adva pii sol

8/20/2019 Adva PII Sol

1/9

FIITJEE L t d . , F I I T JEE Hou se, 2 9 -A , Ka l u Sa r a i , Sa r va p r i y a V i h a r , New De l h i -1 1 00 16 , Ph 461 06000 , 2 65694 93 , Fa x 265139 42 websi t e : ww w . f i i t j e e. c om

ANSWERS, HINTS & SOLUTIONS

FULL TEST – IIPAPER-2

QQ..NNOO PHYSICS CHEMISTRY MATHEMATICS

1. A C A

2. B C A

3. C B B

4. B A A

5. C B A 6. A A A

7. D A B

8. B B A

9. D B A

10. A B A

11. A A B

12. B A C

13. C C C

14. C D B

15. B C C

16. A D C

17. B A D

18. A D B

19. A B A

1.

(A) (q, r), (B) (q, r, s)

(C) (p), (D) (p)

(A) (r), (B) (p)

(C) (s), (D) (q)

(A) (p), (B) (q)

(C) (r), (D) (s)

2.(A) (r), (B) (p, q, r, s)

(C) (q), (D) (s)

(A) (p), (B) (q)

(C) (r), (D) (s)

(A) (r), (B) (s)

(C) (q), (D) (p)

3.

(A) (s)

(B) (r)

(C) (p)

(D) (q)

(A) (p, q, r, s)

(B) (p, q, s)

(C) (r)

(D) (p, r)

(A) (s)

(B) (p)

(C) (r)

(D) (q)

A

L L

I N D I A

T E S T

S E R

I E S

FIITJEE JEE(Advanced)-2014

F r o m C

l a s s r o o m / I n t e g r a t e d

S c h o o l P r o g r a m s 7 i n T o p 2 0 , 2 3

i n T o p 1 0 0 , 5 4 i n T o p 3 0 0 , 1 0 6 i n

T o p 5 0 0 A l l I n d i a R a n k s &

2 3 1 4 S

t u d e n t s f r o m

C l a s s r o o m /

I n t e g r a

t e d S c h o o l P r o g r a m s &

3 7 2 3 S t u

d e n t s f r o m A

l l P r o g r a m s h a v e b e

e n A w a r d e d a R a n k i n J E E

( A d v a

n c e d ) , 2 0 1 3


2/9

AITS-FT-II(Paper-2)-PCM(Sol)-JEE(Advanced)/14


2

P P h h y y s s i i c c s s PART – I

3. According to the question

2 e

e

g hg 1

Rh1

R

2

2

e e

h h1 0

R R

h = e5 1 R

2

4. From figure it is clear180 (i r)

andsini

sinr

sin603

sinr

r = 30 = 90

i = 60 i = 60 r r

r

6. v1

B

mM

v2

C M

16. F = qvB sin = qvB ( = 90)

So, B =F

qv =

20

19 3

3.2 10

1.6 10 4 10

= 5 10

-7 T

Now,p q 70I I

5 102 5 2

I = 4 amp.If the distance of point R from third current carrying current is X, thenBR = 0

70 2 2.5 5 10 04

so x = 1 m

19. P(2r)dr = 2r dvdr

2(r+dr) dvdr

P(2r)dr = 2 dv

dr

dr

P(2r) = 2 02

2v r

R

P = 02

2 v

R


3/9

AITS-FT-II (Paper-2)-PCM(Sol)-JEE(Advanced)/14

FIITJEE L t d . , F I I T JEE Hou se, 2 9 -A , Ka l u Sa r a i , Sa r va p r i y a V i h a r , New De l h i -1 1 00 16 , Ph 461 06000 , 2 65694 93 , Fa x 265139 42

websi t e : ww w . f i i t j e e. c om

3

C C h h e e m m i i s s t t r r y y PART – II

1.

O

H

O

H

2 2OH CH CH OH

OHO+

OH

H

2. (III) is most reactive (resonance activation) followed by N (inductive activation). (II) is moredeactivated (resonance deactivation) followed by (I) (inductive deactivation).

