vlrr gSA (D) f loZ=k lrr~ gSA

4. ;fn 2xtan

)e–1e(nlim

2

xxcosba

0x

c

tgk¡ a, b, c R rFkk c > 0 ;fn

(A) a = 6, b = –6, c = 3 (B) a = 6, b = –6, c = 2 (C) a = 4, b = –4, c = 3 (D) a = 4, b = –4, c = 4

5. ekuk cosf x x vkSj min ,0 ,0 /2

, 32 2

f t t x x

g xx x

ds fy,

ds fy, rc

(A) g x , 0, 3 esa lrr~ vkSj vodyuh; gSA

(B) g x ,2

x

ij lrr gS ijUrq vodyuh; ugha gSA

(C) g x ,2

x

u rks lrr~ gS vkSj u gh vodyuh; gSA

(D) 0, 3 esa g x dk vf/kdre eku 1 gSA







6.

485

2 x3

, ds foLrkj esa x = 3 ij

(A) la[;kRed vf/kdre in 35T vkSj 36T gSA

(B) chtxf.krh; vf/kdre in 35T gSA

(C) chtxf.krh; U;wure in 36T gSA

(D) foLrkj esa inksa dh la[;k 49 gSA

[kaM 2 : (vf/kdre vad : 24) bl [kaM esa vkB (08) iz'u gSaA izR;sd iz'u dk mÙkj 00 ls 99 rd, nksuksa 'kkfey] ds chp dk ,d f} vadh; iw.kk±d gSA

vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

7. ekuk leqPp; A = {x : nx > n(x2 – 3x + 3)} ij Qyu f(x) = 2x3 – 3x2 – 12x + 6 dk U;wure eku m gS

rc |m| dk eku gS -

8. ,d nEifÙk ds dsoy ,d cPpk gS ;k nks cPps gS ;k rhu cPps gksus dh çkf;drk Øe'k% 1

4,

1

2 rFkk

1

4 gSA ;fn

bl çdkj ds lekt esa ,d nEifÙk ds Bhd pkj iksrs&iksrh gksus dh çkf;drk p

q gS rc (q–4p) dk eku gksxkA

(tgk¡ p o q lgHkkT; la[;k,sa gSA)







9. ;fn lery esa 5 o`Ùk vkSj 3 ljy js[kk,¡ gS rc muds vf/kdre izfrPNsn fcUnqvksa dh la[;k gS&

10.

xx

1

x

1

0xecxcosxcosxsinlim cjkcj gS -

11. ;fn f(x), 2 ?kkr dk ,d inh; cgqin ¼ monic polynomial½ bl izdkj gS fd |x2 – x + 2 + f(x)| – |f(x)| = x2 – x +

2 x R rFkk f(1) = 0 rc f(3) dk eku Kkr dhft,A

12. ekuk

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

rc 2012 a dk eku gksxk ;fn 'f', x = 0 ij lrr~ gS -

13. ekuk f : R

2

,0 , f(x) = cosec–1(ax2 – 2px + 4) ls ifjHkkf"kr gS tgk¡ a, p N vkPNknd gS rc 2(a + p)

dk U;wure eku Kkr dhft,A

14. ;fn f : R R, f(x) = x3 – px2 + (q – 10) x, p, q R ,dSdh vkPNknd fo"ke Qyu gS rFkk rc (p + q) dk

U;wure eku cjkcj gS -








[kaM 3 : (vf/kdre vad : 12) bl [kaM esa nks (02) vuqPNsn gSaA izR;sd vuqPNsn ij nks iz'u gaSA izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSA

vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA

iz'u 15 ls 16 ds fy, vuqPNsn

;fn f(x) =

3 3

5

x

3

µ (x – sin x)x 0tan x

x 0

asinx bcosx 1– ce x 0

sin x

, Qyu x = 0 ij lrr~ gS] rks fuEufyf[kr ds mÙkj

nhft,A

15. vUrjky g ' (x) 0 gS, tgk¡ Qyu 2–3µ –2(ax bx)g(x) x e gS&

(A) 1, 1,2

(B)

1,12

(C) 1, 0,2

(D)

1,22

16. 2 2 2 2n

1 2 3 4cnlim ......

n 1 n 2 n 3 n 4cn

=

(A) – 6 – 3µ (B) – – µ (C) + 2µ (D) – 3 – 2µ








fdlh cgqp;ukRed iz'u ds mÙkj fy;s n fodYi fn;s gS ftuesa ls dsoy ,d gh lgh gSA ;fn ,d fo|kFkhZ x`g dk;Z djrk gS rks ;g fuf'pr gS fd og lgh mÙkj dh igpku dj ysxk vU;Fkk mÙkj ;kn`fPNd :i ls nsuk gksxkA ekuk fo|kFkhZ }kjk x`g dk;Z djus dh ?kVuk dks E ls iznf'kZr fd;k x;k gS rFkk P(E) = p ,oa fo|kFkhZ }kjk iz'u dk lgh mÙkj nsus dh ?kVuk dks F ls iznf'kZr fd;k x;k gSA

17. ;fn n = 5, p = 0.75, P(E/F) dk eku gksxk -

(A) 8

16 (B)

10

16 (C)

12

16 (D)

15

16

18. p ds ekuksa dk ,slk lcls cM+k leqPp; ftlds fy;s lEcU/k P(E/F) P(E) oS| gS] gksxk –

(A) [0,1] (B) R (C) 1

0,2

(D) 1

, 12




HkkSfrd foKku

HkkSfrd foK

ku

®





Hkkx-II : HkkSfrd foKku







mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc % dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –2 vad feysaxs

19. ,d xSlh; feJ.k ftlesa xSl A dk 1 g mole (1 = 5/3) vkSj xSl B (2 = 7/5) dks rki T ij ,d ik=k esa j[kk x;k gSA xSl A rFkk B vkil esa vfHkfØ;k ugha djrs gSa rFkk nksuksa xSlksa dks vkn'kZ ekfu,A ;fn xslh; feJ.k dk = 19/13 gS rc

(A) xSl B ds xzke eksyksa dh la[;k 2 gm eksy gksxh

(B) feJ.k ds (CV) dk eku 13

6R ds cjkcj gksxkA

(C) xSl B ds xzke eksyksa dh la[;k 1 gm eksy gksxh

(D) feJ.k ds (CV) dk eku 19

6R ds cjkcj gksxkA




HkkSfrd foKku

HkkSfrd foK

ku

®





20. fp=k esa fn[kk;s x;s ifjiFk esa E1 o E2 vKkr fo- ok- cy ds nks vkn'kZ L=kksr gSaA dqN /kkjk,sa fn[kkbZ x;h gSaA 6izfrjks/k ij foHkokUrj VA – VB = 10V gSA

E1E2

4.00

4.00

3.00 3.00 6.00

3.00A

R2.00A

A

BDC

(A) C o D ds e/; ds 4.00 izfrjks/k esa /kkjk 5A gSA

(B) vKkr fo-ok-cy E1] 36 V gSA

(C) vKkr fo-ok-cy E2] 54 V gSA

(D) izfrjks/k R, 9 ds cjkcj gSA

21. vkos'k– Q rFkk 2Q nks ldsUnzh; xksyksa ds i`"B ij ,d leku forfjr gS] ftudh f=kT;k fp=kkuqlkj Øe'k% ‘R’ o ‘2R’ gSA lgh fodYi@fodYiksa dk p;u dhft,A

–Q +2Q

R2R

(A) fudk; esa lafpr dqy fLFkjoS|qr ÅtkZ 2

0

Q

8 R gSA

(B) nks dks'kksa ds e/; lafpr fLFkjoS|qr ÅtkZ 2

0

Q

16 R gSA

(C) fudk; ds ckgj lafpr fLFkjoS|qr ÅtkZ 2

0

Q

2 R gSA

(D) nks dks'kksa ds e/; lafpr fLFkjoS|qr ÅtkZ 'kwU; gSA




HkkSfrd foKku

HkkSfrd foK

ku

®




22. foLFkkiu fof/k esa oLrq o insZ ds e/; nwjh 96 cm gSA muds e/; j[ks ,d mÙky ySal }kjk cuk;s x;s nks

izfrfcEcksa dh yEckbZ dk vuqikr 4.84 gSA

(A) oLrq o NksVs izfrfcEc dh yEckbZ dk vuqikr 11/5 gSA

(B) ySal dh nks fLFkfr;ksa ds e/; nwjh 36 cm gSA

(C) ySal dh Qksdl nwjh 22.5 cm gSA

(D) NksVs izfrfcEc ls ySal dh nwjh 30 cm gSA

23. leku æO;eku ds nks mixzg o`Ùkh; d{kk esa ?kwe jgs gSaA nksuksa dk vkorZdky 32 fnu rFkk 256 fnu Øe'k% gSA

