xiii rt 6 pcm-16!02!2013 advance paper-2 solution

8/13/2019 XIII RT 6 PCM-16!02!2013 Advance Paper-2 Solution

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XIII (XYZ) RT-6 JEE Advanced [Paper-II] Code-A Page # 1

PHYSICS

PAPER

PART A

[SINGLE CORRECT CHOICE TYPE]

Q.1 to Q.6 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

[6 3 Marks = 18]Q.1 A particle of mass m moving along a straight line is acted on by a retarding force (always directed against

the motion) F = be ! v, where b and ! are positive constants and v is the velocity. At t = 0 it is movingwith velocity v 0. At what time does it come to rest ?

(A) bm

" #0ve1 !$$ (B*) " #1ebm

ov $! (C)bm 0ve! (D)

bm

0ve !$

[Sol. mdt

dv = be+! v

% && $'$!$0

0

t

0

0

v

v dtmb

)v(de

1e 0v $! = mbt

]

Q.2 A pebble is stuck to the edge of the uniform disc of radius R. The masses of both pebble and the disc issame. Initially the pebble is at the topmost point. What is the maximum speed of the centre of disc during

the motion? (here disc does not slip on floor)

C

P

(A)3gR4

(B)3gR2

(C*)3

Rg8(D)

3gR16

[Sol. mg (2R) = 21

23

mR 2( 2

( =R3g8

]

Q.3 The figure shows the elliptical orbit of a planet with the Sun at the focus. The areas of regions 1 and 2 arethe same. If the planet takes times t 1 to go from one end to the other end of region 1 and t 2 to go fromone end to the other end of region 2, then what is the relationship between t 1 and t 2 ?

1 2

(A) t 1 < t 2 (B*) t 1 = t 2 (C) t 1 > t2 (D) Cannot be determined

XIII(XYZ) RT-6 JEE ADV. PAPER 2 DA TE : 16-02-2014


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PHYSICS

Sol.

T

FE

mg

q

,sin+ FE =m(2

mgtan+ 0q4

1- m)sin(

q2

2

+ =(2

+secg +- 330

2

sinmq4

q = ( 2 ]

[PARAGRAPH TYPE]


[2 3 Marks = 6]Paragraph for question nos. 7 to 8

For a given optical system, the principal axis is x-axis and coordinates of object are ( 30, +1, 0) andcoordinates of image are (+20, 2, 0). All coordinates are in cm.

Q.7 If the optical system is a concave mirror, it is located (x-coordinate) at what point ? [3]

(A) origin (B*) 80 cm (C)3

40$ cm (D)3

40cm

Q.8 If the optical system is a convex lens, what is its focal length ? [3]

(A) 12 cm (B) 10 cm (C)3

100cm (D*)

9100

cm


A mass m 1 is attached to the end of a light, rigid rod of length R. The rod is pivoted about its other end,at a point height R above the surface of a horizontal, frictionless table. m 1 is initially held vertically abovethe pivot point. A second mass m 2, of mass equal to m 1 is at rest on the table, directly below m 1, m 1 isreleased from rest and swings down, counterclockwise as shown, making an elastic collision with m 2.

R

m 1

m2

h


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PHYSICS

Column-I Column-II(A) At P such that l = 0.1 mm (P) Intensity is maximum for 01 = 4000 (B) At P such that l = 0.2 mm (Q) Intensity is minimum for 01 = 4000 (C) At P such that l = 0.4 mm (R) Intensity is maximum for 02 = 8000 (D) At P such that l = 0.6 mm (S) Intensity is minimum for 02 = 8000

(T) Intensity is maximum

[Ans. (A) Q (B) PS (C) PRT (D) PS ]PART C

[INTEGER TYPE]

Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are Single digits )

[4 4 Marks = 16]Q.1 In the capillary tube of radius R = 0.5 mm a liquid rose to a height h = 11 mm. Estimate the density in (10 3

kg/m 3) of the liquid, if its surface tension is 0.22 N/m. Take contact angle to be 60 .

