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Preface Page No.
1 . Trigonometric ratio identities & Equations
Exercise 01 - 27
2 . Fundamentals of Mathematics - II
Exercise 28 - 38
3 . Straight Line
Exercise 39 - 70
4 . Circle
Exercise 70 - 92
5 . Mathematical Reasoning, Induction & Statistics
Exercise 92 - 101
6 . Solution of Triangle
Exercise 101 - 125
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MATHEMATICSCLASS : XI
CONTENT
RESONANCE SOLUTIONS (XI) # 1
TRIGONOMETRIC RATIO, IDENTITIES & EQUATIONSEXERCISE # 1
PART - ISection (A) :
A-2. c = 180º
A-4. (a) 3 + 2 + 3 × 31
= 6
(b) 2 × 21
+ 2 × 21
+ 2 × 4 = 10
(c)51
+ 0 = 51
A-6.
2cot)sin(�sin
cos)cos(�
2cot)sin(�sin
cos)cos(�
A-9. tan = � 125
23
< < 2
sin = � 135
and cot = �5
12
LHS =
eccos�eccos�
cot�sin� =
eccos2cotsin
=
513
2�
512
�135
�
= 338181
= RHS
Section (B) :B-4. LHS = cos2 + cos ( + ) { cos cos � sin sin � 2 cos cos }
= cos2 � cos ( + ) . cos ( � ) = cos2 � cos2 + sin2 = sin2 = RHS
B-6. (i)BcosBsinAcosAsin
BsinAsin 22
=
B2sin21
A2sin21
)BAsin()BAsin(
= )BAsin()BAcos(2
)BAsin()BAsin(2
= tan (A + B)
(ii) cot (A + 15º) � tan (A � 15º) = )º15Asin()º15Acos(
� )º15Acos(
)º15�Asin(
= )º15Acos()º15Asin()º15Asin()º15Asin()º15Acos()º15Acos(
=
)º30sinA2(sin21
)º15sinA(sin)º15sinA(cos 2222
=
21
A2sin
A2cos2
= 1A2sin2
A2cos4
B-7 A + B = 45° tan(A + B) = tan(45º)
BtanAtan�1BtanAtan
= 1 tanA + tanB + tanA tanB = 1
(1 + tanA) (1 + tanB) = 2 put A = B = 2º1
22
(1 + tan 2º1
22 )2 = 2 tan 2º1
22 = 1�2
RESONANCE SOLUTIONS (XI) # 2
Section (C) :
C-1. LHS =
4cot2
cos
4�
tan1
4�
tan�1�
2
2
sec 2
9
=
4cot
2cos
2�
cos� sec 2
9 =
4sin
4cos2
cos
2sin� sec
29
= 4sin
1
2sin4sin�
2cos4cos sec
29
= 4sin
1 × cos
29
. sec 2
9 = cosec 4 = RHS
C-3. (ii)AsinAcosAsinAcos
�
AsinAcosAsinAcos
=
Asin�Acos
AcosAsin422 =
A2cosA2sin2
= 2 tan 2A
C-9 tan tan(60° + ) tan(60° � ) = tan 3
LHS = tan
tan3�1
tan3
tan31
tan�3
= tan
2
2
tan3�1
tan�3=
3tan
tan3�1
tan�tan32
3
Put = 20° tan 20°
tan 20° tan80° tan 40° = tan60° = 3
Section (D) :
D-1. Let y = cosx .cos
x
3
2cosx
3
2
y =
x2cos
34
cosxcos21
y =
2x2cos21
xcos21
y = xcosxcosx2cos241
y = 41
[cos 3x + cos x � cosx]
y = 41
cos 3x �1 cos 3x 1
ymin = � 41
and ymax = 41
D-3. (i) y = 10 cos2x � 6 sin cosx + 2 sin2x= 5 (1 + cos 2x) � 3 sin 2x + 1 � cos 2x
= 4 cos 2x � 3 sin 2x + 6 � 22 ba a cos + b sin 22 ba
ymax
= 5 + 6 = 11y
min = � 5 + 6 = 1
(ii) y = 1 + 2 sinx + 3 cos2xy = 1 + 2sinx + 3 � 3 sin2 xy = 1 � (3 sin2x � 2 sinx � 3)
RESONANCE SOLUTIONS (XI) # 3
y = 1 � 3 (sin2x � 32
sinx + 91
� 91
� 1)
y = 1 � 3
910
�31
�xsin2
= � 3
2
3
1�xsin
+
313
ymax
= 3
13 , y
min = � 3
916
+ 3
13 = � 1
(iii) y = 3 cos
3 + 5 cos + 3
y = 3 cos . 21
� 3 23
sin + 5 cos + 3
y = cos23
� sin233
+ 5 cos + 3
y = cos2
13 �
233
sin + 3
ymax
= 4
274
169 + 3 = 7 + 3 = 10
ymin
= � 4
274
169 + 3 = � 7 + 3 = � 4
Section (E) :
E-2. (i)AsinAcos
AsecAsinecAcosAcos
=
AsinAcosAsin�Acos
)AsinA(cosAcosAsinAsin�Acos 22
(ii) cos
1�
tan�sec1
=
cos)sin�1(sin1�cos
cos1
�sin�1
coscos�1
cos 2
=
cossin
cos)sin�1(sin�sin 2
1sincos
�cos
1
=
cossin
cos)sin1()sin1(sin
cos)sin1(cos�1sin 2
(iii)Asin�AcosAsin�cos
AsinAcosAsinAcos 3333
cos2 A + sin2A � sinA cosA + cos2A + sin2 A + sin A cosA = 2
Section (F) :
F-1. (i) LHS = cos 72
cos 74
cos 76
= � cos 7
cos 72
cos 74
= �
7sin.2
78
sin
3
= 81
= RHS
(ii) LHS = cos 11
cos 112
cos 113
. cos 114
. cos 115
= cos 11
cos 112
cos 114
cos 118
cos 11
16
=
11sin.2
1132
sin
5
=
11sin.32
113sin
= 321
= RHS
RESONANCE SOLUTIONS (XI) # 4
F-2. LHS = sin2 + sin2 2 + sin2 3 + ....... + sin2 n
=
2
n2cos1..............
2
4cos1
2
2cos1
= 21
2n [(cos2 + cos4 + cos6 + ........+ cos 2n)]
= 21
2n
2n22
cos.
22
sin
2)2(n
sin
= 21
2n
sin)1ncos(.nsin
= RHS
F-7. cos (S � A) + cos (S � B) + cos (S � C) + cos S
= 2 cos
2BAS2
cos
2AB
+ 2 cos
2CS2
cos
2C
= 2 cos
2
C cos
2
AB + 2 cos
2
BA cos
2C
= 2 cos 2C
2B
cos2A
cos2
Section (G) :
G-4. sin 2 = cos 3 cos
2
2 = cos 3
2
� 2 = 2n ± 3 2
� 2 ± 3 = 2n
= 2n � 2
, 5
n22
= 2n � 2
, 5
n2
21
G-8. tan 2 tan = 1 sin 2 sin = cos 2 cos
0 = cos 3 3 = (2n + 1)2
= (2n + 1)6
.
Section (H) :
H-4. cos2 x + cos2 2 x + cos2 3 x = 1
2
x2cos1+
2x4cos1
+ 2
x6cos1 = 1 cos2x + cos4x + cos6x = � 1
2cos4x cos2x = �2cos22x cos2x = 0 or cos4x + cos2x = 0
2x = (2n +1) 2
or 2cos3x cosx = 0 4
)1n2(x
, (2n + 1) 6
, (2n + 1) 2
Now x = (2n + 1)6
= 3
n +
6
may also be written as
x = (3k + 1)3
+6
,(3k + 2)3
+6
, (3k)3
+6
= 2
k
, 6
5k
,
6k
= (k + 1) 6
, 6
k
( 2
k
is same as (2n + 1) 2
) = 6
m
RESONANCE SOLUTIONS (XI) # 5
H-5. sin2n � sin2(n � 1) = sin2
sin(2n � 1) sin = sin2 sin = 0 or sin (2n � 1) = sin
= m , sin(2n � 1) � sin = 0 2
)2n2(sinncos2
0
2
)1p2(n
, )1n( m ,1n
,
21
p n
Section (I) :
I-1. tan2 � (1 + 3 ) tan + 3 = 0 tan = 1, 3 = n + 4
, n + 3
.
I-3. 4 cos � 3 sec = 2 tan 4 cos � cos
3 =
cossin2
4 cos2 � 3 = 2 sin 4 � 4 sin2 � 3 = 2 sin
4 sin2 + 2 sin � 1 = 0
sin = 8
1642 =
8522
= 4
51
sin =
415
, 4
15 = �cos 36º, sin 18º
= �sin 54º, sin 18º = sin
103
, sin 10
= n + (�1)n 10
or n � (�1)n 103
.
Section (J) :
J-1. 3 sin � cos = 2 2
cos
21
sin23
= 2
2 sin
6 = 2 sin
6 = 2
1 = sin
4
� 6
= n + (�1)n 4
.
J-2. 5 sin + 2 cos = 5
29
5 sin +
29
2 cos =
29
5
sin sin + cos cos = 29
5
cos ( � ) = sin = cos
2 � = 2n ±
2
= 2n ± 2
+
= 2n ± 2
, 2n ± 2
+ 2 = 2n + 2
, 2n � 2
+ 2
For = 2n � 2
+ 2,
RESONANCE SOLUTIONS (XI) # 6
We have = 2n + 2
4 = 2n + 2
1tan25
tan 11
= 2n + 2 tan�1
25
1
125
= 2n + 2
73
tan 1
= 2n + 2
or 2n + 2 where tan�1 73
=
PART - IISection (A) :A-3. 3{cos4 + sin4} � 2{cos6 + sin6}
= 3{1 � 2 sin2 cos2} � 2 {1 × (cos4 + sin4 � sin4 cos2)}= 3 � 6 sin2 cos2 � 2 { 1 � 3 sin2 cos2}= 3 � 6 sin2 cos2 � 2 + 6 sin2 cos2 = 1
A-6.
10cos1
103
cos1
103
cos�1
10cos�1
=
10cos�1 2
10
3cos�1 2
sin2 10
. sin2 103
=
2
415
·4
1�5
= 2
16
4
= 161
Section (B) :B-2. 3 sin = 5 sin
sinsin
= 35
sin�sinsinsin
= 28
2�
tan
2tan
= 4
B-7. cot (A + B) = cot 225° = 1 BcotAcot
1�BcotAcot
= 1
cot A cot B = 1 + cot A + cot B
NowBcotAcotBcotAcot1
Bcot.Acot
= )BcotAcot1(2
BcotAcot1
=
21
Section (C) :
C-3. tanA = 34
A IIIrd quadrant
5 sin 2A + 3 sinA + 4 cosA= 10 sinA cosA + 3 sinA + 4 cosA= 10 sinA cosA + 3 sinA + 4 cosA= 0
RESONANCE SOLUTIONS (XI) # 7
C-6. tan2 = 2 tan2 + 1 ... (i)
cos 2 + sin2 =
2
2
tan1
tan�1 + sin2
= 1tan21
1�tan2�12
2
+ sin2 =
)tan1(2
tan2�2
2
+ sin2
= � sin2 + sin2 = 0which is independent of
C-7*. sin t + cos t = 51
2t
tan1
2t
tan�12t
tan2
2
2
=51
10 tan2 2t
+ 5 � 5 tan2 2t
= 1 + tan2 2t
6 tan2 2t
� 10 tan 2t
� 4 = 0 3 tan2 2t
� 6 tan 2t
+ tan 2t
� 2 = 0
3 tan 2t
2�
2
ttan + 1
2�
2
ttan = 0 tan
2t
= 2 , tan 2t
= � 31
Section (D) :D-1. f() = sin4 + cos2
= sin2 (1 � cos2) + cos2
= sin2 + cos2 � sin2 cos2
f() = 1 � 41
sin22
0 sin22 1f()
max = 1
f()min
= 1 � 41
= 3/4 Range is
1,
43
D-2*. 1 + 4 sin+ 3 cos 4 sin+ 3 cos[� 5, 5]
Max. = 1 + 5 = 6Min. = 1 � 5 = � 4
Section (E) :
E-2. square & adda2 + b2 = 9 + 16 = 25
E-5*. 1 radian ~ 57º (approx.)
sin1 > sin1°
cos1° > cos1
RESONANCE SOLUTIONS (XI) # 8
Section (F) :F-3. A = tan 6° tan 42°
B = cot 66° cot 78°
BA
= tan 6° tan 42° tan 66° tan 78°
BA
=
54tan)660(tan)6�60(tan6tan
. tan 78° tan 42°
BA
=
54tan)1860(tan)18�60(tan.18tan
=
54tan54tan
BA
= 1 A = B
F-5*. cos 10
· cos 102
· cos 104
· cos 108
· cos 10
16
=
10sin2
102sin
5
5
= 321
10sin
1032
sin
= 321
10sin
102
3sin
= � 321
·
10sin
10cos
10sin2
= � 161
cos10
= � 641
5210
F-7*. cos2x + cos2y + cos2z � 2 cosx cosy cosz
(Given x + y = z)= 1 + cos (x + y) cos (x � y) + cos2z � 2 cosx cosy cosz
= 1 + cosz [cos (x � y) + cos (x + y)] � 2 cos x cosy cosz
= 1 + cosz . 2cosx cosy � 2 cosx cosy cosz
= 1= cos (x + y � z)
F-9*. tan A + tan B + tan C = 6, tan A tan B = 2In any ABC,tan A + tan B + tan C = tan A tan B tan C 6 = 2 tan C tan C = 3 tan A + tan B + 3 = 6 tan A + tan B = 3 & tan A tan B = 2Now (tan A � tan B)2 = (tan A + tan B)2 � 4tan A tan B
=9 � 8 = 1
tan A � tan B = ± 1
tan A � tan B = 1 or tan A � tan B = � 1
tan A + tan B = 3 tan A + tan B = 3on solving on solving
tan A = 2 tan A = 1tan B =1 tan B = 2
Section (G) :
G-3. tan x + tan
3x + tan
32
x = 3 3 tan 3x = 3
tan 3x = 1 x = 3
n +
12
, n
RESONANCE SOLUTIONS (XI) # 9
Section (H) :
H-2. sin 7x + sin 4x + sin x = 0
2 sin 4x cos 3x + sin 4x = 0 sin 4x = 0 or(;k) cos 3x = 21
4x = n or 3x = 2n ± 32
x = 4
n,
92
3n2
= 0, 4
, 2
, 92
, 9
4.
H-4.* 2sin2x = sinx + sin3x 2sin2x = 2sinx2x cosx sin2x = 0 or cosx = 1
2x = n or x = 2m x = 2
n, 2m
options (A), (B), (C), (D) are all a part of 2
nx
.
Section (I) :
I-4. cos 2 + 3 cos = 0 2 cos2 + 3 cos � 1 = 0 cos = 4
893 =
4173
As �1 cos 1 a cos = 4
173 only = 2n ± where cos =
4317
I-5. sin + 7 cos = 5
2t1
t2
+
2
2
t1
t17
= 5 where t = tan
2
2t + 7 � 7t2 = 5 + 5t2 tan 2
is root of 12 t2 � 2t � 2 = 0 or 6t2 � t � 1 = 0.
Section (J) :
J-1. tan = �1 4
7,
43
in [0, 2]
cos = 2
1
47
,4
in [0, 2] common value is x =
47
general solution is 4
7n2
, n I.
J-3.* Let E = sin x � cos2 x � 1 E = sin x � 1 + sin2 x � 1 = sin2 x + sin x � 2
=
2
21
xsin
�
49
assumes least value
when sin x = 21
x = n + (�1)n
6 .
EXERCISE # 2
PART - I
3. sinx + sin y = a .....(1)cosx + cos y = b .....(2)
2yx
cos2
yxcos2
2yx
cos2
yxsin2
= ba
RESONANCE SOLUTIONS (XI) # 10
tan
2
yx =
ba
sin
2yx
= 22 ba
a
, cos
2yx
= 22 ba
b
sin (x + y) = 2 sin
2yx
cos
2yx
= 22 ba
ab2
Now for tan
2
yx
(1)2 + (2)2
1 + 1 + 2 cos (x � y) = a2 + b2
cos (x � y) = 2
2ba 22
tan2
2yx
= )yxcos(1)yxcos(1
tan2
2yx
=
22ba
1
22ba
1
22
22
tan
2yx
= ± 22
22
ba
ba4
6. tan = qp
LHS = 21
(p cosec 2 � q sec2) × 22
22
qp
qp
= 21 22
2222qp2sec
qp
q�2eccos
qp
p
22 qp
psin
, 22 qp
qcos
= 22 qp
2cos2sin2sincos�2cossin
21
= 22 qp
4sin)2�sin(
=
22 qp4sin4sin
( = 6)
7. (i) cot 7o
21
= tan 82o
21
=
o
o
21
7sin
21
7cos =
o
o
2
15sin21
7cos2
= )3045sin(
)3045cos(1
=
22
1322
131
= 13
1322
= 13
)13)(1322(
= 6232 = 6432
RESONANCE SOLUTIONS (XI) # 11
= )32( )12(
(ii) tan 142o
21
= � cot 52
o
21
= o
21
52tan
1 =
o
21
745tan
1
= � o
o
21
7tan1
21
7tan1
= � oo
oo
21
7sin21
7cos
21
7sin21
7cos
= �
15cos
21
7sin21
7cos
2oo
= �
15cos15sin1
= �
22
1322
131
= � 2
)13)(1322(
= � 2
])13()13(22[ 2
= � 2
)]324()13(22[
= � )]32()13(2[ = � 3226 = 2 + 632
9. (i) tan9° � tan27° � tan 63° + tan81°
= (tan 9º + tan81º) � (tan 27º + tan 63º)
= º27cosº27sin2
2�
º9cosº9sin22
º63cosº27cosº90sin
�º81cosº9cos
º90sin
=
415
2�
41�5
2º54sin
2�
º18sin2
= 44
)15�15(8
(ii) cosec 10° � 3 sec10° = 2
10sin
23
�10cos21
× 22
º10cosº10sin1
= 4
(iii) 22 sin 10°
35sin2�
5sin40cos
25sec
= 22
º10sinº35sin2�
º5sinº40cosº5cosº5sin2
2º5secº5cosº5sin2
= 22 (sin5º + 2cos45º + cos 35º � cos 25º + cos 45º)
= 22 (sin5º + 2cos45º + 2sin 30º sin (� 5º))
= 22 )2( = 4
(iv) cot 70° + 4 cos 70° = º70sin
º70sinº70cos4º70cosº70cos4
70sin70cos
= º70sin
º140sin2º70cos
= º70sin
º140sin)º140sinº20(sinº70sin
º140sin)º140sinº70(cos
RESONANCE SOLUTIONS (XI) # 12
= º70sin
º20cosº120sin2º70sin
º140sinº60cosº80sin2
= 2 × 323
(v) tan 10º � tan 50º + tan 70º
= tan 10º � tan (60º � 10º) + tan (60º + 10º)
= tan 10º � º10tan31
º10tan�3
º10tan3�1
º10tan3
= º10tan3�1
º10tan3º�10tan92
3
= 3
º10tan3�1
º10tanº�10tan32
3
= 3 tan 30º
= 3
11.
