ENGLISH CLASSROOM CONTACT PROGRAMME (A S : 2018 - …

READ THE INSTRUCTIONS CAREFULLY

GENERAL :

1. This sealed booklet is your Question Paper. Do not break the seal till you are toldto do so.

2. Use the Optical Response sheet (ORS) provided separately for answering the questions.

3. Blank spaces are provided within this booklet for rough work.

4. Write your name, form number and sign in the space provided on the back cover of thisbooklet.

5. After breaking the seal of the booklet, verify that the booklet contains 36 pages andthat all the 18 questions in each subject and along with the options are legible. If not,contact the invigilator for replacement of the booklet.

6. You are allowed to take away the Question Paper at the end of the examination.

OPTICAL RESPONSE SHEET :

7. The ORS will be collected by the invigilator at the end of the examination.

8. Do not tamper with or mutilate the ORS. Do not use the ORS for rough work.

9. Write your name, form number and sign with pen in the space provided for this purposeon the ORS. Do not write any of these details anywhere else on the ORS. Darkenthe appropriate bubble under each digit of your form number.

DARKENING THE BUBBLES ON THE ORS :

10. Use a BLACK BALL POINT PEN to darken the bubbles on the ORS.

11. Darken the bubble COMPLETELY.

12. The correct way of darkening a bubble is as :

13. The ORS is machine-gradable. Ensure that the bubbles are darkened in the correctway.

14. Darken the bubbles ONLY IF you are sure of the answer. There is NO WAY to eraseor "un-darken" a darkened bubble.

15. Take g = 10 m/s2 unless otherwise stated.

Please see the last page of this booklet for rest of the instructions

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)1001CJA102118036)Paper Code

(1001CJA102118036)E

NG

LIS

H

CLASSROOM CONTACT PROGRAMME(Academic Session : 2018 - 2019)

Test Type : UNIT TEST Test Pattern : JEE-Advanced

TEST DATE : 30 - 09 - 2018

PHASE : TOAS & TNASJEE (Main + Advanced) : ENTHUSIAST COURSE

Time : 3 Hours Maximum Marks : 198

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Target : JEE (Main + Advanced) 2019/30-09-2018

1001CJA102118036

Space for Rough Work

SOME USEFUL CONSTANTSAtomic No. : H = 1, B = 5, C = 6, N = 7, O = 8, F = 9, Al = 13, P = 15, S = 16,

Cl = 17, Br = 35, Xe = 54, Ce = 58Atomic masses : H = 1, Li = 7, B = 11, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24,

Al = 27, P = 31, S = 32, Cl = 35.5, Ca=40, Fe = 56, Br = 80, I = 127,Xe = 131, Ba=137, Ce = 140

· Boltzmann constant k = 1.38 × 10–23 JK–1

· Coulomb's law constant pe9

0

1 = 9×104

· Universal gravitational constant G = 6.67259 × 10–11 N–m2 kg–2

· Speed of light in vacuum c = 3 × 108 ms–1

· Stefan–Boltzmann constant s = 5.67 × 10–8 Wm–2–K–4

· Wien's displacement law constant b = 2.89 × 10–3 m–K· Permeability of vacuum µ0 = 4p × 10–7 NA–2

· Permittivity of vacuum Î0 = 20

1cm

· Planck constant h = 6.63 × 10–34 J–s

Enthusiast Course/Phase-TOAS & TNAS/30-09-2018

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BEWARE OF NEGATIVE MARKINGHAVE CONTROL ¾® HAVE PATIENCE ¾® HAVE CONFIDENCE Þ 100% SUCCESS

PART-1 : PHYSICSSECTION–I(i) : (Maximum Marks : 32)

� This section contains EIGHT questions.� Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of

these four option(s) is (are) correct option(s).� For each question, choose the correct option(s) to answer the question.� 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,

both of which are correct options.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 : If first, third and fourth are the ONLY three correct options for a question withsecond option being an incorrect option; selecting only all the three correct options will resultin +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options),without selecting any incorrect option (second option in this case), will result in +2 marks.Selecting only one of the three correct options (either first or third or fourth option), withoutselecting any incorrect option (second option in this case), will result in +1 marks. Selectingany incorrect option(s) (second option in this case), with or without selection of any correctoption(s) will result in –2 marks.

