23 12 2012 JR Solutions Eng

RESONANCE SOLJRMT231212 - 1

PAPER-1PART-I (Physics)

1. Two nonconducting uniformly.....................

Sol. (B)R

KQe

R9KQe

= R

KQe +

R9KQe

+ 21

mV2

R9

KQe2 +

RKQe2

= 21

mV2

0416

R9

eQ =

21

mV2 V = mR9eQ8

0

2. Figure shows a circuit.......................Sol. (C) Pbulb = 80 i = 80 1

3. Circuit contains an uncharged.....................Sol. (D) ii = 1A

After switch is closed if = 68

AA

Then the difference in current is 31

A

4. The shown figure represents.......................Sol. (C) By Snell's law :

1

2

rsinisin

for i = 1, r = 4and 1 = 1.

2 = 4

1

sinsin

.

5. A non relativistic positive....................Ans. (B)

6.7.8.9.10.11.

12. Two thin slabs of refractive...............................Sol. (B,C) 1sin1 = 2sin 2

1 22 ba

a

= 2 22 dc

c

Since 1r

, n and 2r

are coplanar so 21 rn.r

= 0

13.14. The charge Q = C.......................

Ans. 8

Sol. W = QE

R2 = 8J.

15. A point object is placed at .........................Ans. 5

Sol.f1

= (n 1)

21 R1

R1

151

= (1.5 1)

R11

HINTS & SOLUTIONS

MAJOR TEST-1 (MT-1)

(JEE ADVANCE)TARGET : JEE (IITs) 2013

R = 2

15

Equivalent focal length f = n2R

= 5.122

15

=

25

cm

f1

= f2

fm1 =

R)1n(2

R2

f = n2R

system behaves as a concave mirror

u1

v1 =

f1

201

v1

=

52

v = 720

cm

So distance from O = 20 720

= 7

120 =

724

5.

16. In the given circuit.........................Ans. 2

Sol.

i = 5.15.11

412

=

48

= 2A

VEF = 3 2 = 6V 2V + V = 6V = 2

17. A man does work W on ..........................Ans. 6

Sol. Wearth = mgeh = m 34

GdR.h

Wplanet = m. 34

Gd4R

.2h planet

earth

W

W = 2

Wplanet = 2W

= 6W3

.

18.19.20.21.22.23.

PART-II (Chemistry)*24. Which of the following reactions of potassium ..................Ans. (B)

Sol. (2) KO2 + 2H2O KOH + H2O2 + 1/2O2

(3) 4KO2 + 2CO2 2K2CO3 + 3O2

*25. The disintegration rate constant for the ..................Ans.(B)

Sol. t1/2 = 31085.3

693.0

= 180 sec. or 3 min.

DATE : 23-12-2012COURSE NAME : VIJAY(JR)


*26. Which gas is liberated when Al4C3 ..................Ans. (A)Sol. Al4C3 is aluminium carbide, which upon hydrolysis gives CH4.

Al4Cl3 + 2H2O 4Al(OH)3 + 3CH427.

*28. 10 ml of 0.02 M weak monoacidic base ..........................Ans. (B)

Sol. BOH + HCl BCl + H2O

10 0.02 V 0.02

V = 02.02.0

= 10 ml

B+ + H2O BOH + H+

0.01 x x x

Kh = b

w

KK

= ]H[01.0

]H[ 2

12

14

10

10

= ]H[01.0

]H[ 2

[H+] = 6.18 103 M

29.

30.

31.*32. Which of the following are correct statements ................Ans.(AC)Sol. (A) G = H TS < O as S < O so H has to be negative

(B) micelles formation will take place above Tk and above CMC(D) Fe3+ ions will have greater flocculatibility power so smallerflocculating value.

*33. Which is / are the correct statement(s)................Ans. (ACD)

Sol. (A)

COOH

104

OP2CH

COOH

C3O2 (Carbon suboxide) + 2H2O

(B) [B4O5(OH)4]2 + 5H2O 2B(OH)3 + 2 [B(OH)4]

2[B(OH)4] + 2H3O+ 2B(OH)3 + 4H2O

Only [B(OH)4] formed in water reacts with HCl.

