test 3 paper key

PART-1 : MATHEMATICSANSWERKEY : PAPER-1

PART-2 : PHYSICS

PART-3 : CHEMISTRY

NURTURE COURSE : TARGET - JEE (Main + Advanced) 2015MAJOR TEST # 03 Date : 16 - 02 - 2014 Pattern : JEE (Advanced)

PAPER CODETM

Path to success KOTA (RAJASTHAN)0 1 C T 1 1 3 0 5 3

Q. 1 2 3 4 5 6 7 8 9 10A. D C A C A,B,C A,B,D B,C A,B,C B CQ. 11 12A. B BQ. 1 2 3 4 5A. 330 738 020 150 225Q. 1 2 3 4A. 1 6 3 5

SECTION-III

SECTION-IV

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10A. B C C B A,B,C,D B,D A,C B,C D BQ. 11 12A. B DQ. 1 2 3 4 5A. 072 193 706 297 501Q. 1 2 3 4A. 4 2 3 5

SECTION-III

SECTION-IV

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10A. D B A C B,C A,B,C,D A,B,C A,C C AQ. 11 12A. C CQ. 1 2 3 4 5A. 030 038 004 005 002Q. 1 2 3 4A. 1 3 6 3

SECTION-III

SECTION-IV

SECTION-I

PART-1 : MATHEMATICS

PART-3 : CHEMISTRY

PART-2 : PHYSICS

ANSWERKEY : PAPER-2


PAPER CODETM


Q. 1 2 3 4 5 6 7 8 9 10A. C C A B B A,C A,B,C C,D A,B,C A,B,CQ. 11 12 13 14 15 16A. D D A D C C

A B C DP S,T P,Q,R Q,R

Q. 1 2 3 4A. 007 003 007 000

Q.1

SECTION-I

SECTION-III

SECTION-II

Q. 1 2 3 4 5 6 7 8 9 10A. B C D C A A,D A,D A,B,C,D B,D A,B,DQ. 11 12 13 14 15 16A. D A A C C A

A B C DP,Q,S P,Q,R,S,T Q,R P,Q,S

Q. 1 2 3 4A. 066 005 040 040

Q.1

SECTION-I

SECTION-III

SECTION-II

Q. 1 2 3 4 5 6 7 8 9 10A. B B C B C A, C B, C B, C A,B,D A,B,C,DQ. 11 12 13 14 15 16A. C C C C C C

A B C DP,Q,R P,Q,S R P,Q,S

Q. 1 2 3 4A. 004 018 120 002

Q.1

SECTION-I

SECTION-III

SECTION-II

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SECTION-I1. Ans. (B)

(x2 + 3x 15) = (x a) (x b) (1 a) (1 b) = 11& x2 3x + b = (x b) (x g) (2 b) (2 g) = b 2\ (1 a) (2 g) (1 b) (2 b) = 11.(b2)(1 a) (2 g) (b2 3b + 2) = 11(b 2)Now, if b satsisfies x2 3x + b = 0 b2 3b = b b2 3b + 2 = 2 b (1 a) (2 g) = 11

2. Ans. (C)a,b,c are A.P. a = 2b ca,c,b are in G.P. c2 = ab c2 = (2b c).b c2 + bc 2b2 = 0 (c b) (c + 2b) = 0 c = 2b\ a = 4b, b = b, c = 2bQ c > b > a b is negative c is positive.\ Min possible value of c = 2

3. Ans. (C)x [1,0) (x) = 1 + (1) + 1 + (1) = 0x = 0 (x) = 0 + 0 + 1 + 1 = 2x (0,1) (x) = 1 + 1 + 1 + 1 = 4x = 1 (x) = 1 + 1 + 1 + 0 = 3\ sum of all elements = 9

4. Ans. (B)

[cos x] + {sinx} = 13

[ cosx] = 0 & { } 1sin x3

=

Graphically for x [0,2p]

p/2 p 3p2

2p p/2 p 3p2

2p

y=1/3

PART-1 : MATHEMATICS SOLUTIONPAPER-1


PAPER CODETM


5. Ans. (A,B,C,D)A {1,2,......,25}B {1,4,7,.........,25} n(B) = 9C {2,5,.........,23} n(C) = 8D {3,6,.......,24} n(D) = 8If a + b + c is div. by 3 8C3 +

8C3 + 9C3 +

8C1. 8C1.