21AA1
= 31AA
1 +
41AA1
OA1 = OA
2 = OA
3 = OA
4 = r (say)
A1OA
2 =
n2
, AA1OA
3 =
n4
, AA1OA
4 =
n6
nsin
1
=
n2
sin
1
+
n3
sin
1
sin n2
nsin
n3
sin = sin n
3 . sin
n
sin n2
nsin.
n2
cos2 = sin n
3 sin
n
2 sin n2
cos n2
= sin n
3
sin n4
= sin n
3
n4
= � n
3
4 = n � 3 n = 7
13. Pn � P
n�2 = cosn + sinn � cosn�2 � sinn�2
= cosn�2 (cos2 � 1) +sinn�2 (sin2 � 1)
= cosn�2 (�sin2 ) +sinn�2 (� cos2 )= (�sin2 cos2 ) {cosn�4 +sinn�4 }= (�sin2 cos2 ) P
n�4
put n = 4 P
4 � P
2 = (� sin2 cos2) P
0
P4 = P
2 � 2 sin2 cos2
= 1 � 2 sin2 cos2
similarly we can prove the other result also.
15. tan2 + 2 tan . tan 2 = tan2 + 2 tan . tan 2
(tan2 � tan2) + 4 tan tan
22 tan1
1
tan1
1 = 0
(tan2 � tan2) + 4 tan tan )tan1)(tan1(
)tan(tan22
22
= 0
RESONANCE SOLUTIONS (XI) # 13
(tan2 � tan2)
)tan1)(tan1(
tantan41
22 = 0
(tan2 � tan2) (1 � tan 2 . tan 2) = 0 tan2 = tan2 or tan 2 . tan 2 = 1
L.H.S. = tan2+ 2 tan .
2tan1
= tan2 +
tan2tan2
. (1 � tan2) = 1
R.H.S. = tan2 + 2 tan . 2tan
1
= tan2 +
tan2tan2
. (1 � tan2) = 1
19. 13 18 tanx = 6 tan x 3 ................(1)
13 � 18 tanx = 36 tan2x + 9 � 36 tanx tanx = 32
, 61
Put in (1) tanx = 32
is correct x = n + tan�1
32
= n + = , + , � + , �2 + in (�2, 2)
22. tan + sin = 23
...(1)
As tan2 + cos2 = 47
2
sin23
+ cos2 =
47
49
+ sin2 � 3 sin + cos2 = 47
23
= 3 sin sin = 21
= n + (�1)n
6
from (1), tan = 23
� sin = 21
23 = 1 = n +
4
.
23. a cos 2 + b sin 2 = c
2
2
t1
t1a
+
2t1
t2b
= c where t = tan
(c + a)t2 � 2bt + (c � a) = 0 t1 + t
2 =
acb2
, t1t2 =
aca�c
cos2 + cos2 = 2
2cos12cos1 = 1 +
21
[cos 2 + cos 2]
= 1 +
22
22
21
21
t1
t1
t1
t121
simplifying and using values for t1, t
2 we get
cos2 + cos2 = 1 + 22 ba
ac
= 22
22
ba
acba
.
RESONANCE SOLUTIONS (XI) # 14
27. RHS = 3x2 + 2x + 3
Minimum value = )3(4
4)3)(3(4 =
38
> 2
whereas LHS 2 no solution.
PART - II
2. For dodecagon
RR
BA
oA 'OB' = 122
= 30°
OA 'B' = OB 'A' = 75°
75sin
R =
30sin1�3
R =
21
22
)13()1�3(
R = 2
For hexagon AOB = 6
2 = 60°
AOB is equilatecal AB = R = 2
6. A + B + C =
2C
sin
2C
Asin
= 1k
2C
sin2C
Asin
2C
sin�2C
Asin
= 1k1�k
2A
cos2
CAsin2
2A
sin2
CAcos2
= 1k1�k
tan
2B
tan 2A
= 1k1�k
8. 4 cos2
2
x�
4 + xcosxsin4xsin4 224
= 4 cos2
2
x�
4 + | 2 sinx | = 4 cos2
2
x�
4 � 2 sinx
= 2
x�
2cos1 � 2 sinx = 2
10. ecCcosecBcosAcos = CsinBsinAcos
+ CsinAsinBcos
+ BsinAsinCcos
= CsinBsinAsin
CsinCcosBsinBcosAsinAcos
RESONANCE SOLUTIONS (XI) # 15
= CsinBsinAsin2
C2sinB2sinA2sin =
CsinBsinAsin2CsinBsinAsin4
(using conditional identity)
= 2
13.xcos10x3cos5x5cos
)1x2(cos10)x2cosx4(cos5)x4cosx6(cos
= xcos10x3cos5x5cos
xcos210xcosx3cos25xcosx5cos2 2
= 2 cos x xcos10x3cos5x5cos
xcos10x3cos5x5cos
= 2 cos x
15.
cos3cos
= 4 cos2 � 3 = 2 (1 + cos 2) � 3
= 2 cos2 � 1 = 2 cos ( � ) � 1
(cos2 + sin2 ) + (cos2 + sin2 ) + 2 cos ( � ) = a2 + b2
2 cos ( � ) = a2 + b2 � 2
cos3cos
= a2 + b2 � 3
20. sin 3 = 4 sin sin 2 sin 4 sin 3 = (2 sin ) (2 sin 2 sin 4) 3 sin � 4 sin3 = 2 sin (cos 2 � cos 6) 3 � 4 sin2 = 2(cos 2 � cos 6) or sin = 0
3 � 2(1 � cos 2) = 2 cos 2 � 2 cos 6or sin = 0 1 = �2 cos 6 cos 6 = 21
or sin = 0
sin = 0 or cos 6 = 21
= n or = 6
32
n2
= 93
n
= 0, , 9
, 93
, 93
2 ,
9
So eight solutions.
22. 2 cos x = 2 2 2 sin x
x2sin1xcos2 = xcosxsin )xcosx(sin2
1xcos
cosx =
4xsin
see from graph or we can put values given in options to verify..
25. 2 tan2 x � 5 sec x � 1 = 0 2(sec2 x � 1) � 5 sec x � 1 = 0
2 sec2 x � 5 sec x � 3 = 0 sec x = 26
, 21
= 3, 21
RESONANCE SOLUTIONS (XI) # 16
sec x = 3
2
1xsec
cos x = 31
7 solutions in
215
,0 n = 15.
28. 4cos3x � 4cos2x + cos x � 1 = 0
(4 cos2x + 1) (cos x � 1) = 0
cos x = 1x = 2nsolutions in the interval [0, 315] are 0, 2 , 4 , ....., 100
arithmatic mean = 51
100....420 = 50
29. h = 22 )}(sin22sin2{sin)}(cos22cos2{cos
h = [4 cos2 ( + ) (cos ( � ) + 1)2 + 4 sin2 ( + ) (cos ( � ) + 1)2 ]1/2
h = [4{cos ( � ) + 1}2 {cos2 ( + ) + sin2 ( + )}]1/2
h = 2 (1 + cos ( � )) h = 2 × 2 cos2
2
�
h = 4 cos2
2
�
32. y = a cos2 x + 2b sin x cos x + c sin2 x & tan x = ca
b2
z = a sin2 x � 2b sin x cos x + c cos2 x y + z = a + c
and y � z = (a � c) xsinxcos 22 + 4b sin x cos x
= (a � c) cos 2x + 2b sin 2x ( 2b = (a � c) tan x)
= (a � c) [cos 2x + tan x.sin2x] = (a � c)
x2sin
xcos
xsinx2cos2
= xcos
)xx2cos()ca( = (a � c).
33.
n
2BA
sin.2
BAcos2
2BA
cos.2
BAcos2
+
n
2BA
sin.2
BAsin2
2BA
cos.2
BAsin2
= cotn
2BA
+ (�1)n cotn
2BA
RESONANCE SOLUTIONS (XI) # 17
34. sin6x + cos6x = a2
(sin2 x + cos2 x) (sin4 x + cos4 x � sin2 x cos2 x) = a2
(sin2 x + cos2 x)2 � 3sin2 x cos2 x = a2 1 � 3 sin2 x cos2 x = a2
1 � 43
sin2 2x = a2 3
)a�1(4 2 = sin2 2x
0 34
(1 � a2) 1
1 � a2 0 and 4 � 4a2 3
a2 1 and41
a2
� 1 a 1 and a 21
or a � 21
a
21
,�1�
1,
21
38. cos 15x = sin 5x
cos 15x = cos
x5
2 or cos
x5
2
3
15x = 2n ±
x5
2 or 15x = 2n ±
x5
23
x = 10n
+ 40
, n , x = 5
n +
203
, n
and x = 5
n �
20
, n and x = 10n
� 403
, n
40. sin2 x + 2 sin x cos x � 3 cos2 x = 0case-I : cos x 0 tan2 x + 2 tan x � 3 = 0
tan x = 3, 1 x = n + tan�1 (�3), n + 4
case-II : cos x = 0 1 + 0 � 0 = 0 not true.
EXERCISE # 33. (A) sin2 + 3 cos = 3 1 � cos2 + 3cos = 3
cos2 � 3cos + 2 = 0 cos = 1, 2 cos = 1 ( cos 2) = 0 in [�, ] No. of solution = 1
(B) sin x . tan 4x = cos x xcosx4cosx4sin
.xsin
sin4x sinx � cos4x cosx = 0 cos5x = 0 5x = (2n + 1) /2 x = (2n + 1) /10
x = 10
, 103
, 105
, 107
, 109
in (0, )y = 2
xy
x
y = x �12
O 3
So there are five solutions.
(C) (1 � tan2 ) sec2 + 2tan2 = 0
(1 � tan4 ) + 2tan2 = 0
(1 � x2) + 2x = 0 where x = tan2
RESONANCE SOLUTIONS (XI) # 18
2x = x2 � 1 x = 3
tan2 = 3 3tan 3
in
2,
2
Number of solutions = 2
(D) [sin x] + [ 2 cosx] = � 3
[sin x] = � 1 and [ 2 cosx] = � 2 < x < 2 and 1xcos22�
2
1xcos2�
2
1xcos1�
4
5x
for ]2,0[x
45
x
, ]2,0[x
4
5x
25
x22
0 < sin 2x < 1 [sin2x] = 0
4. (A)
Number of solutions = 6
(B) sin x = 2
82 = 1 ± 2
sin x = 1 � 2As sin x takes at least four valuesin [0, n] n 4(C) 1 + sin4 x = cos2 3xL.H.S. 1 and R.H.S. 1 L.H.S. = R.H.S. = 1 sin4 x = 0 and cos2 3x = 1 x = n and 3x = m
x = n and 3x = m
x = n and x = 3
m
x = n
x = �2, �, 0, , 2 in
2
5,
2
5
Number of solutions = 5.(D) A, B, C are in A.P. B = 60º
As sin (2A + B) = 21
2A + B = 30º or 150º
2A = �30º or 90º 2A = 90º A = 45º
C = 180º � A � B = 75º = 125
p = 12.
Comprehension # 1 (5, 6, 7)
5. Given cos + cos = a 2cos
2 cos
2 = a ..... (i)
and sin + sin = b
RESONANCE SOLUTIONS (XI) # 19
2 sin
2 cos
2 = b ..... (ii)
by (i) & (ii)
tan
2 = ab
tan = ab
sin2 + cos2
=
2
2
a
b1
ab
2
+
2
2
2
2
a
b1
a
b1
= 22 ba
ab2
+ 22
22
ba
ba
= 22
222
ba
ab2b2ba
= 1 + 22 ba
)ba(b2
n = 26. sinn A = x
sin2 A = x sin A sin 2A sin 3A sin 4A= sin A (2 sin A cos A) (3 sin A � 4 sin3 A) (4 sin A cos A (1 � 2 sin2 A))= 8 sin4 A (1 � sin2 A) (1 � 2 sin2 A) (3 � 4 sin2 A)If we put sin2 A = x, then given expression is a polynomial of degree 5 in x.
7. If p = 5 sin x + (p � 5), cos x, tan x
sin x, cos x, tan x are in G.P. cos2 x = sin x tan x
cos3 x = sin2 x cos3 x = 1 � cos2 x cos3 x + cos2 x = 1taking cube both sides
cos9 x + cos6 x + 3 cos5 x = 1 cos9 x + cos6 x + 3 cos5 x � 1 = 0
Comprehension # 4
14. sin6x + cos6x < 167
1 � 3sin2x cos2x < 167
sin2x cos2x > 163
sin22x > 43
2
x4cos1 >
43
1 � cos4x > 23
cos4x < 21
Principal is value 4x
34
,32
General value is
3
4n2,
3
2n2x4
n,
32n
,62
nx
RESONANCE SOLUTIONS (XI) # 20
15. cos 2x + 5 cos x + 3 0
2cos2x + 5cosx + 2 0 (cosx + 2)(2 cosx + 1) 0
2cosx + 1 0 ( cosx + 2 > 0)
cosx 21
32
,3
2x
16. 2 sin2
4x + 3 cos 2x 0
1 � cos
2x2 + 0x2cos3
1x2sinx2cos3 21
x2sin21
x2cos23
21
3x2sin
2x +
3
67
n2,6
n2
65
n2,2
n2x2
125
n,4
nx
x
127�
,�
125
,4
�
,
43
in ,
19. Statement-1 : cos A cosec B cosec C
= CsinBsin
Acos =
CsinBsinAsinAcosAsin
= CsinBsinAsin2
A2sin
= CsinBsinAsin2CsinBsinAsin4
= 2
Statement-2 : tan A tan B = 1 iff the triangle ABC is right angled Statment is false
20. y =
3tantan
=
2
3
tan31
tantan3
tan y =
2
2
tan3
tan31
tan2 = 3y1y3
0 y
3
1, (3, )
statement-1 and statement-2 both are true and statement-2 explains statement-1
22. cos x . sin y = 1 Either cosx = 1 and siny = 1 or cosx = �1 and siny = �1
(x,y) =
2,0 ,
25
,0 ,
2,2 ,
25
,2 or (x, y) =
23
, ,
23
,3
Number of pairs = 6
24. log2 ]2cos2cos)(cos)([cos 22
= log2 ]2cos2cos)(sin1)([cos 22
= log2 [1 + cos 2.cos2 � cos 2 cos 2]
= log2 1
= 0
RESONANCE SOLUTIONS (XI) # 21
30.12cos2
3sin
=
21
2(3 sin � 4 sin3 ) = 2(1 � 2 sin2 ) + 1
8 sin3 � 6 sin � 4 sin2 + 3 = 0 (2 sin � 1)(4 sin2 � 3) = 0
sin = 21
, ±23
For sin = ±23
, 2 cos 2 + 1 = 0 so given equation
becomes undefined sin = 21
only = n +(�1)n 6
, n .
32. sin x . xcos8 2 = 1 sin x.|cos x| = 8
1 2 sin x |cos x| =
2
1.
33.
xcos
1xcos
22 y2tan1 2
(3 + sin 3z) = 4
cos2 x + xcos
12 = 2, 1 + tan2 2y = 1, 3 + sin 3z = 2
cos2 x = 1, tan2 2y = 0, sin 3z = �1 x = n, n I.
36. cos 20° + 2 sin2 55° � 2 sin 65°
= cos 20° + 1 � cos 110° � 2 sin 65°
= 1 + 2 sin 65° sin 45° � 2 sin 65°
= 1 + 2 sin 65° 2
1 � 2 sin 65° = 1
41.
cos13sinsin3
+
sin13coscos3
= 4 2 cos
4
cos1sin4 3
+
sin1cos4 3
= 4(cos � sin )
cos1sin3
+ sin = cos �
sin1cos3
cos1cossinsinsin3
=
sin1cossincoscos 3
cos1cos1sinsin 2
=
sin1cossinsincos 2
cos11cossinsin 2
=
sin1)1sin(cossin
either sin = 0 or
cos11cossin2
= � cos
= n or sin2 + cos + 1 = �cos � cos2 cos = �1
= n = 2n or (2n + 1)But at = (2n + 1) , 1 + cos = 0 (2n + 1) = 2x.
42. xsin7xcos6 2 + cos x = 0 xcos77xcos6 2
+ cos x = 0
1xcosxcos7 2 + cos x = 0 1xcosxcos7 2
= � cos x
(so cos x 0) 7 cos2 x � cos x � 1 = cos2 x 6 cos2 x � cos x � 1 = 0
(2 cos x � 1) (3 cos x + 1) = 0 cos x = 21
, 31
RESONANCE SOLUTIONS (XI) # 22
But cos x 0 cos x = 31
cos x = � cos where cos = 31
cos x = cos( � ) x = 2n ± ( � ).
43. x3 + x2 + 4x + 2 sin x = 0 x3 + x2 + 4x = �2 sin x ...(1)when x = 0, 0 = 0 x = 0 is the solutionwhen x [0, ), x3 + x2 + 4x > 0 where as � 2sin x < 0
no solution for x (0, )when x [, 2], x3 + x2 + 4x 3 + 2 + 4 > 2whereas 0 �2 sin x 2 no solution for [, 2]so given equation has only one solution in [0, 2] and that solution is x = 0.
EXERCISE # 4PART - I
1. Clearly = 30º and (60º, 90º)
Hence + lies in (90º, 120º).
2. Let y = 2 sin t
y = 1x2x3
x5x212
2
(3y � 5) x2 � 2x (y � 1) � (y + 1) = 0
x R �
31
,1
D 0 y2 � y � 1 0
y 2
51or y
251
sin t 4
51or sin t
451
range of t is
10,
2
2,
10
3
3. O1BD, DO
BD
1 = cot 30º
BD = 3 similarly EC = 3
BC = AB = AC = 2 + 2 3
area of ABC = 43
2)322( = 43
)3231( 4 = 6 + 34 sq. unit
4.
� = 0, � 2 or 2
� = 0 = cos 2 = e1
RESONANCE SOLUTIONS (XI) # 23
This is true for '4' value of '', ''If � = � 2 = � and = and cos ( + ) = 1 (No solution)similarly if � = 2 = and = again no solution results
5.