34.

*35. In which of the following reactions aromatic ................Ans.(ABD)

Sol. (A) Br

Br

4AgBF2+

+ ++ 6 e aromatic

(B) CH3

CH3 )Metal(K2

-

-

.

.

.

.

CH3

CH3 + H2 - -

10 e, aromatic

(C)

H Br

4AgBF +

H

4 e, antiaromatic

(D) Cl

H

t BuO-

10 e, aromatic

36.

*37. EMF of the following cell is 0.634 volt at .................Ans.6

Sol. At cathode :

21

Hg2Cl2(s) + e Hg(l) + Cl

(aq)

At anode :

21

H2(g) H+(aq) + e

______________________________

21

Hg2Cl2(s) + 21

H2(g) Hg(l) + Cl(aq) + H+(aq)

Ecell = 0cellE 1

059.0 log [Cl][H+]

or 0.634 = (0.28 0) + 0.059 pH

or pH = 059.0

28.0634.0 = 6

*38. How many of the following aqueous solution .................Ans. 5Sol. Acidic, Al2(SO4)3, CoBr2, NH4Cl, CO2, BCl3.

*39. Consider the following chemical reaction .................

Ans. 4

Sol. Assume rate law

r = K[H3AsO4]x [H3O

+]y [I]z

Solving by the help of various experiements

x = 1, y = 2 and z = 1

total order = 4

40.*41. Depending upon the nature of oxides, they .................Ans. 4Sol. P4O6, CrO3, CO2, NO2 are acidic oxides, CO and NO are neutral

oxides and PbO2 and SnO2 are amphoteric oxides.

*42. 1 mole of how many of the following acids .................Ans. 4Sol. H2SO4, H2SO3, H3PO3, H4P2O5 are diprotic. HCl, HNO3, H3PO2,

H3BO3 are monoprotic.

43.

44.

45.

46.PART-III (Mathematics)

47. If a 1x .sin , x 0

f x x

0, x 0

is .....................

Sol. (C)

f(0) = h

)0(f)h(flim

0h

=

hh1

sinhlim

a

0h

h1

sin.hlim 1a0h

This limit will not exist ifa 1 0 a 1.

Now , )x(flim0x

=

x1

sinxlim a0x

= 0 if a > 0.

Thus required set of values of a is (0, 1].

48. Set of all values of x .....................

Sol. (A) 1

x2tan4 1 1

4

tan1 2x 4

1 2x 1

x

21

,21


49. Solution of differential.....................

Sol. (D) dydx

= yx

+ yx

x

1 dydx

+

yx

= y

1

tx

1

2 x dydx

= dydt

dydt.2

yt

= y

1

y2

1t

y21

dydt

I.F. = dyy21

e = ny

21

e = y 1/2

t. y

1= Cdyy

1.

y2

1

Cny.21

y

x

x = y Cyln

50. All possible values of .....................Sol. (B) |A1 kI| = 0

|A||A1 kI| = 0 (|A| 0)|I kA| = 0

Ak

= 0 .k1

A = 0

|A I| = 0 where = k1

=

01

120

201

= 0

= (1 )( )(2 ) + 2(0 (2 )) = 0= 3 + 32 2 4 + 2= 0 = 3 32 + 4 = 0

= 2,2,1 k = 1, 21

51. Set of all values of .....................

Sol. (C)

xcos

21

21

xcos > 0

21

cos x < 0 i.e cos x > 21

i.e x

3,

3

I

n3

n2,3

n2

52.53.54.

55. Let f :

2,0 [0, 1] be .....................

