Answerkey ADV Score -3

8/10/2019 Answerkey ADV Score -3

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Corporate Office : ALLEN CAREER INSTITUTE, SANKALP, CP-6, INDRA VIHAR, KOTA-324005

PHONE : +91 - 744 - 2436001, Fax : +91-744-2435003, E-mail: [email protected]. in Website: www.a llen.ac.inKOTA / HS - 1

PAPER CODE

TM

Path to success KOTA(RAJASTHAN)0 1 C T 3 1 3 0 7 1

PATTERN : JEE (Advanced)LEADER & ENTHUSIAST COURSE

SCORE-I

Date : 08 - 03 - 2014TARGET : JEE (Main + Advanced) 2014

TEST # 03

SECTION-I

1. Ans. (A)

Sol. Let g(x) = 1 + x + ....+

nx

n

g'(x) = 1 + x + ....+

n 1x

n 1

-

-xn= n (g(x) - g'(x))

n

n

x g(x) g '(x)I dx n dx

x g(x)1 x ......

n

-= =

+ + [ ]en x log (g(x)) C= - + .

2 nx xf(x) n 1 x .....

2 n

= + + + + l

2. Ans. (D)

Sol. Put z = 1, sum of all coefficients= (1 + 2 + 3 + .........n)2

= (Sn)2= Sn3

3. Ans. (B)

Sol.

6 / 7 2 /3

x 0

1 1(cos x) ( sin x) (cos x) ( sin x)

7 3lim 2x

- -

- - -

l= =

1 1 1 2

2 3 7 21

- = .4. Ans. (B)

Sol : x y y x 5+ + - =Q .....(1)

2x

5x y y x

=+ - -

2xx y y x

5+ - - = .....(2)

(1) + (2) 2x

2 x y 55

+ = +

squaring, 4y = 25 +24x

1524x

15

2

2

d y 2

dx 25=

Q. 1 2 3 4 5 6 7 8 9 10

A. A D B B D C A,B A,D A,B,C A,B

Q. 11 12 13

A. B B B

A B C D

P Q T S

Q. 1 2 3 4 5 6

A. 5 5 7 5 6 4

Q.1

SECTION-I

SECTION-IV

SECTION-II

SOLUTION

PART-1 : MATHEMATICS ANSWER KEY


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01CT313071KOTA / HS - 2

08-03-2014TARGET : JEE (Main + Advanced) 2014

TM

Path to success KOTA(RAJASTHAN)

5. Ans. (D)

Sol. No. of non decreasing onto functions is the number of positive integral solutions of

x1+ x2+ ....x5= 10. .. . (1). Which is9C5

And number of total non decreasing functions is number of non negative integral solutions of (1) which

is 14C4

. So the required number of function is 14C4

9C5

= 875

6. Ans. (C)

Sol. 1 1r

r 1 r r 1 rI sin tan

(r 1)(r 2) 1 r 1 r

- - + - + -

= = + + + + 1 1tan r 1 tan r- -= + -

1nS tan n 14

- p= + - .

7. Ans. (A,B)

Sol :Let ( )n

n( 2 1) I f, 2 1 f '+ = + - =

where I = Integer and 0 < f , f ' < 1Now ( ) ( ) ( )

n n n 1n n

0 12 1 C 2 C 2-

+ = + + ( )n 2

n n2 nC 2 ..... C

-

+ +

( ) ( ) ( )n n n 1

n n0 12 1 C 2 C 2

-

- = - + ( ) ( )n 2 nn n

2 nC 2 ..... 1 . C-

+ + -

If n is even, then f + f ' = 1 and I = odd integer f = 1 f '

\ Given limit = 1 . (1)odd int = 1 ( )nlim f ' 0\ =If n is odd, the f = f ' and I = even int.

\ Given limit = 0 (1)even int.= 08. Ans. (A,D)

Sol. Number of such numbers of k digits is the number of solutions of the equation x1+ x

2+x

3...+ x

k= 8

subject to {x1 1 and x

2, x

3, ...., x

k 0}which is 6 +kC

k1.

such numbers are 1 + 8 + 36 + 120 =165. (1 digit) (2 digit) (3 digit) (4 digit)

So 170thnumber is 10043.