9C1= 56 + 56 + 84 + 576 = 772 \ (A)abc is div. by 3 25C3

17C3 = 2300 680 = 1620 \ (D)a + b + c is even 12C3 +

13C2. 12C1 = 220 + 78 12

= 220 + 936 = 1156 \ (B)abc is even 25C3

13C3 = 2300 286 = 2014 \ (C)6. Ans. (B,D)

(A) coeff. x24 = (1 + 2 + ...+ 25)25.26

3252

= - = -

(B) coeff of x23 =

2 2

1 . 11.2

2

=

( ) ( )2 225 13 25 13 17

2

- =

( )325 325 17 325500050

2 2

-= = =

(C) constant term = (25)!(D) sum all coeff we (obtain by putting x = 1)i.e. 0

7. Ans. (A,C)c2 = a2 + b2 2ab cosC= 13 6 = 7

2 3

C

B ANow, length of internalangle bisector through vertex

01CT113053KOTA / HS - 2

16-02-2014TARGET : JEE (Main + Advanced) 2015TM

Path to success KOTA (RAJASTHAN)

PAPER 1

2ab C 12 3 6 3C cos

a b 2 5 2 5= = =

+\ (A)

& length of median through vertex C is

2 2 21 1 192a 2b c 26 72 2 2

+ - = - = \ (C)8. Ans. (B,C)

(xy) = (x).(y)Put x = 2 & y = 1 (1) = 1(2) = 0 ; put x =2, y = 2 (4) = 102 ; similarly(8) = 103 & (16) = 104

\ ( )4 rr 0

2 11111=

= & ( )4 1 rr 0

10 31-=

=Paragraph for Question 9 & 10

9. Ans. (D)

(x) + g(x) = |x 1| + 2|x 2| + 3|x 3| + 4|x

4| + 5|x 5|.

15x+551 2 3 4 513x+53 9x+45 3x+27 5x+55 15x55

20

clearly minimum is achieved at x = 4 & ((x) +

g(x))min = 15

10. Ans. (B)(x) = g(x) has two intersection points. Plot thegraphs of (x) & g(x)

Paragraph for Question 11 & 12

11. Ans. (B)

(1,1)

L = 0

S =01

(a,b)

L = 0 is common chord of (a,b) w.r.t. circle S1 = 0\ ax + by + (x + a) (y + b) 2 = 0

(a + 1) x + (b 1)y + a b 2 = 0& 3x + 4y + 4 = 0comparing, we geta 1 b 1 a b 2

3 4 4+ - - -= =

\ a + b = 285

-12. Ans. (D)

Assume S2 = 0 as (x a)2 + (y b)2 = 1

applying condition of orhtogonality we getr1

2 + r22 = d2 (a + 1)2 + (b 1)2 = 4

Also, 3a + 4b = 1 or 3a + 4b = 9\ 4 circles possible.

SECTION III1. Ans. 072

12! = 210. 35.52.71.111

Now, divisors of 4k + 2 will have exactly 1 two\ (5 + 1) (2 + 1) (1 + 1) (1 + 1) = 72

2. Ans. 193

A B

C

PQ

Let the side lengths of square P & Q be x & yrespectively.\ AB = x + y + x cotA + ycotB= x + y + x cotA + y tanA

Now, 4

tan A3

=

\ 5 = x + y + 3x 4y x y 54 3 4 3 7

+ + =Now, we have to minimize x2 + y2 which issquare of perpendicular distance from (0,0) to

x y 54 3 7

+ = \ 14449

\ r + s = 1933. Ans. 706

AXB is maximized. If we draw a family ofcircle passing through A & B and it will touchx-axis.

C(3,0)

A(0,4)

B(3,8)

X( ,0)a

\ Using power of point

KOTA / HS - 301CT113053

NURTURE COURSE 16-02-2014TM


PAPER 1

(CX)2 = CA CB

(a + 3)2 = 5 10 a + 3 = 5 2

5 2 3a = -\ a = 5, b = 3 a4 + b4 = 625 + 81 = 706

4. Ans. 297

Total number of rectangles

= 7C2 7C2 4(27+9)

= 441 144 = 297

5. Ans. 501

2004 (x3 3xy2) 2005 (y3 3x2y) = 0

Dividing by y3, we get

3 2

3 2

x x x2004 3 2005 1 3

y y y

- - -

31 2

1 2 3

xx x x x x2004

y y y y y y

= - - -

where 31 2

1 2 3

xx x, ,

y y y are roots of equation. Put

x1

y= , we get

31 2

1 2 3

xx x1 1 1

y y y

- - -

( ) ( )2004 2 2005 2 12004 1002

- - -= =

\ ( ) ( )( )1 2 3

1 1 2 2 3 3

y y y501

2 y x y x y x=- - -

SECTION IV

1. Ans. 4

a cosB b cosA = 35

c

a cosB + b cosA = c

Adding we get 2a cosB = 8

c5

& 2b cosA =2

c5

\ a cos B tan A4 4bcos A tan B

= =

2. Ans. 2(x) = x2 sgnx

Now, (x + a) > 2 (x) " x [a,a + 2]Case: I If a + 2 < 0 a < 2 (x + a)2 sgn(x + a) > 2x2sgn xNow, sgnx < 0 & sgn(x + a) < 0(x + a)2 < 2x2 x2 2ax a2 > 0which is not possible " x [a, a + 2]Case : II a < 0 < a + 2