4,0
tan in
4,0 and 0 < tan < 1
cot in
4,0 and cot > 1
Let tan = 1 � 1 and cot = 1 +
2 where
1 and
2 are very small and positive, then
t1 = 11
1)1( , t
2 = 21
1)1( , t
3 = 11
2 )1( , t
4 = )1(
22)1(
t4 > t
3 > t
1 > t
2
6. 2sin2 � 5sin + 2 > 0 (sin � 2)(2sin �1) > 0
1/2
O
y
sin < 21
[ �1 1sin ]
From graph, we get
2,
65
6,0
7. 2sin2 � cos2 = 0 ............(i)
sin = 21
2cos2 � 3sin = 0 ............(ii)�2sin2 � 3sin + 2 = 0
sin = 21
, �2
So sin = 21
is the only solution
at = 6
5,
6
8.*2
xsin4
+ 3
xcos4
= 51
2
xsin4
+ 3
)xsin1( 22 =
51
2
xsin4
+ 3
xsin2xsin1 24 =
51
5 sin4x � 4 sin2x + 2 = 56
25 sin4x � 20 sin2x + 4 = 0
(5 sin2x � 2)2 = 0 sin2x = 52
, cos2x = 53
tan2x = 3
2and
8
xsin8 +
27
xcos8 = 125
1
9. f() =
22 cos5cossin3sin
1 =
2)2cos1(5
2sin23
22cos1
1
= 2cos42sin36
2
f()max
= 56
2
= 2
RESONANCE SOLUTIONS (XI) # 24
10.
nsin
1
�
n3
sin
1
=
n2
sin
1
n3
sinn
sin
nsin
n2
cos2
= n2
sin
1
sin n4
= sin n
3
n4
= (�1)k n
3 + k , k
If k = 2m n
= 2m
n1
= 2m , not possible
If k = 2m + 1 n
7 = (2m + 1) n = 7, m = 0
11. tan = cot 5
cossin
=
5sin5cos
cos 6= 0
6 = (2n + 1) 2
= (2n + 1) 12
; n
= 125
�
, �4
, 12
�
, 12
, 4
, 125
.........(1)
sin2 = cos4 sin2 = 1 � 2 sin2 2 2sin22 + sin2� 1 = 0
sin2 = � 1, 21
2= (4m � 1)2
, p + (�1)p 6
= (4m � 1)4
, 2
p+ (�1)p
12
; m, p I
= �4
, 12
, 125
...........(2)
From (1) & (2)
125
,12
,4
�
Number of solution is 3.
12. P = {: sin � cos = 2 cos }
sin = ( 2 + 1) cos tan = 2 + 1
= n+ 8
3 ; n I
Q = {: sin + cos = 2 sin } cos = ( 2 � 1) sin
tan = 1�2
1 = 2 + 1 = n+
83
; n I
P = Q
RESONANCE SOLUTIONS (XI) # 25
13. As tan(2 � ) > 0, � 1 < sin < �23
, [0, 2]
2
3 < <
35
Now 2cos(1 � sin) = sin2( tan /2 + cot /2)cos � 1
2cos(1 � sin) = 2sin cos � 1 2cos + 1 = 2sin( + )
As
35
,2
3 2cos + 1 (1, 2) 1 < 2sin( + ) < 2
21
< sin( + ) < 1
As + [0, 4] +
6
5,
6 or +
6
17,
6
13
6
� < < 6
5 � or
613
� < < 6
17 �
67
,32
32�
,2
3�
35
,2
3
correct option is (A, C, D)
PART - II
1. u = 2222 sinbcosa +
2222 cosbsina
u2 = a2 cos2 + b2 sin2 + a2 sin2 + b2 cos2 + 2 2222 sinbcosa ×
2222 cosbsina
u2 = (a2 + b2) + 2 22222222 cosabasina�ba
u2 = 22 ba + 2 222222224 cossinababaa
u2 = 22 ba + 2
2sin
2
abba 2
22222
.
min(u2) = a2 + b2 + 2ab = 2ba
and max(u2) = a2 + b2 + 22 ba = 2 22 ba
Now, max(u2) � min(u2) = (a � b)2
2. sin + sin = � 6521
and cos + cos = � 6527
squaring and adding, we getsin2 + sin2 + 2 sin sin + cos2 + cos2 + 2 cos . cos
=
2
6521
�
+
2
6527
�
2 + 2 cos ( � ) = 42251170
cos2
2�
= 42254
1170
= 130
9
cos
2
� =
130
3� ( < � < 3
2
<
2
� <
23
)
RESONANCE SOLUTIONS (XI) # 26
3. tan 2P
and tan 2Q
are the roots of equation ax2 + bx + c = 0
tan2P
+ tan 2Q
= �ab
and tan2P
tan 2Q
= ac
2P
+ 2Q
+ 2R
= 2
( P + Q + R = )
2
QP =
2
� 2R
2
QP =
4
( R = 2
)
tan
2QP
= 1
2Q
tan.2P
tan�1
2Q
tan2P
tan
= 1 a/c�1a/b�
= 1 c = a + b
4. cos x + sin x = 21
2/xtan1
2/xtan�12
2
+
2/xtan1
2/xtan22
=
21
, Let tan 2x
= t
2
2
t1
t�1
+ 2t1
t2
=
21
3t2 � 4t � 1 = 0 t = 3
72
as 0 < x < 0 < 2x
< 2
tan 2x
is positive
t = tan 2x
= 3
72
Now tan x = 2/xtan�1
2/xtan22 = 2t�1
t2 tan x = 2
372
�1
372
2
=
374
�
5. Given equation is 2 sin2 x + 5 sin x � 3 = 0
(2 sin x � 1) (sin x + 3) = 0
sin x = 21
( sin x �3)
It is clear from figure that the curve intersect the line at four points in the given interval.Hence, number of solutions are 4.
6. Given, cos x + sin x = 21
2x
tan1
2x
tan1
2
2
+
2x
tan1
2x
tan2
2
= 21
RESONANCE SOLUTIONS (XI) # 27
Let tan 2x
= t 2
2
t1
t1
+ 2t1
t2
=
21
3t2 � 4t � 1 = 0 t = 3
72
As 0 < x < 0 < 2x
< 2
tan 2x
is positive.
t = tan 2x
= 3
72
Now, tan x =
2x
tan1
2x
tan2
2
= 2t1
t2
tan x = 2
372
1
372
2
tan x = �
721
723
×
721
721
tan x = �
374
.
7. 2{cos ( � ) + cos ( � ) + cos ( � )} + 3 = 0(cos + cos + cos )2 + (sin + sin + sin )2 = 0 cos = 0 = sin
8. tan 2 = tan (( + ) + ( � )) = )tan()tan(1)tan()tan(
=
125
.43
1
125
43
= 1548
4)59(
=
33414
= 3356
Hence correct option is (1)
9. A = sin2x + cos4x = sin2x + (1 � sin2 x)2
= sin4x � sin2x + 1
= 2
2
21
�xsin
+
43
= 43 A 1
10. 3sin P + 4 cos Q = 6 ...(i)4 sin Q + 3cos P = 1 ...(ii)
Squaring and adding (i) & (ii) we get sin (P + Q) = 21
P + Q = 6
or 6
5 R =
65
or 6
If R = 6
5 then 0 < P, Q <
6
cos Q < 1 and sin P < 21
3 sinP + 4 cosQ < 211
So R = 6
RESONANCE SOLUTIONS (XI) # 28
Fundamentals of mathematics - II
OBJECTIVE QUESTIONSSection (A) :
A-5. logp log
p np
1
)p(
=
n
p p
1log
= �log
p pn = �n
independent of p.
A-7*. N = 3log
135log
15
3 � 3log
5log
405
3
= 5log27log 33 15log3 � log3 5.log3 405
= 5log3 3 5log1 3 � log35 log3(81 × 5)
= (3 + log3 5) (1 + log3 5) � log3 5(4 + log3 5)= 3.
A-8*. log23 > 1, log
1210 < 1 log
23 > log
1210
log65 < 1, log
78 > 1 log
65 < log
78
log326 < 3, log
29 > 3 log
326 < log
29
log16
15 < 1, log10
11 > 1 log16
15 < log10
11
Section (B) :
B-4*. (log5x)2 + log
5x x5 = 1
(log5x)2 + log
5x5 � log
5xx = 1 (log
5 x)2 + xlog5log
5log
55
5
� xlog5log
xlog
55
5
= 1
(log5x)2 + xlog1
1
5 � xlog1
xlog
5
5
= 1 Let log
5x = t
t2 + t1
1
� t1
t
= 1 t1
t1)t1(t2
= 1
t3 + t2 + 1 � t = 1 + t
t3 + t2 � 2t = 0
t(t2 + t � 2) = 0
t(t � 1) (t + 2) = 0
t = 0, 1, � 2 log5x = 0, 1, �2
x = 1, 5, 251
B-6*.
5xlog
29
xlog 32
3
x = 3 3 33 xlog �
29
log3 x + 5 = logx 3 3
23 xlog � 29
log3 x + 5 = 23
logx 3 Let log3 x = t
t2 � 29
t + 5 = t2
3 2t3 � 9t2 + 10t � 3 = 0
t = 1 satisfies it2t3 � 9t2 + 10t � 3 = 2t2(t � 1) � 7t(t � 1) + 3(t � 1)
= (t � 1) 3t7t2 2
= (t � 1) (2t � 1) (t � 3)
RESONANCE SOLUTIONS (XI) # 29
t = 1 t = 21
t = 3
log3 x = 1 log3 x = 21
log3 x = 3
x = 3 x = 31/2 x = 27.
B-9. Number of digits in integral part = number of digit in 6012 before decimalP = 6012
logP = log 6012 = 12 log 60 = 12[log 6 + 1] = 12 [log 2 + log 3 + 1]= 12 [.3030 + .4771 + 1] = 12 [1.7801] = 21.3612
number of digits in integral part = 22
B-10. log16 x = 43
x = 163/4 x = 8.
Section (C) :
C-3. log1 � x
(x � 2) 1x > 2 ..................(1)(i) When 0 < 1 �x < 1 0 < x < 1
So no common range comes out.(ii) When 1 � x > 1 x < 0 but x > 2
here, also no common range comes out. , hence no solution.Finally, no solution
C-6.
)15(log7
10log
)x2)(8x(
23.0
0
For )x2)(8x( to be defined
(i) (x � 8) (2 � x) 0(x � 2) (x � 8) 0 2 x 8
Now Let say y = log0.3
7
10(log
25 � log
22) = log
0.3
710
(log25/2)
Let y < 0 (assume)
then log0.3
7
10 (log
2 5/2) < 0
7
10 log
2 5/2 > 1 log
2 5/2 >
107
25
> 2(7/10) which is true
So y < 0so denominator is � ve and numerator is +ve, so inequality is not satisfied,
thus )x2)(8x( = 0
x = 2, 8 .....(i)Now 2x � 3 > 31 (x � 3) > log
2 31 x > 3 + log
224.9 (approx) x > 7.9
x = 8
C-8. Domain x2 + 4x � 5 0 x (� , � 5] [1, )Case I :x (� , � 5] [1, 3)� ve < + ve alsways true x (� , � 5] [1, 3) ... (1)Case II :x [3, ) .. (i)
RESONANCE SOLUTIONS (XI) # 30
x � 3 < 5�x4x2
x2 � 6x + 9 < x2 + 4x � 5 x > 57
... (ii)
(i) (ii) x [3, ) ... (2)(1) (2) x (� , � 5] [1, ) Ans. (A)
Section (D) :
D-3. z = 4)i1(4
)i1()i(
11= 4)i1(
4
i2
= 2/i
i4
e
e42i
)i1(2
= 2/ie2
|z| = 2 amp (z) = 2
2
2)z(amp
|z|
= 4 (D)
D-6*. |z1 + z
2|2 = |z
1|2 + |z
2|2
0zzzz 2121 2
1
2
1
zz
�zz
0zz
zz
2
1
2
1
2
1
zz
is purely imaginary
so amp
2
1
zz
is may be 2
or � 2
D-8.
D-9. z1/3 = a � ib
z = (a � ib)3
x + iy = (a3 � 3ab2) + i(b3 � 3a2b)
22 b3�a
ax
by
= b2 � 3a2
by
�ax
= 4(a2 � b2) k = 4
EXERCISE # 2
1. (i)
3/1
10
7log
1.01
log
15 5
= (7 + 1)1/3 = 2
(ii) log3/4
log2 2/12/18 = log
3/4 log
2 (2)3/4 = 1
RESONANCE SOLUTIONS (XI) # 31
(iii)
27log1
491
=
2log7log2 77)7( = 147log2 )7(
= 27 )14(log7
= 196
1 & 7log 5/15 = 7log55 = 7
7 + 196
1
(iv) 5log37 + 7log53 � 7log35 � 3log57
= 5log37 + 7log53 � 5log37 � 7log53
{using property blogca = alogcb }
= 0
5. (i) log10 (x2 � 12x + 36) = 2
(i) x2 � 12x + 36 > 0 (x � 6)2 > 0 x R � {6}
(ii) x2 � 12x + 36 = 100 x2 � 12x � 64 = 0
(x � 16) (x + 4) = 0 x = 16, �4.
(ii) log4 log3 log2 x = 0 log3 log2 x = 1 log2 x = 3 x = 23 x = 8.
(iii)
x93 9
21
xloglog = 2x log9 x + 21
+ 9x = 32x
log9 x + 21
+ 9x = 9x log9 x = �21
x = 9�1/2 x = 31
(iv) 2 log4 (4 � x) = 4 � log2 (�2 � x)
(i) 4 � x > 0 x < 4(ii) �2 � x > 0 x < �2
(iii) log2 (4 � x) = 4 � log2 (�2 � x) log2 (4 � x) (�2 � x) = 4
(4 � x) (�2 � x) = 16 �8 � 2x + x2 = 16 x2 � 2x � 24 = 0 (x � 6) (x + 4) = 0
x = 6 (not possible) , x = �4.
(v) log10
2 x + log10
x2 = log10
2 2 1
log10
2 x + 2 log10
x + 1 = log10
2 2 (log10
x + 1)2 = log10
2 2log
10x + 1 = ± log
10 2
x = 201
and 51
(vi) log4(log2 x) + log2(log4 x) = 2 21
log2 (log2 x) + log2 (log4 x) = 2
21
log2 (2 log4x) + log2 (log4x) = 2
21
log2 2 + 21
log2 (log4 x) + log2 (log4 x) = 2
23
log2(log4 x) = 23
log2 (log4 x) = 1
log4 x = 2 x = 42 x = 16.
(vii) 35xlog
x
= xlog510
3
5xlog log x = 5 + log x
RESONANCE SOLUTIONS (XI) # 32
log2 x + 2 log x � 15 = 0
(log x + 5)(log x � 3) = 0
log x = �5, log x = 3
x = 10�5 , x = 103.(viii) Domain x � 1 > 0 and x + 1 > 0 and y � x > 0
x > 1 x > � 1 x < 7 x (1, 7) .........(i)
� log2 (x � 1) � log2 (x + 1) = 1 + )x7(log21
2
� log2 (x2 � 1) + log2 (7 � x)2 = 1
log2 1x
)x7(2
2
= 1
1x
)x7(2
2
= 2
x2 + 14x � 51 = 0
(x + 17) (x � 3) = 0
x = 3 or x = � 17 (rejected)
x = 3
6. (a) log10 2 = 0.3010 . log10 3 = 0.4771let x = 615
log10 x = 15 log10 6= 15(log10 2 + log10 3)= 15(0.3010 + 0.4771)= 11.6715
characteristic of 615 is 11 number of digits in 615 is 12.(b) let x = 3�100
log10 x = �100 log10 3= � 47.71
number of zeroes immediately after the decimal in 3�100 is 47.
10. (i) 0x
6x4log
51
x6x4
> 0 x ),0(23
,�
....(i)
& 1x
6x4
0
x2x
x (�, � 2] (0, ) ....(ii)
(i) (ii) x
23�
,2�
(ii) log2 (4x � 2.2x + 17) > 54x � 2.2x + 17 > 0(2x)2 � 2.2x + 17 > 0 x R and 4x � 2.2x + 17 > 32
(2x)2 � 2.2x � 15 > 0 (2x + 3) (2x � 5) > 0
2x < � 3 or 2x > 5 x or x > log2 5 x (log2 5, )
(iii) (logx)2 � logx � 2 0x > 0 ....(i)(log x � 2) (log x +1) 0
log x � 1 or log x 2
x 101
or x 100 ....(ii)
(i) (ii) x
,100
101
,0
RESONANCE SOLUTIONS (XI) # 33
(iv) log0.5(x + 5)2 > log1/2 (3x � 1)2
(x + 5)2 > 0 x R � {� 5} ........(i)
(3x � 1)2 > 0 x R �
31
........(ii)
(x + 5)2 < (3x � 1)2
8x2 � 16 x � 24 > 0 x2 � 2x � 3 > 0
(x � 3) (x + 1) > 0 x (�, � 1) (3, ) ........(iii)(i) (ii) (iii) gives
(�, �5) (�5, �1) (3, )
(v)21
2log12x3
3x2 + 1 > 1 x2 > 0 x R � {0}
2 < (3x2 + 1)1/2
3x2 + 1 > 4 (x � 1) (x + 1) > 0
x (�, �1) (1, )
(vi) 2xlog (x + 2) < 1 x + 2 > 0 x > � 2
Case-I : when 0 < x2 < 1 x (�1, 0) (0, 1)then x + 2 > x2 x2 � x � 2 < 0
}0{�)1,1(x
Case-II : x2 > 1 |x| > 1x + 2 < x2 x2 � x � 2 > 0
),2()1,2(x
Hence , ),2()1,0()0,1()1,2(x
11. (i)2x
1x2
< 1
Case-I :x � 2 < 0 x < 2 ........(i)2x � 1 (x � 2)2
x (�, 1) (5, ) ........(ii)x (i) (ii) x (�, 1) .......(A)
Case-II : x � 2 > 0 x > 2 ........(iii)2x � 1 (x � 2)2
2x � 1 < x2 � 4x + 4
x2 � 6x + 5 > 0
x (�, 1) (5, ) ........(iv)x (iii) (iv)x (5, ) .......(B)x (A) (B)x (�, 1) (5, )
(ii) x < |x|1
Case-I : x < 0 ......(i)1 � |x| 0 1 + x 0 x � 1 .........(ii)x (i) (ii) x [�1, 0) .......(A)
Case-II : x 0 .....(i)1 � x 0 x 1 .......(ii)x2 < 1 � x
x2 + x < 1 x2 + x + 41
< 45
2
21
x
<
45
251
< x < 2
15 ......(iii)
RESONANCE SOLUTIONS (XI) # 34
x (i) (ii) (iii) x
215
,0 .......(B)
x (A) (B) x
215
,1
(iii) 8x6x2 1x
Domain x + 1 0 x �1
x2 � 6x + 8 0 (x � 2) (x � 4) 0 x 2 or x 4 Domain x [�1, 2] [4, )squaring x2 � 6x + 8 x + 1 x2 � 7x + 7 0
2
27
x
�
421
0 x
2217
,2
217
(iv) 2x�x28 > 6 � 3x
(a) 8 + 2x � x2 0 x [�2, 4] .... (i)case - Iwhen (i) 6 � 3x 0 x 2 ... (ii)so 8 + 2x � x2 > 36 + 9x2 � 36 x
10x2 � 38x + 28 < 0
5x2 � 19x + 14 < 0
(5x � 14) (x � 1)< 0
x
514
,1 .... (iii)
by (1) and (2) and (3)x (1, 2]Case -
6 � 3x < 0 x > 2+ ve > �ve
so x > 2 .... (iv)by (1) and (4)x (2, 4]so by case (1) and (2) x (1, 4]
(v) x2 � 7x + 10 0 and 14x � 20 � 2x2 0(x � 2) (x � 5) 0 and (x � 2) (x � 5) 0 ...........(i)so x = 2 or x = 5now check for x = 2
9 log4
4
1 � 9
� 9 � 9
which is true hence x = 2 is a solutionnow check x = 5
29
log
8
5 � 3
log2
85
� 32
(1.6)3 44.096 4which is falseso only solution is x = 2
RESONANCE SOLUTIONS (XI) # 35
(vi) Domain x > 0log2
2x + 2 log
2x 0
log2x (log
2x + 2) 0
log2x � 2 or log
2x 0
0 < x 41
or x 1
x
41
,0 [1, ) .......(i)
Case-I 4 � log2x < 0
positive < negative (false)Case-II 4 � log
2x 0 log
2x 4
log2
2x 2 log2x < 2 (4 � log
2x)2
Let log2x = t
t2 + 2t < 2 (4 � t)2
t2 � 18t + 32 > 0
(t � 16) (t � 2) > 0 t < 2 t > 16log
2x < 2 log
2x > 16 (Rejected)
log2x < 2
x < 4 .........(ii)by (i) and (ii)
x
41
,0 [1, 4)
13. Square root of 7 + 2i = ±
2725
i2
725 = ±(4 + 3i)
where |7 + 24 i| = 25
15. (i) z = R3 � (3 + i)+m + 2i = 03 � 3 + m = 0 & � + 2 = 0
= 28 � 6 + m = 0 m = � 2
(ii) If one root is i then other is � i
Let forth root is .