1. A current source carrying current i = i0sinwt is connected to primary of an ideal transformer.The primary has a self inductance 4L and secondary has self inductance L. There is no fluxleakage. The resistance of primary is zero but resistance of secondary circuit is R as shown.(Take: wL = R) :-

~R

4L L

(A) The current through secondary is 02i cos t4pæ öw -ç ÷

è ø.

(B) The voltage drop across resistance is 02 i R cos t4pæ öw -ç ÷

è ø.

(C) The emf across primary is 02i R sin t4pæ öw -ç ÷

è ø.

(D) The emf across primary is 2i0R cos t4pæ öw -ç ÷

è øSpace for Rough Work

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2. The intensity versus 1l

graph for x rays is shown here. Two lines of characteristic x-rayspectra (ka & kb) are also shown.

I

4.34 5.56 (Å )–18

(A) The accelerating potential difference across the x-ray tube is 99.2 kV(B) The atomic member of target is Z = 74.

(C) 1l

= 5.56 (Å)–1 corresponds to Ka x ray & ( ) 11 4.34 Å-

=l

corresponds to Kb x ray

(D) 1l

= 4.34 (Å)–1 corresponds to Ka x ray & 1l

= 5.56 (Å)–1 corresponds to Kb x ray.

3. Assume that an electron is moving in a circular path due to magnetic field. Since it is a closedloop, the electron must form a standing wave. The magnetic field is B, charge on the electronis q :-

(A) The minimum flux of the external magnetic field can be hq

(B) The minimum flux of the external magnetic field can be h2q .

(C) At the minimum radius, the magnetic field due to moving electron can nullify the externalmagnetic field at the centre of the circle.

(D) The average current due to movement of electron is quantized such that i = ni0.(i0 is average current for circle of smallest radius).



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4. The cross section of a prism is a regular hexagon. A narrow beam of light strikes a face of theprism just below the midpoint M of edge AB. The beam is parallel to the top and bottom facesof the prism. Find the value(s) of refractive index of the prism for which emergent beam willbe parallel to the incident beam :-

A F

EM

B

C D

(A) 3 (B) 1.5 (C) 2 (D) 2.55. In a tube designed for photoelectric effect, the potential difference across the tube is twice

the stopping potential :-

L

(A) No electrons are emitted.

(B) Electrons are emitted but they travel a maximum of L2 before turning back.

(C) If the distance is reduced to L2 keeping the potential difference same, the electrons will

be captured by the collector plate.(D) If the polarity of cells is reversed, all the emitted electrons will be captured by the collector

plate.

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6. A conducting disc of radius a is at the centre of a fixed flat circular coil of radiusb (b >> a) and n turns. Sliding contacts are made at the centre and edge of the disc andconnected to ends of outer coil. Total resistance of circuit is R. There is a certain minimumangular velocity w0 of disc such that if current somehow flows will continue to flow incircuit :-

ab

n

w0

(A) 0 20

4bRna

w =m

(B) w0 = 20

2aRnbm

(C) If w < w0, current decrease to zero(D) If R = constant & w > w0, current goes on increasing


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7. In a large sample of radioactive nuclei we choose 4 active nuclei :-

(A) The probability that in 1 half life exactly two nuclei would have decayed is 38 .

(B) The probability that in 1 half life 2 or more than 2 nuclei would have decayed is 1116 .

(C) The probability that in 1 half life no nuclei would have decayed is 1

16 .