Put x = 0 (a) > 2(0)

a2 > 0 Not possible

Case : III a > 0 \ (x + a)2 sgn(x + a) > 2x2 sgnx x2 2ax a2 < 0\ (a) < 0 2a2 < 0 true (a + 2) < 0

a2 + 4a + 4 2a2 4a a2 < 0

a2 > 2 a 2, 3. Ans. 3

a2 + b2 = c2 ; Also ( )1 ab 2 a b c2

= + +(a + b)2 c2 = 2ab

(a + b + c) (a + b c = 2ab\ a + b c = 8 Now a2 + b2 = (a + b 8)2= a2 + b2 + 2ab + 64 16(a + b)

(a 8) (b 8) = 32

a 8 = 2 b 8 = 16

a 8 = 4 b 8 = 8

a 8 = 1 b 8 = 32 \ 3 triangles4. Ans. 5

2(x) = ((x)) = a(ax + b) + b3 (x) = a(a2x + ab + b) + b

n(x) = anx + an1b + an2b+.....+b

7(x) = a7x + b(a6 + a5 +....+1) = 128x + 381

a = 2 & 7a 1

b 381a 1

- = - 127b = 381 b = 3\ a + b = 5

01CT113053KOTA / HS - 4



PAPER 1

PART2 : PHYSICS SOLUTIONSECTIONI

1. Ans. (D)

Sol. v = RTM

g

vHe

= 2H

v

( )5 7RT R 3003 5

4 2=

T = 504 K = 231C2. Ans. (C)

Sol. Pavg

= 2 21 A v

2mw

v = Tm = fl, w = 2pf

3. Ans. (A)

Sol. F' + 32

r g F3

p r =

F = [(3rrg) + p0]pr2

4. Ans. (C)

Sol. Fext

= 0 so COMv 0=r

vboard

= vdisc

= vRw v = v

5. Ans. (A, B, C)Sol. P V r6. Ans. (A, B, D)Sol. Q = W + DE

int

7. Ans. (B, C)

Sol. L/4L/4

CM

Mv = 2Mv' v' = v2

22ML L L2 M Mv

12 42 2

+ w =

8. Ans. (A, B, C)

Sol. T = m

2k

p

2 2max

1 1mv kA

2 2=

amax

= w2A9. Ans. (B)

Sol. N kx = 200

m(g + a)

Maximum upward acceleration is 5m/s2

N + 200= 60(10 + 5) = 900

N = 700 N Reading = 70 kg

10. Ans. (C)Sol. Till 12 sec. block remain contact with weight

machine so elongation in spring balance will notchange.

11. Ans. (B)Sol. Nm : Total mass of a gas = nM so F =

2(n1 2OM )

2Ov =

2

2

11 O

O

8RT2n M

Mp12. Ans. (B)Sol. T

1 > T

2 so (momentum)

1 > (momentum)

2

So net force on membrane will be towards right.SECTIONIII

1. Ans. 330Sol. Since volume of pipe is negligible, thus V of air

remains 7l.By mole conservation,

1 1 2 2 m m

1 2

P V P V g 700 7 g 770 7T T 1000 300 1000 T

r r = = T 330 = [Dh = 35 2 mm]

KOTA / HS - 501CT113053



PAPER 1

2. Ans. 738

Sol. Constant velocity achieved = 1.22

= 0.6 m/sec

\ we have,v2 = u2 + 2as (0.6)2 = 02 + 2 a 1

\ a = 0.18 m/sec2

From Newton's 2nd law of motion

0.05M M ga g g

2M 0.05M 41M 41 = = = +

\ g = 0.18 41 = 7.38 m/sec2 = 738 cm/sec2.3. Ans. 020

Sol. rl

ghA F = Vrl

g = Mg

1.2 103 10 .1 103 104 F = 10 10

F = 20N4. Ans. 150

Sol. CP = C

V + R = 3R

T0 = 0 0

P VnR

Tf = 0 f 0 0

P V P VnR 2nR

=

0 0f 0

P VT T T

2nR-D = - =

\ DQ = nCPDT = n 3R 0 0P V

2nR = 1.5 P

0V

0 =

1.5 105 1 J

5. Ans. 225

Sol. Net force on liquid = mliquid

a = 1.8 1000

1.25 = 2250 N

SECTION-IV1. Ans. 1

Sol.hl

q

21 gsin t2

= ql

2gh tsin 2

= =q

l

t = 2hg

so time does not depend upon inclination q. Itonly depends upon vertical height.

