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23 12 2012 JR Solutions Eng
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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)
RESONANCE SOLJRMT231212 - 2
*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
RESONANCE SOLJRMT231212 - 3
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)
RESONANCE SOLJRMT231212 - 4
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
RESONANCE SOLJRMT231212 - 5
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.
RESONANCE SOLJRMT231212 - 6
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
RESONANCE SOLJRMT231212 - 7
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.
RESONANCE SOLJRMT231212 - 8
*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]
RESONANCE SOLJRMT231212 - 9
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
RESONANCE SOLJRMT231212 - 10
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
RESONANCE SOLJRMT231212 - 11
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.