(D) The probability that in 1 half life exactly 1 nuclei would have decayed is 14

8. A converging and a diverging lens in contact produce an image of a distant object on a screen30 cm away. If the diverging lens is moved 5 cm towards the screen, the screen has to bemoved 15 cm to keep the image sharp. Choose correct option(s) :-(A) The screen has to be moved away from lens.(B) The screen has to be moved towards the lens.(C) Focal length of converging lens is 10 cm.(D) Focal length of diverging lens is – 15 cm.


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SECTION–I(ii) : (Maximum Marks : 18)� This section contains THREE paragraphs.� Based on each paragraph, there are TWO questions.� Each question has FOUR options (A), (B), (C) and (D) ONLY ONE of these four options

is correct.� For each question, darken the bubble corresponding to the correct option in the ORS.� For each question, marks will be awarded in one of the following categories :

Full Marks : +3 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 If none of the bubbles is darkened.Negative Marks : –1 In all other cases.

Paragraph for Question No. 9 and 10It is known that accelerating charges radiate electromagnetic waves. This is the method ofpower being radiated by a cellphone. A cellphone has an antenna in which we can assumethat charges undergo oscillation with an amplitude x0 and angular frequency w. This averagepower can be assumed to depend on charge Q, x0, w, t0 and velocity of light c.

9. The formula for power P is (a is dimensionless constant) :-

(A) 2 4 2

02

0

aq xc

wÎ (B)

2 4 20

30

aq xc

wÎ (C)

4 20

2 60

aq xc

wÎ (D)

4 4 20

2 40

aq xc

wÎ

10. A cell phone radiates 3W power at 300 MHz. If a = 13 , find the number of photons being

emitted per unit time ?(A) 1.5 × 1025 /sec (B) 8 × 1023/sec (C) 4 × 1019 /sec (D) Information insufficient



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Paragraph for Question No. 11 and 12Spectroscopic analysis is one way to find the energy levels of atom. Another way is findingenergy lost by inelastically scattered particles. The figure shows the arrangement formeasuring energy levels of H atom using inelastically scattered electrons. A beam of electronall with the same energy is fired through a container of H gas. The electrons scattered atsame convenient angle are sent through a magnetic field which bends them in circular path.A photographic film records the number of electrons striking and their radii in magneticfield this tells us the energy lost in collision.

Photographic plateH gas

electronbeam

11. The graph of number of electrons versus energy loss looks like :-

(A)

No. ofparticles

O 10.2 12.1 13.6 (eV)energy loss

(B) 0 0.54 0.85 1.51 (eV)

energy loss

(C)0 1.89 3.4

energy loss(eV)

(D) 0 13.6 (eV)

energy loss12. Consider an electron scattered at right angle to beam direction. If the maximum radius of

electron in the magnetic field is 1 cm, what is the smallest radius ? The accelerating voltageis 20 V :-

(A) 2 2 cm5

(B) 0.7 cm (C) 7.9cm20 (D) 17 cm

5Space for Rough Work

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Paragraph for Question No. 13 and 14

23192 U decays to 231

91 Pa . The proposed mode is b+ decay or K shell capture. The masses are

MU = 231.03621 amu, MPa = 231.03588 amu. mass of electron = 0.51 MeV/C2, 1 amu = 931 MeV/C2.Neglect the mass of neutrino emitted.

13. 231U can decay by :-

(A) K shell capture & b+ decay both. (B) K shell capture but not b+ decay

(C) b+ decay but not K shell capture (D) Neither b+ decay nor K shell capture.

14. Assuming neutrino to be like a photon, what is the recoil velocity of 231Pa atom in K capture.1 amu = 1.6 × 10–27 kg :-

(A) 443.3 m/s (B) 4.43 m/s (C) 44.3 m/s (D) 4433.3 m/sSpace for Rough Work


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SECTION–II : (Maximum Marks : 16)MATCHING LIST TYPE QUESTION (ANSWER AS FOUR DIGIT INTEGER)

� This section contains FOUR questions.� The answer to each question is a FOUR DIGIT INTEGER ranging from 1111 to 4444, both

inclusive� This section contains four questions, each having two matching lists. Choices for the correct

combination of elements form List-I (P), (Q), (R), (S) and List-II (1), (2), (3), (4)� Given matching type questions in which column-I (PQRS) match with column-II (1, 2, 3, 4)

and represent your answer as Four digit integer for example correct match forP ® 2; Q ® 3; R ® 1, S ® 4. Then you have fill the answer in section-II as 2314.Each matching in column-II may be/may not be correct/can be used more than onetimes for matching in column-I. For example your answer can be 2233, 1333, 1433, 4321 etc...