9. Ans. (A,B,C)

Sol.

x2

2

0

f(t)f(x) (1 x ) 1 dt

1 t

= + + +

on differenting we get2

2

f '(x ) (x 1)

f (x) 1 x

+=

+logef(x) = x + log

e(1 + x2) + c. Since (f(0) = 1)

f(x) = ex(1 + x2)10. Ans. (A,B)

Sol. Since sum of the greatest and the smallest terms is zero x < 0 and n 1x

112

+

+

is integer.

that x can be 10 and 21.11. Ans. (B)

Sol. Let the az2+ bz + c = 0 has roots z1

and z2

and |z1

| = 1

z1+z

2=

b

a

-and z

1z

2=

c

a |z

1z

2| =

c

a= 1 |z

1| = |z

2| = 1


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TM


|z1+ z

2| =

b1

a

-= 1 2

1 2

1 1(z z ) 1

z z

+ + =

(z

1+ z

2)2= z

1z

2 b2= ac.

12. Ans. (B)

Sol. Since each equation has a root having modulus b2 = ca and c2= ab a3= b3= c3

a, b, c are z, zw, zw2

a, b, c are the vertics of an equilaterial triangle.13. Ans. (B)

Sol. az2+ bz + c = 0 and bz2+ cz + a = 0 have a root of modulesa, b, c are (z, zw, zw2).ratio of cofficients of the two equatins are sameboth the roots are same.

SECTION II

1. Ans. (A) (P); (B) (Q); (C) (T); (D) (S)

Sol. (A) P(A) =7

28

2

C 3

C 4=

(B) P(B) =1

7+

6 1 1. .

7 4 3+

6 1 1 1. . .

7 4 3 4 =

1

4(Playing (Playing (Playing

in I round) in II round) in round III)

(C) P(C) = 8 44 2

1 1 1 1. .

C C 2 840=

(D) P(D) =

6

37

3

C1 1.8 C 14

=

SECTIONIV

1. Ans. 5

Sol Q sin1(sin12) + cos1(cos12)= 12 4p+ 4p 12 = 0\(n 2)x2+ 8x + n + 4 > 0 "xn 2 > 0 n > 2 (i)D < 0 64 4(n 2)(n + 4) < 0n2+ 2n 24 > 0

(n + 6)(n 4) > 0n > 4 (ii)2. Ans. 5

Sol :

( )

2

2if x , 1

x

f(x) a bx if 1 x 1

2if x 1

x

- - -

= + -

>

Q continuity at x = 1 a + b = 2 and differentiability at x = 1

2b = 2 b = 1 and a = 3 \ 2a + b = 53. Ans. 7

2 2 2

a b c: : 3 : 4 : 5

a 1 b 1 c 1=

+ + +


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01CT313071KOTA / HS - 4


TM


letA B

a tan ,b tan2 2

= = andC

c tan2

=

Since ab + bc + ca = 1A + B + C = 180.

sinA : sinB: sinC = 3 : 4 : 5.

So2 2 2

2 2 21 a 1 b 1 c51 a 1 b 1 c

- - -+ + = + + + 5 [cosA + cosB + cosC] = 7

4. Ans. 5

MB

C 4 A

90 - aa q

a

3

In DAMC AM = 2 R sin a= c sin a

Also in DABM, BM = 2 R sin (90 - a)= 2R cos aInD ACM CM = 2R sin (90 - a + q)

= cos(q - a)

AM + BM + CM = 2R [sin a+ cos a+ cos (q - a)]= c [sin a(1 + sin q)] + cos a[1 + cos q]= sin a[c + a] + cos a[c + b]

Maximum value = 2 2(a c) (b c)+ + +

= 2 29 8 145+ =

5. Ans. 6

Sol. f(x)n 2 n 2

n nn

sin x cos xlim

sin x cos x

+ +

++

=( )