[Ans. 4]

[Sol. h = grm2 00

1-

1 = 332

101110105.02

110222$$

$

22222222

1 = 4 103 ]

Q.2 Conductive rod OA rotates about the point O in the plane perpendicular to the magnetic field B = 1 T,with an angular velocity ( = 300 rad/s. The free end of the rod slides along a circular arc of radiusR = 0.1 m (Fig). The arc between point C and point mounting rod included battery with EMF ) and theinternal resistance R, the direction of rotation rod and the direction of the magnetic induction are shown.Resistance of rod, arc and contact is not taken into account. Determine the voltage V (in Volts) at theterminals of the battery. (Write '2V' in OMR Sheet)

B1A+

I2 O

*3 rC

R

5

[Ans. 3 ][Ans. 3 ]

[Sol.21

B( R2 ]

Q.3 A pendulum consists of a disk of mass M and radius R and a massless rod of length l. Find approximatetime period (in sec) of the system if the disk is mounted to the rod by a frictionless bearing so that it is


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PHYSICS

perfectly free to spin? Assume small amplitude. (Take : = 0.993 m, g = 9.8 m/s 2)

R

M

0

[Ans. 2]

[Sol. g2T -' ]

For Y 1 , Y 2 & Y 3

Q.3 ,d vkn'kZ xSl pkj "ek xfrdh f;kvks a ds }kjk pf; ize dks iw jk djrk gSA bu lHkh izeks a ls lEcaf/kr "ek,s ae'k%Q1 = 5960 J, Q 2 = - 5585 J, Q 3 = - 2980 J vkS j Q 4 = 3645 J gSA rFkk xSl }kjk fd;k x;k dk;Z W 1 = 2200 J, W 2 =

825 J, W 3 = 1100 J vkSjW 4 gSA rks ph; ize dh n{krk dk izfr'kr yxHkx eku

1 X izkIr gks rk gSA rks X dk eku fudVLFk iw .kkZd ds :i es a OMR Sheet es a Hkfj;sA[Ans : 1]

Q.4 The concave side of a thin plano-concave lens with a radius of curvature of 50 cm is silver-plated toobtain a convex mirror. An object is located at a distance of 100 cm in front of this mirror. If focal lengthof equivalent mirror is 50/x cm. Fill the value of x in OMR. [Ans. 3]

[Sol. )15.1(f 1 $' 6 7 89: ; $ 50

1 =100

1$

f 1

= f 2

+mf 1

= 6 7 89

: ; $$

1001

2 +502

1m

0

< =1.5

R=50cm=

10042 .

=503

F =3

50 ]


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CHEMISTRY

PAPER

PART A



[6 3 Marks = 18]

Q.1 For the given conversion :

Me C OH

O

Me C H

O

Suitable options are :(A) Reduction by LiAlH 4(B*) Reaction with SOCl 2 followed by H 2 / Pd-BaSO 4(C) Catalytic reduction by H 2 / Ni(D) Reaction with PCl 5 followed by H 2O

Q.2 Which of the following complex will give white ppt. with BaCl 2(aq)?(A) [Co(NH 3)5SO 4]NO 2 (B) [Cr(NH 3)5SO 4]Cl(C*) [Cr(NH 3)5Cl]SO 4 (D) Both (B) & (C)

Q.3 An element crystallizes in FCC with edge length equal to 1600 pm. Calculate maximum radius of anatom which can fill the tetrahedral void without distorting the lattice.

(A) pm245 (B*) pm290 (C) pm2180 (D) 290Sol. For FCC

4r = a2

r =4

a2= pm2400

416002 !"

To fill the tetrahedral void without distorting the lattice,

radius of atom = 0.225 2400 = pm290

Q.43AlCl

MeCl # # $ # AA

A is

(A) (B) (C) (D*)

[Sol. Nitrobenzene do not react with an electrophile at room temperature. It is used as solvent for FCreactions. ]



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CHEMISTRY

Q.5 X + H 2SO4 #$ # % Y a colourless gas with irritating smell.