� For each question, marks will be awarded in one of the following categories :Full Marks : +4 If only the bubble corresponding to the correct answer is darkened.Zero Marks : 0 In all other cases.

1. A conducting wire of linear mass density 10 gm/m and tension of 100 N is being considered. Inlist-I, the wire is subjected to certain wave pattern. The whole wire is in a magnetic field of0.5 T in a direction perpendicular to plane of motion of string. In list-II, the net emf acrosscertain section at certain time is recorded. Match them. (x in m, t in sec)

List–I List–II

(P) y = 2cm sin ( )2 x vt50

pæ ö-ç ÷è ø

emf between (1) 0 V

x = 0 and x = 25 m at t = 0 sec.

(Q) y = 1 cm sin ( )2 x vt100

pæ ö+ç ÷è ø

emf between (2) 0.5 V

x = 0 and x = 100 m at t = 0 sec.

(R) y = 2 cm sin ( )2 x sin 2 t100

pæ ö pç ÷è ø

emf between (3) 1 V

x = 0 and x = 100 m at t = 1 sec.

(S) y = 1 cm cos ( )2 x sin 2 t100

pæ ö pç ÷è ø

emf between (4) 2 V

x = 0 and x = 50 m at t = 0 sec


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2. In Young's double slit experiment set-up with light of wavelength l = 6000 Å, distance betweenthe two slits is 2 mm and distance between the plane of slits and the screen is 2m.y1 = 0.075 mm, y2 = 0.15 mm, y3 = 0.2 mm and y4 = 0.36 mm. I0 is the intensity of light fromeach slit. Then match the intensity at different points.

2m

ab

c

de

2mm

S2

S1y1

y3

y2

y4

List-I List-II

(P) b (1) 2I0

(Q) c (2) 0.38I0

(R) d (3) 3.414I0

(S) e (4) I0



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3. In the list-I, we have certain optical system and object. Comment on the nature of imageformed :-

List–I List–II

(P)

f = 10cm

15cm

(1) Magnified and virtual image.

(Q)

f = –10cm

15cm

(2) Diminished and virtual image

(R)

R = 20 cm

15cm

µ = 1.5

air

(3) Magnified and real image

(S)

15 cm

µ = 1.5air

R = 20 cm

(4) Diminished and real image.


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4. In a discharge tube filled with hydrogen, electrons strike the hydrogen atoms. Due to this,the electron in the hydrogen atom goes to 4th excited state. Match the correct order ofmagnitude.

List–I List–II

(P) The wavelength of electron in excited state. (1) 10–7 m

(Q) The maximum wavelength of the striking (2) 10–12 m

electron

(R) The possible wavelength of the emitted (3) 10–10m

photon while hydrogen atom desxcites

(S) The radius of electron orbit in excited state. (4) 10–9 m


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PART-2 : CHEMISTRYSECTION–I(i) : (Maximum Marks : 32)







1. The following graph is experimentally obtained for the reaction A ¾® 2B at 25ºC

–1.4

–0.7

0.0 10 17 24 (minute)

(A)M

ln [A]

The correct statement (s) for the reaction is/are : (Given ln 2 = 0.7)(A) Time for 87.5% reaction of A is 21 min.(B) The initial concentration of A was e M.(C) The time at which concentration of A and B become equal is 7 minute.

(D) The rate of appearance of B is d[B]dt = (0.2 min–1) [A]


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2. 1 M HCl solution was slowly added in to one litre of 0.37 kg of impure lime [Ca(OH)2] solutionand the following plot was obtained.