2. Ans. 6

Sol. 11

Tv = m ,

12

2

T vv

2= =m

At = 2

1 2

2vv v+ Ai =

2 v / 29

v v / 2 +

= 6 mm

3. Ans. 3

Sol.

qvp/g

vt/g

vp/t

t / g p / gv v cos= q

q

30

vt/g

vp/gvp/t

( )t / g p /gv v

sin 30 sin 30=

q - 4. Ans. 5

01CT113053KOTA / HS - 6



PAPER 1

PART-3 : CHEMISTRY SOLUTIONSECTION-I

1. Ans. (D)

2. Ans. (B)

Sol. 1.5 10-3 (10-2x) = 0.03 5 40

1000

x = 3

3. Ans. (A)

4. Ans. (C)

5. Ans. (B,C)

6. Ans. (A,B,C,D)

7. Ans. (A,B,C)

8. Ans. (A, C)

9. Ans. (C)

10. Ans. (A)

Q En2 > En1 n2 > n1( ) ( )2 1 2 1n n n n 1

2

- - + = 6 & 2 2

n (n 1)15

2- =

n2 = 6 & n1 = 3.No. of lines in Balmer series = 6 2 = 4

En2 En1 = 4.53 = E6 E3

13.6 2 2Z Z

13.6 4.5336 9

+ =

Z = 2\ species is He+ ion.

11 Ans. (C)

12. Ans.(C)

SECTION-III

1. Ans. 030

5ml10 ml

2 210ml

1CO O CO

2+

Vf = 2COV + Volume of remaining air.

= 10 + 20 = 30 ml

2. Ans. 038

6BrO3 + 10Cr3+ + 22H2O 3Br2 + 10HCrO4

+ 34H+

x = 6, y = 10, z = 22

(x + y + z) = 38

3. Ans. (004)

4. Ans. (005)

5. Ans. (002)

SECTION-IV

1. Ans. 1

P=3atm

N (g)+3H (g) 2NH (g)

2 2

3

T=300KV=24.63 L

N2(g) + 3H2(g) 2NH3(g) 1 3

1x 33x 2x

Total moles at equilibrium

= 1 x + 3 3x + 2x

= 4 2x

Now from ideal gas equation -

PV = nRT

3 24.63 = n 0.0821 300

n 3=\ 4 2x = 3

2x 1=\ Moles of NH3 produced = 2x = 1 mol.2. Ans. 3

3. Ans. 6

4. Ans. 3

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KOTA / HS - 7

PART-1 : MATHEMATICS SOLUTIONSECTION-I

1. Ans. (C)(1000 1 10)2 10 = 19,990

2. Ans. (C)((x)) = 2 (x) = 0 or (x) = 3

1 solution two solutions

Number of solutions = 3.3. Ans. (A)

a=6 a=10

a b > 4

a b < 4

a + b = 16

a b

7 9

8 8

9 7

Number of solution = 3.

4. Ans. (B)y=sin(cos x)1

1 1

y = x + 1

5. Ans. (B)1

h(x) 33

1

g 13

= - & g(3) = 1a

b 13

+ = - and 3a + b = 1

3 5a , b4 4

= = - 1a b+ = 2

PAPER-2


PAPER CODETM


6. Ans.(A,C)DOA'A

1cos15

OA = 45

45

BAy=(2 3)x

0 A' x=115

OA = sec15OB = OA cos45

23 1

3 1= = -

+ = AB

( ) ( )( )B 3 1 cos60 , 3 1 sin 60 - - 3 1 3 3

,2 2

- - 7. Ans. (A,B,C)

a2 + 1 = sinq cosq + a.sinq + a.cosqa2 + sin2q + cos2q = sinq cosq + asinq + acosq sinq = cosq = a

tanq = 1 & 1a2

= or 1a2

= -

sum of possible values of 'a' is zero.tanq = 1 has 4 solution in [0,4p].