;fn igys mixzg dh f=kT;k R gS rks

(A) nwljs mixzg dh d{kk dh f=kT;k 8R gksxhA

(B) nwljs mixzg dh d{kk dh f=kT;k 4R gksxhA

(C) nwljs mixzg dh dqy ;kaf=kd ÅtkZ igys ds lkis{k vf/kd gksxh A

(D) nwljs mixzg dh xfrt ÅtkZ igys ls de gksxhA

24. fn;s x;s ifjiFk esa fcUnq A dk foHko fcUnq B ds foHko ls 9V mPp gSA lgh fodYiksa dk p;u djksA

(A) izfrjks/k R dk eku 1 gS

(B) izfrjk/sk R dk eku 7 gS

(C) B rFkk D ds e/; foHkokUrj dk ifjek.k 30V gSA

(D) B rFkk C ds e/; foHkokUrj dk ifjek.k 15V gSA




HkkSfrd foKku

HkkSfrd foK

ku

®





vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

25. ,d ik=k (bldk ty rqY;kad ux.; gS) esa 0ºC ij 50gm ty Hkjk gSA ;fn Qqgkj dks cUn j[krs gq, ty ls çfr lSd.M Å"ek dh ek=kk H fu"dkflr dh tk,] rks 10 sec esa iwjk ty te tkrk gSA ;fn Qqgkj tkjh j[ks tks 0ºC ij 2 gm/sec ty Qqgkjrh gS] rks iwoksZDr Å"ek fu"dklu nj leku j[krs gq;s] ty ds iwjk teus esa x

3 lsd.M dk foyEc gksrk gSA 'x' dk eku Kkr dhft,A (cQZ dh xqIr m"ek /kkfjrk = 80 cal/gm)

26. ,d 'kadq dqpkyd inkFkZ dk cuk gS bldh frjNh lrg ij dqy vkos'k Q = 1 C ,dleku :i ls forfjr gSA vuUr ls 'kadq ds 'kh"kZ rd q = 1 mC ifj{k.k vkos'k dks ykus esa fd;k x;k dk;Z (twy esa) Kkr dht,A 'kadq dh

frjNh yEckbZ = 1m gSA




HkkSfrd foKku

HkkSfrd foK

ku

®




27. ikuh ds 4

3

vUnj fp=kkuqlkj 4 cm f=kT;k dk ,d ok;q dk cqycqyk fLFkr gSA ,d izs{kd O cqycqys dh

O;klr% v{k AB dks izsf{kr djrk gSA ikuh esa v{k ij fcUnq A ls ml fcUnq fcEc dh nwjh feseh0 esa Kkr dhft;s tks izs{kd }kjk A ij fn[krk gqvk izrhr gksrk gSA

O

gok dk cqycqyk

A B

28. ,d xzg ] nks inkFkksZ ls cuk gS ftuds ?kuRo n'kkZ,uqlkj 1 rFkk 2 gS

xzg dh lrg ij rFkk xgjkbZ ‘R’ ij xq:Roh; Roj.k leku gSA 1

2

vuqikr dk eku x

6 gSA x dk eku Kkr

dhft,A

29. ,d vkn'kZ ,d ijek.koh; xSl TV3/2 = fu;rkad çØe dh ikyuk djrh gS

bl çØe ds fy, eksyj fof'k"V Å"ek 5R

y

gS tgk¡ R xSl fu;rkad gSA '2y' dk eku Kkr dhft,A




HkkSfrd foKku

HkkSfrd foK

ku

®




30. ltZ lizslj (surge suppressor) ,d midj.k gS rks ifjiFk esa vpkud vf/kd /kkjk ds izokg dks jksdrk

gSA ;g ,sls pkydh; inkFkZ dk cuk gksrk gS ftlls izokfgr /kkjk blds fljksa ij vkjksfir foHkokUrj dh

r`rh; ?kkr ds lekuqikrh gksrk gSA tc vkjksfir foHkokUrj 200V gksrk gS rks lizslj ls O;f;r ÅtkZ 10 okWV

gksrh gSA tc foHkokUrj 400V rd c<+k;k tkrk gS rks O;f;r ÅtkZ 8X okWV gksrh gSA X dh x.kuk dhft;sA

31. fp=k esa mÙky niZ.k dh lrg ls 25 cm nwjh ij ,d fcEc j[kk gS vkSj ,d lery niZ.k bl çdkj j[kk tkrk

gS rkfd nksuksa niZ.kksa ls cus çfrfcEc ,d gh ry esa ikl&ikl gksaA lery niZ.k fcEc ls 20 cm nwj j[kk gSA

mÙky niZ.k dh oØrk f=kT;k D;k gS ?

32. rkEcs dh ,d vk;rkdkj pknj 2.00 mm eksVh vkSj i`"B foek gS 8.0 cm x 24 cm gSA ;fn yEcs fdukjksa dks ,d

24 cm yEch V~;wc cukus ds fy;s tksM+k tkrk gSA V~;wc ds fljksa ds e/; V~;wc dk izfrjks/k ekbØks vkse esa Kkr

djsaA rk¡cs dh izfrjks/kdrk = 1.6 x 10–8 -m.




HkkSfrd foKku

HkkSfrd foK

ku

®





bl [kaM esa nks (02) vuqPNsn gSaA izR;sd vuqPNsn ij nks iz'u gaSA

izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSA

vadu ;kstuk :

iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A

_.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA


rkjksa dk ,d ;qXe ,d mHk;fu"B nzO;eku dsUnz ds ifjr% ?kwerk gSA ,d rkjs dk nzO;eku M o nwljs dk nzO;eku

m bl izdkj gS M = 2m rFkk rkjksa ds dsUnzksa ds e/; nwjh d gSA (d izR;sd rkjs ds vkdkj ls cgqr vf/kd gSA)

33. rkjksa dk mHk;fu"B nzO;eku dsUnz ds ifjr% ?kw.kZu dky gS (d, m, G ds inksa esa)

(A) 2

34d

Gm

(B)

238

dGm

(C)

232

d3Gm

(D)

234

d3Gm

34. nks xzgksa dh xfrt ÅtkZ dk vuqikr (Km/KM) gS &

(A) 1 (B) 2 (C) 4 (D) 9




HkkSfrd foKku

HkkSfrd foK

ku

®





çkjEHk esa ,d vkn'kZ ,d ijek.kqd xSl ds T1 rki ij n eksy gSA nkc rFkk vk;ru dks /khjs&/khjs bl izdkj

nqxquk djrs gSa fd P-V fp=k ij ,d ljy js[kk vkysf[kr gksrh gS &

35. bl izfØ;k ds fy,] vuqikr 1

W

nRT cjkcj gS (tgk¡ W xSl }kjk fd;k x;k dk;Z gS) :

(A) 1.5 (B) 3 (C) 4.5 (D) 6

36. ;fn izfØ;k ds fy, vkSlr eksyj fof'k"V Å"ek C ls ifjHkkf"kr gks] rks C

R dk eku gS &

(A) 1.5

(B) 2

(C) 3

(D) 6



CH

EM

IST

RY

CHEMISTRY






Hkkx : III jlk;u foKku

Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al = 27, Si = 28,

P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5, Zn = 65, As = 75,

Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]







mnkgj.k% ;fn fdlh iz'u ds fy, dsoy fodYi (A),(B) vkSj (D) lgh fodYi gS] rc % dsoy fodYi (A),(B) vkSj (D) pquus ij +4 vad feysaxs ’ dsoy fodYi (A) vkSj (B) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (B) vkSj (D) pquus ij +2 vad feysaxs ’ dsoy fodYi (A) pquus ij +1 vad feysaxs ’ dsoy fodYi (B) pquus ij +1 vad feysaxs ’ dsoy fodYi (D) pquus ij +1 vad feysaxs ’ dksbZ Hkh fodYi u pquus ij ¼vFkkZr~ iz'u vuqÙkfjr jgus ij½ 0 vad feysaxs vkSj vU; fdlh fodYiksa ds la;kstu dks pquus ij –2 vad feysaxs



CH

EM

IST

RY

CHEMISTRY





37. 0.1 eksy K ;qDr ,d yhVj tyh; foy;u esa Bksl HgI2 dh c<+rh gq;h ek=kk feyk;h tkrh gSA fuEu esa ls

dkSulk@dkSuls xzkQ feyk;h x;h Hg2 dh ek=kk ds lkFk lgh ifjorZu iznf'kZr djrs gS\

(TB : DoFkukad, TF : fgekad ) (Hg2 vfoys; Bksl gS tks fd vkf/kD; – esa ?kqy tkrk gS)

(A)

TF

0.05 0.10

HgI2 ds eksy

(B)

TF

0.05 0.10

HgI2 ds eksy

(C)

TB

0.05 0.10

HgI2 ds eksy

(D)

TB

0.05 0.10

HgI2 ds eksy



CH

EM

IST

RY

CHEMISTRY





38. fuEu esa ls dkSulk@dkSuls vk;ju dk v;Ld gS@gSa \ (A) gsesVkbV (B) esXusVkbV (C) flMsjkbV (D) dsykekbu