3.1

KE

1 2

2 1

KEKE

1

1

0.99

2

1 2KE 0.99 KE

KE2 1.02 KE1

% change is KE = 2 1

1

KE KE100

KE

2%

4. H 0, S 0 Reaction may be non-spontaneous at 25oC

G H T S = 180 – 298 × 150 × 10

-3

= 135.3 > 0= Non-spontaneous

To make it spontaneous G 0 . We have to increase the temperature.3H 180 10

TS 150

= 1200 K = 927oC.

11. Above critical temperature (TC), gas can not be liquified on cooling, the average energy ofmolecule decreases.

12. Positive charge on nitrogen of diazo group is stabilized by electron releasing group.

13. Within amino acid, proton is accepted from COOH group by NH2 group to from

3" COO R NH "

.

14. More the number of alkyl substitute at double bond, greater its thermodynamic stability.

15. C – H bond is broken in non rate determining step, therefore, substitution of -H by deuteriumdoesn’t affect the rate of reaction.


4/9



4

16. There are total four types of -H and two types of carbonyl, hence a total of eight aldol would beformed.

17. ocellE 2.37 0.8

= 3.17 V

o c0.059E logK2

clogK 107.45

18. Maximum work = nFE= 6 × 10

2 kJ


5/9




5

M M a a t t h h e e m m a a t t i i c c s s PART – III

1. Perfect square = 100 – 1 = 9(excluding one)

Perfect cubes = 31100 3/1

Perfect 4th powers = 1/ 4100 1 2

Perfect 5th powers = 11100 5/1 Perfect 6

th powers = 11100 6/1

Now, perfect 4th

powers have already been counted in perfect squares and perfect 6th powers

have been counted with perfect squares as well as with perfect cubesHence the total ways = 9 + 3 + 1 – 1 = 12

2. [sin –1

x] > [cos –1

x] x > 0

cos1-1 sin1

1

x

y

y =1/2

O

Clearly [cos –1

x] =

]1,1(cosx,0

1cos,0x,1

[sin –1x] =

1,1sinx,1

)1sin,0[x,0

Hence [sin –1x] > [cos –1x] x [sin 1, 1].

3. As roots are of opposite sign, product of roots < 0 a2 + b2 – 1 < 0 a2 + b2 < 1 |a + ib| < 1So, the point a + ib l ies inside a circle of centre (0, 0) and radius 1.


6/9



6

4. |z – 2 + 3i| = 6 is a circle with centre (2, –3) and radius 6and |z – 4 – i| = |z – 12 – i| is the perpendicular bisectorof (4, 1) and (12, 1) is a line parallel to axis imaginary.Now this lineThe line is tangent to circle at complex number (8 – 3i).Hence only one complex number satisfies the above

equation.

(8, 1)

1

(2, -3)(8, -3)

(4, 1) (12, 1)

6. Coordinates of point T (a cos , 0) so distance from focus of the point T is a (e cos )

7. There are 8 even and 9 odd numbers. So probabilities of getting first even number is8

17 and

probabilities of getting second odd number =9

17

, so required probabilities =8 9 72

17 17 729

8. BI = r cosecB

2 = 4R sin

A

2 sin

C

2

BI1 = r 1 secB

2 = 4R sin

A

2 cos

C

2

II1 =2 2(BI) (BI ) = 4R sin

A

2

1II = 4R Asin 2 R =15

8

A

B C

I1

I

90

9. yr =1 2 3 n 1

1 1 1 ....... 1r r r r

log y =n 1

p 1

1 plog 1

r r

nlim logy

=

k

0

log(1 x) dx = (k + 1) log(1 + k) – k

11. px2 + 4xy + qy2 + 4a(x + y + 1) = 0 represents pair of straight lines iff

4apq + 16a2 – 4a2p – 4a2q – 16 a2 = 0

42 – 4a + ap + aq – pq = 0.For real , 16a2 – 4.4(ap + aq – pq) 0 (a – p)(a – q) 0 a p or a q

12. f(x) = ax2 – bx + 2

f(0) = 2f(– 1) = a + b + 2 < 0 ( a + b + 4 < 0)


7/9




7

f(0) f(– 1) < 0 One root lie between (– 1, 0)Nothing can be said about ab

13. Here only one condition is given. So degree is 2.

14. Any point on the parabola is (t2, t). Shift the origin to (–4, 0) so that the line becomes X + y = 0and the point (t2, t) becomes (4 + t2, t) where X = x + 4. If (X1, y1) is the image of (4 + t

2, t) in

X + y = 0, then2

1

1

X 0 1 4 t

y 1 0 t

X1 = –t, y1 = –(4 + t2) and in the original coordinates x1 = –4 – t, y1 = –4 – t

2

the equation of image is (x + 4)2 = –(y + 4).