1 2 3 4

1

2

3

4

Volume of HCladded (litre)

[OH ](Mol/lt)

–

Now select the correct statements among the following :-(A) Lime sample is 50% pure(B) It 2 litres of 4M CH3COOH (Ka = 10–5) is mixed with above lime sample solution

(freshly prepared), pH of solution is 5 at 25ºC(C) Above lime sample contains 1.204 × 1024 calcium atoms.(D) Lime sample is 40% pure.

3. Select reaction in which chemical composition of product precipitate is changed when exposedto air for some time :-(A) BaCl2(aq.) + Na2SO3 (aq.) ® ppt - 1

(B) FeSO4 (aq.) + NaOH (aq.) ® ppt - 2

(C) BaS + ZnSO4 (aq.) ® ppt - 3

(D) MnSO4 (aq.) + NaOH (aq.) ® ppt - 4



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4. Which of the following reaction proceed by decrease in oxidation number of underlineNitrogen.

(A) 2 3KNO KI CH COOH+ + ¾¾®

(B) 2NO H- ++ + NH –C2 –NH2

O(C) 2 4 2 4NO MnO dil. H SO- -+ + ¾¾®

(D) 2NaNO Metallic zinc Castic soda+ + ¾¾®

5. The purpose of addition of NH4Cl solution in group-V radical is(A) to decrease concentration of 2

3CO - ion by acid-base reaction

(B) to decrease concentration of 23CO - ion by common-ion effect

(C) to increase concentration of 23CO - ion by common-ion effect

(D) to increase concentration of 23CO - ion by acid-base reaction

6. 4KMnO / H¾¾¾¾¾¾®+

(A) Δ¾¾® (B) 4LiAlH¾¾¾¾® (C) H – C – HH

O

(D)

For the above reaction sequence correct statement(s) is/are :

(A) Compound D is CH2O

O

(B) Compound C gives red colour victor meyer test(C) Compound A gives brisk effervescence on treatment with NaHCO3

(D) Compound D is a cyclic acetal

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7. Which of the following compounds show faster reaction with (CH3)2NH as compared to

Br

NO2

?

(A)

F

NO2

(B)

Cl

NO2

(C)

Br

NO2

NO2(D)

I

NO2

8. Which of the following compounds can show Fehling solution test after treatment with aqueousKOH?

(A)

O

CH3

C – H(B)

H HO

HOH

HOH

OHO – C– CH3

O

O

(C) H

ClCl(D)

HO OPh


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Paragraph for Questions 9 and 10The pH of blood is 7.4. pH of blood should be maintained constant, otherwise it may causeillness. The buffer in the blood is formed by CO2 and 3HCO- ion. The reaction available in theblood is a reversible reaction

CO2 + H2O � H+ + 3HCO- K = 4.5 × 10–7

[At constant body temperature 37ºC]9. Calculate the ratio of conjugate base to acid necessary to maintain blood at its proper pH

(A) 4.5 (B) 3.75 (C) 11.25 (D) 1410. The pH of blood streams is maintained by a proper balance of H2CO3 and NaHCO3

concentration. What volume of 5 M NaHCO3 solution should be mixed with 10 mL sampleof blood which is 2M is H2CO3 in order to maintain a pH of 7.4.[Antilog of 1.0532 = 11.31 and log 4.5 = 0.6532](A) 0.07836 Litre (B) 0.04524 Litre (C) 0.07836 mL (D) 45.24 Litre


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Paragraph for Questions 11 and 12MCl3.6H2O is exist in different solvated isomers. 0.1 M, 1 litre solution of its hydrate isomer(X)when react with AgNO3(Excess) it produce 14.35 gm of white ppt.Crystals of (X) when added in dil. NaOH produce coloured ppt (Y). Bright yellow residueis obtain on fusion of yellow ppt (Y) with Na2CO3 & Air.(given that atomic weight of O = 16, H = 1, Cl = 35.5, Ag = 108)

11. Which is CORRECT option.(A) (X) is paramagnetic, inner orbital octahedral complex(B) (X) is diamagnetic, inner orbital octahedral complex(C) None of hydrate isomer of (X) can show geometrical isomer(D) All hydrate isomers of MCl3.6H2O have different water molecules but same Cl– inside

the coordination sphear12. Oxidation number of metal(M) is change when :-

(A) Crystals of (X) react with AgNO3(Aq.)