8. Ans. (C,D)

( )( )( )50

rr 1

C r 1T 1

r 1 r 2 r 3 r+

+= + + +

51 r 1r 3

- + , 48 > 2r, r < 24

T24 & T25 are numerically grestest term.9. Ans. (A,B,C)

ab = 1 Also1b =a a

2 = 3a 1

3 1a -a = a & 1

3 1a= b =

a a -10. Ans. (A,B,C)

Obuiously all options are correct except (D)

01CT113054KOTA / HS - 8



PAPER 2

11. Ans. (D)Let N = 10a + b10a + b = ab + a + b10a = ab + a b = 9

12. Ans. (D)S(S(N)) = 3 S(N) = 12 or S(N) = 21 or S(N) = 3 or S(N) = 30

N 93

39

84

48

75

57

66

= N 3012

21

=

10 possible values of N.13. Ans. (A)

(x) = cos1(2x2 1)Put x = cosq\ q = cos1x, 0 < q < p\ 1y cos cos2 , 0 2-= q q p

2 , 0 2y

2 2 , 2 2

q q p= p - q p < q p1

1

2 cos x, 0 x 1y

2 2 cos x, 1 x 0

-

-

= p - - r re r2 2> r re1 1

22 0 DT > 0

9. Ans. (B, D)

10. Ans. (A, B, D)

11. Ans. (D)

Sol. 1m4l = l = 4m

12. Ans. (A)

Sol. v = fl f = v 1 0.25Hz4

= =l

13. Ans. (A)

14. Ans. (C)

15. Ans. (C)

Sol. P P0 =

2

20

0

1 vv 1

2 v

r -

=

22 200 0

A1 1v 1 v

2 A 2

r - r

v0 =

( )02 P P-r

01CT113054KOTA / HS - 10



PAPER 2

16. Ans. (A)

Sol. F = vrel

dmdt

= 0dV

vdt

r = v0(rA0v0) = rA0v02

= rA0 ( )02 P P - r

= 2A0(P P

0)

SECTIONII

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

Sol. For (A) :

tnet

about OO' = 0 L is conserved (P)

t about O on A of mg 0. (Q)

Since KE of system increase

work done by internal force is not zero.(S)

For (B) :

tnet

about OO' = 0 L is conserved (P)

telec

about O on A is zero, but tmg

about O is

non-zero on A. (Q)

Fext

is horizontal direction is zero. (R)

Since KE of system changes work is doneby internal forces (S).

Fexternal net

= 0 COM is at rest (T)

For (C) : Since there is torque of mg about

OO' L is not conserved.

Fext

in horizontal direction is zero.

For (D) : Fext

(friction) is present between

boy & ground in horizontal direction.

(Not R & T)

SECTIONIII

1. Ans. 066

Sol.2

2 11

I10 log

Ib - b =

b2 60 = 10 log 4 = 10 2 0.3010 b

2 = 66

dB

2. Ans. 005

Sol. P0 + r

mgH r

mgy = P

air

rmg (0.8 + 1) r

mgy = P

air

Pair

= rmg (1.8 y) ...(i)

Also, P0(pr2)h = P

air(pr2) (h y)

( )( )

mair

0.8 hgP

h y

r = - ...(ii)

Solving (i) & (ii)

y 0.25 m

3. Ans. 040

Sol.x

30m 30m

2 260 2 x 30300 1000

+- = 0.1

4. Ans. 040

Sol.dp dm 10

F v 1000 2ghdt dt 1000

= = =

10 2 10 0.8 = 40 N.

KOTA / HS - 1101CT113054



PAPER 2

PART-3 : CHEMISTRY SOLUTIONSECTIONI

1. Ans. (B)

2O gas

gas

r M3r 2 32

= = Mgas = 48

Vm, real = 1000800

48 l = 60 l

Vm, ideal = RT 0.08 500

40P 1

= = l

Z = 60 340 2

=

2. Ans. (B)

3. Ans. (C)

4. Ans. (B)

5. Ans. (C)

6. Ans. (A, C)

7. Ans. (B, C)

8. Ans. (B,C)

9. Ans. (A,B,D)

10. Ans. (A,B,C,D)

11 Ans. (C)

12. Ans. (C)

Sol. H2 (g) + I2 (g) 2HI (g)

I Eqm 1 x/2 3x/2 x

II Eqmx x32 2

- - x x32 2

- - x + x

or (3x) 3 x 2x

2 2

2

x (2x)x x (3 x)1 32 2

= - - - x = 1.5 Ans.

KC =2x

x x1 32 2

- -

=2(1.5)

(1 0.75) (3 0.75)- - =4 Ans.]

13. Ans. (C)

14. Ans. (C)

15. Ans. (C)

16. Ans. (C)

SECTIONII

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

SECTIONIII

1. Ans. 004

2. Ans. 018

3. Ans. 120

4. Ans. 002

test 3 paper key

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