15. Shift the origin to the point (0, 3) so that any point (x1, y1) on the reflected line is given by

1

1

1 q(h 3)x 1 0

py 3 0 1

h

(since m = 0)

1 11 q(h 3)x , y 3 h

p px1 = qy1 + 1 – 6q

and hence the reflected line is px qy = 1 6q.

16. Equation of the tangent to the given circle at (2, – 1) is 2x + y – 3 = 0.Shift the origin to the point (–2, 0) so that the two lines becomes y = X and 2X + y – 7 = 0. Any

point on the line is (h, 7 – 2h) and its reflection in y = X is given by 1

1

X 0 1 h

y 1 0 y 2h

X1 = 7 – 2h, y1 = h X1 = 7 – 2y1 x1 + 2 = 7 – 2y1 and hence the reflection of the tangent is

2y + x – 5 = 0 or y =x 5

( 2)2 4

which touch the parabola y2 = – 5x.

17. In the new definition l =x

x y z , etc.

18. d(O, P) = k |x| + |y| + |z| = kwhich represents a set of 9 planes makingintercepts of lengths k on positive as wellnegative sides of all three axes. See theadjacent figure.

ZX

Y

(0, 0, k)(k, 0, 0)

(0, k, 0)

(0, 0, k)

Z

X

Y

(0, k, 0)

19. The maximum value of d(O, P) = circumradius of sphere = a.


8/9



8

SECTION – B

1. (A) I =

2 2 2

2 2 2 2 2 2 2 22 2 2 2

dx x 2xdx x a x ax 2 dx

a x a x a xa x a x

=

2

2 2 22 2

x dx

2I 2aa x a x

2 2 2 22 2

dx 1 dx

2a a xa x

= 2 2 2

xc

2a a x

.

(B) Put x = a sec dx = a sec tan d

I =2 2

2 2 2 2

asec tan d 1 x asin c c

a sec a tan a a x

.(C)

Put x = a sin dx = a cos d

I = 22 2

a cos acos dcot d

a sin

= –cot – + c

= – 2 2 2 2

1 1a x x a x xsin c cos cx a x a

.

(D) Put x = a sec dx = a sec tan d

I = 2a tan asec tan d

a tan dasec

= a tan – a + c

= 2 2 1 xx a asec ca

.

2. (A) Required area =

4

0

12 xdx 4 4

2

= 4

3 / 20

4 8x 8

3 3

(B) / 2

5 5

0

sin x cos x dx

=

/ 2

5

0

4 2 162 sin xdx 2 5 3 15

(C) For equation S + k = 0 to represent pair of lines

1 2 2

2 3 1 0

2 1 1 k

3(1 + k) – 1 – 2(2 + 2k + 2) – 2(2 + 6) = 0 k = –22


9/9




9

(D) Let p.v. of given points be ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ A i j k , B 2i 2 j k and C 3i k , so that two vectors in

the plane may be ˆ ˆ AB i j

and ˆ ˆ ˆ AC 2i j 2k

Thus,

2 1

1 1 0 0

2 1 2

–2 – 2(–2) + (–1 – 2) = 0

=1

2

3. (A) The equation will have roots of opposite sign if it has real roots and product of roots is

negative 4 (b2 + 1)

2 12 (b2 3b + 2) 0 and

2b 3b 20

3

1 < b < 2

(B) The probability of problem being solved is 1 P A P B P C

= 1

1 1

1 1 12 3

=2 2

, 13 3 3

(C) x = 5 – (y + z)yz + x (y + z) = 8yz + (y + z) (5 – (y + z)) – 8 = 0y2 + y (z – 5) + (z2 – 5z + 8) = 0For real solution, (z – 5)

2 – 4 (z

2 – 5z + 8) 0

(z – 1)7

z 03

1 z 7

3

adva pii sol

Documents