Y + K 2Cr2O7 + H 2SO 4 #$ Green solution.[X] & [Y] are(A*) SO 3

2 , SO 2 (B) Cl , HCl (C) CO 32 , CO 2 (D) All of these

Q.6 The value of Henry's constant for three different gases A, B & C at 298 K are 40.3 K bar, 1.67 K bar& 1.83 10 5

K bar. Which of the following order is correct increasing order of solubility of the gases

in water?(A) B, A, C (B*) A, B, C (C) C, B, A (D) C, A, B

Sol. As value of henry's constant K h increases, solubility of gases in water decreases. So order of solubilityA < B < C

[PARAGRAPH TYPE]



In order to explain extent of adsorption theoretically Langmuir proposed an isotherm on the basis of following assumptions.(a) The adsorbed gas behaves ideally.(b) Only a monolayer is formed by the adsorbed gas.(c) Every adsorption site is equivalent and the ability of a particle to bind is independent of whatever

or not nearby sites are occupied.(d) The adsorbed gas molecules are localised, i.e., they do not move around on the surface.By considering equilibrium between the two opposite processes of adsorption & desorption, Langmuirobtained that extent of adsorption (x/m) can be obtained as.

bp1

p.a

m

x

&! where 'a' & 'b' are constants.

Based on this info, answer the questions that follow.Q.7 Which of the following options is correct for Langmuir isotherm.

(A) At very low pressuremx

will become constant.

(B) At moderate pressuresmx

will be linearly dependent on pressure.

(C*) At high pressuresmx

should be constant independent of pressure.

(D) At high pressuresmx

should be varying linearly with pressure.

Sol. bp1p.a

mx

&!

At very high pressureba

mx ! which is constant independent of pressure.


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CHEMISTRY

Q.10 A carbohydrate undergoes the following conversion.

D-LyxoseHCl

NaCN # # $ # 'A' # # # # # $ # ( 42 BaSOPd / H 'B' # # $ # )OH 3 'C'

'C' can be(A) D-Galactose (B) D-Talose (C*) Both (D) None of these

Q.11 Which of the following is correct presentation of L-Gulose.

(A)

OHCH|

HCHO|

HCHO|

HCHO|

OHCH|CHO

2

((

((

((

((

(B)

OHCH|

OHCH|

HCHO|

HCHO|

OHCH|CHO

2

((

((

((

((

(C*)

OHCH|

HCHO|

OHCH|

HCHO|

HCHO|CHO

2

((

((

((

((

(D)

OHCH|

OHCH|

HCHO|

OHCH|

HCHO|CHO

2

((

((

((

((

[Sol. D-Gulose and L-gulose are enantiomers. ]

PART B

[MATRIX TYPE]

Q.1 has four statements (A, B, C, D) given in Column-I and five statements (P, Q, R, S, T) given in Column-II .Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II .

[1 3+3+3+3 Marks = 12]Q.1 Column I Column II

(Name of the compound) (Formula)(A) Dead burnt plaster (P) CaCl 2(B) Quicklime (Q) Ca(OH) 2(C) Slaked lime (R) CaO(D) Gypsum (S) CaSO 4

(T) CaSO 4.2H 2O[Ans. (A) S (B) R (C) Q (D) T]

PART C

[INTEGER TYPE]Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are Single digits )

[4 4 Marks = 16]Q.1 Number of compounds, which can evolve CO 2 on heating, are.......

OCOOH H C CH COOH3

NO 2

H C C CH C O H3 2

NH O


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CHEMISTRY

O

OH

H C CH COOH3

CH COOH2

COOHCOOH

Ph CH CH COOH2

OH

HO C C OH

O OH

3C CH

2 COOH

[Ans.6 ]

Sol. (i)

OCOOH

*-Keto acid gives CO 2 +

(ii) H C CH COOH3

NO 2

*-Nitro acid gives CO 2 +

(iii)H C C CH C O H3 2

NH O*-imino acid gives CO 2 +

(iv)

O

OH*-, unsaturated CO 2 +

(v)COOHCOOH

1,6 dioic acid gives CO 2 +

(vi) HO C C OH

O O

1,2,dioic acid gives CO2 +

Q.2 Calculate at what approximate pH will 10 2 moles of Al(OH) 3(s) will dissolve in 1 litre of solution if

Ksp Al(OH) 3 = 8 10 33 & K f Al(OH) 4 = 8

10 36&. [Ans. 9]

Sol. Al(OH) 3(s) ! Al3+

(aq) + 3OH (aq) ; K sp = 8 10 33

Al3+(aq) + 4 OH (aq) ! Al(OH) 4 ; K f = 810 36

On adding two equations,Al(OH) 3(s) + OH (aq) ! Al(OH) 4 (aq) , K = K sp K f

= 8 10 33 8

10 36

K = 10 3

As K is very large Al(OH) 4 = 10 2 moles

Al(OH) 3(s) + OH (aq) ! Al(OH) 4 (aq) ; K = 10 3x M 10 2 M

103 =x

10 2( - x = 10 5 M

' pOH = 5 - pH = 9


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CHEMISTRY

[Sol.

