IIT Study Circle Paper1

8/8/2019 IIT Study Circle Paper1

1/14

xa/2

a, d

4a2

yx2d

F F

f

f

MOCK TEST PAPER I

SOLUTION

1. (B) f F ma 0 = > W. D. = F x f x 0 <

2. (A) ( ) ( )y yB ATcos

a g a gm

= > =

( ) ( )B AB A

y h y

t t

= =

W X Z + ( )120 7.5 90 8 30 5 > +

W 2y ( )120 7.5 2 60 8.5 < energy released

X y Z + ( )90 8 60 83.5 30 5 > +


3/14

V

dqu

ql

V

d

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

( )

0 0

eq

0

A Ak 2 2C

d dk 1 A

2d

= +

+ =

0Aq k v2d = l

0u

Aq v

2d =

11. (A) (B), (C)

A

B

C

2

A

2

B

2

C

va i

R

va a i a j

Rv

a 2ai jR

= =

=

r$

r$ $

r$ $

12. (A), (B)

E.dl 0

g.dl 0

=

=

ur ur

r ur

13. (B)

( )F f r 0F f

= =

( )

Fsin f mgsin

mgsinF

1 sin

+ =

= +


4/14

airmedium

S1

S2D

O

14. (C) ( ) ( )F N mg F cos = /

1 mgsintan ; F1 sin3

= = + Q

15. (A)

( )

( )

mgsin Fsin f

F f mgsin

F1 sin

= + +

=

= + +

for, ( )minF , sin 1 + =

2 6 3

= =

16. (A)

Path difference = xd

D

at Pfor maxima.

medium

xdn

D=

( )0nD nD

xd d t

= = +

( )20

Speed of nth order maxima dx nDat t dt d t

= = +

17. (B) fringe width( )0

D Dd d t = =

+

18.( )0

1t

+

decreases with time

hence pattern gets narrower central maxima in always at 0.


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C

AD

12

3 4

y

x

19. (A) (q)(B) (p)(C) (r)(D) (q)

As water flows out,

PC decrease as volume increase.

Dp also decrease.

2 2A A ctm A

A

1 h 1p v g p v O2 2 2p conct

+ + = + +

=

( )2 2B A A AB

1 h 1p 4v pg p v2 2 2p decrease

+ + = +

20. (A) (p),(q)(B) (p),(r)(C) (q),(s)


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H OH

H OH

COOH

COOH

H OH

H OH

COOH

OH

H 3

C 4

C5CH

2CH =

1CH

CH3

(D) (q), (s)

1 2 3 4

1 2 3 4

1 2 3 4

y 0 y 0 y 0 y 0

v 0 v 0 v 0 v 0

a 0 a 0 a 0 a 0

< > > > >

21. (B) For ( )4Zn Hg NCS we have K sp = S2

7S 2.2 10 0.000469= =

22. (B) [ ]10

14

max

10Ag 10 M1

= =

12

10remaining

10Br 0.01M

10

= =

Hence 99% of Br have get removed23. (B) The p ka of CH 3COOH is 4.74 hence mixture

(B) is best so that the ratio of the buffer components will be 1 resulting into abuffer with a high buffer capacity

24. (C) 3-methyl-pent-1-en-4-yne

25. (C) C = C C C C C C C C C

1-deceneC C = C C C C C C C C 2-deceneC C C = C C C C C C C 3-deceneC C C C = C C C C C C 4-deceneC C C C C = C C C C C 5-decene

So 4 of these can show cis trans isomerism26. (B) it has to be the meso compound which is

Which can be drawn as

27. (C) This is because CH 3 shows +I effect.

28. (B) There is no loss of C on ozonolysis


7/14

29. (A), (B), (C) Factual question

30.

31. (B), (C) There is lp lp repulsion in F 2 so the bond strength is lesser than Cl 2 .

32. (D) Carbon has small size so form bonds readily

33. (C) 2 3 2 3Na CO 2HCl 2NaCl H CO+ +

34. (A)[ ]HIn

H InK

HIn

+ = , so that ratio depends on HInK and pH .

35. (C) HPh will not exhibit color change during this titration.

36. (C) Since it contains the metal-carbon bond.

37. (C) Protic solvent cannot be used

38. (D) aldehyde + RMgX gives after hydrolysis a secondary alcohol

39. (A) (q)(B) (s)(C) (r)(D) (p)These are as per the VSEPR theory

40. (A) - (s)(B) - (r) (q) (s)( C) (q) (s)(D) - (s)

Assuming Neglection of No in s option

MOCK TEST PAPER

SOLUTION I

41.(C) The domain of this function ( ) [ ]xf x log x

= is ( )1,

For 1 < x < 2, f(x) = 0 and for n x n 1 < + where n I + and n 2 , the function f(x) is decreasing. Therefore, the maximumvalue of f(x) is 1.A = 1

( )g x sinx cosx

1 sinx cosx 2

= + +

minimum value of g(x) = 1, B = 1


8/14

A = B

42.(A) Let ( ) ( )x 2 x 3f x e x and g x e x = =

( ) ( ) ( )x 2 x x

f ' x e 2x x e e .x x 2

= + =

0

1

0 2 +

2

So, f(x) has one solution.