Sol. (ABCD) Applying LMVT on y = f(x) for x

2,0

f (x) =

22/01

02

)0(f2

f

Hence , (B)

Let g(x) = (f(x))2

g(0) = (f(0))2 = 0 ; g

2 =

2

2f

= 1, By LMVT

g(x) =

2

02

)0(g2

g

2f(x) f (x) =

2 f(x)f (x) =

1Hence , (C)

Let g(x) = f(x) 2

2x4

g(0) = 0 ; g

2 = 0 By Rolles theorem g (c) = 0 for some

c

2,0

i.e. f (c) 2c8

= 0 f (c) = 2

c8

Hence , (D).

56. If f(x) = 1947 + .....................

Sol. (ABD) f(x) = 1947 +

x

1t1

1.)t(f dt

by Lebnitz theorem.

f(x) = f(x)

x1

1

So )x(f)x('f

dx =

x

11 dx

ln(f(x)) = x lnx + c ...(1)at x = 1, f(1) = 1947 c = ln(1947) 1By equation (1)

then F(x) = x

e 1x.1947

Option (A) is correctFor option (B)

f(1947) = 1947e 946

.1947 = e1946

f(2013) = 2013e2013

.1947

obvious f(2013) . > f(1947)for option (D)


2013

1947r1947

)r(ef =

2013

1947r

r

re

= 1947e1947

+

2012

1948r

r

re

+ 2013e2013

Then

2012

1948r

r

re

<

2012

1947r

r

re

<

2013

1947r1947

)r(ef

57.

58. An extreme value of .....................

Sol. (BC) 4 sin2x + 3 cos2x 24 (sin 2x

+ cos 2x

)

= 3 + sin2x 24 (sin 2x

+ cos 2x

)

Let sin 2x

+ cos 2x

= t.

Then t2 = 1 + sin x, 1 t 2 and the expression becomes3 + (t2 1)2 24t = t4 2t2 24t + 4

Let f(t) = t4 2t2 24t + 4f(t) = 4t3 4t 24 = 4(t 2) (t2 + 2t + 3) < 0

{ 1 t 2 } f(t) is a decreasing function

minimum value of f(t) is at t = 2 and minimum value

= 4 (1 6 2 )maximum value of f(t) is at t = 1 and maximum value

= 21

59. An intercept made by a.....................Sol. (AB) y = |x| = 2 x = 2

y = 2

0

dt|t| =

2

0

2

2t

= 2

Hence point of contact (2, 2)Equation of tangent (y 2) = 2(x 2)Intercept on x-axis = 1

For x = 2, y = 2

0

dt|t| = 2

0

tdt =

2

0

2

2t

= 2.

Point of contact ( 2, 2) hence intercept = 1.

60. Following usual notations .....................

Sol. (2) )CcosBcosA(cosR2Atan

a

2

Csin21

2

BAcos.

2

BAcos2R2 2

2BA

cos2

BAcos

2C

sin21R2

2C

sin.2B

sin.2A

sin41R2 = 2R + 2r = 22

61. Find the area of the .....................

Sol. (4) required area = 4 1

0

dx)nx(

= 1

0)xnxx(4 = 4 1 = 4

62.

63. If function f(x) = .....................Sol. (6) 3 = a , 2 + a = m + b , 5 = m + b

f(x) =

2x1m

1x03x2

m = 1, b = 4

f(x) =

2x14x

1x03x3x

0x32

, f(x) =

2x11

1x03x2

f(c) = 23

0236

, f (c) = 2c + 3

23

= 2c + 3 c = 43

c(a + m + b) = 6

64. If

dx)xx4(e 22x2

.....................

Sol. (1)

dx)xx4(e 22x2

= dxx2.x2.e2x

+ dxx.e 2x

2

= 2x 2xe dxe2

2x +

dxx.e 2x2

= 2x 2xe

dxx2ex2x1

+ dxx.e 2x

2

= 2x 2xe x1

2xe dxx.e 2x

2

+ dxx.e 2x

2

= 2x 2xe x1

2xe + c = c)xx2(e 1x2

a = 2, b = 1 a + b = 1

65.

66. If I1 =

1

0x )x1(e

dx .....................