39. fuEu esa ls dkSulk@dkSuls dFku xyr gS@gSa \

(A) Å"ekjks/kh (insulated ) ik=k esa vkn'kZ xSl dk eqDr izlkj ,d mRØe.kh; izØe gSA

(B) tc (Gfudk;)T,P < 0, rc vfHkfØ;k Å"ek{ksih gksuh pkfg,A (C) lHkh :)ks"eh; izØe vkblks,UVªkWfid ¼isoentropic½ gksrs gSA (D) STP ij 1 eksy N2 1 atm vkSj 273 K ij 1 eksy He xSl dh vis{kk vf/kd ,UVªkWih j[krh gSA

40. ikSVsf'k;e lYQsV dh 20g ek=kk dks 150 cm3 ty es ?kksyk x;kA ckn es foy;u dks oS|qr vi?kfVr fd;k x;kA oS|qr vi?kVu ds i'pkr~ foy;u esa ikSVsf'k;e lYQsV dk vo;o nzO;eku ds vuqlkj 15% FkkA NTP ij izkIr xSlksa dk vk;ru Kkr dhft, \ ( OH2

d = 1 g cm–3)

(A) 45.6L H2 xSl (B) 22.8L O2 xSl (C) 42 L H2 xSl (D) buesa ls dksbZ ugha

41.

Cl

P3AlCl

QNaH

O O R

C115100

CO)NH( 324

O

SHCl

P, Q, R vkSj S esa ,sjkseSfVd ;kSfxd gS@gSa & (A) P (B) Q (C) R (D) S



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42. nh xbZ vfHkfØ;k inksa ds fy, lgh dFku igpkfu;s

3AlClVkWyqbZu eq[; mRikn

(A) R =

(B) R =

(C) fn;s x;s rhuksa vfHkfØ;k inksa esa ls nks bysDVªkWuLusgh izfrLFkkiu in gSaA (D) (i) MeMgBr (ii) aq.NH4Cl ls vfHkfØ;k djus ij ;kSfxd P jsflfed feJ.k nsrk gSA



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vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA

'kwU; vad % 0 vU; lHkh ifjfLFkfr;ksa esaA

43. ,d vKkr /kkrq bysDVªksM X ds lkFk ,d ekud xsYosfud ¼oksfYVd½ lSy dk lSy ladsru uhps n'kkZ;k x;k gS &

X(s) | X3+ (1 mol dm–3) | | Pb2+ (1 mol dm–3) | Pb (s)

mijksDr ifjfLFkfr ls lEcfU/kr fdrus dFku lgh gSa \

(a) mijksDr lSy ladsru esa f}d mnxz js[kkvksa (II) }kjk iznf'kZr lSy dk ?kVd yo.k lsrq dgykrk gSA

(b) mijksDr lSy es Pb(s) vkWDlhdkjd ¼oxidizing agent½ gSA

(c) mijksDr lSy ds bysDVªksMksa ls tqM+s oksYVehVj ij izkjfEHkd ikB~;kad 1.53 V gSA

[ o

Pb|Pb2E = – 0.13 V] rc vKkr /kkrq X dk ekud vip;u foHko 1.66 V gSA

(d) bl vfHkfØ;k esa gksus okyh dqy vfHkfØ;k ds fy, larqfyr lehdj.k gS %

2X(s) + 3Pb2+ 2X3+ + 3Pb(s)

(e) ;fn X(s) | X3+ (aq) v)Z lSy esa oS|qr vi?kV; dh lkUnzrk c<+rh gS rks oksYVehVj dk izkjfEHkd ikB~;kad

?kVsxkA



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44. 50 mL, 0.1 M CuSO4 foy;u dks 200 lSd.M dh vof/k ds fy, 0.965 A dh /kkjk ds lkFk oS|qrvi?kfVr fd;k tkrk gSA bysDVªkWM ij gksus okyh vfHkfØ;k,¡ gS &

dSFkksM : Cu2+ + 2e– Cu(s)

,uksM : 2H2O O2 + 4H+ + 4e–

oS|qrvi?kVu ds nkSjku vk;ru es ifjorZu ugh ekurs gq,] oS|qrvi?kVu ds vUr ij –24SO dh eksyj lkUnzrk

ifjdfyr dhft,A viuk mÙkj 100 ls xq.kk djus ds i'pkr~ nhft,A

45. fuEu esa ls fdrus izØe esa v;Ld ls /kkrq ds fu"d"kZ.k ds nkSjku dkcZu vip;u gksrk gS\

ckWDlkbV Al

dkWij ikbjkbV Cu

gsesVkbV Fe

xsysuk Pb

dkusZykbV Mg

flYokbu K

dsflVsjkbV Sn

ftad Cys.M Zn

46. 0.01 eksyy tyh; CH3COOH ds fgekad es voueu 0.0198ºC gSA 1 eksyy ;wfj;k foy;u –1.8ºC ij

terk gSA eksyjrk dks eksyyrk ds cjkcj ekurs gq,] CH3COOH foy;u dh pH gS &



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47. vk/kqfud vkoZr lkj.kh esa ekuk, a = ijek.kq Øekad 'z' = 120 j[kus okys dkYifud rRo dh lewg la[;k gSA

b = 'Pd' rRo dh vkoZr la[;k

c = ysUFksukbM Js.kh esa vfUre rRo dk ijek.kq Øekad

Kkr dhft, (a + b + c)

48. C8H18 ds dqy lajpuk leko;oh;ksa dh la[;k gS (x) rFkk bues ls (y) leko;oh; 1 eksuksDyksjks mRikn nsrs gSA viuk mÙkj (x+y) ds :i es nsosA

49. fuEu vfHkfØ;k esa fufeZr mRiknksa dh dqy la[;k gS &

50. C6H

2Br

4 ds fdrus leko;oh Br

2/FeBr

3 ds lkFk czksehuhdj.k djus ij dsoy ,d izdkj ds mRikn cuk;sxsA



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[kaM 3 : (vf/kdre vad : 12) bl [kaM esa nks (02) vuqPNsn gSaA izR;sd vuqPNsn ij nks iz'u gaSA izR;sd iz'u esa pkj fodYi (A), (B), (C) rFkk (D) gSaA bu pkj fodYiksa esa ls dsoy ,d fodYi lgh gSA

vadu ;kstuk : iw.kZ vad % +3 ;fn flQZ lgh fodYi gh pquk x;k gSA 'kwU; vad % 0 ;fn fdlh Hkh fodYi dks ugha pquk x;k gS ¼vFkkZr~ iz'u vuqÙkfjr gS½A _.k vad % –1 vU; lHkh ifjfLFkfr;ksa esaA


dkWij izFke iafDr laØe.k /kkrqvksa esa lokZf/kd mRd`"V gksrk gS rFkk vusd ns'kks es lw{e ek=kk es izkIr gksrk gSA

dkWij ds v;Ld es pkYdsUFkkbV (CuSO4.5H2O), ,VsdsekbV (Cu2Cl(OH)3), D;wizkbV (Cu2O), dkWij XykUl

(Cu2S) rFkk esysdkbV (Cu2(OH)2CO3) fufgr gSA ;|fi fo'o dkWij mRiknu dk 80% pkYdksikbjkbV

(CuFeS2) v;Ld ls vkrk gSA pkYdksikbjkbV ls dkWij ds fu"d"kZ.k es vkaf'kd HktZu] vk;ju dk fu"dklu rFkk

Lor% vip;u gksrk gSA

51. feyk;h x;h flfydk ds lkFk pkYdksikbjkbV ds vkaf'kd HktZu ls fuEu mRikfnr ugh gksrk gS &

(A) FeSiO3 (B) FeS (C) Cu2S (D) CuO

52. Lor% vip;u esa vipk;d Lih'kht ……… gksrh gSA

fjDr LFkku ds fy, lgh fodYi dk p;u dhft,A

(A) Cu+ (B) O–2 (C) S–2 (D) SiO2



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csUthu ,d vlUr`Ir gkbMªksdkcZu gS ysfdu lkekU; ifjfLFkfr;ksa ds vUrxZr ;ksxkRed vfHkfØ;k,sa ugha nsrk gSA ;g eq[;r% bysDVªkWu Lusgh izfrLFkkiu vfHkfØ;k nsrk gSA csUthu ds izfrLFkkih O;qRiéksa esa] vkxs izfrLFkkiu dk vfHkfoU;kl vuqukn izHkko] f=kfoe dkjdksa bR;kfn }kjk fu/kkZfjr gksrk gSA blds vk/kkj ij fuEu iz'uksa ds mÙkj nhft, &

53. fuEu esa ls dkSulh vfHkfØ;k esa fn;k x;k mRikn eq[; mRikn gS &

(A)

Br

NO2 lYQksuhdj.k

Br

NO2

SO3H

(B)