CH=CH 2

# # # # # $ # (( 3AlCl / ClCMe

||O

C Me

# # # # $ # &HClZnHg

CH CH2 3

# # $ # NBS

CH CH 3

Br

Me CO Na3! )

/ %

CH=CH 2

]

Q.2 (A)dil.HCl

HgCl 2

excess (B)

P h e n o p

h t h e

l e n e

Phenophthelene

(B) Colourlessodourless gas

+ + H O (C) 2givesgoldenyellowflame

Ca(OH)Lime water

2

(E)

(F)PinkcolourBrowncompound

White ppt.

(D)

No change

+ H O2

Find :Number of / bond in (A)Number of 0 bond in (B)Number of 0 bond in (A)Number of / bond in (B) respectively

[Ans. 3212]Sol. (A) = Na 2CO 3

(B) = CO 2(C) = NaCl(D) = Ca(HCO 3)2(F) = HgCO 3.3HgONumber of / bond in (A) = 3Number of 0 bond in (B) = 2Number of 0 bond in (A) = 1Number of / bond in (B) = 2


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CHEMISTRY

Q.3 At 298 K, oCombustionH% (Sucrose) = 5737 kJ/mole & oCombustionG% (Sucrose)= 6333 kJ/mole.Estimate additional non PV work that is obtained if temperature is raised to 310 K. Assume %(nCP)reaction= 0 for given temperature range. Express answer in kJ. [Ans. 0024]

Sol. At 298 K,%G = %H T%S 6333 = 5737 298 %S298 %S = 596' % S = 2 kJ/molIf (%CP)r = 0

12 TT SS %!% & 12 TT HH %!%' At 310 K%G = %H T%S= 5737 310 2= 6357 kJ / mol' Additional non PV work = 6357 6333

= 24 kJ /mol

Only For Y 3Q.3 Born-Haber p dk mi;ksx vk;fud ;kS fxdks a ds lEHkou ,UFkSYih dh x.kuk es a fd;k tkrk gSA ;fn /oZikru

"ek Na = 26 kcal/g. atom, cU/k fo;kstu tkZ Cl2 = 54 kcal/mole dh] Na (g) dh vk;uu tkZ =117 kcal/ mol ] Cl (g) dh bySDV kWu xzg.k ,UFkSYih = 84 kcal/g atom] NaCl dh lEHkou ,UFkSYih= 99 kcal/mole gS rks NaCl ds tkyd tkZ dk ifjek.k (K. cal/mole) es a D;k gksxkA

[Ans : 0185]

[Sol LE54

99 27 117 84 H2

( ! & & ( & %

LEH% = 185]


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XIII_(XYZ)_RT-6_JEE A DVANCE _P APER -II CODE -A P AGE # 17

MATHEMATICS

PAPER

PART A



Q.1 The equation of common tangent to the curves y 2 = 20x and120y

25x 22 ! = 1 having positive gradient

is

(A*) 5 y = x + 25 (B) 5 y = x 25 (C) 5 y = x + 5 (D) 5 y = x 5

[Sol. y = mx +m5

(m > 0) .(1)

" 2m25 = 25m 2 + 120 (using condition of tangency) # m = 5

1 . ]

Q.2 If the equationx1

xpx1

x2

!!$ % &'

( ) * + 29 = 0 (p + R) has exactly four distinct real solutions,

then the true set of values of p is

(A*) $ % &'

( ) *,*

229

, - { 10} (B) { 10}

(C) $ % &

'( ) *

,* 229

, (D) $ % &

'( )

,*

,229

[Sol. Putx1

x ! = t, we get, f(t) = t 2 + pt + 25

Case-I: Now, one root is greater than 2 and other is less than 2.