(B) ppt (Y) react with conc. NaOH(C) Bright yellow residue react with conc. H2SO4 and acetic acid(D) Bright yellow residue react with mixture of dil. H2SO4 and ethanol



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Paragraph for Questions 13 and 14The R1 and R2 for five peptides are listed in the table given below.

NH – CH – CO – NH – CH – CO – NH – CH – COO 3

R1 R2H

+

Peptide R1 R2

(I) –H –CH2CONH2

(II) –CH3 –CH3

(III) –CONH2 –CH2NH2

(IV) –COOH –COOH(V) –COOH –CONH2

13. How many of these peptides are positively charged at pH = 7:(A) 1 (B) 2 (C) 3 (D) 4

14. How many of these peptides are negatively charged at pH = 7 :(A) 1 (B) 2 (C) 3 (D) 4


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1. List-I List-II

(P) Electron moving in second orbit of He+ ion (1) Radius of orbit in which electron

is moving is 0.529 Aº.

(Q) Electron moving in third orbit of H-atom. (2) Total energy of electron is

– 13.6 × 9 eV

(R) Electron moving in first orbit of Li2+ ion. (3) Velocity of electron is62.18 10

3´ m/sec.

(S) Electron moving in second orbit of Be3+ ion. (4) de-Broglie wavelength of

electron is 15013.6 Aº



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2. One mole of N2(g) is taken in 1 litre empty container fitted with movable piston at 300 K. Ifit is heated to 1200 K at constant pressure then match the following parameters

List-I List-II

(P) Z1 (1) 1/8 times

(Q) Z11 (2) 2 times

(R) l (3) 1/2 times

(S) urms (4) 4 times


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3. Consider given reaction sequence given in list-I and identify colour of final product inList-II

List-(I) List-(II)(REACTION SEQUENCE) (COLOURED PRODUCT)

(P) Hg + dil HNO3 ¾® Aq. solution + gas (1) Dark metallic pptAq. solution + Na2CO3 ® coloured product

(Q) Fe + dil H2SO4 ¾® Aq. solution + gas (2) Blue colouration

Aq. solution 2 2

3 6

( i ) H O / H ,( ii ) K [Fe(CN ) ]

+ D¾¾¾¾¾¾® coloured product

(R) Bi + conc. HNO3 ¾® Aq. solution–1 + gas (3) Yellow oxo salt

SnCl2.2H2O Aq.(excess) KOH¾¾¾¾¾¾® Aq. solution–2

Aq. solution–1 + Aq. solution–2 ® coloured product

(S) R2CHOH 2Red P Cl- +¾¾¾¾¾® (X) 2

2

( i)NaNO HCl( ii ) HNO

+¾¾¾¾¾¾® Aq. solution (4) Red-Brown complex

Aq. solution + NaOH ¾¾® coloured productSpace for Rough Work


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4. List-I List-IIReactions Most appropriate reagent

for conversion

(P) C – H

O

HO HO

(1) NH2 – NH2 / KOH / D

(Q)

O

N+ N+(2) Zn/Hg/HCl

(R) O O

O

O O(3) NH2 – NH2 / H2O2

(S) Ph – CH = CH – CH3

O

Ph – CH = CH – C – H (4) (i) SeO2 / (ii) MnO2


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PART-3 : MATHEMATICSSECTION–I(i) : (Maximum Marks : 32)







1. Let ( )/ 2

13 11

0

I cosx dxp

+= ò and ( )