CN

NO2

2Br

Fe

CN

NO2

Br

(C)

2 3Cl /FeCl

Cl

(D)

3 2

3

CH CO O

AlCl

COCH3

54. bysDVªkWuLusgh izfrLFkkiu fØ;kfof/k }kjk csUthu ds ,fYdyhdj.k ds fy, iz;qDr fuEu esa ls dkSuls vfHkdeZ Zd leku mRikn nsxsa \

(p) futZy AlCl3 dh mifLFkfr esa

CH3–CH–CH2–Cl

CH3

(q) H2SO4 dh mifLFkfr esa (CH3)3 COH

(r) H2SO4 dh mifLFkfr esa (CH3)2 CH-CH2-OH (s) H2SO4 dh mifLFkfr esa

CH3–C–CH2–OH

CH3

CH3 (A) p, q, r (B) q, r, s (C) p, q, r, s (D) p, q, s






ADVANCED PATTERN



ladsr ,oa gy

PAPER-1


1. ;fn f(x) = sin(tan–1(cos(cot–1x))) ……………

Sol. 2

1)x(flim

x

;

2

1–)x(flim

x

f(x) = xcotcostansin 1–1–

f(x) dk ifjlj

2

1,

2

1– gSA

f(x) = sin(tan–1(cos(cot–1x)))

=

2

1–

x1

xtansin

ekuk

2

1–

x1

xtan

tan = 2x1

x

cot–1x = (0, )

x = cot

(i) ekuk

2

,0

cos = xsin

1

= 2

2

2 x1

x

x

x1

1

(ii) ;fn

,2

x < 0

cos = xsin

1=

2t1

1

= –

2x

11

1

= 2x1

|x|

=

2x1

x

x < 0

f(x) = sin() = ecxcos

1

=

2cot1

1– =

2

2

x

x11

1–

=

1x2

|x|

2

f(x) = 2x21

x

(x > 0 ds fy, blhizdkj)

f(x) = 2x21

x

2. ekukfd Qyu f(x) = (log3x)4 ……………

Sol. ekuk log3x = t x [1, 729]

t [0, 6]

f(x) = t4 + 12t2(3 – t)

= (t(t – 6)2 = = ((t – 3)2 – 9)2 t [0, 6]

M = 81, m = 0

3. ekukfd f : R [0, 1……………

Sol.

vkjs[k ls fodYi ls igpkfu,A

4. ;fn ……………

Sol. 2

x.x

xtan

)e–e1(nlim

2

2

2

xxcosba

0x

c

2x

)e–e1(nlim

2

xxcosba

0x

c

a + b = 0 b = –

a

2x

)e–e(

)e–e(

)e–e1(nlim

2

x)xcos–1(a

x)xcos–1(a

x)xcos–1(a

0x

c

c

c

2x

e–elim

2

x)xcos–1(a

0x

c

2

x

2

)xcos–1(a

0x x

1–e–x

1–elim

c

= 2






2x

1–e–x

)xcos–1(alim

2

x

20x

c

2x.x

1–e–x

)xcos–1(alim 2c

c

x

20x

c

(i) c = 2 ds fy,, 2

a– 1 = 2 a = 6 b = – 6

a + b + c = 2

(ii) c > 2 ds fy,, 2

a– 0 = 2 a = 4

a + b + c > 2

(iii) c < 2 ds fy, lhek fo|eku ugh gSA

5. ekuk cosf x x ………

Sol.

6.

485

2 x3

, ds foLrkj esa ………

Sol.

4934 r

21

5

35 r 1 36

7. ekuk leqPp; ………

Sol. A : x > x2 – 3x + 3 x2 – 4x + 3 < 0

x (1, 3)

f (x) = 6x2 – 6x + 12

= 6(x – 2)(x + 1)

x = 2 ds fy, f(x) U;wure gSA

f(2) = –14

8. ,d nEifÙk ds dsoy ………

Sol. A : Bhd ,d cPpk ; B : Bhd nks cPps ; C : Bhd rhu cPps

P(A) =1

4 P(B) =

1

2 P(C) =

1

4 ; E : nEifÙk ds Bhd

pkj iksrs&iksrh gSA

P(E) = P(A).PE

A

+ P(B).PE

B

+ P(C).PE

C

P(E) =1

4.0 +

1

2

21 1 1

. .22 4 4

+ 1

4

1 1 13. . .

4 4 2

=27

128

9. ;fn lery esa 5 o`Ùk ………

Sol. 5

2C .2 + 3

2C + 5

1C .3

1C 2 = 20 + 3 + 30 = 53

10.

xx

1

x

1

0xecxcosxcosxsinlim ………

Sol. x1

0xxsinlim

= 0

x1

0xxcoslim

(1)0 = 1ee 0x

1–xcoslim

0x

x0x

ecxcoslim

= 0

=

)ecx(cosnxlim0xe

= x

1

)x(sinnlim–

0x

e

= 2

0x

x

1–

xcotlim–

e

= e0 =1

11. ;fn f(x), 2 ?kkr dk ,d inh; cgqin ………

Sol. |x2 – x + 2 + f(x)| = |x2 – x + 2| + |f(x)| x R

(x2 – x + 2) f(x) 0 x R

f(x) 0 x R but f(1) = 0

f(x) = (x – 1)2 (pwfd f(x) ,d inh; cgqin gS)

f(3) = 4

12. ekuk

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

………

Sol. ekuk

503a

|sin x|

tan2x

tan8x

(1 | sinx |) ; x 06

f(x) b ; x 0

e ; 0 x6

f(x) ds fy, x 0 ij lrr~ gS

limf(x) f(0) limf(x)

x 0 x 0

503a

lim lim |sin x|

x 0 x 0f(x) (1 | sinx |)

limx 0

503a|sinx|

|sinx| 503ae e

vc

tan2xlim lim tan8xx 0 x 0

f(x) e

2

8e

lehdj.k (i) esa j[kus ij

1503a 4e b e

1503a 4

1 1e e 503a a

4 2012 vkSj

1

4b e






1

a2012

vkSj 1

a2012

13. ekuk f : R

2

,0 , ………

Sol. f(x) vkPNknd gksxk tcfd ax2 – 2px + 4 dk U;wure eku 1

gSA

a4

–= 1 4p2 = 12a

p2 = 3a a, p N

U;wure eku 2(a + p) ds fy, a = 3, p = 3

2(a + p) = 12

15. vUrjky g ' (x) 0 gS, , ………………

Sol. LHL = h 0lim

3 3

5

h sin h

tan h

= h 0lim

3

h sinh

h

2 2

2

h sin h hsinh

h

=.1

6 (3) =

2

RHL

= h 0lim

3 2 2

3

h h h ha h ..... b 1 ...... 1 c 1 ........

3! 2! 1! 2!

h

b + 1 – c = 0 a – c = 0 b + c = 0

b

2

– c

2 = 0 b – c = –1

1 1

b c2 2

1a

2

RHL = a

6 –

c

6

= 1

6

(1) =

1

6

2

=

1

6

= 1

6

,

1

3

g(x) = x.

21 12 x x

2 2e

= x.

2x xe

g'(x) = x. 2x xe

(1 – 2x) +2x xe .1

= 2x xe

(x – 2x2 + 1)

g'(x)= – 2x xe

(2x2 –x – 1)

16. 2 2 2 2n

1 2 3 4cnlim ......

n 1 n 2 n 3 n 4cn

=

Sol. 2c = 1

2

1

n 2n<

2

1

n r<

2

1

n 1

2

r

n 2n<

2

r

n r<

2

r

n 1

2n

2r 1

r

n 2n <

2n

2r 1

r

n r <

2n

2r 1

r

n 1

nlim 2

1

n 2n×

2n(2n 1)

2

<

nlim

2n

2r 1

r

n r <

2n

1lim

n 1 ×

2n(2n 1)

2

nlim

2n

2r 1

r

n r = nlim 2

n(2n 1)

(n 2n)

= 2 – 6 – 3µ

17. ;fn n = 5, p = 0.75, ………………

18. p ds ekuksa dk ,slk lcls cM+k ………………

Sol. P(F / E).P(E)

P(E / F)P(F / E).P(E) P(F / E).P(E)

= p

1p (1 p)

n

(17) For p = 0.75 and n = 5

P(E/F) = 15/16

(18) P(E/F) = P(E)

p

p1

p (1 p)n

1

p (1 p) 1n

n 1 1

p 1n n

p 1

p [0,1]

PART : II PHYSICS

19. ,d xSlh; feJ.k ………………………. Sol. vkn'kZ xSl ds fy;s CP – Cv = R rFkk = (Cp/Cv),

– 1 =

V

R

C or CV =

R

1

(CV)1 = = R

(CV)2 = =

rFkk (CV)mix = = R

ÅtkZ laj{k.k ls e., U = U1 + U2,

(1 + 2) (CV)mixT = [1 (CV)1 + 2 (Cv)2] T

(CV)mix =

R = =

13 + 132= 9 + 152, i.e., 2 = 2 g mole






20. fp=k esa fn[kk;s ……………………….