2 = 23
= 43
2a�
= 2 + i + (� i) + 43
= 411
a = 211�
20. (i) z = 1 + 2518
ie
= 259
ie
259
i�259
iee
z =
259
cos2 259
ie
|z| =
259
cos2 Arg z = 259
(ii) z = 6/5i�6/ii e2ee2
|z| = 2 Arg z = 6
5�
.
RESONANCE SOLUTIONS (XI) # 36
(iii) |z| = 2
2 1tan1
= sec21Arg z = 2 Arg(tan 1 � i)
= 2
2�1 = 2 �
(iv) z =
5cosi
5sin
5sin2
)1�i(
|z| =
5eccos
2
1
5sin2
2
Arg(z) = � 5
�4
=
2011
EXERCISE # 3PART - I
1. 1x � 1x = 1x4 .....(i)
squaring both sides
(x + 1) + (x � 1) � 2 1x2 = 4x � 1
(1 � 2x) = 2 1x2 .....(ii)squaring both sides1 + 4x2 � 4x = 4x2 � 4
4x = 5 x = 45
does not satisfy equation (i)
No solution
2. 2 log2 log
2 x + log
1/2 log
2 x22 = 1
log2 (log2 x)2 � log2 log2 x22 = 1 log2
x22log
xlog
2
22
= 1
xlog23
xlog
2
22
= 2 Let log2 x = y
y2 � 2y � 3 = 0 (y � 3) (y + 1) = 0
y = 3, �1 log2 x = 3, �1,
but log2 x > 0 log2 x = � 1 is not possible x = 8
3. (a) z1 = z2 = z3 = 1
11zz = 22zz = 33zz = 1
Given 1 = 321 z1
z1
z1
= 321 zzz = 321 zzz = 1
1 = 321 zzz
(b) � = arg (z) < 0arg (�z) = � arg (�z) � arg (z) = � � (�) = Hence (A)
RESONANCE SOLUTIONS (XI) # 37
4. log3/4 log8 (x2 + 7) + log1/2 log1/4 (x2 + 7) 1 = 2
log3/4 31
log2 (x2 + 7) � log2 2
)7x(log 22
= 2
let log2 (x2 + 7) = t
log3/4 3t
� log2 2t
+ 2 = 0 log3/4 3t
+ 1 �
1
2t
log2 = 0
log3/4 4t
= log2 4t
4t
= 1 t = 4
log2 (x2 + 7) = 4this gives x = ± 3.
5.21
log2(x � 1) = log2 (x � 3)
1x = x � 3
(x � 1) = x2 � 6x + 9
x2 � 7x + 10 = 0
(x � 5) (x � 2) = 0 but x 2 x = 5
6. Let 1zz
zz1
21
21
|1 � z
1 2z | < |z2 � z
1|
(1 � z1 2z ) (1 � 21zz ) < (z
2 � z
1) ( 12 zz )
1 + |z1|2 |z
2|2 � |z
1|2 � |z
2|2 < 0 (1 � |z
1|2) + (|z
1|2 � 1) |z
2|2 < 0
(1 � |z1|2) (1 � |z
2|2) < 0
which is true because of |z1| < 1 < |z
2| .
7 . (2x)n2 = (3y)n3
n2 n(2x) = n3 n(3y) = n3 (n3 + ny) ......... (1)also 3nx = 2ny
nx n3 = ny n2 ......... (2)
by (1) n2 n(2x) = n3 (n3 + ny) n 2 . n (2x) = n3
2n
3nnx3n
n22 n2x = n23 (n2 + nx) 3n2n 22 (n2x) = 0
n2x = 0 x = 21
8. Let .......23
1�4
23
1�4 = t
t23
1�4 = t
4 � 23
1t = t2
t2 + 23
1t � 4 = 0 23 t2 + t � 212 = 0
t = 232
21223411�
=
232
171�
t = 26
16,
26
18�
RESONANCE SOLUTIONS (XI) # 38
t = 23
8,
2
3� and
2
3� is rejected
so 6 + log3/2
23
8
23
1 = 6 + log
3/2
94
= 6 + log3/2
2
32
= 6 � 2 = 4
PART - II
1. Let z = r1 ei and w = r
2 ei z = r
1 e�i
Given, |z| = 1 i2
i1 er.er = 1
r1r
2 = 1 ...(i)
and arg (z) � arg () = 2
� = 2
Then,
i2
i1 er.erz
= r1r
2 )(ie
From Eqs. (i) and (ii), we get
z = 1. 2/ie
= cos 2
� i sin 2
z = �i.
2.x
i1i1
=
x
)i1()i1()i1()i1(
=
x
2
2
i1
)i1(
=
x
2i211
x
i1i1
= (i)x = 1 (given)
(i)x = (i)4n ,where n is any positive integer. x = 4n.
3. Since , z + i w = 0 z = �i w z = iw w = �iz
Also, arg(zw) = arg(�iz2) = arg(�i) + 2 arg(z) =
�2
+ 2 arg (z) =
2)iarg(
2 arg (z) = 2
3 arg (z) =
43
.
4. Let z = 1i
1
z =
1i1
= 1i
1
= �1i
1
.
5. Let roots be p + iq and p � iq p, q Rroot lie on line Re(z) = 1 p = 1product of roots = p2 + q2 = = 1 + q2
(1, (q 0, roots are distinct) Ans.
RESONANCE SOLUTIONS (XI) # 39
STRAIGHT LINE
EXERCISE # 1PART - I
Section (A) :
A-3._ (i) centroid
312120
,3
1650 (7, 8)
A(0,0)
B(5,12) C(16,12)
13 20
11
(ii) Let coordinates of circumcentre is O (x, y).Therefore OA = OB = OC
x2 + y2 = (x � 5)2 + (y � 12)2 = (x � 16)2 + (y � 12)2
x2 + y2 = (x � 5)2 + (y � 12)2 10x + 24y = 16g(x � 5)2 + (y � 12)2 = (x � 16)2 + (y � 12)2
2x = 21 x = 221
, y = 38
(iii)
112013
12132012110,
112013
1316205110 (7, 9)
(iv) 2 =
11132012131102012
,111320
13111613205 (27, � 21)
A-4. Let coordinates of P(x,y)given PA = PB (x � 3)2 + (y � 4)2 = (x � 5)2 + (y + 2)2
4x � 12y = 4
x � 3y = 1 ...(i)
125
143
1yx
21
= 10
6x + 2y � 26 = ± 20 3x + y � 13 = ± 10
3x + y = 23 ...(ii) 3x + y = 3 ...(iii)Solving (i) and (ii) we get (7, 2)Solving (i) and (iii) we get (1, 0)
Section (B) :B-2. Let equation of line is x + my + n = 0 ...(i)
given
1a3a
,1a
a 23
,
1b3b
,1b
b 23
and
1c3c
,1c
c 23
are collinear
1t3t
,1t
t 23
is general point which satisfies line (i)
1tt3
+ m
1t3t2
+ n = 0
t3 + m t2 + nt � (3m + n) = 0
RESONANCE SOLUTIONS (XI) # 40
a + b + c = �
m
ab + bc + ac =
n
abc =
nm3
Now LHS = abc � (ab + bc + ac) + 3 (a + b + c)
=
)nm3( �
n + 3
m = 0
B-5. Let point is P(x, y) and A(ae, 0) and B(�ae,0)
Given |PA � PB| = 2a 2222 y)aex(y)aex( = 2a
Let 22 y)aex( = A, 22 y)aex( = B
Hence A � B = 2a
A2 � B2 = (A + B) (A � B) A + B = �2xe
Hence 2A = 2a �2xe
A = a � ex
(x � ae)2 + y2 = (a � ex)2 2
2
a
x �
)1e(a
y22
2
= 1
Section (C) :
C-3. ObviousC-4. By parametric form 11
2 2
p(4, 1) 3x � y = 0
Q
sin
22
111,cos
22
114
it lies on 3x � y = 0
12 + 22
33 cos � 1 �
22
11 sin = 0
1 + 22
3 cos �
22
sin = 0 3cos � sin = � 22
squaring both sides9cos2 + sin2 � 6 sin cos = 8(sin2 + cos2)cos2 � 6sin cos � 7 sin2 = 07tan2 + 6tan � 1 = 0
tan = � 1, 71
.
Hence required line are x + y = 5, x � 7y + 3 = 0
Section (D) :
D-2. foot of perpendicular
32x
= 13y
= � 22 )1(3
)4323(
(x, y)
1029
,1023
image
32x
= 13y
= � 2 22 )1(3
)4323(
(x, y)
514
,5
13
RESONANCE SOLUTIONS (XI) # 41
slope of line perpendicular to the line y = 3x � 4 is � 31
hence its equation
y � 3 = � 31
(x � 2) x + 3y � 11 = 0
D-5. L1 : 4x + 3y � 7 = 0
L2 : 24x + 7y � 31 = 0
a1 a
2 + b
1b
2 = 4 × 24 + 3 × 7 > 0
Hence + sign gives obtuse angle bisector and � sign gives acute angle bisector
Now, put origin in both 4 × 0 + 3 × 0 � 7 < 0
24 × 0 + 7(0) � 31 < 0
Hence sign gives that bisector in which origin lies.Hence origin lies in obtuse angle bisector
Now, equation of bisector
57y3x4
= ±
2531y7x24
+ sign x � 2y + 1 = 0
� sign 2x + y � 3 = 0
Section (E) :
E-4. 12x2 � 10xy + 2y2 + 11x � 5y + = 0This represents pair of straight lines if = abc + 2fgh � af2 � bg2 � ch2 = 0
we get = 2Now12x2 � 10xy + 2y2 + 11x � 5y + 2 = (6x � 2y + p) (2x � y + q)
compair both sides 2p + 6q = 11�p � 2q = �5
solving both we get p = 4, q = 21
Hence required lines are 6x � 2y + 4 = 0 3x � y + 2 = 0
2x � y + 21
= 0 4x � 2y + 1 = 0
solving both equations we get point of intersection
2
5,
2
3
Now angle between both lines
tan = 21
21
mm1
mm
= 231
23
=
71
= tan�1 71
Now equation of pair of angle bisector
21225
y23
x22
= 525
y23
x
2x2 + 4xy � 2y2 + 16x � 4y + 7 = 0
E-5. Homogenize x2 + y2 = a2 by y = mx + c
we get x2 + y2 = a2
2
cmxy
This equation represents pair of lines passing through origin.That will be right angle ifcoeff. x2 + coeff. y2 = 0 2c2 = a2 (1 + m2)
Section (F) %
F-1_. (i) (2, 5, 8) (ii) (�5, �4, �3) (iii) (�3, 0, 7) (iv) (8, 2, 5)
RESONANCE SOLUTIONS (XI) # 42
F-3_. (i)
(ii)
PART - IISection (A) :
A-1*. AB = 94 = 13
BC = 1636 = 2 13
CD = 94 = 13
AD = 1636 = 2 13
AC = 164 = 65
BD = 4916 = 65its rectangle
A-4. If H is orthocentre of triangle ABC, then orthocentre of triangle BCH is point A
Section (B) :
B-2*. Since A, B, C are coffe. nearSlope of AB = Slope of BC A(k, 2 � 2k) B(1 � k, 2k) C(� k� 4, 6 � 2k)
k1kk2k22
=
4kk1k26k2
1k2k42
=
56k4
10 � 20k = (4k � 6)(2k � 1)
(4k � 6)(2k � 1) + 10(2k � 1) = 0 k = 21
, � 1
RESONANCE SOLUTIONS (XI) # 43
B-3. AP = 22 )4�y(x
BP = 22 )4y(x |AP � BP| = 6
AP � BP = ± 6
22 )4�y(x � 22 )4y(x = ± 6
On squaring we get the locus of P9x2 7y2 + 63 = 0
Section (C) :
C-2. x1 + y
1 = 5 ... (i)
x2 = 4 ... (ii)
co - ordinates of G are (4, 1)
3
xx1 21 = 4 ....(iii)
and ¼rFkk½3
2yy 21 = 1 ... (iv)
solving above equations, we get B & C.
C-4.
Let coordinates of point P by parametricP(2 + r cos 45º, 3 + r sin 45º)
It satisfies the line 2x � 3y + 9 = 0
2
2
r2 � 3
2
r3 + 9 = 0 r = 24
Section (D) :
D-1. a2x + a by + 1 = 0origin and (1, 1) lies on same side.a2 + ab + 1 > 0 a RD < 0 b2 � 4 < 0
b (�2, 2)
but b > 0 b (0, 2)
D-4. p = 22 )16(2
56422
=
26091
q(�11, 4)
64x+8y+35 = 0
1
2
pB : 2x � 16y � 5 = 01
q = 22 864
35481164
p < q Hence 2x � 16y � 5 = - is acute angle besectory
RESONANCE SOLUTIONS (XI) # 44
Section (E) :
E-2. m1 + m
2= � 10
m1m
2 =
1a
given m1 = 4m
2 m
2 = � 2, m
1 = � 8,
a = 16
E-5. Homogenize given curve with given line
3x + 4xy � 4x + 1 = 02
2x + y = 1
3x2 + 4xy � 4x(2x + y) + 1(2x + y)2 = 03x2 + 4xy � 8x2 � 4xy + 4x2 + y2 + 4xy = 0� x2 + 4xy + y2 =coeff. x2 + coeff. y2 = 0Hence angle is 90º
Section (F) :
F-3_. x2 + y2 + y2 + z2 + z2 + x2 = 362(x2 + y2 + z2) = 36
23zyx 222
F-4_. The two numbers are x and x + 2(a) x > 10(b) x + 2 > 10 x > 8(c) x + x + 2 < 34
2x < 32 X < 16Now x must be between 10 < x < 16
x (11, 13), (13, 15)
F-6_. Let the third PH reading is x
7.4 < 3
x42.848.7 < 8.2
22.2 < 15.90 + x < 24.66.3 < x < 8.7PH range should be in between 6.3 to 8.7
F-8_. Standard result.
EXERCISE # 2PART - I
3.
A, S, B are collinear
0
112
1xxx
100
121
RESONANCE SOLUTIONS (XI) # 45
3x1 = 2x
2.... (1)
B, R, C are collinear
0
103
1xxx
112
122 x1 � 2x
2 + 3 = 0 ... (2)
Solving (i) and (ii) we get x1 =
23
x2 =
49
Hence
0,
49
Q,0,23
P ,
43
,49
S,43
,23
R
6. (i) D is mid point of BC
Hence co-ordinates of D are
2
yy,
2
xx 3232
Therefore, equation of the median AD is
12
yy
2
xx1yx
1yx
3232
11 = 0
Applying R3 2R
3
2yyxx
1yx
1yx
3232
11
= 0
1yx
1yx
1yx
22
11 + 1yx
1yx
1yx
33
11 = 0
(using the addition property of determinats)(ii) Let P(x, y) be any point on the line parallel to BC
Area of ABP = Area of ACP
1yx
1yx
1yx
22
11 = 1yx
1yx
1yx
33
11
1yx
1yx
1yx
22
11 � 1yx
1yx
1yx
33
11 = 0
This gives the equation of line AP.
(iii) Let AD be the internal bisector of angle A,
DCBD
= CABA
= bc
D
bc
bycy,
bc
bxcx 2323
RESONANCE SOLUTIONS (XI) # 46
Let P(x,y) be any point on AD then P,A,D are collinear
1
cb
bycy
cb
bxcx1yx
1yx
2323
11
= 0
R3 (b + c) R
3
cbbycybxcx
1yx
1yx
2323
11
= 0
ccycx
1yx
1yx
33
11 + bbybx
1yx
1yx
22
11 = 0 (Addtion property)
c 1yx
1yx
1yx
33
11 + b 1yx
1yx
1yx
22
11 = 0
This is the equation of AD.
9. equation of line L1 is
y � 25
= 2 . (x � 23
)
or 2x � y � 21
= 0
or 4x � 2y � 1 = 0
equation of line L2 is
y � 25
= 1 (x � 23
) or x � y + 1 = 0
Point C is mirror image of point A w.r.t line L1
4)2(��x
= 2�
)3(�y =
20)1�6�8(�2�
C(4, 0)similarly B is mirror image of A in line L
2 = 0
1�)3�y(
1)2(��x =
2)13�2(�2�
B(2, � 1) D(1, 23
) ; E (0, 1)
median through B is
(y + 1) = 1�2/5
(x � 2) 5x + 2y = 8
median through C is
(y � 1) = )0�x(4�
1 x + 4y = 4
11. a2 + b2 = c2 .... (i)Let L is (x
1, y
1)
L is foot of perpendicular from point P(a, b) on line ABequation of AB is bx + ay � ab = 0
2211
ba
)ab�abab(�a
b�yb
a�x
RESONANCE SOLUTIONS (XI) # 47
211
c
ab�
ab�y
ba�x
x1 = a � 2
22
2
2
c
)b�c(a
c
ab = a3/c2
a3 = c2x1
.... (ii)similarly
b3 = c2y1
....... (iii)using these relations (ii) & (iii) in equation (i), we getrequired locus.
14. Given pair of lines are a2x2 +2h(a + b)xy + b2y2 = 0 ax2 +2hxy + by2 = 0Equation of pair of bisectors of first pair is
22
22
b�a
y�x = bah
xy
b�ay�x 22
= hxy
Which is also bisector of second pair.Hence both pair are equally inclined.