( ) ( )x 2 3 x x 2g ' x e .3x x e e x x 3 = =

( )3

3g 3 1

e = >

0 3

0 3

+ +1

Therefore. g(x) has two solutions

43.(D) Let ( )1 1x , y be the foot of the r from focus upon the tangent.( )

( ) ( )

1 12 2

1 1

3 1 8x 1 y 11

3 1 3 1x , y 2, 2

+ + += = = + =

With respect to (2, 2), images of (7, 13) and (1, 1) will lie on the axis and thedirectrix of the parabola. The two points are (3, 17) and (5, 3)

slope of the axis1 17

81 3

+= = +

equation of the directrix is, ( )1y 3 x 58

+ = x 8y 19 0 + + =

44.(B)

2

2

2 1 dxx x x

I 1 11

xx

+ = + +

Let 1 11 txx

+ + =

2

dt 2I 2 c

t t= = +

2xc

x x 1= +

+ +


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45.(D) For n 3 , 1 2 3 n 1na a a ...... a

an 1

+ + + +=

( ) n 1 2 n 1n 1 a a a ......... a = + + +

( )

1 2 nn 1

n n

nn

a a .......... aa

nn 1 a a

nna

an

++ + +=

+=

= =

n 1 n 9 3

1 2 23

2

a a a a 99

a a 19 aa 992 2a 179

+ = = =+ +

= ==

46.(B)

Total words =2 3 2 2 3

1 1 2 1 2

4! 4! 4! 4!C C C C C 4!

3! 2!2! 2! 4! + + + +

= 24 + 6 + 72 + 1 + 24 = 127

Probability that it has all letters distinct = 24

127.

47.(B) ( )( )3

2 22 2

1 1f g x x 3 x

x x = + +

( )( ) ( )3

g x 3g x= ( ) 3f x x 3x =

( ) 2f ' x 3x 3=

48.(A) Let the roots be k ,k,kr

r

31 a 1 1k r 1 , k k r 8 8 2 + + = = =

2

2

arr r 18

ar r 1 1 0

4

+ + =

+ + =

a a1 2 and 1 2

4 4


10/14

a 12, a 4

MORE THAN ONE CORRECT

49.(A),(B), (D) Let ( ) ( )( )g x 4 lnx lnx ..........= +

( )4, 1 x e

g x 5, x e

, x e

<

50.(B), (C) ( )f x 3sinx cosx 2 2sin x 26 = + = +

Since f(x) is one to one and onto, therefore f 1 exists.

( )( ) ( )1 1fof x x 2sin f x 2 x6

= + =

( )1 1 xf x sin 12 6

= +

x1 1

2 for all [ ]x 0,4

( )1 1

1

x 2f x cos

2 2 6

2 x 2cos3 2

= +

=

51.(A),(C) ( )( )

( ) ( )( )p x,y ,z

f x,y,zx y y z z x

= + + +

The polynomial ( )p x,y,z has degree 3and ( ) ( ) ( )p x,x,z p x,y,y p x,y,x 0= = = Thus p has factors x y, y z, z x upto a constant which turns out to be 1.

( )( ) ( )( )

( ) ( )( )

x y y z z xf x,y,z

x y y z z x

=+ + +

Since, x y x y < + y z y z < + z x z x < +

we get ( )f x,y,z 1< Using x y z, y z x, z x y < < <

we get ( ) z x yf x,y,zx y y z z x

< + + +


11/14

O

P (x , y )1 1

Q12

xy yz zx 1

x y y z z x 8=

The equation of the circle is x 2 + y2 = 10

1 1y 1, x 3= = ( ) ( )1 1P x ,y P 3,1=

OP 3i j= +uuur

Q lies on AP, A < Q < PQ (1, 1) or (2, 1) as abscises a is an integer

There are two values of OQuuur

OQ i j and 2i j= + +uuur


12/14

OP,OQ 3 1 4 and 6 1 7= + = + =uuur uuur

PQ OQ OP 2i j 3i j i

Also i j 3i j 2i

= = + = + =

uuur uuur uuur

Max and min of PQ 2=uuur

and 1 respectively.

PARAGRAPH II

Slope of tangent at P(x, y) =dy

Mdx

= Equation of tangent is y = mx mX + Y

The coordinate of Q Y YX ,2m 2

Y2tan

YX

2m

=

56.(B)If r rN and D are of same sign i.e., tan 0 >

Y tan

Xm Y2tan

X

=

Solving the differential equation the curve is found to be 2X Y 3Y= and tan 1 n

4 = = + , nI

but since4 = would be sufficient.

4

=

57.(D)When r rN and D are of opp sign, tan 0 <

2

Y tanXtan X Y 3Y

Y2tan

X

= =

and tan 1 =


13/14

AML

S

B N

T

K

O

4 =

, for Y 04

for Y 04

= > = =

(B) For f(x) to be have two equal roots,( )k 2 and f ( x ) f 2 0, k 1 = =

(C) For f(x) to be invertible ( )f ' x 0


14/14

x R k 2 = (D) For f(x) to have three real and distinct roots, ( ) ( )k 2, f x f 2 0 <

( )( )( )2

k 1 k 7k 7 4k 5 0k 4, 3, 2,6

+ + > =

60. (A) (r), (B) (s), (C) (q), (D) (q)

( ) ( )( )

1 11 1f x sin x tan x4x 1

x 1

= + + + + +

Domain of f(x) is [1, 1] a + b = 0

maximum value of g(x) is g(1) =

3

4 maximum value of h(x) occurs

When ( ) 4x 1x 1

+ + +is minimum

max value of ( ) 1h x4

=

Rang of f(x) is 3 ,14

3m 1

4 + = l 1

4=

IIT Study Circle Paper1

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