So, f(2) < 0. Hence, p + $ % &'

( ) *,*

229

,

Case-II: Both roots greater than 2 and equal.

Hence, a2b*

> 2 and D = 0, p > 2" p < 2 and D = 0 # p2 100 = 0 # p = 10So, p = 10

" The range of p is $ % &'

( ) *,*

229

, - { 10}.

Hence, the greatest integral value of p is 10. Ans. ]



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MATHEMATICS

Q.3 In triangle ABC, if (sin B + sin C) 2 sin 2A = 2R

4., then tan A is equal to

[Note: All symbols used have usual meaning in triangle ABC.]

(A)23

(B)21

(C)178

(D*)158

[Sol. We have, .* )as(s = 4 #

2Acot = 4. So, tan A =

158

. Ans. ]

Q.4 The total number of words with three letters which are formed by using the letters of word IITJEEsuch that no two vowels are together, is(A) 12 (B) 16 (C) 18 (D*) 20

[Sol.vowels

2E,2I // ; T = 1, J = 1

" Number of words formed = ( 2C1 3!) + 2 (1 + 1) + ( 2C1 2!) = 20. Ans. ]

Q.5 Ifdxdy

= 2x23

yey!

and y (0) = 1, then

(A*) y 2 = e2x 2e2x lny (B) y 2 = e2x + 2e 2x lny

(C) y 2 = e 2x 21

e2x lny (D) y 2 = e 2x +21

e2x lny

[Sol. dydx

= 3

2x2

y

ye ! # e 2x

dydx

=x2

3 ey1

y

1 *!

put e 2x = t # yt2dydt ! = 3y2 (L.D.E.)

t y2 = 0 dyyy

2 23 + C

e 2x y2 = 2 ln y + C Now y (0) = 1# C = 1" e 2x y2 = 2 ln y 1 # y2 = e 2x 2e 2x ln y. Ans. ]

Q.6 Let P (x, y) be a variable point on the curve 4x 2 + 9y 2 8x 36y + 15 = 0,

then min. (x2 2x + y

2 4y + 5) + max. (x

2 2x + y

2 4y + 5) is equal to

(A)32536

(B)1325

(C)2513

(D*)36

325

[Sol. We have,

925

)2y(

425

)1x( 22 *!* = 1

So, min. ((x 1)2 + (y 2)2) =9

25 and max. ((x 1)2 + (y 2)2) =

425

. ]


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MATHEMATICS

[PARAGRAPH TYPE]


Paragraph for question nos. 7 & 8

Let PQ be a variable focal chord of the parabola C 1 : y2 8x = 0 and A be its vertex.

The locus of centroid of triangle APQ is another parabola C 2.

Q.7 The length of latus-rectum of parabola C 2 is

(A)32

(B)34

(C*)38

(D)3

16

Q.8 The area of quadrilateral formed by tangents and normals at ends of latus-rectum of parabola C 2, is

(A)3

32(B*)

932

(C)3

16(D)

916

[Sol.(i) Centroid (h, k)

" 2h3

= $$ %

&''(

) !

21

21

t

1t .(1)

and4k 3

= $$ %

&''(

) *

11 t

1t .(2)

y

A x

Q

(h,k) S(2, 0)G Focus

(0, 0)

P(2t , 4t )2 11

$$''(

) *

121 t

4,

t

2

# Locus of G (h, k) is C 2 : y2 = $ % &'

( ) *

34

x38

(another parabola)

(ii) Area = 82

32 $ % &'

( )

=9

48 1 =

932

(square units). ]

Paragraph for question nos. 9 to 11

Consider a plane 2 : x + y z = 1 and the point A (1, 2, 3). A line L has equationsx = 1 + 3r, y = 2 r, z = 3 + 4r, where r is parameter.