15. Let equation of chords hx + ky = 1By homogenisation
3x2 � y2 � 2x (hx + ky) + 4y (hx + ky) = 0
it makes 90º. Hence
coeff. x2 + coeff. y2 = 03 � 2h � 1 + 4k = 0 h � 2k = 1
Hence all chords are concurrent at (1, � 2)
Similarly homogenize 3x2 � y2 � 2x + 4y = 0
3x2 + 3y2 � 2x(hx + ky) + 4y (hx + ky) = 0
again coeff. x2 + coeff. y2 = 0
3 + 3 � 2h + 4k = 0 h � 2k = 3 3h
� 3k2
= 1
Hence, all chords passes though
32
,31
.
PART - II
1. here tan = 51
tan 2 = 2
51
�1
51
2
= 125
required line y = 12
x5
4. p = 5
a00 =
5
a
tan 45º = xp
p = x
Hence area = 21
(2x)(p) = px = p2 = a/5
RESONANCE SOLUTIONS (XI) # 48
8. Image of A(3, 10) in 2x + y � 6 = 0
23x
= 110y
= � 2
22 12
6106
23x
= 110y
= � 4
A' = (� 5, 6)
Equation of A'B is y � 3 =
4536
(x � 4)
y � 3 = � 31
(x � 4)
3y � 9 = � x + 4
x + 3y � 13 = 0
10. By geometrya2 + b2 = (a + b)2 ....(i)
By section formula
h = ba
=
a)ba(n
k = ba
=
b)ba(k
Put value of and in (i)
2
22
a
)ba(h + 2
22
b
)ba(k = (a + b)2
2
2
a
h + 2
2
b
k = 1
Locus of (opchim) is 2
2
a
x + 2
2
b
y = 1
12. x2 � 2pxy � y2 = 0
pair of angle bisector of this pair )1(1
yx 22
= p
xy
x2 � y2 + p2
xy = 0
compare this bisector pair with x2 � 2qxy � y2 = 0
p2
= �2q pq = �1.
14. 1 = 2x � 3y � 6 = 0
2 = 3x � y + 3 = 0
3 = 3x + 4y � 12 = 0
Hence [� 1, 3]
[� 2, 3]
RESONANCE SOLUTIONS (XI) # 49
16. Both A & B are same side of line 2x � 3y � 9 = 0
Now, perimeter of APB will be least when points A, P, B were collinear. Let B' is image of B
Then 2
0x =
34y
= � 2
22 )3(2
9120
B'
1374
,1384
Now equation of AB' is y = 110
74(x + 2)
point of intersection of given line & Q is P
1737
,1721
.
EXERCISE # 3
1. (A) Slope of such line is ± 1
(B) area of OAB = 21
× 3 × 4 = 6 sq. units
y
xO
B
A(0,�3)
(�4,0)
(C) To represent pair of straight lines c33
311
312
= 0 c = 3
(D) Lines represented by given equation are x + y + a = 0 and x + y � 9a = 0
distance between these parallel lines is = 2
a10 = a25
Comprehension # 2 (5, 6, 7)Slopes of the lines
3x + 4y = 5 is m1 = �
43
and 4x � 3y = 15 is m2 =
34
m1 m
2 = � 1
given lines are perpendicular and A = 2
Now required equation of BC is
(y � 2) = )4/tan(m1)4/tan(m
(x � 1) ......(1)
where m = slope of AB = � 43
equation of BC is (on solving (1))x � 7y + 13 = 0 and 7x + y � 9 = 0
L1 x � 7y + 13 = 0
L2 7x + y � 9 = 0
5. c + f = 46. Equation of a straight line
through (2, 3) and inclined at an angle of (/3) with y-axis ((/6) with x-axis) is
)6/cos(2x
= )6/sin(
3y
x � y3 = 2 � 33
Points at a distance c + f = 4 units from point P are
(2 + 4 cos (/6), 3 + 4 sin (/6)) (2 + 32 , 5)
RESONANCE SOLUTIONS (XI) # 50
and (2 � 4 cos (/6), 3 � 4 sin (/6)) (2 � 32 , 1)only (A) is true out of given options
7. Let required line be x + y = a
which is at |b � 2a � 1| = |5 � 4 � 34 � 1| = 34 units from origin
required line is x + y � 64 = 0 (since intercepts are on positive axes only)
8. ax3 + bx2y + cxy2 + dy3 = 0since this is homogeneous pair represent there straight lines passing through origin
ax3 + bx2y + cxy2 + dy3 = (y � m1x)(y � m
2x)(y � m
3x)
or put y = mx in given equation we getm3d + cm2 + bm + a = 0
m1 + m
2 + m
3 =
dc
m1m
2 + m
2m
3 + m
3m
1 =
db
m1m
2m
3 =
da
given two lines + hence m1m
2 = � 1 m
3 = a/d
eliminate m3 from remaining equation
10. S2 is standard result.
equation of angle bisectors of lines given in S1 are
52y4x3
= ± 5
2y3x4 x � y = 0 and 7x + 7y � 24 = 0
14._ Let R(5, 1) divides line segment joining P(2,10) and Q(6, � 2) in : 1
5 = 126
= 3
Hence Harmonic conjugate divides in 3 : 1 externaly
Hence required part is
13106
,13218
= (8, -8)
19. Required point is foot of perpendicular from (0, 0) on the given line which is 3
0 =
40
= 25
)1(
EXERCISE # 4PART - I
1. A(x1,y
1), B(x
2, y
2), C(x
3, y
3)
x1, x
2 , x
3 and y
1 , y
2, y
3 are in G.P. of common ratio r.
x2
= x1r , x
3 = x
1r2 , y
2 = y
1r, y
3 = y
1r2
Area of triangle ABC = 21
111
ryryy
rxrxx2
111
2111
= 0 A, B & C are collinear..
2. Let m be the slope of PQ then
tan 450 = )2(m1)2(m
1 = m212m
RESONANCE SOLUTIONS (XI) # 51
± 1 = m212m
m + 2 = 1 � 2m or �1 + 2m = m + 2
m = 31
or m = 3
PR makes 450 with PQ
equation of PQ y �1 = 31
(x � 2)
x + 3y � 5 = 0
equation of PR is y � 1 = 3(x � 2)
3x � y � 5 = 0
combined equation of PQ and PR (x + 3y � 5) (3x � y � 5) = 0
3x2 � 3y2 + 8xy � 20x � 10y + 25 = 0
3. S is the mid point of Q and R
S
213
,2
67 =
1,
213
slope of PS = m = 2/132
12
=
92
equation of line passing through (1, �1) and parallel to PS is
y + 1 = 92
(x � 1) 2x + 9y + 7 = 0
4.
BC = 2AB = 2AC = 2Hence ABC is an equilateral triangle. In equilateral triangle incentre coincides with centroid. Thus
3300
,3
120
3
1,1
5. p (h, k) be a general point in the first quadrent such that d(P, A) = d(p, O) |h � 3| + |k � 2| = |h| + |k| = h + k
[h and k are positive point (h, k) being in quadrent]If h < 3, k < 2, then (h, k) lie in region If h > 3, k < 2, then (h, k) lie in region If h > 3, k > 2, then (h, k) lie in
y = 2 A(3,2)IIIIV
I IIx+y=5/2
(0,0) (5/2,0)x = 3
O
x = 1/2
x
If h < 3, k > 2, then (h, k) lie in IV
In region 3 � h + 2 � k = h + k h + k = 25
In region h � 3 + 2 � k = h + k k = 21
not
possible
RESONANCE SOLUTIONS (XI) # 52
In region h � 3 + k � 2 = h + k �5 = 0 Not
possible
In region IV 3 � h + k � 2 = h + k h = 21
Set consist line segment x + y = 25
of finite length
In Ist region and x = 21
in the IV region.
6. c1 ac
1
= a1
cbyaxbcyacxa
bcycbyaxyabxa
acxaybxacabyxa
2
2
2
c1
c
1 + bc
2 + cc
3
= a1
cbyaxcyb)cba(
bcyaxcbyy)cba(
acxbxayx)cba(
222
222
222
= a1
byaxccyb1
cybaxcbyy
acxbxayx
as a2 + b2 + c2 = 1c
2 c
2 � bc
1 , c
3 c
3 � cc
1
= a1
byaxcy1
baxcy
aayx
R1 x R
1
= ax1
byaxcy1
baxcy
axaxyx2
R1 R
1 + yR
2 + R
3
= ax1
byaxcy1
baxcy
001yx 22
= (x2 + y2 + 1) (ax + by + c)Given = 0 ax + by + c = 0 which represent astraight line
7. The x-coordinate of intersection of lines 3x + 4y = 9 and y = mx + 1 is x = m43
5
For x being an integer 3 + 4m should be divisor of 5i.e. 1, �1, 5 or �5
3 + 4m = 1 m = 21
(Not integer)
4m + 3 = � 1 m = � 1 (Interger)
3 + 4m = 5 m = 21
(Not an integer)
3 + 4m = � 5 m = � 2 (integer)
there are two integral value of m
RESONANCE SOLUTIONS (XI) # 53
8.
In parallelogram OABCB(0,1) and point A in the point of intersection of y = mx and y = nx + 1
nm
1x
and y =
nmm
Now area of parallelogram = 2 (OAB)
=
nm1
121
2
= |nm|1
9. y = |x| � 1
y = �|x| + 1
Region is clearly square with vertices at thepoint (1,0), (0,1), (�1,0), (0,�1). So,
its area = 22 = 2.
10. Let XOS = and XOT = 2
let p(cos , sin ), then TOP = � 2
let Q be the image of P in OT. Then QOT = � 2
QOX = �
XOQ = �
Q is image of P in the line whose slope is tan 2
11.
The line segment QR makes an angle 60º with the positive direction of x-axis.
hence bisector of angle PQR will make 120º with +ve direction of x-axis.
Its equationy � 0 = tan120º (x � 0)
y = � x3
0y3x
RESONANCE SOLUTIONS (XI) # 54
12.
as OPA ~ OQC
OQOP
= OCOA
= 3
4/9 =
43
13. The line y = mx meets the given lines in P
1mm
,1m
1 and Q
1mm3
,1m
3. Hence equation of L
1 is
y � 1m
m
= 2
1m1
x y � 2x � 1 = � 1m
3
.........(i)
and that of L2 is y �
1mm3
= � 3
1m3
x y + 3x � 3 = 1m
6
.........(ii)
Form (i) and (ii) 3x3y1x2y
= �
21
x � 3y + 5 = 0; which is a straight line
14. equation of line y � 2 = m(x � 8) where m < 0
P
0,
m2
8 and Q (0, 2 � 8m)
Now OP + OQ = m
28 + |2 � 8m|
= 10 + )m(�2
+ 8(�m) 10 + 2 )m(8m2
18
15. The number of integral points that lie in the interior of square OABC is 20 × 20. These points are (x, y) where x,
y = 1, 2, ........., 20. Out of these 400 points 20 lie on the line AC. Out of the remaining exactly half lie in ABC.
number of integeral point in the triangle OAC = 21
[20 × 20 � 20] = 190
Alternative SolutionThere are 19 points that lie in the interior of ABC and on the line x = 1, 18 point that lie on the line x = 2 and soon. Thus, the number of desired points is
19 + 18 + 17 + .... + 2 + 1 = 2
1920 = 190.
RESONANCE SOLUTIONS (XI) # 55
16. Refer Figure Equation of altitude BD is x = 3.
slope of AB is 4304
= � 4
slope of OE is 1/4Equation of OE is
y = 41
x.
Lines BD and OE meets at (3, 3/4)
17. The lines given by x2 � 8x + 12 = 0 are x = 2 and x = 6.
The lines given by y2 � 14y + 45 = 0 are y = 5 and y = 9
Centre of the required circle is the centre of the square. Required centre is
295
,2
62 = (4, 7).
18. x2 � y2 + 2y = 1x = ± (y � 1)
Bisector of above lines are x = 0, y = 1so Area between x = 0, y = 1 and x + y = 3
= 21
× 2 × 2 = 2 squ. units
19. A line passing through P(h, k) and parallel to x-axis is y = k the other lines given are y = x and y + x = 2Let ABC be the formed by the points of intersection of the lines (i) , (ii) and (iii) be A(k, k) , B(1, 1), C (2�k, k)
Area of ABC = 21
1kk�2
111
1kk
= 4h2
C1 C
1 � C
2 21
1kk2�2
110
1k0
= 4h2
21
|(2 � 2k) (k � 1)| = 4h2 (k � 1)2 = 4h2
k � 1 = 2h, k � 1 = � 2h k = 2h + 1 k = � 2h + 1
locus of (h, k) is y = 2x + 1 y = � 2x + 1
20.
R is centroid hence R
34
,3
RESONANCE SOLUTIONS (XI) # 56
21.RQPR
= OQOP
OQOP
RQPR
= 5
22
but statement � 2 is false
Ans. (C)
22. P (� sin ( � ) , � cos )Q (cos ( � ), sin )
R (cos ( � + ) , sin ( � )) 0 < , , < 4
xR = cos ( � ) cos � sin ( � ) sin
xR = x
Q . cos + x
P . sin
yR = sin cos � cos sin
yR = y
Q . cos + y
P . sin
For P, Q, R to be collinear sin + cos = 1
sin
4 = 2
1 not possible for the given interval
4,0
non collinear
23. (1 + p) x � py + p (1 + p) = 0 ......(1)(1 + q) x � qy + q(1 + q) = 0 ......(2)on solving (1) and (2), we get C(pq, (1 + p) (1 + q)) equation of altitude CM passing through C and perpendicular to AB is x = pq .......(3)
slope of line (2) is =
q
q1
slope of altitude BN (as shown in figure) is = q1q
equation of BN is y � 0 = q1q
(x + p)
y = )q1(q
(x + p) ....... (4)
Let orthocentre of triangle be H(h, k) which is the point of intersection of (3) and (4) on solving (3) and (4), we getx = pq and y = � pq h = pq and k = �pq
h + k = 0 locus of H(h, k) is x + y = 0
24. Let slope of line L = m
)3(�m1
)3(��m
= tan 60º = 3
m31
3m
= 3
taking positive sign, m + 3 = 3 � 3m
m = 0taking negative sign
m + 3 + 3 � 3m = 0
m = 3
RESONANCE SOLUTIONS (XI) # 57
As L cuts x-axis m = 3
so L is y + 2 = 3 (x �3)
PART - II
1. (h � a1)2 + (k � b
1)2 = (h � a
2)2 + (k � b
2)2
2h(a1 � a
2) + 2k(b
1 � b
2) + 2
121
22
22 baba = 0
compare with (a1 � a
2)x + (b
1 � b
2) y + c = 0
c =
2baba 2
121
22
22
.
2. 3h � 1 = a cos t + b sin t
3k = a sin t � b cos t
squaring and add. (Locus)(3x � 1)2 + 9y2 = a2 + b2
3. x2 � 2pxy � y2 = 0
pair of angle bisector of this pair )1(1
yx 22
= p
xy
x2 � y2 + p2
xy = 0
compare this bisector pair with x2 � 2qxy � y2 = 0
p2
= �2q pq = �1.
4. Equation of AC
y � a sin =
sincoscossin
(x � acos )
y(cos + sin ) + x(cos � sin ) = a(sin cos + sin2 � sin cos + cos2 )y(cos + sin ) + x(cos � sin ) = a.
5. G
32k
,3h
3h2
+ (k � 2) = 1 2h + 3k = 9
Locus 2x + 3y = 9.
6. Let equation of line isax
+ by
= 1
it passes through (4, 3)a4
+ b3
= 1
RESONANCE SOLUTIONS (XI) # 58
sum of intercepts is �1 a + b = �1 a = �1 � b
b1
4
+ b3
= 1
4b � 3 � 3b = �b � b2
b2 + 2b � 3 = 0
b = �3, 1
b = 1, a = �22
x
+ 1y
= 1
b = �3, a = 22x
+ 3
y
= 1.
7. x2 � 2cxy � 7y2 = 0
sum of the slopes m1 + m
2 =
7c2
Product of slopes m1m
2 =
71
given m1 + m
2 = 4m
1m
2
7c2
=
74
c = 2.
8. Pair 6x2 � xy + 4cy2 = 0 has its one line 3x + 4y = 0
y = 4
x36x2 +
4x3 2
+ 4c 16x9 2
= 0
24x2 + 3x2 + 9cx2 = 0 c = �3.
9. ax + 2by + 3b = 0bx � 2ay � 3a = 0
ab6ab6x
= 22 a3b3
y
= 22 b2a2
1
Hence point of intersection (0, �3/2)
Line parallel to x-axis y = �3/2.
10. a, b, c are in H.P. b2
= a1
+ c1
a1
� b2
+ c1
= 0
given lineax
+ by
+ c1
= 0
Clearly line passes through (1, �2).
11. Centroid is
3
7 ,1
RESONANCE SOLUTIONS (XI) # 59
12. Pair of lines ax2 + 2(a + b)xy + by2 = 0
Area of sector A1 = 1
2r21
A2 = 2
2r21
1 +
2 = 180º
given A1 = 3A
2
1 = 3
2
2 = 45º ,
1 = 135º
Angle between lines is = ba
ab)ba(2 2
= 1
4 abba 22 = a2 + b2 + 2ab
3a2 + 3b2 + 2ab = 0.
13. Let equation of line is
by
ax = 1.
By section formula
2a
= 3 a = 6
2b
= 4 b = 8
6x
+ 8y
= 1 4x + 3y = 24.
14. Since (1, 1) and (a, a2) Both lies same side with respect to both linesa � 2a2 < 0 2a2 � a > 0
a(2a � 1) > 0
a (�, 0)
,
2
1
3a � a2 > 0 a2 � 3a < 0 a (0, 3)
Hence after taking intersection a
3,
21
.
15. AB = 22 )1k()1h(
BC = 1
AC = 22 )1k()2h(
AB2 + BC2 = AC2 (h � 1)2 + (k � 1)2 + 1 = (h � 2)2 + (k � 1)2
2h = 2 h = 1
Area of ABC = 22 )1k()1h(21
× 1 = 1
(K � 1)2 = 4 k � 1 = ±2 k = 3, �1.
16.
RESONANCE SOLUTIONS (XI) # 60
The line segment QR makes an angle 60º with the positive direction of x-axis.
hence bisector of angle PQR will make 120º with +ve direction of x-axis.
Its equationy � 0 = tan120º (x � 0)
y = � x3
0y3x
17. Bisector of x = 0 and y = 0 is either y = x or y = �x
If y = x is Bisector, thenmx2 + (1 � m2)x2 � mx2 = 0 m + 1 � m2 � m = 0 m2 = 1 m = ±1.
18. Slope of PQ = k1
1
Hence equation of to line PQ
y � 27
= (k � 1)
2
)k1(x
Put x = 0
y = 27
+ 2
)k1()k1( = �4
7 + (1 � k2) = �8 k2 = 16 k = ±4.
Hence possible answer = �4.
19. p(p2 + 1) x � y + q = 0
(p2 + 1)2 x + (p2 + 1) y + 2q = 0 are perpendicularfor a common line lines are parallel slopes are equal
1
)1p(p 2
= � )1p(
)1p(2
22
p = � 1
20. BPAP
=
13
(x + 1)2 + y2 = 9((x � 1)2 + y2)x2 + 2x + 1 + y2 = 9x2 + 9y2 � 18x + 9
8x2 + 8y2 � 20x + 8 = 0
x2 + y2 � 4
10x + 1 = 0
circumcentre
0,
45
.