Sol.

ifjiFk dks iqu% [khpus ds i'pkr~

(a) 4 = 5A ,

(b), (c) ywi (1) ls – 8(3) + E1 – 4(3) = 0

E1 = 36 volt

ywi (2) ls + 4(5) + 5(2) – E2 + 8(3) = 0

E2 = 54 volt

(d) ywi (3) ls – 2R – E1 + E2 = 0

R = 2 1E E

2

=54 36

2

= 9

Ans. (a) 5.00 A (b) 36.0 V, (c) 54.0 V (d) 9.00 .

21. vkos'k– Q rFkk ………………………. Sol.

–Q +2Q

R2R

fudk; dh dqy ÅtkZ

=

2 2

0 0 0

Q (2Q) 1 Q2Q

8 R 8 2R 4 2R

=

2

0

Q

8 R

nks dks'kksa ds e/; ÅtkZ

= dr.r4r4

Q

2

1R2

R

2

2

20

0

=

2

0

Q

16 R

22. foLFkkiu fof/k ………………………. Sol.

Position 1

Principle axis

O

u v

v u

1

2

Position 2

izFke o f}rh; fLFkfr ds fy, v

u =

1

O

, u

v=

2

O

2

2

v

u =

1

2

= 4.84

v

2.2u and v + u = 96 v = 66 , u = 30

2

O

= v 11

2.2u 5 A lgh gSA

ySUl dh nks fLFkfr;ksa ds e/; nwjh = v - u = 36 cm B lgh gS

ySUl dh Qksdl nwjh f = uv 66 30

u v 66 30

= 20.63

C xyr gS

NksVs izfrfcEc ls ySal dh nwjh = u = 30 cm D lgh gS

23. leku æO;eku ……………………….

Sol.

3 / 2

1 1

2 2

T R

T R

3 / 2

2

32 R

256 R

R2 = 4R

(dqy ;kaf=kd ÅtkZ)1 = GM

8R

(dqy ;kaf=kd ÅtkZ)2 = GM

2R

(dqy ;kaf=kd ÅtkZ)2 > (dqy ;kaf=kd ÅtkZ)1

(xfrt ÅtkZ)1 = GM

8R

(xfrt ÅtkZ)2 = GM

2R

(xfrt ÅtkZ)2 < (xfrt ÅtkZ)1

24. fn;s x;s ifjiFk ……………………….

Sol. 24 15 6

IR 1 2 1

A BV V 6 Ir

33

9 6R 4

R = 7

VBC = 15 – 3(2) = 9V

VBD = 30V

25. ,d ik=k ………………………. Sol. H × 10 = 50 × 80 H = 400 cal/sec.

tc Qqgkj dk;Z'khy gS

(50 + 2t)80 = 400 t 4000 + 160 t = 400 t 4000 = 240 t t = 400/24 t = 50/3

t = 50/3 – 10 = 20/3 sec.

So vr% x = 20.

26. ,d 'kadq dqpkyd ……………………….

Sol. 'kadq ds fy, = Q

R






dx x

oy; dk {kS=kQy = rdx = R

x dx

oy; ds dkj.k 'kh"kZ ij foHko =

Rk x dx

x

dqy foHko =

Rk x dx

x

=

2kQ

W = qV = q ·

0

Q

2

=

– 3 – 6 910 10 18 10

1

= 18

27. ikuh ds 4

3

……………………….

Sol. nh xbZ fLFkfr ds fy, fdj.k vkjs[k ;gk¡ fn;k x;k gS &

;gk¡ fuxZr fdj.k ER izs{kd }kjk fcUnq A ls vkrh gqbZ izsf{kr gksxhA

x dk eku fudkyus ds fy, tks A ls fcEc dh nwjh gS] ge fuxZr fdj.kksa ER dks vkifrr fdj.ksa ekurs gq;s izdk'k fdj.kksa dks foijhr fn'kk esa ysrs gS rFkk izfrfcEc dh fLFkfr nks viorZu ds ckn Kkr djrs gSaA

I viorZu ds fy, ge iz;ksx djrs gSaA

431

v 2R =

431

R

v =– 3R

II viorZu ds fy, ge iz;ksx djrs gSa &

43 1

x R

=43 1

R

x = – 2R

bl izdkj OA= 2× 4 = 8 cm.

28. ,d xzg] nks inkFkksZ ……………………….

Sol. 2

GM

(2R) =

2

GM'

R

M

4 = M

4

3R3 1 +

4

3(8R3 – R3)2 =

31

44 . R .

3

1 + 72 = 41

1

2

= 7

3

29. ,d vkn'kZ ,d ……………………….

Sol. TV3/2 = fu;rkad

3 / 2(PV)V

nR = fu;rkad

PV5/2 = fu;rkad

PVx = fu;rkad

x = 5/2

C = Cv + R

1 x

cgqijek.koh; çØe ds fy,

C = 3

2R +

R

51

2

=5R

6.

30. ltZ lizslj ……………………….

Sol. P = V = KV3 P = KV4

2

1

P

P =

4

2

1

V

V

2P

10 =

4

400

200

P2 = 160 W

X = 20

31. fp=k esa mÙky ………………………. Sol. lery niZ.k }kjk cuk izfrfcEc niZ.k ds ck;h vksj 20 cm ij gksxk pqfd nksuk fejj dk izfrfcEc ,d gh txg ij gS mry inZ.k ds fy, izfrfcEc dh fLFkfr 15 cm ck;h vksj gksxk

u = – 25 cm v = + 15 cm

mi;ksx ls 1

v + 1

u = 1

f = 2

R

1

15 – 1

25 = 2

R

R = 75 cm. Ans. R = 75 cm.

32. rkEcs dh ,d ……………………….

Sol. R = A

A = 8 cm x 2 mm = 16 x 10–5 m2

R = 5

28

10x16

10x24x10x6.1

= 24 x 10–6 = 24 µ Ans. 24

34. nks xzgksa dh ………………………. Sol. ekuk mHk;fu"B dsUnz ds lkis{k nksuksa rkjksa dh dks.kh; pky gSA

mHk;fu"B dsUnz budk nzO;eku dsUnz gSA






m nzO;eku ds rkjs ij vfHkdsfUnz; cy

m22d

3 =

2

Gm(2m)

d

gy djus ij T =

234

d3Gm

blh izdkj xfrt ÅtkZ dk vuqikr Hkh tM+Ro vk?kw.kksZ ¼fudk; ds nzO;eku dsUnz ds ifjr%½ ds vuqikr ds cjkcj gksxkA

+

2

2m

m

22M

M

2dm

K 32

K d2m

3

1I

21I

2

35. bl izfØ;k ds fy, ……………………….

Sol. W = oØ ls f?kjk {ks=kQy = 3

2 P1V1

2 1

2 1

V 2V

P 2P

vkSj P1V1 = nRT1

vr%

1

w

nRT.

1 1

1 1

3. P V

2

P V

36. ;fn izfØ;k ds fy, ……………………….

Sol. nC T = Q nCT = 6n RT1

dT = 4T1 – T1 = 3T1

n . C . 3T1 = 6nRT1

C

R = 2


37. 0.1 eksy K ;qDr ,d yhVj tyh; ……………….

Sol. HgI2 + 2KI(aq) K2[HgI4] 2K+ + [HgI4]2–

ladqy fuekZ.k ds dkj.k d.kksa dh la[;k esa deh vk jgh gSA

38. fuEu esa ls dkSulk@dkSuls vk;ju ………………. Sol. gsesVkbV – Fe2O3

esXusVkbV – Fe3O4

flMsjkbV – FeCO3

dsykekbu – ZnCO3

39. fuEu esa ls dkSulk@dkSuls dFku ………………. Sol. (B) Lor% vfHkfØ;k Å"ek{ksih gksus dh vko';drk ugha gSA (C) mRØe.kh; :)ks"eh; izØe vkblks,UVªkWfid gksrk gSA (D) de vk;ru esa xSl dh leku ek=kk ds fy, U;wu ,UVªkWih gksrh gSA

40. ikSVsf'k;e lYQsV dh 20g ek=kk dks ………………. Sol. oS|qrvi?kVu ds igys nzO;eku (H2O) = 150 g

oS|qrvi?kVu ds i'pkr~ m (H2O) = 150 – 113.3 = 36.7g

2H O

n = 2.04 mol

pwafd

2H2O 2H2 + O2

2H

n = 2.04 mol V = 45.6L

2O

n = 1.02 mol. V = 22.8L

41.

Cl

P3AlCl ……………….

Sol. ;






42. nh xbZ vfHkfØ;k inksa ds fy, lgh ……………….

Sol.