Sol. (0) I1 =

1

0x )x1(e

dx; I2 =

4

032

tan

cos

sin.

tan2

e2

d

= 21

1

0

x

x2e

dx = 21

1

0

x1

)x1(2e

dx = 2e

1

0x )x1(e

dx

2

1

I

I =

e2


67. Find the number of .....................

Sol. (1) f(x) = 3x4 4x3 + 6x2 + ax + b

f (x) = 12x3 12x2 + 12x + a and f (x) = 36x2 24x + 12

= 12(3x2 2x + 1) > 0 x

f (x) = 0 can have only one real root

f(x) has only one critical point.

68. Let f(x) = x3 + x2 10x + 11.....................

Sol. (3) g (f(x)) = x g(f(x)) f (x) = 1

for g(3) we require f(x) = 3.

x3 + x2 10x + 8 = 0

(x 1) (x 2) (x + 4) = 0

x = 4 ( x < 0).

f (x) = 3x2 + 2x 10

f (4) = 48 8 10 = 30.

g(f(4)) . f ( 4) = 1 = 3

69.

PAPER-2PART-I (Physics)

1. Two concentric conducting...............................

Sol. (A) r < a E = 0

a < r < b E = 2r

KQ

r > b E = 2r

KQ r

2. In a meter bridge experiment..........................

Sol. (B)XR

= BCAB

= 41

; X = 4R

R8R

=

1 = 9

8m

3. Calculate the energy stored..........................

Ans. (D)

4.

5.

6.

7. A liquid of refractive index................................

Sol. (C)eqf1

= 1f1

+ 2f1

+ 3f1

= (1.5 1)

R11

+ (1.33 1)

R1

R1

(1.5 1)

1R1

= R5.0

R165.0

+ R5.0

eqf1

= R835.0

feq = 835.0R

> 0

So converging

Similarly Q is also converging

8.

9. A converging mirror..........................

Ans. (D)

10.

11. Calculate maximum velocity..........................

Sol. (C) In triangle PMC

cos53 = MCMP

53

= R4

R

12 = 8R

R = 23

m (R is the maximum radius of halfcircle)

Rmax = qB

mumax Umax = 3 m/s.


12. In previous question, .....................

Sol. (B) R = qBmu

= 24 m

Let, MPQ =

By geometry,

CPO = (37 )

In CPO,

)CPOsin(OC

= )PCOsin(OP

)37sin(20

= )37180sin(

24

)37sin(5

=

365

sin(37 ) = 21

= 1807

rad.

= mqB

= 2 rad/sec. t = 3607

sec.

13. Calculate emf of .......................... Ans. (C)

14. Calculate the internal........................ Ans. (B)

Sol.(13 to 14)

case-I S2 is open

Potential gradient = LE

so 6 = LE

2L

E = 12 V

case -II S2 is closed

i = r10

6

6 ir = LE

12L5

6 r10

6

r = 5

10rr6

= 1 6r = r + 10 r = 2

15. Charge on plate 1, when......................

Sol. (C) At point P, Enet = 0

q3 = 2q

So, charge on outer surfaces of

(1) and (2) is 2q

and it will not

change with time

(1) and (2) have same potential

0 + 0

1

0

1

A)qq(

A)vt(q

= 0

q1( + vt) = (q q1) q1 = )vt2(q

So, at t = v

q1 = )2(q

=

3q

So, charge on plate (1) at that time = 6q

3q

2q

.

16. Current in wire when..............................

Sol. (B) As charge on outer surface of plate (1) doesn't changes

with time. So only charge on inner surface will change

i = dtd

(q1) =

vt2q

dtd

i = 2)vt2(

vq

At t = v2

, i = 16

qv.

17.

18.