21.5x
+ by
= 1
513
+ b
32 = 1
b32
= � 58
b = � 20
5x
� 20y
= 1 4x � y = 20
Line K has same slope � c3
= 4
c = � 43
4x � y = � 3
distance = 17
23
Hence correct option is (3)
RESONANCE SOLUTIONS (XI) # 61
22.
AD : DB = 5:22 OD is angle bisectorof angle AOB St : 1 true
St. 2 false (obvious) Ans.
23. x + y = |a|ax � y = 1
if a > 0x + y = aax � y = 1
------------------------------------x(1 + a) = 1 + a as x = 1y = a � 1
It is in the first quadrantso a � 1 0a 1a [1, )If a < 0x + y = � a
ax � y = 1
+-------------------------------x(1 + a) = 1 � a
x = a1a�1
> 0
1a1�a
< 0 .............(1)
y = � a � a1a�1
= a1
a1�a�a� 2
> 0
�
1a1a2
> 0 1a1a2
< 0 .............(2)
from (1) and (2) a {}
24. = 3h � 2 = 3k
= 3k +2third vertex on the line 2x + 3y = 92 + 3 = 92(3h) + 3(3k + 2) = 92h + 3k = 12x + 3y � 1 = 0
RESONANCE SOLUTIONS (XI) # 62
25.
C
514
,58
Line 2x + y = k passes C
514
,58
514
582
= k
k = 6
26. (y � 2) = m(x � 1)
OP = 1 � m2
OQ = 2 � m
Area of POQ = 21
(OP)(OQ) = 21
m
21 (2 � m)
= 21
2
m4
m2
=
m4
m421
m = �2
ADVANCE LEVEL PROBLEMSPART - I
1. Condition for concurrency
cc41
bb31
aa21
= 0 b2
= a1
+ c1
So a, b, c are in H.P.
2. x2(sec2� sin2) � 2xy tan + y2sin2= 0
|m1 � m
2| = 21
221 mm4)mm(
2
222
2 sin
sinsec4
sin
tan2 = 2
3. Equation of family of curves passing through intersection of C1 & C
2 is
x2 + 4y2 � 2xy � 9x + 3 + (2x2 + 3y2 � 4xy + 3x � 1)= 0 .............(i)
It will give the joint equation of pair of lines passing through origin,if coefficient of x = 0 & Constant = 0 = 3put = 3 in equation (i), we getx2 + 4y2 � 2xy + 6x2 + 9y2 � 12xy = 0
It will subtend 90º at origin if coeff. of x2 + coeff. of y2 = 0= �19
RESONANCE SOLUTIONS (XI) # 63
4.
5
A
(3, 2)
C
5
B
O
For B and C apply Parametric form
cos3x
=
sin2y
= ± 5
Points are (7, 5) & (�1, �1)
5.
From figure it is clear that A is orthocentre of ABC
6. px2 � qxy � y2 = 0m
1 = tan m
2 = tan
m1 + m
2 = � q , m
1 m
2 = � p
tan (+ ) = p1q
7. (2 + ) x + (1 � 2) y + (4 � 3) = 0
distance from point A is = 22 )21()2(
)34()21(32)2(
= 10
= 1Hence, the required line is 3x � y + 1 = 0
8.
To find equations of AB and CD AB and CD are parallel to 3x � 4y = 0 and at a distance of 2 units from (1, 1)
3x � 4y + k = 0
and5
k4�3 = 2 k � 1 = 10
k = 11, � 9
equations of two sides of the square which are parallel to 3x � 4y = 0 are
3x � 4y + 11 = 0 and 3x � 4y � 9 = 0
Now the remaining two sides will be perpendicular to3x � 4y = 0 and at a distance of 2 unit from (1, 1) 4x + 3y + k = 0
and5
k34 = 2 k + 7 = 10
k = 3, � 17 remaining two sides are4x + 3y + 3 = 0 and 4x + 3y � 17 = 0
9. Givenx cos + y sin = a .......(1)x sin � y cos = b .......(2)square (1) and (2) then add them.x2 + y2 = a2 + b2
RESONANCE SOLUTIONS (XI) # 64
10. Let point of concurrency of given family of lines is Q and it can be obtained by solving3x + 4y + 6 = 0and x + y + 2 = 0 Q (� 2, 0)
Now required line will pass through Q(� 2, 0) and perpendicular to PQ.
Equation of required line is
y � 0 = 34�
(x + 2) 4x + 3y + 8 = 0
11. (i) After reflection about line y = x position of point will be (1, 4)(ii) After this step (3, 4)
(iii) (h + ki) = (3 + 4i) ei/4 = (3 + 4i)
i
2
1
2
1 h = �
2
1, k =
2
7
Hence the final position will be
2
7,
2
1
(h, k)
(3, 4)
12. Let the point P be (x, y)|x| + |y| = 3 .......(i)
�P(x, y)Case - 1 x > 0, y > 0Equation (i) will become :
x + y = 3
B
O A
Area (OAB) = 29
Similarly for each quadrant , a triangle will be formed. Hence area enclosed will be 18.
13.
P (� 4, � 2)
and Q (� 2, � 6)
Let slopes of PM and QM be m1 and m
2 respectively.
m1 = 3 and m
2 =
21
.
Let �� be the acute angle between PM and QM
tan = 21
21
mm1
m�m
tan = 1 =
4
RESONANCE SOLUTIONS (XI) # 65
14. For collinearity of 3 points
1sincos
13
11
102
= 0
3 sin � cos = 2 = 32
15. For ABCa + b > c, b + c > a, c + a > bx2 + 4x + 3 > x2 + 3x + 8 x > 5
x2 + 5x + 11 > x2 + 2x x > 311
2x2 + 5x + 8 > 2x + 3 2x2 + 3x + 5 > 0 x RCommon to all is x > 5.
16.
point of intersection of the two ray is P(0, 2)
Point A is
0,
3
2 or
0,
3
2
and PO is bisector of the angle between two rays required point is (0, 0)
17. Slope BD = �1,
A
B C
D
(x , y )2 2
(x , y )1 1
(x, y)
(a, b)
O
b2
b2
a+ , ,
Equation of BD, x + y = a + bequation of AC x � y = a
On solving, we get O
2
b,
2
ba
B (a + b, 0)m
AC = 1 = tan
OA = OD = 2
b
Apply parametric form for finding A & C
2
12b
ax
=
2
12b
y
= ± 2
b
A & C are (a, 0) and (a + b, b)
18.qpr
prq
rqp
= 0
p3 + q3 + r3 � 3pqr = 0 (p + q + r) (p2 + q2 + r2 � pq � qr � rp) = 0
RESONANCE SOLUTIONS (XI) # 66
19. k1u � k
2v = 0 .... (i)
k1u + k
2v = 0 .... (ii)
equations of bisectors of the angles formed by lines (i) and (ii) are
221
221
21
)akbk()bk�ak(
vk�uk
=
221
221
21
)ak�bk()bkak(
)vkuk(
k1u � k
2v = (k
1u + k
2v) .... (iii)
(i) by taking positive sign in (iii), we getk
1u � k
2v = k
1u + k
2v.
2k2v = 0 v = 0
(ii) by taking negative sing in (iii), we getu = 0
PART - II
1.
cos1x1 =
sin2y1 = ±
36
( (1, 2) lie below the line)
x1 = 1 +
36
cos, y1 = 2 +
36
sin
(x1, y
1) lies on x + y = 4
3 + 36
(sin + cos) = 4
sin + cos = 6
3 =
2·3
3
2
1(sin + cos) =
2
3 ·
2
1
sin( + 4
) = 23
= sin 60º or sin 120º
= 12
, 125
rejectedis
36
2.
tan =
43
·34
1
43
34
= 247
257
24
cosec = 725
RESONANCE SOLUTIONS (XI) # 67
P1 is from AB to CD, P
2 is from AD to BC
for finding P1 choose arbitrary point (a, a) on AB
P1=
5a2
5
a3a3a4
for P2 choose arbitrary point (a/2, 0) on AD
P2 =
5a
5
aa20
Area = P
1P
2 cosec =
7a2 2
3. Slope of BC is
= )3(1)1(3
=
22
= � 1
Equation of a line parallel to BC isy = � x + c i.e. x + y � c = 0
its distnace from the origin is
21
2
c c =
2
1
Equations of the lines are
x + y ± 2
1 = 0
Since the required line intersects OB and OC, therefore, it is the line whose y intercept is negative. H e n c e
the required line is x + y + 2
1 = 0.
4. AB = 36164 = 425
BC = 36576 = 612
AC = 169256 = 425AB = AC BC, triangle is isoscelesand in isosceles triangle O, H, I, G are collinear
5. D is mid point of AB and lies on the line 3 x + y = 6
3 · 2
12
+ 2
12 = 6
32 � 7 + 2 = 0 ........(1)
= 31
, 2
multiplication of slope of AB & line = �1
1
12
(�3) = �1
2 � � 2 = 0 ........(2) = �1, 2
= 2 satisfies both (1) & (2)
6. AB = AC
A
B C(1, -10)
x+y-3 = 07x-y+3 = 0
m = mBC
RESONANCE SOLUTIONS (XI) # 68
The bisectors are 25
3yx7 = ±
2
)3yx(
Their slopes are 31
, � 3
Required lines are y + 10 = 31
(x � 1) and y + 10 = � 3(x � 1)
i.e. x � 3y � 31 = 0 and 3x + y + 7 = 0
7.
AB = 2d OAPB is a cyclic quadrilateral and OP will be diameter of the circumcircle of this quadrilateralLet Q be the centre of the circle in AQT
sin= 2
12
1 yx
d2
..... (i)
tan = ba
ab�h2 2
.... (ii)
from (i) and (ii), we get
221
21
2
d4�yx
d2
ba
ab�h2
Locus of P(x1, y
1) is
(x2 + y2) (h2 � ab) = d2 {(a � b)2 + 4h2}
8. The slopes of the lines AB, BC and CA are �1, �71
and �7 respectively
Let m1 = �
71
, m2 = �1, m
3 = � 7
m1 > m
2 > m
3
tangent of internal angles of the triangle are
A
B
x +
y �
5 =
0
7x + y + 14 = 0
Cx + 7y � 7 = 0
tan A = 43
, tan B = 43
and tan C = � 724
interior angles A and B are acute and interior angle C is obtuse internal bisector of B = acute bisector of B 3x + 6y � 16 = 0
External bisector of C acute bisector of C 8x + 8y + 7 = 0Internal bisector of A acute bisector of A 12x + 6y � 11 = 0
9. Equation of line passing through P(�1, 2) making angle with + ve direction of x-axis is given by
cos1x
=
sin2y
= r1 , r
2 , r
3(parametric form)
where r1, r
2, r
3 are distances of points A, Q, B from point P respectively.
Hence coordinates of A(r1 cos � 1, r
1 sin + 2)
RESONANCE SOLUTIONS (XI) # 69
But A lies on x-axis
Hence r1 sin + 2 = 0 sin = �
1r2
coordinates of point B (r3 cos � 1, r
3 sin + 2)
Point B lies on y-axis hence
r3 cos � 1 = 0 cos =
3r1
Coordinates of point Q (r2cos � 1, r
2 sin + 2)
Hence h = r2 cos � 1 cos =
2r1h
and k = r2 sin + 2 sin =
2r2k
Now given that r1 , r
2, r
3 are in H.P.
2r2
= 1r1
+ 3r1
2r2
= � 2
sin + cos
2r2
= � 21
2r2k
+ 2r
)1h( 2 = �
21
(k � 2) + (h + 1)
4 = � k + 2 + 2h + 2 2h = klocus y = 2xAlt : Use P and Q are harmonic conjugates with respect to A and B.
10. Let P(h, k) be a variable point on the lines passing through the origin.
22
11
kh
hykx
=
(kx1 � hy
1)2 = 2 (h2 + k2)
locus of P(h, k) is (x1y � xy
1)2 = 2 (x2 + y2)
solving it, we get(y
12 � 2) x2 � 2x
1y
1 xy + (x
12 � 2) y2 = 0.
11. Let the line (L) through the origin isx = r cosy = r sin
as L intersects L1 at Q and OQ = r
1
r1 sin = m
1r
1 cos + c
1..............(1)
similarly, L intersects L2 at R and OR = r
2
r2 sin = m
2r
2 cos + c
2..............(2)
Let P (h, k) & OP = r r2 = r
1 r
2..............(3)
& h = r cos ..............(4)k = r sin ..............(5)
putting the values of r1 and r
2 from (1) and (2) in (3)
r2 = )cosm(sinc
1
1
. )cosm(sin
c
2
2
..............(6)
putting the value of cos and sinfrom (4) and (5) in (6), we get
r2 =
rh
mrk
rh
mrk
cc
21
21 (k � m
1h) (k � m
2h) = c
1c
2
replacing (h, k) by (x, y) we get the desired locusas (y � m
1x) ( y � m
2x) = c
1c
2
RESONANCE SOLUTIONS (XI) # 70
12. take any point on line3x + 2y + 4 = 0
2x + 3y + 1 = 0
3x + 2y + 4 = 0
134
,1320
(0, �2)
(�2,1)
put x = 0, we get y = � 2
Now image of (0, �2)
in line 2x + 3y + 1 = 0
20x
= 3
2y = � 2
94160
= 1310
Hence x = 1320
and y = 1330
� 2 = 134
Point of intersection of 2x + 3y + 1 = 0 and 3x + 2y + 4 = 0 is (�2, 1)
Hence equation of other line y � 134
= 213/20113/4
1320
x
After simplification, we get 9x + 46y = 28
CIRCLE
EXERCISE # 1PART - I
Section (A) :
A-2. Since BD is diameter of circleHence (x � a) (x � 0) + (y � 0) (y � a) = 0
x2 + y2 = a (x + y)
A-6. x = �3 + 2sin x + 3 = 2 sin y = 4 + 2cos y � 4 = 2 cos
Squarring and add (x + 3)2 + (y � 4)2 = 4
Section (B) :
B-4. S1 (9)2 + (0)2 � 16 = 65 > 0
Since (9, 0) lies outside the circle. Hence two real tangents can be drawn.Now S x2 + y2 � 16
S1 9x � 16
Hence pair of tangents SS1 = T2
(x2 + y2 � 16) (65) = (9x � 16)2
65x2 + 65y2 � 1040 = 81 x2 + 256 � 288 x
16x2 � 65y2 � 288x + 1296 = 0
Angle between these tangents = )ba(abh2 2
= 6516
651602
=
49658
B-5. given 6gf 22 = 2 f3g3gf 22
3g2 + 3f2 + 12g + 12f + 6 = 0 g2 + f2 + 4g + 4f + 2 = 0
Section (C) :
C-4. Area of triangle formed by pair of tangents & chord of contact is = 22
3
LR
RL
Here R = a
L = 222 akh
Hence Area =
22
2/3222
kh
akha
RESONANCE SOLUTIONS (XI) # 71
C-7. T = S1
�2x � 3y +3(x �2) +4 (y � 3) + 9
= 4 + 9 � 12 � 24 + 9
x + y + 5 = 0
Section (D) :D-1. S
1 : x2 + y2 � 2x � 6y + 9 = 0 C
1(1, 3), r
1 = 1
S2 : x2 + y2 + 6x � 2y + 1 = 0 C
2(�3, 1), r
2 = 3
C1C
2 = 416 = 20
n + r2 = 4
Hence C1C
2 > r
1 + r
2Both circles are non-intersecting.
Hence there are four common tangents.
Transverse common tangents :
coordinate of P
31
91,
31
33
2
5,0
Let slope of these tangents is m
y � 25
= m(x � 0) mx � y + 25
= 0
Now 2m1
25
3m
= 1 21
m = 2m1
m2 + 41
� m = 1 + m2 m = �43
, other tangents is vertical
Equation of tangents x = 0
�43
x � y + 25
= 0 �3x � 4y + 10 = 0 3x + 3y = 10
Direct common tangents
coordinate of Q
3191
,3133
Q(3, 4)
Hence equations y � 4 = m(x � 3) mx � y + (4 � 3m) = 0
2m1
m343m
= 1
|1 � 2m| = 2m1
1 + 4m2 � 4m = 1 + m2 3m2 � 4m = 0 m = 0, 34
Hence equation y � 4 = 0(x � 3) y = 4
y � 4 = 34
(x � 3) 4x � 3y = 0
D-3. Equation of circle passing through origin is x2 + y2 + 2gx + 2fy = 0This circle cuts the circle x2 + y2 � 4x + 6y + 10 = 0 orthogonally
2g(�2) + 2f(3) = 0 + 10
�2g + 3f � 5 = 0 ...(1)& x2 + y2 + 12y + 6 = 0 also
2g(0) + 2f(6) = 6 + 0 f = 21
RESONANCE SOLUTIONS (XI) # 72
�2g + 23
�5 = 0 2g = �27
g = �47
Hence circle x2 + y2 + 2
4
7x + 2
2
1y = 0
2x2 + 2y2 � 7x + 2y = 0
Section (E) :E-1. Equation of circumcircle of this triangle
L1L
2 + L
2L
3 + L
3L
1 = 0
(x + 2y � 5)(x + y � 6)+(x + y � 6)(2x + y � 4)+(x + 2y � 5)(2x + y � 4) = 0
coef. of xy = 0 3 + 3 + 5 = 0 3 + 5 + 3 = 0 ...(1)coef. x2 = coef. y2 1 + 2 + 2 = 2 + + 2
= 1 = �56
Hence (x + 2y � 5) (x + y � 6) + (x + y � 6)(2x + y � 4) �56
(x + 2y � 5) (2x + y � 4) = 0
x2 + y2 � 17x � 19y + 50 = 0
E-2. x2 + y2 � 10x + (2x � y) = 0 ....(i)x2 + y2 + 2x ( � 5) � y = 0Centre (� ( � 5) , /2)Using on y = 2x
)5(22
102
5
Putting = 4x2 + y2 � 2x � 4y = 0
PART - II
Section (A) :
A-1.
diameter = 24
r = 22
A-6.* Let equation of required circle isx2 + y2 + 2gx + 2fy + c = 0it passes through (1, �2) & (3, �4)
2g � 4f + c = �5
6g � 8f + c = �25
4g � 8f + 2c = �10
6g � 8f + c = �25
�2g + c = 15
circle touches x-axis g2 = c g2 � 2g � 15 = 0
g = 5, � 3
g = 5, c = 25, f = 10 x2 + y2 + 10x + 20y + 25 = 0g = �3, c = 9, f= 2 x2 + y2 � 6x + 4y + 9 = 0
RESONANCE SOLUTIONS (XI) # 73
Section (B) :B-1. Point on the line x + y + 13 = 0 nearest to the circle x2 + y2 + 4x + 6y � 5 = 0 is foot of from centre
12x
= 1
3y = �
22 11
1332 = �4
x = �6 y = �7
B-3. Let slope of required line is my � 3 = m(x � 2)
mx � y +(3 � 2m) = 0
length of from origin = 3 9 + 4m2 � 12m = 9 + 9m2
5m2 + 12m = 0 m = 0, �5
12
Hence lines are y � 3 = 0 y = 3
y � 3 = �5
12(x � 2) 5y � 15 = �12x + 24 12x + 5y = 39.