43. ,d vKkr /kkrq bysDVªksM X ds lkFk ,d ………………. Sol. EºlSy = Eºvip;u (c) – Eºvip;u (a)

Eºvip;u (a) = – 1.66 V

45. fuEu esa ls fdrus izØe esa v;Ld ls /kkrq ……………….

Sol. gsesVkbV Fe

xsysuk Pb

dsflVsjkbV Sn

ftad Cys.M Zn

46. 0.01 eksyy tyh; CH3COOH ds fgekad es ……………….

Sol. ;wfj;k ds fy,, Tf = Kf × m Kf = m

Tf =

1.8

1 = 1.8

vc] CH3COOH ds fy,

Tf = i × Kf × m i = 1.1

vr%, i = 1 + blfy, = 0.1

vc, [H+] = C = 0.01 × 0.1 = 10–3 M

pH = 3

47. vk/kqfud vkoZr lkj.kh esa ekuk, ……………….

Sol. a = 2, b = 5, c = 71

a + b + c = 2 + 5 + 71 = 78.

49. fuEu vfHkfØ;k esa fufeZr ……………….

Sol.

50. C6H2Br4 ds fdrus leko;oh Br2/FeBr3 ds ……………….

Sol. 32 FeBr/Br

32 FeBr/Br

32 FeBr/Br

51. feyk;h x;h flfydk ds lkFk pkYdksikbjkbV ……………….

52. Lor% vip;u esa vipk;d ………………. Sol. (Q.51 & Q.52)

2CuFeS2 + 4O2 Cu2S + 2FeO + 3SO2

Cu2S + FeO + SiO2 FeSiO3 ¼xyuh; /kkrqey½ + Cu2S

¼esVh½

53. fuEu esa ls dkSulh vfHkfØ;k esa fn;k ……………….

54. bysDVªkWuLusgh izfrLFkkiu fØ;kfof/k ………………. Sol. (Q.53 & Q.54)

3AlCl

6 5 3 318 C to 80 C

( )

C H C(CH ) r̀rh;d C;wfVy csUthu

dsoy mRikn






3BF

6 5 3 2 2 360 C( )

C H C(CH ) CH CHdsoy mRikn

r̀rh;d isfUVy csUthu

2 4H SO

6 6 3 6 5 3 33C H CH COH C H C(CH )

r̀rh;d C;wfVy csUthu

2 4H SO

6 6 3 2 6 5 3 32C H CH C CH C H C(CH )

r̀rh;d C;wfVy csUthu

PAPER-2


1. LrEHk–I esa fn;s x;s ……….

Sol. f(x) = 2

1

x 7x 12 vlrr~ fcUnq x = 3, 4 gSA

(A) vc g(x) = 4x – 3 – x2 = 3 x2 – 4x + 6 = 0

dk dksbZ gy ugha gSA

rFkk g(x) = 4x – 3 – x2 = 4

x2 – 4x + 7 = 0 dk dksbZ gy ugh gSA

fog dk dksbZ vlrr~ fcUnq ugha gSA

(B) Qyu f(x) = x2 lHkh txg lrr~ gSA

g(x) = x 1 x 0

x 1 x 0

g(0) = – 1 x 0Lim

g(x) = 1 rFkk

x 0Lim

g(x) = – 1

g(x) , x = 0 ij lrr~ ugha gSA

ysfdu x 0Lim

fog(x) =

x 1Lim

x2 = 1

x 0Lim

fog (x) =

x 1Lim

x2 = 1, fog (0) = f(–1) = 1

fog , x = 0 ij lrr~ gSA

vlrr~ fcUnqvksa dh la[;k 0 gSA

(C) f(x) = 2

1

15x 8x 1 , x = –

1

3, –

1

5 ij vlrr~ gSA

g(x) = 2

1

x 3x 2 , x = 1, 2 ij vlrr~ gSA

1, 2 vlrr~ fcUnq gSA

vr% g(x) = 2

1

x 3x 2 = –

1

3 x2 – 3x + 5 = 0

(dk dksbZ gy ugha gS)

g(x) = 2

1

x 3x 2 =

1

5 x2 – 3x + 7 = 0 (dk dksbZ

gy ugha gS) vr% 2 fcUnqvksa ij vlrr~ gSA

(D) f(x) = 2

1

6x 25x 14 , x = –

2

3, –

7

2 ij vlrr~

gSA

g(x) = 2

1

x x 2 , x = –1, 2 ij vlrr~ gSA

vc g(x) = 2

1

x x 2 = –

2

3 2x2 – 2x – 4 + 3 = 0

2x2 – 2x – 1 = 0

x = 2 4 8

4

=

1 3

2

g(x) = 2

1

x x 2 = –

7

2 7x2 – 7x – 12 = 0

;fn x =7 385

14

vr% 6 vlrr~ fcUnq gSA

2. lwph I dks lwph II ……….

Sol. (P) 0 < 16

x ; 2

6

x1 ; 3

6

x2 ;

46

x3 ; 5

6

x4 ; 5

6

x

f(x) = 0 f(x) = –1 f(x) = –2 f(x) = –3 f(x) = – 4 f(x) = –5

6 elements vo;o

(Q) sin3 = 3sin – 4sin3

4sin3 = 3sin – sin3

E=3sin

1

)34sin(–3

4sin3)32sin(–

3

2sin33sin–sin3

=

)3sin(

1

3sin–3

4sin33sin–

3

2sin3)3sin–sin3(

=

3sin3–3

4sin

3

2sin(sin3

)3sin(

1

=

3sin3–3

cossin2(sin3)3sin(

1

=

3sin3–)sin–(sin3)3sin(

1 = –3

(R) sinx cosx – 3cosx + 4sinx – 12 – 1 > 0

sinx(cosx + 4) – 3(cosx + 4) – 1 > 0

(cosx + 4)(sinx – 3) > 1

lnSo _.kkRed gSA

dksbZ gy ugha

(S) sin2x + sinx cosx = n

2sin2x + sin2x = 2n

1 – cos2x + sin2x = 2n

sin2x – cos2x = 2n – 1

gy ds fo|eku gksus ds fy,

21–n22–






12n212–

2

1

2

1n

2

1

2

1–

n ds iw.kk±d ekuksa dh la[;k 2 gSA

3. ekukfd f(x) =

2x

2x

,x2–3

,3x4–x2

, ……….

Sol. y = x2 – 4x + 3; n 2 ; y = 3 – 2x ; x > 2

y = x2 – 4x + 3; (–, 2] [–1, )

at x = 2; y = –1

f ,dSdh gS rFkk vkPNknd gSA

f(x)

2

3

–1

x2 – 4x + 3 – y = 0

x = 2

)y–3(14–164

x = 2

y444

x = y12 at x = 2, y = –1 ij

x = 0, y = 3 ij _.kkRed fpUg ysus ij

f–1 = 2 – x1 ; x –1 y = 3 – 2x, x > 2

x = 2

y–3, x > 2

f–1 =2

x–3, x < – 1

f(f(1)) = 3 vkSj f(f(2)) = 8

sin–1(sin3) + cos–1(cos8) = ( – 3) + (8 – 2) = 5 –

4. ekuk f(x) =

4x

4x2

2x

,

,

,

3bx6

x–a

1bx2

……….

Sol.

4

x

lim f(g(x)) = a – 4

vkSj –

4x

lim

f(g(x)) = 4b + 1

f(g(x)), x = 4

lrr~ gSA

a – 4 = 4b + 1 a – 4b – 5 = 0 …….(1)

ax + by + 1 = 0, (–1, 3) ls xqtjrh gSA

–a + 3b + 1 = 0 ………(2)

(1) o (2) dks gy djus ij

a = –11, b = – 4

5. ;fn = 2

2

0x x

x4x–2–x2cosbxsinalim

……….

Sol. = 2

223

0x x

x4x–2–.....!2

)x2(–1b.....!3

x–xa

lim

= 2

32

0x x

.......6

a–x)b2–4(x)1–a(x)2–b(

lim

lhek fo|eku gksus ds fy,

b = 2, a = 1 vkSj = 0

6. ;fn 2)y(sin2–)y(sin2 21–41–)x(cos 21– = 1 ……….

Sol. 1)1–)y((sin2 221–)x(cos 21– = 1 ……….(1)

lehdj.k (1) laHko gS ;fn

21– )x(cos2 = 1 vkSj 01–)y(sin 21–

cos–1x = 0 vkSj sin–1y = ±1

x = 1 vkSj y = ±sin1

x – y = 1 + sin1;k x – y = 1 – sin1

7. sindk eku Kkr dhft,A ……….

Sol. sin(4) = 2sin2 cos2 ; = 3

1tan 1–

2 = 2tan–1

3

1= tan–1

4

3

= 2

4

3tancos

4

3tansin 1–1–

= 2 × 25

24

5

4

5

3

24

7tancos

7

1tan2cos 1–1–

= 25

24

8. ,d iafDr esa 15 dqflZ;ka ……….