19. In the given circuit, the potential...........................

Ans. 15

Sol.R12

= R

V12

= 24 V

RV

+R

V12 +

RV0

= 0

3V + 12 = 0

3 (24 ) + 12 = 0

72 + 3 + 12 = 0

4 = 60 = 15 V


20. A ray of light is incident ................................

Ans. 25

Sol. = ( 1) A

= (1.5 1) 1 + (2 1) 2 = 0.5 + 2 = 2.5

21. A battery of internal resistance.............................

Ans. 19

22. A point chare 9C is placed..........................

Ans. 30

Sol. V = CBkq

= 27

109109 69 =

27100081

= 3000 volt

23.

PART-II (Chemistry)

*24. When H2O2(aq) is allowed to decompose ..............

Ans. (A)

Sol. H2O2 2O + 21

O2

In R1, R2 and R3 moles of H2O2 taken initially are 0.2, 0.5, 0.3

therefore moles {or volume} of oxygen produced will also follow

order R2 > R3 > R1

Graph : (1) (2) (3)

*25. Consider the following nuclear reactions..................

Ans. (E)

Sol. He2NM 42xy

23892

where NN 23088xy

2LN AB

xy

LL 23086AB

Total number of neutrons in L23086 is = 230 86 = 144 Ans.

*26. Which of the following statements are correct

(I) Boiling point of H2O is higher than H2Te.....................

Ans. (D)

Sol. (I) On account of Hbonding H2O has higher melting and boiling

points.

H2O H2Te

m.p/K 273 222

b.p/K 373 269

(II)

(IV) B2B.O = 1 and B2+ B.O =

21

; C2 B.O. = 2 and C2+ B.O. = 1.5

N2 B.O = 2.5 and N2 B.O = 3 ; O2 B.O. = 2 and O2

+ B.O. = 2.5

Bond strength bond order.

27.

28.

29.

30.

31.

32.

33.

*34. After passing 20 amp current from battery ...................

Ans. (D)

Sol. Deposition order Cu > M > Zn

Cu required 96500

t =

Ew

20 t = 1 2 2 96500

t = 19300 sec.

In 28950 19300 = 9650 sec M2+, will deposit as M.

mole of M 2 = 96500

965020

mole of M deposit = 1

Initial mole of M2+ = 1 2 ; remain M2+ moles = 2 1 = 1 ;

[M2+] = 21

= 0.5

*35. What will be the volume of gas formed at ...................

Ans. (A)

Sol. Total charge = 96500

2895020 = 6F

Mole of Br2 formed = 2

Mole of Cl2 formed = 1.

36.

37.

38.

39.


*40. Match the reactions in Column-I with nature ..............

Ans. (A) q, r, s ; (B) p ; (C) q, r ; (D) p, q, r

Sol. (A) B2H6(g) + 6H2O() 2H3BO3(aq) + 6H2(g)

B(OH)3 + 2H2O [B(OH)4] + H3O

+ (monobasic Lewis

acid)

trigonal planar

(B) 3MnO42 + 3H2O

ionationdisproport 2MnO4

(aq) +

MnO(OH)2 (s) + 4OH (aq).

MnO4 as well as MnO(OH)2 both act as oxidising agents.

OH, acts as ligand in many coordination complexes like

[Cr(OH)4] etc.

(C) CaCN2(s) + 3H2O() CaCO3 (s) + 2NH3(g)

pyramidal

(D) 2CsO2 + 2H2O 2CsOH + H2O2 + O2

*41. The rate of change of ln k with temperature ..............

Ans. 50 kJ/mol

Sol. K = AeEq/RT

or nK = nA 2RT

Ea

Given 2RT

Ea = 0.0668

or Ea = 0.0668 8.314 103 3002 = mol

kJ 50

*42. How many next nearest neighbours are present ..............

Ans. 12

Sol. Next nearest neighbour of Zn+2 would be = no of nearest sur-

rounded Zn2+ ions

next nearest neighbour of S2 would be = no of nearest S2 ions

= 12 (due to FCC)

and their number of neighbour ratio is 1 : 1 and that make Zn2+

neighbour are 12.

*43. The number of six membered carbon rings in the ..............

Ans. 20

Sol. It contains twenty sixmembered rings and twelve five

membered rings.