Q.9 The coordinates of a point B on line L such that AB is parallel to the plane 2 is(A) (10, 1, 15) (B) ( 5, 4, 5) (C) (4, 1, 7) (D*) ( 8, 5, 9)

Q.10 The equation of plane containing the line L and the point A has the equation(A) x 3y + 5 = 0 (B) 3x y 1 = 0 (C*) x + 3y 7 = 0 (D) 3x + y 5 = 0

Q.11 The distance between the points on the line L which are at a distance of 34

from the plane 2 is

(A*) 4 26 (B) 20 (C) 10 13 (D) 5 13


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MATHEMATICS

[Sol.(i) A (1, 2, 3); B (1 + 3r, 2 r, 3 + 4r)

" 1 (3r) + 1 ( r) 1 (4r + 6) = 0 [As AB is parallel to plane 2 .]# r = 3. So, B ( 8, 5, 9)

(ii) The equation of plane is413600

)3z()2y()1x(

*

*** = 0

A ( 1

, 2 , 3 ) L

< 3, 1,

4 >

( 1 , 2

, 3 )

# x + 3y = 7. Ans.(iii) 2 : x + y z 1 = 0

P (1 + 3r, 2 r, 3 + 4r)

Now, 3|1r2| ! = 3

4 # r =

23

or25*

. So, points are $ % &'

( )

9,21

,2

11 and $ % &'

( ) ** 7,

29

,213

.

# Distance between them = 4 26 . ]

PART B

[MATRIX TYPE]

Q.1 has four statements (A, B, C, D) given in Column-I and five statements (P, Q, R, S, T) given in Column-II .Any given statement in Column-I can have correct matching with one or more statement(s) given in Column-II .

Q.1 Column-I Column-II(A) If 5x 2 + 5y 2 + ax + by c = 0 is the equation of smallest circle (P) 1

which is passing through points of intersection of circle x 2 + y 2 = 4and line 3x + 4y = 5, then the value of (2c + a + b) is equal to (Q) 2

(B) Number of solutions of the equation sin x +2

1cos x = $

%

&'(

) 3!4

xsin 2

in x + (0, 3) is equal to (R) 4

(C) Suppose n is a natural number such that

| i + 2i 2 + 3i 3 + . + ni n | = 18 2 where i = 1* , (S) 5

then n is equal to

(D) If S = 4 56,

/

*!* !1k

k 1k 1 22cot , then cot (S) is equal to (T) 6

[Ans. (A) T ; (B) Q ; (C) T ; (D) Q][Sol.

(A) S + 7L = 0 # (x2 + y2 4) + 7 (3x + 4y 5) = 0; Centre $ % &'

( ) 7*7* 2,

23

Now, 3 $ % &'

( ) 7*

23

+ 4 ( 27) = 5 # 7 =52*

" The equation of required circle is 5x 2 + 5y 2 6x 8y = 10.

(B) sin x +2

9 cos x = $

%

&'(

) 3!4

xsin 2 # (1 cos x) (1 2sin x) = 0 # cos x = 1 or sin x =2

1


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MATHEMATICS

" x =6

3,

653

+ (0, 3).

(C) Let z = (i + 2i 2 + 3i 3 + 4i 4 + . + ni n.)As, (i + 2i 2 + 3i 3 + 4i 4) = 2 2i = 2 (1 i)

# n = 36. So, n = 6

Aliter: | z | =i1

ni2

1i 1nn

**$$ %

&''( ) * !

(D) S n = 6/ !!

*$$ %

&''(

) !

*n

1k k 1k

k 1k 1

221

22tan = 4 56

/

*!* *n

1k

k 11k 1 2tan)2(tan = (tan 1 (2n+1) tan 12)

" S = nn

SLim,:

=2

3 tan 12 = cot 12 # cot (S) = 2. Ans. ]

PART C[INTEGER TYPE]

Q.1 to Q.4 are "Integer Type" questions. (The answer to each of the questions are Single digits )

Q.1 Find the number of solutions of equation sin 1(4sin 2; + sin ; ) + cos 1( 1 + 6sin ; ) =2

3,

in ; + [0, 5 3]. [Ans. 0006]

[Sol. We must have, 4sin 2; + sin ; = 1 + 6sin ; # sin ; = 1,41

# 6 solutions.]

Q.2 If k )z4( jyi)x(a !!7/ , k )y3( jxiyb !!/ and

4 5k x)1( j)z2(izc !7***/ are sides of the triangle as shown infigure, then find the value of 7 (where x, y, z are not all zero.)

a

bc

[Ans. 0000]

[Sol. As, cba /!So, 7x + y + z = 0

x + y + 2z = 0(7 + 1) x + 3y + 4z = 0

" For non-trivial solution, put . = 0

# 43121111

!7

7 = 0 # 7 = 0. Ans. ]


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MATHEMATICS

Q.3 If 03

!