B-5. Line parallel to given line 4x + 3y + 5 = 0 is 4x + 3y + k = 0This is tangent to x2 + y2 � 6x + 4y � 12 = 0
5k612
= 5
6 + k = ±25 k = 19, �31
Hence required line 4x + 3y � 31 = 0, 4x + 3y + 19 = 0
B-9. As we knowPA.PB = PT2 = (Length of tangent)2
Length of tangent = 916 = 12
B-10. Let any point on the circle x2 + y2 + 2gx + 2fy + p = 0 (, )This point satisfies 2 + 2 + 2g + 2f + p = 0Length of tangent from this point to circle x2 + y2 + 2gx + 2fy + q = 0
length = 1S = qf2g222 = pq
Section (C) :C-2. Required point is foot of
23x
= 51y
= �
254856
= �1
x = 1, y = 4
C-4.* Let point on line be(h, 4 � 2h) (chord of contact)
hx + y (4 � 2h) = 1
h(x � 2y) + 4y � 1 = 0 Point
4
1,
2
1
Section (E) :E-1. Let required circle is x2 + y2 + 2gx + 2fy + c = 0
Hence common chord with x2 + y2 � 4 = 0
is 2gx + 2fy + c + y = 0This is diameter of circle x2 + y2 = 4 hence c = �4.
Now again common chord with other circle2x(g + 1) + 2y(f � 3) + (c � 1) = 0
This is diameter of x2 + y2 � 2x + 6y + 1 = 0
2(g + 1) � 6(f � 3) + 5 = 0
2g � 6f + 15 = 0
locus 2x � 3y � 15 = 0 which is st. line.
RESONANCE SOLUTIONS (XI) # 74
E-2. Common chord of given circle6x + 4y + (p + q) = 0This is diameter of x2 + y2 � 2x + 8y � q = 0
centre (1, �4)
6 � 16 + (p + q) = 0
p + q = 10
EXERCISE # 2PART - I
3. Equation of circle whose diameter's end points are (a, b) and (h, k)(x � a) (x � h) + (y � b) (y � k) = 0
x2 + y2 � x(a + h) � y(b + k) + ah + bk = 0
it touches x-axis.
Hence g2 = c
2
2ha
= ah + bk
(h � a)2 = 4bk Locus of (h, k) is (x � a)2 = 4by.
5. As we know if two lines are m
1m
2 = �1
hk
= �1
2 � k = �2 + hLocus of (, ) is x2 + y2 = xh + yk
6.
2b
,ah2 lies on circle
2(2h � a)(2h � 2a) 2b
(�2b) = 0
4(h �a)(2h � a) + b2 = 08h2 � 12ah + 4a2 + b2 = 0D > 0144a2 � 4 × 8 (4a2 + b2) = 09a2 � 8a2 � 2b2 > 0 a2 > 2b2
9.
Let required equation of circle is x2 + y2 + 2gx + 2fy + c = 0Now common chord of given circle with required circle areCommon chord 2gx + 2fy + (c + 4) = 0 it is also diameter of circle x2 + y2 = 4. Hence c = �4
similarly with x2 + y2 � 6x � 8y + 10 = 0 2x(g + 3) + 2y(f + 4) � 14 = 0
6(g + 3) + 8(f + 4) � 14 = 0
RESONANCE SOLUTIONS (XI) # 75
6g + 8f + 36 = 0 3g + 4f + 18 = 0
With circle x2 + y2 + 2x � 4y � 2 = 0 2x (g � 1) + 2y(f + 2) � 2 = 0
�2(g � 1) + 4(f + 2) � 2 = 0
�2g + 4f + 8 = 0
2g � 4f � 8 = 0
after simplification g = �2, f = �3, c = �4
Hence circle x2 + y2 � 4x � 6y � 4 = 0
13. 42 � 5m2 + 6 + 1 = 0
(3 + 1)2 = 5(2 + m2) 22 m
1m.03
= 5
Hence centre (3, 0), radius = 5
15. Equation of circle having centre (x1, y
1) and radius 'd'
(x � x1)2 + (y � y
1)2 = d2
x2 + y2 = a2
Equation of common chord2xx
1 + 2yy
1 � x
12 � y
12 � a2 + d2 = 0
2xx1 + 2yy
1 � 2a2 + d2 = 0
PART - II
1. Point
t1
,t lies on x2 + y2 = 16
t2 + 2t
1 = 16 t4 �16t2 + 1 = 0 ........(i)
If roots are t1, t2, t3, t4 then
t1t2t3t4 = 1 .........(ii)
5.2
c01 = 2 c � 1 = ±2 c = �1, 3
But c = �1 common point is one
c = 3 common point is infiniteHence c = �1 is Answer.
8. Equation of chords of contact from (0, 0) & (g, f)gx + fy + c = 0gx + fy + g(x + g) + f(y + f) + c = 0
gx + fy +
2cfg 22
= 0
Distance between these parallel lines = 22
22
fg2
cfg
11.
(x + g)(x � 2) + (y + f)(y �1) = 0
RESONANCE SOLUTIONS (XI) # 76
12. cos /3 = 5
)3k()2h( 22
Locus (x + 2)2 + (y � 3)2 = 6.25
14.
slope of C1C
2 is tan = �
34
By using parametric coordinates C2 (± 3 cos , ± 3 sin )
C2 (± 3 (�3/5) , ± 3 (4/5)
C2 (± 9/5 , 12/5)
20. (x2 + y2 � 6x � 4y � 12) + (4x + 3y � 6) = 0
This is family of circle passing through points of intersection of circlex2 + y2 � 6x � 4y � 12 = 0 and line 4x + 3y � 6 = 0
other family will cut this family at A & B.Hence locus of centre of circle of other family is thiscommon chord 4x + 3y � 6 = 0
22. Let any point P(x1, y
1) to the circle x2 + y2 �
5x16
+ 15
y64 = 0
x1
2 + y1
2 � 5
16x
1 +
1564
y1 = 0
Length of tangent from P(x1, y
1) to the circle are in ration
2
1
S
S =
60y564
x548
yx
15y532
x524
yx
1121
21
1121
21
= 60y
564
x5
48y
1564
x5
16
15y5
32x
524
y1564
x5
16
1111
1111
= 900y128x96
225y32x24
11
11
= )225y32x24(4
225y32x24
11
11
=
21
24. Two fixed pts. are point of intersection ofx2 + y2 �2x �2 = 0 & y = 0
RESONANCE SOLUTIONS (XI) # 77
Point x2 � 2x � 2 = 0
(x � 1)2 � 3 = 0
x �1 = 3 , x � 1 = 3
)0,31( )0,31(
25. 512C3C4
= C C = 1, 6
EXERCISE # 3
Match the column :
1. (A) S1 � S
2 = 0 is the required common chord i.e 2x = a
Make homogeneous, we get x2 + y2 � 8.4 2
2
a
x = 0
As pair of lines substending angle of 90° at origin
coefficient of x2 + coefficient of y2 = 0 a = ± 4
(B) y = 22 3 (x � 1) passes through centre (1, 0) of circle
(C) Three lines are parallel
(D) 2(r1 + r
2) = 4
r1 + r
2 = 2
( 3 , 2)
2rr 21 = 1
Comprehension # 2 (6 to 8)6. PQC
1 and PRC
2 are similar
2
1
PRCofAreaPQCofArea
= 2
2
21
r
r =
259
7. Let mid point m(h, k). Now equation of chordT = S
1
hx + ky + 3(x + h) = h2 + k2 + 6hit passes through (1, 0)h + 3(1 + h) = h2 + k2 + 6hlocus x2 + y2 + 2x � 3 = 0
But clear from Geometry it will be arc of BC
8. Common chord of S1 & answer of 7
4x + 3 = 0 x = �3/4
RESONANCE SOLUTIONS (XI) # 78
at x = �3/4
2
343
+ y2 = 9 y2 = 9 �
1681
y2 = 1663
y = ± 473
Hence tan = )4/31(
473
=
773
tan = 7
3
10. Statement-1 is true and statement-2 is false as radius = 21
22
11. Statement-1 : There is exactly one circle whose centre is the radical centre and the radius equal to the lengthof tangent drawn from the radical centre to any of the given circles.
Statement-2 is True But does not explain Statement-1.
13. (0, 0) & (8, 6) lie on the director circle of x2 + y2 � 14x + 2y + 25 = 0
so � = 0
16. P = 29
1856 = 29
r2 = p2 + 32 = 38 r = 38
19. x2 + y2 � 8x � 12y + p = 0
Power of (2, 5) is S1 = 4 + 25 � 16 � 60 + P = P � 47 < 0 P < 47
Circle neither touches nor cuts coordinate axesg2 � c < 0 16 � p < 0 p > 16f2 � c < 0 36 � p < 0 p > 36
taking intersection P (36, 47)
EXERCISE # 4PART - I
1. The lines given by x2 � 8x + 12 = 0 are x = 2 and x = 6.
The lines given by y2 � 14y + 45 = 0 are y = 5 and y = 9
Centre of the required circle is the centre of the square. Required centre is
2
95,
2
62 = (4, 7).
2. Clearly from the figure the radius of bigger circle
r2 = 22 + {(2 � 1)2 + (1 � 3)2}
r2 = 9 or r = 3
3. The equation of circle having tangent 2x + 3y + 1 = 0 at (1, � 1)
(x � 1)2 + (y + 1)2 + (2x + 3y + 1) = 0x2 + y2 + 2x( � 1) + y(3 + 2) + ( + 2) = 0 ... (i)
RESONANCE SOLUTIONS (XI) # 79
equation of circle having end points of diameter (0, � 1) and (�2, 3) is
x(x + 2) + (y + 1) (y � 3) = 0
or x2 + y2 + 2x � 2y � 3 = 0 ... (ii)since (i) & (ii) cut orthogonally
2
)23(21.
2)2�2(2
(� 1) = + 2 � 3
2 � 2 � 3 � 2 = � 1
2 = � 3 = � 3/2
from equation (i), equation of required circle is2x2 + 2y2 � 10x� 5y + 1 = 0
4. 22 )1k()0h( = 1 + |k|
or
h2 + k2 � 2k + 1 = 1 + 2|k| + k2
h2 = 2|k| + 2k x2 = 4y if y > 0 & x = 0 if y 0
5. Clearly P is the incentre of triangle ABC.
r = s
= s
)cs)(bs)(as(
Here 2s = 7 + 8 + 9 s = 12
Here r = 12
3.4.5 = 5
6. Statement-1 is true because point (17, 7) lies on the director circle and Statement-2 is equation of director circleof given circle.
7. )r2)(3(21
18 r = 6
Line, y =
r2
(x � 2) is tangent to circle
(x � r)2 + (y � r)2 = r2
2 = 3r and r = 6r = 2
8. (ax2 + by2 + c) (x2 � 5xy + 6y2) = 0 x = 3y or x = 2y or ax2 + by2 + c = 0If a = b and c is of opposite sign, then it will represent a circleHence (B) is correct option.
9*. PS . ST = QS . SR
R
T
Q
P
s
Now HM < GM
ST1
PS1
2
< ST.PS
RESONANCE SOLUTIONS (XI) # 80
PS1
+ ST1
> SR.QS
2 B is correct and A is wrong.
Now QR = QS + SRApplying AM > GM
2SRQS
> SR.QS
QR > 2 SR.QS ST.PS
2QR4
ST1
PS1
> ST.PS
2 >
QR4
D is correct and C is wrong
�B� and �D� are correct.
10.
Let G () be the centre of C
= 233
� 1. cos 30 = 3
= 23
� 1 . sin 30 = 1
equation of C is
(x � 3 )2 + (y � 1)2 = 1
11.
FGD = DGE = 120° F = ( 3 , 0) and
GF = GE = GD = 1 E =
23
,23
12. Slope QR = 3 equation of QR is y � 23
= 3
23
x
y = 3 x and slope of RP = 0 equation RP is y = 0
13. The distance between L1 and L
2 is
13
6 < 2
Statement �1� is True because distance between lines is less than radius but L2 need not be a diameter.
Statement �2� is False because if
L1 is diameter then L
2 has to be a chord of circle
Thus �C� is correct
RESONANCE SOLUTIONS (XI) # 81
14. For required circle, P(1, 8) and O(3, 2) will be the end points of its diameter.
(x � 1) (x � 3) + (y � 8) (y � 2) = 0 x2 + y2 � 4x � 10y + 19 = 0
15. (r + 1)2 = 2 + 9r2 + 8 = 2
r2 + 2r + 1 = r2 + 8 + 92r = 16r = 8
16. Since distance between parallel chords is greater than radius, therefore both chords lie on opposite side ofcentre.
2 cos k2
+ 2 cos k
= 3 + 1
Let k2
=
2 cos + 2 cos 2 = 3 + 1
2 cos + 2 (2 cos2 � 1) = 3 + 1 4 cos2 + 2 cos � (3 + 3 ) = 0
cos = )4(2
)33(1642 =
)4(23412122
= 4
11212
=
4)132(1
cos k2
= 23
, 2
)13( Rejected
k2
= 6
k = 3 [k] = 3
17. Let equation of circle isx2 + y2 + 2gx + 2 fy + c = 0
as it passes through (-1,0) & (0,2) 1 � 2g + c = 0
and 4 + 4 f+ c =0
also f2 = c f = �2, c= 4 ; g = 25
equation of circle isx2 + y2 + 5x � 4y + 4 =0
which passes through (�4, 0)
18. 2x � 3y = 1, x2 + y2 6
S
)V()()()(
4
1,
8
1,
4
1,
4
1,
4
3,
2
5,
4
3,2
Plot the two curvesI, III, IV will lie inside the circle and point (I, III, IV) will lie on the P regionif (0, 0) and the given point will lie opposite to the line 2x � 3y � 1 = 0
P(0, 0) = negative, P
43
,2 = positive, P
4
1,
4
1 = positive P
4
1,
8
1 = negative
RESONANCE SOLUTIONS (XI) # 82
P
4
3,
2
5 = positive , but it will not lie in the given circle
so point
43
,2 and
41
,41
will lie on the opp side of the line
so two point
43
,2 and
41
,41
Further
4
3,2 and
4
1,
4
1 satisfy S
1 < 0
19. Circle x2 + y2 = 9line 4x � 5y = 20
P
520�t4
,t
equation of chord AB whose mid point is M (h, k)T = S
1
hx + ky = h2 + k2 ........(1)equation of chord of contact AB with respect to P.T = 0
tx +
520�t4
y = 9 ........(2)
comparing equation (1) and (2)
9kh
20�t4k5
th 22
on solving45k = 36h � 20h2 � 20k2 Locus is 20(x2 +y2) � 36 x+ 45y = 0
Sol. 20 to 21
20.
B divides C1 C
2 in 2 : 1 externally
B(6, 0)Hence let equation of common tangent isy � 0 = m(x � 6)
mx � y � 6m = 0
length of r dropped from center (0, 0) = radius
2m1
m6
= 2 m = ±
22
1
equation is x + y = 6 or x � y = 6
21. Equation of L is
x � + c = 0
length of perpendicular dropped from centre = radius of circle
RESONANCE SOLUTIONS (XI) # 83
= 1 C = �1, �5
x � y = 1 or x � y = 5
PART - II
1. S1 : (x � 1)2 + (y � 3)2 = r2 C
1 (1, 3), r
1 = r
S2 : x2 + y2 � 8x + 2y + 8 = 0 C
2 (4, �1), r
2 = 3
circles intersect |r1 � r
2| < C
1 C
2 < r
1 + r
2
|r � 3| < 5 < r + 3
|r � 3| < 5 �5 < r � 3 < 5 �2 < r < 8
5 < r + 3 r > 2After intersection 2 < r < 8.
2. Point of intersection of 2x � 3y = 5
3x � 4y = 7 is (1, � 1)
Hence centre (1, �1), Area = 154 = r = 7
equation of circle (x � 1)2 + (y + 1)2 = 72
x2 + y2 � 2x + 2y = 47.
3. Let centre of circle is (h, k) and it passes through (a, b)equation of circle is (x � h)2 + (y � k)2 = (h � a)2 + (k � b)2
This circle cuts x2 + y2 � 4 = 0 orthogonally2g
1g
2 + 2f
1f2 = c
1 + c
2
2g1(0) + 2f
1(0) = �(h � a)2 � (k � b)2 + h2 + k2 � 4
2ah + 2kb �
Hence locus of (h, k) is 2ax + 2by � = 0.
4. Equation of circle(x � p) (x � h) + (y � q) (y � k) = 0
x2 + y2 � x(h + p) � y(q + k) + (ph + qk) = 0
This circle touches x-axis g2 = c
= ph + qk
Locus of (h, k) is (x � p)2 = 4qy.
5. Point of intersection of 2x + 3y + 1 = 0 3x � y � 4 = 0 is (1, �1)
and circumference of circle = 2r = 10 r = 5Hence equation of circle (x � 1)2 + (y + 1)2 = 25
x2 + y2 � 2x + 2y � 23 = 0.
6. By family of circle x2 + y2 � 2x + (x � y) = 0
centre of this circle
lies on y = x = = 1
Hence x2 + y2 � x � y = 0.
7. Let S1 : x2 + y2 + 2ax + cy + a = 0
S1 : x2 + y2 � 3ax + dy � 1 = 0
common chord S1 � S
2 = 0 5ax + y(c � d) + (a + 1) = 0
given line is 5x + by � a = 0
RESONANCE SOLUTIONS (XI) # 84
compare both = =
a = = �1 �
(i) (ii) (iii)From (i) & (iii) a2 + a + 1 = 0 a = , 2 no real a.
8. Draw a line parallel to x-axis at a distance 2 unit.Now by definition of parabolalocus of a point whose distance from a fixed point (0, 3)is equal to its distance from a fixed line is a parabola.
9. Point of intersectionof lines3x � 4y � 7 = 0
2x � 3y � 5 = 0 is (1, � 1)
Area of circle = r2 = 49 r = 7Hence equation of circle (x � 1)2 + (y + 1)2 = 72 x2 + y2 � 2x + 2y = 47
10. cos = =
Locus of (h, k) is x2 + y2 = .
11. Let equation of circle is (x � h)2 + (y � k)2 = (h + 1)2 + (k � 1)2
it touches x-axis g2 = c
h2 = 2k � 2h � 2 k =
k k .
12.
= � 1 h = �3
= �2 k = �4 Hence Q(�3, �4).