Sol. Case-I : 2 left to 6 and 1 right to 6 two can sit in the ways (1, 3),

(1, 4), (2, 4) and one who sit right to 6 can sit in 8 ways

total number of ways = 24 Case-II : 1 left to 6 and 2 right to 6 the person left to 6 in 4 ways Two right to 6 can be done in following way x1 + x2 + x3 = 7

x1 1, x2 1, x3 0

Number of integral solution = 21 total ways 21 × 4 = 84 ways Case-III : All 3 right to 6

x1 + x2 + x3 + x4 = 6(x1 1, x2 1, x3 1, x4 1)

coefficient t3 in (1 – t)–4 6C3 = 20

Case-I+ Case-II + Case-III = 128

Total arrangements = 128 × 4! = 384Hence = 8






9. ekuk ,d ik=k esa ………. Sol. og rhu ckj mBkrk gS rFkk 6 vkrk gSA

1, 2, 3 1, 3, 2 2, 3, 1 2, 1, 3 2, 2, 2 3, 1, 2 3, 2, 1

;s lHkh ?kVuka, lelEHkkoh gS rFkk ijLij viothZ gS muesa ls dsoy 7 fLFkfr;ksa es ls ,d

blfy, 1/7 izkf;drk gSA

10. lehdj.k ……….

Sol. {sinx} + {–sinx} =

xsinif1

xsinif0

{sinx} + {–sinx} =

xsin1

xsin0

fn;

fn;

sin x x = 0, 2

, ,

2

3, 2,

2

5,3

11. Find the number of integral values of x in the domain of f(x)

= cos–1[2 – 4x2]. ……….

f(x) = cos–1[2 – 4x2] ds izkUr esa x ds iw.kk±d ekuksa dh la[;k

Kkr dhft,A………. Ans. 0

Sol. –1 [2 – 4x2] 1

–1 2 – 4x2 <2

–3 – 4x2 <0

0 < x2 4

3

x

2

3,

2

3– – {0}

12. Find the number of ………. lehdj.k ………. Ans. 2

Sol. x = 0 and x = 1

13. Let f(x) =

x2cos–x2sin1

x2cosx2sin1cot 1–

……….

ekukfd f(x) =

x2cos–x2sin1

x2cosx2sin1cot 1–

……….

Ans. 5

Sol. f(x) = cot–1(cotx)

5

1x

)x(f = cot–1 = (cot1) + cot–1(cot2) + cot–1(cot3) + cot–

1(cot4) + cot–1(cot5)

= 1 + 2 + 3 + 4 – + 5 – = 15 – 2

14. If

0x|2–x|

0xx–)x(f , ……….

;fn

0x|2–x|

0xx–)x(f , ……….

Ans. 5

Sol.

f(x) = –x 2

–1

1

1 2 3 x

y

lehdj.k f(x) = 1 ds gyksa dh la[;k x = –1, 1, 3 gSA

f(f(x)) = 1 ds fy, ekuk f(x) = p

f(p) = 1 f(p) = –1,1,3

(i) p = – 1 ; f(x) = –1; dksbZ gy ugha

(ii) p = 1 ; f(x) = 1 ; 3 gy

(iii) p = 3 ; f(x) = 3 ; 2 gy

Total 5 dqy 5 gy gSA

15. ekuk f(x), 6 ?kkr dk ,d inh; cgqin ……….

Sol. f(x) = (x2 – 1)(x2 – 4)(x2 – 9) + x2 + 4

2f = 20 and vkSj 3f = 19

2f – 3f = 1

16.

xtan2–x1

x2sin

)1–x(4lim

1–2

1–1x

Sol. xtan2–xtan2–

)1–x(4lim

1–1–1x

=

xtan–4

4

)1–x(4lim

1–1x

apply LHR

=

2

1x

x1

1–4

4lim = –2

17. ekuk f(x) =

3x,}x{–

}x{–sinb

3x,a

3x

23

4–e

3–x

1

3–x

1

……….






Sol.

3–x

1

3–x

1

3–x

1

3x3x

3

21

3

4–3

e

lim)x(flim

= 01

0–0

= 0

)3x(–

)3xsin(–blim)x(flim

–– 3x3x

= b

x = 3 ij Qyu lrr~ gksus ds fy,

f(3+) = f(3) = f(3–) = 0 = a = b

a = b = 0

18. ;fn (a + b)n ds izlkj ……….

Sol. t2 = nC

1 an – 1b = 135

t3 = nC

2 an – 2b2 = 30

t4 = nC

3 an – 3 b3 = 3

10

, rc

nC1 . nC

3 a2n – 4 b4 = 135 × 3

10

2

2n

3n

1n

)C(

C.C

= 2

1

n 3

)2n)(1n(n

= 4

)1n(n 22

3

)2n(

= 4

1n

n = 5

3

2

t

t

= 25

15

C

C

b

a

b

a

= 2

9

× 2.1

45

. 5

1

a = 9b

5a4b = 135 9

a5 5

= 135

a5 = 27 × 9 a = 3

b = 3

1

PART : II PHYSICS

19. 1P rFkk 2P f}/kqo ………………………

Sol. cyk?kw.kZ = P E

= 0 b/c = 0

[fp=k (i) esa] fp=k figure (ii)

= 2

13

2KPP

d =

1 2

3

2KPP

d

fp=k (i) esa 1P ij cy

1

dP

dr

=

1 2

4

6KPP

d

fp=k (ii) esa fØ;k&izfrfØ;k }kjk 1P ij cy

F =1 2

4

3KPP

d

20. i`Foh vkSj mixzg ………………………

Sol. (A) vkorZdky T gS : 2

2 3

earth

4T R

GM

(B) v d{kh; pky 2 earthGM

vR

(C) dqy ÅtkZ = earth sGM m

2R

(D) mixzg ds dsUnz ij xq:Roh; {ks=k dk ifjek.k = earth

2

GM

R

21. nks xksyh; xzg ………………………

Sol. Ves = 2GM

R =

342.G . R

3

R

=

4G

3

R

Ves R

P dk i`"Bh; {ks=kQy = A = 4RP2

Q dk i`"Bh; {ks=kQy = 4A = 4 RQ2

RQ = 2Rp

R dk nzO;eku MR = MP + MQ

3

R

4R

3 =

3

P

4R

3 +

3

Q

4R

3

RR3 = RP

3 + RQ3

= 9RP3

RR = 91/3 RP RR > RQ > RP

blfy, VR > VQ > VP

R

P

V

V = 91/3 rFkk P

Q

V

V=

1

2

22. iznf'kZr fp=k esa ………………………

Sol.

130 cm

23. nks d.k A rFkk ………………………

Sol. xq:Roh; cy ds dkj.k] nksauks d.k leku le; esa leku Å¡pkbZ rd fxjsaxsA vr% fLFkj oS|qr cy dk Å/okZ/kj fn'kk esa ?kVd ugha gksxkA

vr% /kjkry ij igq¡pus esa yxk le; 2h

g , v0 ls LorU=k gksxkA

24. ;fn E ,d leku ………………………

Sol. (A, B, D)

i = ne A Vd i Vd (A)

P =

2V

R=

22E

R P E2 (B)

P = i2R P i2 (D)






25. ,d vkn'kZ f}ijek.kqd ………………………

Sol. ljy js[kk AB dh lehdj.k P = 0

0

0

2PV 5P

V

ljy js[kk AB dk <ky = 0

0

2P

V

tc izØe dks Å"ek'kks"kh ls Å"ek{ksih esa ifjofrZr fd;k tkrk gS rks <ky] :)ks"e P-V oØ ds <ky ds leku gksrk gS vFkkZr~ <ky =

P 7 P

V 5 V

0

0

2P 7P

V 5V

P = 0

0

P10V

7 V

10

7

0

0

PV

V =

00

0

2PV 5P

V

0

35V V

24

0

0

P10P

7 V

35

24 V0 = 0

25P

12

3P0V0 = nRT0

0

25P

12

0

35V

24

= 825

288 P0 V0 = nRT

0

T 875

T 288(3) =

875

864

T = 875

864 T0

26. 0.8 gm/cm3 ?kuRo ………………………

Sol. 0.8 × 5 × S × 15 = 210

4.2

S = 5

6cal/gm-°C

27. R = 2 cm f=kT;k ………………………

Sol. = jdA = EdA =

R

0

kr . 2r dr =

3k2 R

3

= 3

3

2 kR

=

6

38

3 16 10

22 10

100

= 3 × 10–8

28. Hkqtk a = 0.1 m ………………………

Sol. ABCD ls gksdj fo|qr ¶yDl

A B

C D z

G

F E x

y

H

1 =EA

= (2ˆ ˆx i y j ). (

2 â i )

= 0 as x = 0

EFGH ls gksdj fo|qr ¶yDl

2 = (2ˆ ˆx i y j ). (

2 â i )