/ 213 1

20

I cosx dxp

-= ò , then value of 2

1

II

é ùê úë û

is equal to (where [.]

greatest integer function)(A) 0 (B) 1 (C) 2 (D) greater than 2

2. Let 1 x

1 xg(x) t f '(t) dt

+

-= ò where f(x) does not behave like a constant function in any interval

(a, b) and the graph of y = f '(x) is symmetrical about the line x = 1 then

(A) g(x) is increasing " x Î R (B) g(x) is increasing only for x > 1

(C) g'(x) is an even function (D) g'(x) is symmetrical about the line x = 2



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3. If ( ) ( )( )xe e e xsin x log x cot x log log x log e dx+ +ò = f(x) + C ; and f(1) = 0, (where C is the integration

constant) then the value of f(e) is less than

(A) 12

(B) –1 (C) 3

2-

(D) –2

4. If = Îòl

1 n

n0

( nx)I dx,n Nx

, then

(A) In has finite value for all natural values of 'n'.

(B) + + = -32 4

1 2 3

II I 4I I I

(C) I4 = –9(D) I8 = 40331

5. Let ƒ : R ® R be a differentiable function which satisfies ( ) ( )= - " Îòx

0

ƒ x x ƒ x dx x R , then-

(A) ƒ(x) is a monotonic function (B) ƒ(x) is an onto function

(C) ƒ(x) is an odd function (D) y = x is tangent to ƒ(x)


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6. The value of ( )21

4 1cos 1 cosn

n k

n k klt Cn n a®¥ =

-æ ö = -ç ÷è ø

å , then

(A) a + c = 20 (B) 4ac

= (C) a – c = 12 (D) ac = 16

7. Consider two events while throwing a pair of diceA : Getting a sum which is a prime numberB : Getting a sum less than or equal to 6, then events defined are(A) Mutually exclusive(B) Exhaustive(C) Equally likely(D) Mutually exclusive, equally likely & exhaustive

8. If x

2 t

0

ƒ(x) 3x e ƒ(x t)dt-= + -ò , then

(A) ƒ(x) = 0 has 3 solutions (B) ƒ(x) is monotonic increasing

(C) ƒ(x) = 4 has 2 solutions (D) 1

1

ƒ(x)dx 2-

=ò


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Paragraph for Questions 9 and 10Let A,B,C be three sets of complex numbers as defined below.

{ } ( )2z zA : z :| z | 5 , B z : 2 z z

2iì ü-ï ïæ ö£ = ³ +ç ÷í ýè øï ïî þ

,C = {z : Im(z) < 2}.

On the basis of above information, answer the following questions :9. The number of complex number (whose both real part and imaginary part are integers) in the

set AÇBÇC is-(A) 12 (B) 13 (C) 14 (D) 15

10. The least positive argument of the complex number in the set A Ç B ÇC whose both real part

& imaginary parts are integers is 1 atanb

- æ öç ÷è ø

(where, a & b relatively prime numbers),

then (a + b) is -(A) 1 (B) 2 (C) 3 (D) 4

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Paragraph for Questions 11 and 12Given planes : P1 : 4x + 3y + z – 8 = 0

P2 : x + 4y – 3z + 6 = 0P3 : Acute angle bisector of P1 and P2.

P1 ® P1' [after 90° rotation about line of intersection of P1 and P2 such that it crosses P2]

On the basis of above information, answer the following questions :11. If equation of acute angle bisector of P1' and P3 is ax + by + cz – 2 = 0, then value of a + b + c is -

(A) 13 (B) 10 (C) 8 (D) 1412. The distance of plane from origin which is perpendicular to P3 and passing through

intersecting line of P1 and P2.

(A) 1426 (B)

278 (C)

226 (D)

1450

Paragraph for Questions 13 and 14A curve passing through origin is such that slope of tangent at any point is reciprocal of sumof coordinates of point of contact.