*44. Let MgTiO3 exists in pervoskite structure in which..............

Ans. 36

Sol. When all the atoms of face diagonal of a face are removed.

81

4Mg

25TiO

d = 24323 10)

65

2(106

25

16482421

1

=

333555

1.6

100

= 36 gm/cc

45.

46.

PART-III (Mathematics)

47. The value of .............................

Sol. (C)

I = 4

2

0sin x cos x dx

=

2

0 4xcosxsin

dx =

22

40

sec xdx

tanx 1

tanx = t2

041t

tdt2

= 2

0 43dt

)1t(

1

)1t(

1= 2

032 )1t(3

1

)1t(2

1

31

21

02 = 31

48. Range of the function.............................

Sol. (D) f(x) = 2 + 23x4x7x

324

Let h(x) = x4 7x2 4x + 23

= (x2 4)2 + (x 2)2 + 3

Range of h(x) [3, ) Range of 'f ' (2, 3]


49. If the slope of tangent to.............................

Sol. (B) dxdy

= xcos

xsiny 2

dxdy

+

xcos

1 y =

xcosxsin2

I.F. = xdxsece = )xtanx(secne = xtanxsec

1

y. xtanxsec

1

= dxxcosxsin

.xsin1

xcos 2 = dxxsin1

xsin2

y. xsin1

xcos

=

dx

xsin11

1xsin

=

dxxcos

xsin1xxcos

2

= x cosx + tanx secx + c

xsin1xcos.y

= x cosx + tanx secx + c at x = 0, y = 0 , c = 2

1y

= + 1 + 1 + 2 y 4

50. The number of integral values .......................

Sol. (D) Let f(x) = 2tanx + 2x + sin2x

f (x) = 2sec2x + 2 + 2cos2x > 0 x

4,

4 f is M. I.

a

4f,

4

f

f

4

= 2 2

1 = 3 2

f

4 = 2 + 2

+ 1 = 3 + 2

Number of integral values of a is '9'

51.

52. Let f(x) =

1|x|2x

cos

1|x|)x(g1|x|

,....................

Sol. (A) LHL at x = 1 = h 0lim f(1 h)

=

1 h 1

h 0lim cos (1 h

2

; p =

h

0h 2h

sinlim

lnp = 2h

sinnhlim0h

=

h/12h

sinnlim

0h

= 2/hsin2/hcos

lim0h

2

. (h2)

=

2h

tan

h2

.hlim0h

= 0

f(1) = 1 g(1) = 1

RHL at x = 1

RHL =

1h1

0hh1(

2coslim

=

h0

0h 2h

sinlim

= 1

53.

54. If f(x) = cosec 2x + cosec 22x .........................

Sol. (B) g(x) = cosec 2x + cosec22x + ....... + cosec 2nx + cot 2nx

since cosec + cot = cot 2

g(x) = cot x

0xlim (cos x)cotx + (sec x)cosecx

= xcot)1x(coslim

0xe

+ ecxcos)1x(seclim

0xe

= xcos1xsinxcos

lim0xe

+ )1x(secxsin

xtanlim

2

0xe = e0 + e0 = 2


55. The least value of.............................

Sol. (A) If x (1, 2) (3, 4), then

tan1 )4x()2x()3x()1x(

cot1 )3x()1x(

)4x()2x(

= tan1 )4x()2x()3x()1x(

)4x()2x()3x()1x(

tan 1 =

If x (, 1) (2, 3) (4, ), then

tan1 )4x()2x()3x()1x(

cot1 )3x()1x(

)4x()2x(

= tan1 )4x()2x()3x()1x(

)4x()2x()3x()1x(

tan 1

= 0

least value =

56. Let f : R [1, 1] defined .............................

Sol. (D) y = f (x) = 2

2

x 1

x 1

= 1 2

2

x 1

f ' (x) = 22 )1x(

x4

x > 0, f is increasing and x < 0, f is decreasing

range is [1, 1) into

57.

58.

59.

60.

61. cosec1 5 + cosec1 65 .............................