= x2.a2 = a4 = 1.0 × 10–4 Nm2/C

BCGF ls gksdj fo|qr ¶yDl

3 = (2ˆ ˆx i y j ). (

2â j )

= a3 = 1.0 × 10–3 Nm2/C

EADH ls gksdj fo|qr ¶yDl

4 = (2ˆ ˆx i y j ). (

2â j ) = 0 as y = 0

ABFE ls gksdj fo|qr ¶yDl

5 = (2ˆ ˆx i y j ). (

2 â k ) = 0

CDHG ls gksdj fo|qr ¶yDl

6 = 0

usV ¶yDl = (1.0 × 10–4 + 1.0 × 10–3) N-m2/C = 11 × 10–4 N-

m2/C

29. nks irys le:i ………………………

Sol. 1

43

11

f

2

R

+ 2

43

1

2

R

=1

24

f = 24

30. = 1 × 10–5 -m ………………………

Sol. | J | = 1

|E | sin30°

| J | =106 × 10–4 × 1

2= 100

2A/m2 = 50 A/m2

31. ,d :)ks"e ik=k ………………………

Sol. Å"ekxfrdh ds çFke fu;e ls Q = W + U

0 = fan

f( W ) P v n R T

2

Wfan = n RT + nf

2RT

Wfan = n Cp T

Wfan = (1) f

R R2

(500 k) (pwafd xSl /khjs&/khjs çlkfjr

gksrh gS vr% p = fu;r] vr% T v)

Wfan = 14 kJ






32. ,d ik=k esa 0.014 kg ………………………

Sol. ik=k esa Hkjh x;h xSl, V = fu;r

(Q)v = nCvT

;gk¡, n = (0.014 103)/28 = (1/2) mol

ukbVªkstu f}ijek.kqd xSl gS, Cv = (5/2) R,

iz'ukuqlkj,

rms 2 2

rms 11

V T

V T = 2, i.e., T2 = 4T1

T = 4T1 – T1 = 3T1 = 3 300 = 900 K

(Q)v = 1 5

2 2 2 × 900 = 2250 cal

33. izfrjks/kdrk 3 m – m ………………………

Sol. R = A

10 = A

103 3

A

=

3

104

fo|qr ¶yDl = E.A

AV

= x × 10–3

310

204 = x × 10–3

x = 6

34. nks ladsfUnz; [kks[kys ………………………

Sol. ekus ys fd vkUrfjd dks'k ij vfFkZx ds ckn vkos'k q’ gks tkrk gS] rks blds foHko dks 'kwU; gksus ds fy,

3

q'q0

R3

Kq

r

'Kq

vr% /kjkry esa cgus okyk vkos'k 3

q gSA

35. fuEu fn;s x;s ………………………

Sol. ikLdy ds fu;e ls ge izkIr djrs gSaA

P0 = P0 + AghA – BghB + CghC – DghD

95°C g(52.8 cm) – 5°g (49 cm) + 95°Cg(49 cm) –

5°Cg(51) = 0

5 C

95 C

52.8 49

49 51

= 1.018 ......(i)

vk;ru izlkj xq.kkad

y = 95 C 5 C

5 C

v v

V (95 C 5 C)

=5 C 95 C

95 C(90 C)

y = 5 C

95 C

11

90 C

lehdj.k (i) ls

y = (1.018 – 1) 1

90 =

90

018.0=

90000

18 = 2 × 10–4

js[kh; izlkj xq.kkad =

4y 2 10

3 3

36. fp=kkuqlkj ,d vuar ………………………

Sol. lEiw.kZ csyukdkj lrg (dsk.k = 2) ls ikfjr ¶yDl

30° 30°

aA B

O

= in

0

Q

csyukdkj lrg AB ls ikfjr ¶yDl = nh xbZ lrg ls ikfjr ¶yDl

= in

0

Q

6=

06

n = 6

PART : III CHEMISTRY 40. VkbVsfu;e /kkrq varfj{k izkS|ksfxdh es ……………….

Sol. ºrH = 140.5 kJ

ºrS = – 0.058 kJ

ºrG 0

41. fuEufyf[kr vfHkfØ;kvksa esa ……………….

Sol.

42. fuEufyf[kr ;kSfxdksa (P, Q rFkk S) dk ………………. Sol.

3 2 4

2

HNO /H SO

(NO )

(–OH o/p funsZ'kd gSA)






3 2 4

2

HNO /H SO

(NO )

(–OCH3 izcy lfØ; dkjd gSA)

3 2 4

2

HNO /H SO

(NO )

(izfrLFkkiu vfHkfØ;k lfØ; oy; esa p-fLFkfr ij gksxhA)

43. M(s) + xH+ Mx+ + x

2H2 vfHkfØ;k ……………….

Sol. Elsy = XM|ME +

2H |HE

= X

0

M|ME –

0.059

x log (MX+) +

2H |HE –

0.059

2 log

2H

2

P

[H ]

= XM|ME –

0 . 0

1log (M+x)1/x –

0.059

1log

2

1/2

HP

[H ]

= X

0

M|ME – 0.0591 log

2

1/2x 1/x

H(M ) . P

[H ]

x = 3

44. AgX(Ksp = a × 10–b M2) ds 1 yhVj lar`Ir ………………. Sol. AgX Ag+ + X– 10–6 + S S

27 × 10–6 = 6 (10–6 + S) + 8S + 7 × 10–6 S = 10–6 Ksp = (10–6 + S) S = 2 × 10–12 M2

45. ,d vkn'kZ vjs[kh; cgqijek.oh; xSl ds ,d ………………. Sol. W = –Pext (V2 – V1) = –1 (8 – 2) = – 6 L atm

D;ksafd q = 0 E = W

3(8Pf – 12) = – 6

Pf = 4

5 atm

i

f

T

T =

58

46 2

=

12

10

S = 3 × 300

12 ln

12

10 +

300

12ln

8

2

= 3.34 J/K

46. MX3 fo;kstu dh ek=kk () 0.33 ds ………………. Sol. MX3 M3+ + 3X–

1 – 3 i = 1 + 3 = 2

47. fuEu esa ls fdrus A ds ……………….

Sol. Ads v;Ld

(i) ckWDlkbV A2O3.2H2O

(ii) Øk;ksykbV Na3 AF6

48. ;fn vkorZ lkj.kh ds nh?kZ ………………. Sol. (i) 9F (ii) 17Cl (iii) 2He (iv) 3Li

49. uhps fn;s gq;s ;kSfxdks vFkok vk;uksa es ………………. Sol. H2SO3 Sulfurous acid ; HClO Hypochlorous acid

HClO2 Chlorous acid ; Sn2+ Stannous ion ;

Fe+2 Ferrous ion

50. fuEu es ls fdruh vfHkfØ;kvksa es eq[;………………. Sol. (6), (7)

51. fuEu vfHkfØ;k esa cuus okys ……………….

Sol. HCN

52. uhps fn;s x;s dkcZ/kuvk;uksa es] ……………….

Sol. (i), (iii), (iv), (v), (vi), (vii), (viii)

54. viuk vfUre mÙkj (X) rFkk (Y) ds ………………. Sol. (4 + 3) = 7






ADVANCED PATTERN



ANSWER KEY

PAPER-1


1. (B) 2. (D) 3. (AD) 4. (BC) 5. (BD) 6. (CD) 7. 0

8. 8 9. 8 10. 7 11. 0 12. 2 13. 5 14. 5

15. 1 16. 2 17. 6 18. 1

PART : II PHYSICS

19. (AB) 20. (ABCD) 21. (AB) 22. (ABD) 23. (BCD) 24. (BC) 25. (20)

26. (18) 27. (80) 28. (14) 29. (12) 30. (20) 31. (75) 32. (24)

33. (D) 34. (B) 35. (A) 36. (B)


37. (AD)

38. (ABC) 39. (ABC) 40. (AB) 41. (ABCD) 42. (BCD) 43. (03)

44. (10) 45. (04) 46. (03) 47. (78) 48. (19) 49. (02) 50. (03)

51. (D) 52. (C) 53. (A)

54. (A)

PAPER-2


1. (B) 2. (D) 3. (AD) 4. (BC) 5. (BD) 6. (CD) 7. 0

8. 8 9. 8 10. 7 11. 0 12. 2 13. 5 14. 5

15. 1 16. 2 17. 6 18. 1

PART : II PHYSICS

19. (A) 20. (C) 21. (BD) 22. (ABD) 23. (AB) 24. (ABD) 25. (5)

26. (5) 27. (3) 28. (1) 29. (8) 30. (5) 31. (7) 32. (5)

33. (6) 34. (3) 35. (5) 36. (6)


37. (B) 38. (D) 39. (AC) 40. (ACD) 41. (BCD) 42. (ABD) 43. (3)

44. (2) 45. (3) 46. (2) 47. (2) 48. (9) 49. (5) 50. (2)

51. (2) 52. (7) 53. (5) 54. (7)



Doubtion...Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA,...

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