13. Slope of tangent at point with ordinate ln3 is

(A) 1 (B) 13 (C)

12 (D) – 2

14. If I =[ ]a + a

-ò

sin cos

1

x dy (where [.] denotes the greatest integer function and a Î R), then I is

(A) 1 1

ee 3

- - (B) 1e 2

e- - (C) 1

ee

- (D) 1 1e

e 3+ +


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1. List-I List-II

(P) If the line x 4 y 2 z 2k1 1 2- - -

= = lies in the plane 2x – 4y + z = 3, (1) 92

then k is equal to

(Q) If the lines x 1 y 1 z 12 3 4- + -

= = and x 3 y k z1 2 1- -

= = intersect, (2) 32

then value of (k – 1) is

(R) A plane passes through the point (1, 1, 1) cuts the coordinate axes at (3) 52

A, B & C respectively such that OA = OB = OC where O is the origin,then volume of tetrahedron OABC is equal to

(S) A plane P passes through (1, –2, 1) is perpendicular to two planes (4) 72

2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane P from

the point 5 3

1, ,2 2

æ ö-ç ÷

è ø is

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2. Let 'g' be a derivable function on [–7, 5] such that g(0) = 5. The graph of y = g'(x) is shownbelow which consists of three line segments and a semicircle centred at origin andradius 2.

1 2 3 4

5

–6 –3 –2 –1–1

(3,3)y=g'(x)

2

(5,–1)

List-I List-II

(P) g(–7) (1) p + 5

(Q) g(2) (2)32

(R) g(3) – g(2) (3)152

- p

(S)9g g(2)2

æ ö -ç ÷è ø

(4)154


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3. Let ( )p pæ ö æ ö= +ç ÷ ç ÷

è ø è øn,k2k 2kz cos i sinn n , n Î N

[Note : where = -i 1 ]List-I List-II

(P) ( )10

10,kk 1

2 z=

å (1) –1

(Q) ( )=

- å10

11,kk 1

2 z (2) 0

(R) ( )10

10,kk 1

z=

Õ (3) 1

(S) ( )10

11,kk 1

z=

Õ (4) 2

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4. List-I List-II(P) If the angle between the tangents from (1, 3) to the (1) 0

curve y2 = 4a(x – 1) is 56p

, then a2/3 is

(Q)x 2 3

2 3x 0

log(e 3x 5x 9x )2 limlog(cos x x 3x 6x )®

é ù+ - +ê ú- + -ë û

is (2) 8

(where [.] denotes greatest integer function)

(R) If x

x x

(1 xe )dxh(x)xe e x 1

-=

+ + +ò (3) 4

& h(0) = 1n2

æ öç ÷è ø

l , then h(x )

xlim e

®¥ is

(S) Sum of intercepts made by any tangent of (4) 32y ( x 2) 2 x(2 x y)= + - + +

on coordinate axes is


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QUESTION PAPER FORMAT AND MARKING SCHEME :16. The question paper has three parts : Physics, Chemistry and Mathematics.17. Each part has two sections as detailed in the following table.

I have read all the instructionsand shall abide by them.

____________________________Signature of the Candidate

I have verified the identity, name and Formnumber of the candidate, and that questionpaper and ORS codes are the same.

____________________________Signature of the Invigilator

NAME OF THE CANDIDATE ................................................................................................

FORM NO. .............................................

Que. No. Category-wise Marks for Each Question MaximumSection Type of Full Partial Zero Negative Marks of the

Que. Marks Marks Marks Marks section+4 +1 0 –2

One or more If only the bubble(s) For darkening a bubble If none In all I(i) correct 8 corresponding corresponding to each of the other 32option(s) to all the correct correct option, provided bubbles is cases

option(s) is(are) NO incorrect option darkeneddarkened darkened

Paragraph +3 0 –1Based If only the bubble If none In all

I(ii) (Single 6 corresponding to — of the other 18correct the correct option bubbles is casesoption) is darkened darkened

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II ( Matching 4 corresponding to — other — 16Lists Type) the correct option cases

is darkened

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