Sol. (D) cosec1 x = tan11x

12

T1 = cosec1 5 = tan1 2

1 = tan1

42

= tan1

3.11

13 = tan1 3

tan1 1

T2 = cosec1 65 = tan1 8

1 = tan1

162

= tan1

3.51

35= tan

1 5 tan1 3

T3 = cosec1 325 = tan1 18

1 = tan1

362

= tan1

7.51

57

= tan1 7 tan1 5

Similarly Tn = tan1(2n + 1) tan1 (2n 1)

Adding all terms we get

sum = tan1 (2n + 1) tan1 1 = tan1

1nn

as n sum = tan1 (1) = 4

62.

n

1r42

1

n rr2

r2tanlim .............................

Sol. (B)

n

1r42

142

n

1r

1

)rr1(1

r2tan

rr2

r2tan

n

1r22

221

)rr1)(rr1(1

)rr1()rr1(tan

n

1r

2121 rr1tanrr1tan

tan1 (1 + n + n2) tan1 1

1tannn1tanlimrr2r2

tanlim 121n

n

1r42

1

n

442

63.

64. Let f(x), g(x) be non .....................

Ans. 01

Sol. Diff. w.r.t. y.

f(x + y) = f(x) f(y) - g(x) g(y)

g(x+y) = g(x) f(y) + f(x) g(y)

Put y = 0 f(x) = g(0) g(x)

g(x) = f(x) g(0)

g(0) = )x(g)x('f

= )x(f)x('g

f(x) f(x) + g(x) g(x) = 0


Integrating

f2(x) + g2(x) = k (let)

f2(x + y) + g2(x + y) = f2(x) f2(y) + g2(x) g2(y) + g2(x) f2(y) + f2(x)

g2(y)

= f2(x) (f2(y) + g2(y)) + g2(x) (g2(y) + f2(y))

= (f2(y) + g2(y)) (f2(x) + g2(x))

k = k . k

k = 1( k = 0 f(x) = 0, g(x) = 0)

f2(x) + g2(x) = 1.

65. The equations of tangents .....................

Ans. 41

Sol. y2 2x2 4y+ 8 = 0

0dxdy

4x4dxdy

y2 4y2x4

dxdy

4k2h4

dxdy

)k,h(

Equation of tangent is (y k) = 4k2h4

(x h).

It passes through (1, 2)

(2 k)(2k 4) = 4h (1 h)

or, 4k 2k2 8 + 4k = 4h 4h2

or, 4k k2 4 = 2h 2h2

or, 2h2 + k2 + 2h 4k + 4 = 0

k2 2h2 4k + 8 = 0

2h 4 = 0 or h = 2 k = 0 or 4

Equation of tangent at (2,0); y = )4(8

(x 2)

or y = 2x + 4 or y + 2x = 4

and equation of tangent at (2, 4) is y = 2x

66. If y = tan1

)ex(n

)x/e(n2

2

.....................

Ans. 00

Sol. Let y1 = tan1

)ex(n

)x/e(n2

2

and y2 = tan1

nx61nx23

Let a = 2 n x,

then y1 = tan1

2

2

nxne

nxne

= tan1

a1a1

= tan1

4tan =

4

, where a = tan

Similarly y2 = tan1

a31a3

= + , where tan = 3

y =

4 + ( + ) = 4

+ tan1 3 = constant

dxdy

= 0 2

2

dx

yd = 0

67.

68. If 2

6 5 4 3 2

3x 2x

x 2x x 2x 2x 5

dx ............

Ans. 00

Sol.

5x2x2xx2x

x2x323456

2

dx

=

4)1xx(

x2x3223

2

dx = 21

tan1

21xx 23

+ C

F(x) = 21

tan1

21xx 23

+ C

F(1) F(0) = 21

21

tan23

tan 11 = 21

tan1 74

0 < F(1) F(0) < 21

. 2

< 1

[F(1) F(0)] = 0

69.

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23 12 2012 JR Solutions Eng