2

044 dxxcosxsin

x2sinx = k

2

45

44 dxxcosxsin

x2sin

$$$$

%

&

''''

(

)

!03

3, then find the value of k. [Ans. 0002]

[Sol. Let I 1 = 03

!

2

044 dxxcosxsin

x2sinx =

8

23(Using King and Queen property)

Also, I 2 = k

2

periodwithPeriodic

45

44 dxxcosxsin

x2sin

$$$$$

%

&

'''''

(

)

$ % &'

( )

!3

3

30 = k

2

4$ % &'

( ) 3

=16k 23

So, k = 2. Ans. ]

Q.4 Let a < b < c be three integers such that a, b, c is an arithmetic progression and a, c, b is a geometricprogression, then find the smallest positive value of c. [Ans. 0002]

[Sol. a = A d,b = A Id,Awhere +

c = A + dc 2 = ab(A + d) 2 = (A d) AA2 + d 2 + 2Ad = A 2 Add2 = 3Add = 0, d = 3Ac = A + d = 2A" cmin = 2. ]

PART D

[INTEGER TYPE]

Q.1 to Q.3 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits )

Q.1 Let A and B be two points on the major axis of the ellipse16y

25x

22

! = 1, which are equidistant from

the centre. If C and D are the images of these points in the line mirror y = mx, m ? 0 then find themaximum area of quadrilateral ACBD . [Ans. 50]

[Sol. Image of A(h, 0) in the line mirror mx y = 0

$ % &'

( )

!*/

**/*

1m

mh2

10y

mhx

2

x = 22

m1

)m1(h!*

, y =1m

mh22 ! AB

D

C

(h, 0)( h,0)

y

x

y = mx


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MATHEMATICS

" C @ $$ %

&''(

) !!

*1m

mh2,

m1

)m1(h22

2

Similarly D @ $$ %

&''(

) !

*!

**1m

mh2,

m1)m1(h

22

2

AC =22

222

2

2

)m1(hm4

m1)m1(h

h!

!$$ %

&''(

) !** = 22

22

22

24

)m1(

hm4

)m1(

hm4!

!!

=)m1(m1mh2

2

2

!!

AD = 22222

2

2

)m1(

hm4

m1

)m1(hh

!!$$ %

&''(

) !*! = 22

22

22

2

)m1(

hm4

)m1(

h4!

!! =

22 m1)m1(

h2 !!

" Area = AC AD = 2222

)m1(

)m1(mh4

!

! =

m1

m

h4 2

!

Area| max = 2254 1

= 50 Ans. ]

Q.2 Consider the curves C 1 : | z 2 | = 2 + Re(z) and C 2 : | z | = 3 (where z = x + iy, x, y + R and i = 1* ).They intersect at P and Q in the first and fourth quadrants respectively. Tangents to the C 1 at P and Qintersect the x-axis at R and tangents to the C 2 at P and Q intersect the x-axis at S. If the area of . QRS

is 27 , then find the value of 7. [Ans. 0010]

[Sol. Let z = x + iyThen C 1 : | z 2 | = 2 + Re(z)# | (x 2) + iy | = 2 + x# (x 2)2 + y2 = (x + 2) 2

R

P ( 1, 2

2 )

(3,0)O

(0,0)

Q ( 1 , 2 2 )

S x

y

# y2 = 8xand C 2 : x

2 + y2 = 9For point of intersections of C 1 and C 2

x2 + 8x = 9# x2 + 8x 9 = 0#

(x + 9) (x 1) = 0

# x=1(" y2 = 8 # y = 22" P = 522,1 and Q 522,1 *Equation of tangent at P on parabola is y 22 = 4 (x + 1) ....(1)Equation of tangent at Q on parabola is

y 22 = 4 (x + 1) ....(2)Solving (1) and (2) , R = ( 1, 0)Equation of tangent at P on circle is

9y22x /! ......(3)


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xiii rt 6 pcm-16!02!2013 advance paper-2 solution

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