( )

n2

2

nn

tan x tan x 1cos x lim

tan x 1

+

+

=

2

2

cos x if 0 x4

sin x if x4 2

p

p p < Q at t = 2sec ; i > 1 amp

3. Ans. (A)

Sol. If mass is more radius will be moremv

RqB

=

, ( )F q v B=

rr

4. Ans. (A)

Sol. 11 2 2

R 1 R R 2,R R 3 R 1

+= =+

5. Ans. (D)

Sol.t

R C1 11 11 1

1 1 1 1

tI e nI

R R R C

e e= = -l

Similarly 22

2 2 2

tnI

R R C

e= =l

From figure 3, it is clear that

1 2

1 2R R

e e

= ...(i)

and1 1 2 2

1 1

R C R C> ...(ii)

6. Ans. (D)

Sol. Since final medium is same so ray is parallel to each other.7. Ans. (A,B,D)Sol. An indoergic reaction is one in which net energy has to be supplied.8. Ans. (B,D)9. Ans. (A,D)

10. Ans. (A,B)11. Ans. (D)12. Ans. (B)

SOLUTION

PART-2 : PHYSICS ANSWER KEYQ. 1 2 3 4 5 6 7 8 9 10

A. A C A A D D A,B,D B,D A,D A,B

Q. 11 12 13

A. D B C

A B C D

P,S,T P,Q,R,T R,S,T P,R,S,T

Q. 1 2 3 4 5 6

A. 6 3 9 4 5 5

Q.1

SECTION-I

SECTION-IV

SECTION-II


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13. Ans. (C)

SECTIONII

1. Ans. (A) (P,S,T); (B) (P,Q,R,T); (C) (R,S,T); (D) (P,R,S,T)

For (A): The motion of the particle is on a straight line with constant speed.

For (B): The motion of the particle is a helix with constant pitch. Projection on x and y axis are a0coswt& a0sinwt respectively.For (C): The motion of the particle is elliptical path and origin is not at the focus. Projection on x & y

axis are a0sinwt & b0coswt respectively.For (D) :The motion of the particle is on circular path with origin as center. Projection of particles

motion on x & y axis are a0coswt & a0sinwt respectively.

SECTION-IV

1. Ans. 6

Sol. KaX-ray results from the transition of an electron from L shell to K shell. If the energy of the atom with

a vacancy in the K shell is EKand the energy with a vacancy in the L shell is E

L, the energy of the photon

emitted is EK EL. the energy of the 71 pm photon is

hcE =

l 31242eVnm

17.5keV71 10 nm-

= =

Thus, EK E

L= 17.5 keV

EL= 23.32 keV 17.5 keV = 5.82 keV.

2. Ans. 3

Sol. E0=

( )2k 4k

d / 2 d

l l=

Electric field of r >>l.

E =( )

2 2

kkQ

r r

l=

l

02

E dE

4r

=

l

3. Ans. 94. Ans. 4

Sol. As the material is of refractive index 1, qris negative and q'

rpositive. Now t r rq = q = q

The total deviation of the outcoming ray from the incoming ray is 4qi10

4 410 = .5. Ans. 5

Sol.

2BrV

2

w=

2BrI

R 2R

w= =

2 3B r rmg 0

2R 2

w - r =

1

2 4

4mg R100s

B r

-rw = =

6. Ans. 5Sol. I = 2pe

0aEv = 0.5 mA.


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TM


SECTIONI

1. Ans. (C)

2. Ans. (C)

Sol. + CH CH=CH3 2H PO3 4

CH CHCH3 3

O ,D2H O3

+

OH

+CH COCH3 3

A B C

NH OH/H2+

(CH ) C=NOH3 2Acidic reagent

CH CONHCH3 33. Ans. (B)

Sol. Polymerization takes place by formation of following stable reaction intermediate

CH CH3+

CH CH3

CH CH3

4. Ans. (A) ; 5. Ans. (C) 6. Ans. (B) ; 7. Ans. (A,C)8 Ans. (A,B,C)Sol. When vicinal di carboxylic acid is heated anhydride formation takes place to produce a stable compound.9. Ans. (A,B,C) ; 10. Ans. (B,C) ; 11. Ans. (B) ; 12. Ans. (C)13. Ans. (A)

Sol. For the visibility1.55 eV < DE

Answerkey ADV Score -3

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