13. S1 + S
2 = 0 should satisfy (1, 1)
(2 + 3 + 7 + 2p � 5) + (1 + 1 + 2 + 2 � p2) = 0
= �
p2 6 p ±
but at p = ± the 2nd circle is
x2 + y2 + 2x + 2y � 6 = 0
satisfies (1, 1) and obviously P and Q
so p = ± is also acceptable
RESONANCE SOLUTIONS (XI) # 85
� 1 1 7 + 2p 6 � p2
p2 + 2p + 1 0p � 1
14. r = = 5
< 5
� 25 < m + 10 < 25 � 35 < m < 15
Hence correct option is (1)
15. x2 + y2 = ax ...........(1)
centre c1 and radius r
1 =
x2 + y2 = c2 .........(2) centre c
2 (0, 0) and radius r
2 = c
both touch each other iff|c
1c
2| = r
1 ± r
2
=
= ± |a| c + c2 |a| = c
16. Circle whose diametric end points are (1, 0) and (0, 1) will be of smallest radius.(x � 1)(x � 0) + (y � 0) (y � 1) = 0
x2 + y2 � x � y = 0
17. Nowh2 = (1 � 2)2 + (h � 3)2
0 = 1 � 6h + 9
6h = 10
h =
Now diameter is 2h =
ADVANCE LEVEL PROBLEMPART - I
1. x2 + y2 � 5x + 2y � 5 = 0 + (y + 1)2 � 5 � � 1 = 0
+ (y + 1)2 =
So the axes are shifted to
New equation of circle must be x2 + y2 =
RESONANCE SOLUTIONS (XI) # 86
2.
Equation of circum circle of triangle OAB x2 + y2 � ax � by = 0.
Equation of tangent at origin ax + by = 0.
d1 = and d2 =
d1 + d2 = = diameter
3. Ler r be the radius of new circle
C1C2 = .
So r = 2
Slope of line joining C1 and C2 i.e. tan = 2 Equation of line joining C1 and C2 is
= = 2 + =
x = 2 and y = 5 Centre (2, 5)
4.
Area of ABCD = 4 .
5. Equations of two circles touching both the axes arex2 + y2 � 2c1 x � 2c1y + c1
2 = 0 .....(i)x2 + y2 � 2c2x � 2c2y + c2
2 = 0 .....(ii) (i) & (ii) are orthogonal also 2c1c2 + 2c1c2 = c1
2 + c22
or 6c1 c2 = (c1 + c2)2 ....(iii)Now point P(a, b) lies over the circle
x2 + y2 � 2cx � 2 cy + c2 = 0.so c2 � 2c(a + b) + a2 + b2 = 0 c1 & c2 are roots of this equationso c1 + c2 = 2(a + b) ....(iv)and c1 c2 = a2 + b2 ....(v)from (iii), (iv) & (v), we get
6(a2 + b2) = 4(a + b)2.
6. Let two circles are S = 0 and S = 0having radius r1 and r2 respectively.
=
S r12 = r2
2 S1
S1 � S = 0 Locus of P(h,k)
S � S = 0 which represents the equation of a circle.
RESONANCE SOLUTIONS (XI) # 87
7. tan 60º = =
and
sin 60º = =
Let coordinates of any point P on the circle be P (r cos , r sin)
PA2 = + (r sin)2
PB2 = (r cos )2 + (1 � r sin)2
PC2 = (r cos + )2 + (r sin)2
and PD2 = (r cos)2 + (r sin + 1)2
PA2 + PB2 + PC2 + PD2 = 4r2 + 8 = 11 r =
8. = tan�1 tan =
sin = and cos =
A (OA cos , OA sin ) A (3, 2)Similarly B (OB cos , OB sin ) (6, 4)Now it can be checked that circles C1 and C2 touch each other.Let the point of contact be C.
C
required radical axis is a line perpendicular toAB and passing through point C
y � = � (x � 5)
9. Equation of circle (x � 2)2 + (y + 2)2 + (x + y) = 0 ........(i) Centre lies on the x-axis = � 4 put in (i)
equation of circle is x2 + y2 � 8x + 8 = 0
(, ) lies on it2 = � 2 + 8 � 8 0
greatest value of �� is 4 + 2
10. Let �d� be the common difference
the radii of the three circles be 1 � 2d, 1 � d, 1
equation of smallest circle is x2 + y2 = (1 � 2d)2 ........(i) y = x + 1 intersect (i) at real and distinct points x2 + x + 2d � 2d2 = 0 ....(ii) D > 0 8d2 � 8d + 1 > 0
d > or d <
but d can not be greater than
d
RESONANCE SOLUTIONS (XI) # 88
11. Let the coordinates of P and Q are (a, 0) and (0, b) respectively equation of PQ is bx + ay � ab = 0 .......(i) a2 + b2 = 4r2 .....(ii) OM PQ equation of OM is ax � by = 0 .......(iii)Let M(h, k) bh + ak � ab = 0 ........(iv) and ah � bk = 0 .......(v)
On solving equations (iv) and (v), we get
a = and b =
put a and b in (ii), we get(h2 + k2)2 (h�2 + k�2) = 4r2
locus of M(h, k) is (x2 + y2)2 (x�2 + y�2) = 4r2
12. Equation of circle passing through (0, 0) and (1, 0) isx2 + y2 � x + 2fy = 0 .......(i)
x2 + y2 = 9 ......(ii)(i) & (ii) touch each other.so equation of Radical axis is x = 2fy + 9 ......(iii)line (iii) is also tangent to the circle (ii) on solving (ii) & (iii), we get(1 + 4f2)y2 + 36fy + 72 = 0 .......(iv)
D = 0 f = ± .
13. a2 � bm2 + 2d + 1 = 0 ......(1)and a + b = d2 .......(2)Put a = d2 � b in equation (1), we get
(d + 1)2 = b(2 + m2)
= ......(3)
From (3) we can say that the line x + my + 1 = 0 touches a fixed circle having centre at (d,0) and radius
=
PART - II1. Let the circumcentre be P(h, k)
Equation of AB is
a =
on solving
= 2(h + k) = + a
locus of circumcentre P(h,k) is
2 (x + y) a =
2. S1 x2 + y2 = a2
S2 x2 + y2 = b2
S3 x2 + y2 = c2
equation of 1 is ax cos + ay sin = b2
1 is tangent to circle S3
RESONANCE SOLUTIONS (XI) # 89
c = ca = b2 Hence a,b,c are in G.P.
3. Equation of circle touching y - axis isx2 + y2 + 2gx + 2fy + f2 = 0 it passes through (4, 3) & (2, 5)so 25 + 8g + 6f + f2 = 0
29 + 4g + 10f + f2 = 0solving above two equations, we get(g, f) (�2, � 3) & (� 10, � 11).
So equations of circles are x2 + y2 � 4x � 6y + 9 = 0 and x2 + y2 � 20x � 22y + 121 = 0
for circle x2 + y2 � 4x � 6y + 9 = 0.
tan = =
=
So tan is max at k = 3.at k = 3, tan = 1 = 45°
4. 1 4x + 3y = 10 2 3x � 4y = � 5
Let be the inclination of 2
tan =
equation of 2 in parametric form
= = ± 5
co-ordinates of centres are (5, 5), (�3, �1)
5. centre lies over the line 2x � 2y + 9 = 0
So let coordinate of centre be
Let the radius of circle be 'r'So equation of circle is
(x � h)2 + = r2
x2 + y2 � 2hx � y(2h + 9) + 2h2 + 9h � r2 + = 0
given circle cuts orthogonally to x2 + y2 = 4
so 2h2 + 9h + � r2 = 0 or 2h2 + 9h � r2 = �
so equation of required circle can be written as x2 + y2 � 2hx � y (2h + 9) + 4 = 0
(x2 + y2 � 9y + 4) + h (�2y � 2x) = 0
so this circle always passes through points of intersection of x2 + y2 � 9y + 4 = 0 and x + y = 0
so fixed points are (�4, 4) and
6. Centre of C1 lies over angle bisector of 1 & 2Equations of angle bisectors are
= ±
RESONANCE SOLUTIONS (XI) # 90
x = 5 or y = �
Since centre lies in first quadrantso it should be on x = 5.So let centre be (5, )
3 = = 2, � From the figure r =
But � so = 2.
So equation of cirlce C2 is(x � 5)2 + (y � 2)2 = 52
x2 + y2 � 10x � 4y + 4 = 0.
7. OA = a and AQ = QP = QR
OQ =
AQ = = PQ
(OA)2 = (OQ)2 + (AQ)2
a2 = 2 + 2 + (p � )2 + (q � )2
22 + 22 � 2p� 2q+p2+q2 � a2 = 0.Locus of the middle point Q (, ) is2 x
2 + 2y2 2 p x 2 q y + p2 + q2 a2 = 0
8. Let the equation of required straight line be y = mx + c.
= .....(i)
For PCM = tan 2.
PM = 5cot 2 .....(ii)
For PQM = PM sin (90 � )
= cos
on solving, we get = 30°
Equation of tangent at P(� 2, � 2) is
3x + 4y + 14 = 0.
tan 60° =
=
m =
Now on substituting value of 'm' in equation (i), we get
c = or
but c should be (�ve)
So equation of line y = x +
RESONANCE SOLUTIONS (XI) # 91
9. Let the centre of the circle be (h, k) and radius equal to �r� h2 + k2 = r2 ......(i)
and = r
2 � h � k = r .....(ii)and h = 1 � r .......(iii)
put h = 1 � r in (ii), we get k = r (1 � ) + 1Now put the values of h and k in (i), we get
(r (1 � ) + 1)2 + (1 � r)2 = r2
r2 (3 � 2 ) � 2 r + 2 = 0
hence radius i.e. r is the root of the equation (3 � 2 ) t2 � 2 t + 2 = 0
10. Let the equation of the circles be x2 + y2 + 2gx + 2fy + d = 0 .......(i) these circles pass through (0, a) and (0, �a)
a2 + 2fa + d = 0 ......(ii)and a2 � 2fa + d = 0 ......(iii)solving (ii) and (iii), we get f = 0, d = � a2
put these value of f and d in (i), we getx2 + y2 + 2gx � a2 = 0 ......(iv)
y = mx + c touch these circles =
g2 + (2cm) g + a2 (1 + m2) � c2 = 0 ......(v)equation (v) is quadratic in 'g' Let g1 and g2 are its two roots g1g2 = a2 (1 + m2) � c2
the two circles represented by (iv) are orthogonal 2g1g2 + 0 = � a2 � a2 g1g2 = �a2
a2 (1 + m2) � c2 = � a2
c2 = a2 (2 + m2) Hence proved
11. Let OAB = and OAB =
+ = and OBA =
length of AB is �a� and length of AB is �b� from the figure
A (b cos , 0) and A(a cos , 0)similarly B(0, a sin ) and B (0, b sin )Let c(h, k) be the centre of circle 2h = a cos + b cos
= �
2h = a cos + b sin ........(i)and 2k = a sin + b sin
= �
2k = a sin + b cos ........(ii)
on solving (i) and (ii), we get cos = and sin =
sin2 + cos2 = 1 locus of C(h, k) is (2ax 2by)² + (2bx 2ay)² = (a² b²)²
RESONANCE SOLUTIONS (XI) # 92
12. One circle lies within the other circle C1C2 < |r1 � r2|
<
squaring both sides, we get
� 2gg1 < � 2 � 2c
gg1 > c + .
gg1 � c > . ......(i)
gg1 � c > 0 gg1 > cagain squaring both sides of (i), we get�2cgg1 > � c (g2 + g1
2) c(g � g1)2 > 0 c > 0 and from (i), we can say that gg1 will also be > 0
MATHEMATICAL REASONING, INDUCTION & STATISTICSEXERCISE # 1
PART - I
Section (A) :
A-1. By definition of 'statement'.
A-7. The negation of ��Everyone in Germany speaks German�� is - there is at least one person in Germany who doesnot speak German.
A-10. Statement (A) All prime numbers are even.Statement (B) All prime numbers are odd.Both false
A-12. If it is a holiday as well as sunday than also the office can be closed.
A-13*. Polygon cannot be both concave and convex
A-16. Obvious
A-19. (~ T F) ~T T (F F) F T F F T F T
A-21. p q means(i) p is sufficient for q (ii) q is necessary for q (iii) p implies q (iv) if p then q (v) p only if q
Section (B) :
B-3. Contrapositive of (p q) r is ~ r (p q )
B-4.
RESONANCE SOLUTIONS (XI) # 93
Section (C) :
C-2. x1 + x
2 + ...... x
n = nM
(x1 + x
2 ..... + x
n) � x
n + x = nM � x
n + x
so
average = .
C-5. Average speed over the entire distance =
= =
C-7. =
C-10. � , � 3, � , � 2, � , + , + 4, + 5 ( > 0)
C-12. =
=
.(2n�1) =
C-14 Frequency of f = 10C5 which has maximum value
Section (D) :
D-2. 34, 38, 42, 44, 46, 48, 54, 55, 63, 70
median = = 47
= 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86
so mean diviation about median = = 8.6
D-5. new
=
= = odd
D-8. = = 60
= = = 10.4
RESONANCE SOLUTIONS (XI) # 94
D-9. = Now =
new
= = .
D-11. = 250
= = 5
coeff. of variation = = 10%
Section (E) :E-4*. Let n = 1 then p(A) = 64
Let p(k) is divisible by 6432k + 2 � 8k � 9 is divisible by 64
Now,P(k + 1) = 32(k + 1) + 2 � 8(k + 1) � 9
= 32k + 2 × 9 � 8 × k × 9 � 9 × 9 � 8 + 72 + 64 k
= 9(32k + 2 � 8k � 9) + 64 (k + 1)
Which is divisible by 64
E-6. Let p(n) = n3 + (n + 1)3 + (n + 2)3 , p(A) = 36, p(B) = 99 both are divisible by 99Let it is true for n = kk3 + (k + 1)3 + (k + 2)3 = 9q ; q adding 9k2 + 27k + 27 both sidesk3 + (k + 1)3 + (k + 2)3 + 9k2 + 27k + 27 = 9q + 9k2 + 27 k + 27(k + 1)3 + (k + 2)3 + (k + 3)3 = 9r ; r
Comprehension # 1 (1 to 3)1. If p then q means p only if q
2. If p then q p is sufficient for q
3. p is false, q is false so p q is true.
Comprehension # 2 (4 to 6)
A.M. = = 12
= = = 3
coeff. of variation = = × 100
MATCH THE COLUMN
1. (A) = 5 + = 15
Md = x
11 + 10 = = 10 + 10 = 20
variance remains unaffected on addition of a constant
RESONANCE SOLUTIONS (XI) # 95
(B) = 5 + = 5 + 10 = 15
Md = x
11 = 10
(C) Mean and median get multiplied by 2 and variance by 22
(D) = = 16
Md = x
11 + 11 = 10 + 11 = 21
variance remains unaffected on addition of a constant
EXERCISE # 2
4.
7.
9. (p q) [~ p (p ~ q)]= (p q) [(~ p p) (~ p ~ q)]= (p q) [t (~ p ~ q)]= (p q) (~ p ~ q)= (p q) [ ~ ( p q)] = talso (~p q) t = t
12. (p q) ~p = (p ~ q) (q ~p)= t (q ~p) = q ~p = ~p q
15. p : it rainsq : crops will be goodS
1 : p q , S
2 : ~p S : ~q
Not valid
17. p : it rains tomorrowq : I shall carry my umbrellar : cloth is mendedP : p (r q)Q : p ~rS : ~qP : T, Q : T S : TT S not valid
RESONANCE SOLUTIONS (XI) # 96
18. S.D.(xi) = S.D. (x
i � 8) = = = 2
20. n = 200, mean = 40 = = 8000
correct = 8000 � 34 + 43 = 8009 correct mean = = 40.045
Also 2 = � 225 = � 1600
x2 = 36500
correct = 36500 � (34)2 + (43)2 = 365693
correct 2 = � (40.045)2
= 1828.465 � 1603.603 = 14.995
21.
2 =
= = 1.2
so variance of A = 1.2 < 1.25 = variance of Bso more consistent team = A
22. 2 = = = 9
coefficient of variation = = 25
26. (i) Given statement is true for n = 1(ii) Let us assume that the statement is true for n = k
i.e. 1.3 + 2.32 + 3.33 +.......+ k.3k =
(iii) For n = k + 1,L.H.S. = 1.3 + 2.32 + 3.33 +.......+ k.3k + (k + 1) 3k+1
= + (k + 1) 3k+1 = = R.H.S.
so by principle of mathematical induction the statement is true for all n N
RESONANCE SOLUTIONS (XI) # 97
EXERCISE # 3PART - I
MATHEMATICAL REASONING :
1. r : x is a rational number iff y is a transcendental number r = ~p q
Statement-1 is false and Statement-2 is false.
2.
3. Statement-1 :
Statement-2 : False.
4. Negation of Q is ~
It may also be written as ~
5.
6. Let p : I become a teacherq : I will open a schoolNegation of p q is ~ (p q) = p ^ ~ qi.e. I will become a teacher and I will not open a school.
STATISTICS :
7. Let average marks of the girls = x
= 72 x = 65
8. No change median is 5th observation (If observation are in asending order)
RESONANCE SOLUTIONS (XI) # 98
9. Correct variance = �
= 222 � 144 = 78.00
10. If we change scale by using x + h then median increases by h. so median is not independent of change of scale.From histtegranm we can see highest frequency so made.
11. = 0
= a2 S.D. = |a| = 2
12.
so median = 22 = , mode = 24
13. 2 0
0 0 n 16
14. Variances remain uneffected by adding some constant to all observationsso V
A = V
Bso V
A/V
B = 1
15. Let no. of student = 100 number of boys = n,
= 50 n = 80
so 80%
16. = 6 a + b = 7 ...(1)
= 6.80 (a � 6)2 + (b � 6)2 = 13
solve a = 3, b = 4
17. Statement-1 : � = �
= (2n + 1 � 3)
Statement-2 : Obvious
18. = = 1 + 50d
Mean deviation = =
= = 225 = 225
.d = 255 d = 10.1
RESONANCE SOLUTIONS (XI) # 99
19. x2 = 4 � = 4
� (2)2 = 4 = 40
similarly = 105
2 = � = � = 5.5
20. Median = 25.5 a
Mean deviation about median = 50 = 50
24.5 a + 23.5a + ..... + 0.5a + 0.5a + .... + 24.5a = 2500
a + 3a + 5a + ..... + 49a = 2500 (50a) = 2500 a = 4
21. Correct mean = observed mean + 230 + 2 = 32Correct S.D. = observed S.D. = 2
22. A.M. of 2x1, 2x
2 ..... 2x
n is
= =
So statement-2 is falsevariance (2x
i) = 22 variance (x
i) = 42
so statement-1 is true.
MATHEMATICAL INDUCTION :
23. Put k = 1LHS 1 RHS = 4LHR RHSLet S(k) is truethen 1 + 3 + 5 +....(2k � 1)
= 3 + k2
add (2k + 1) both the side1 + 3 + 5 +.... + (2k � 1) + (2k + 1)
= 3 + k2 + 2k + 1S(k + 1) = 3 + (k + 1)2
then if S(k) is true S(k + 1) is also true.
24. For n 2n2 + n < n2 + n + n + 1n2 + n < (n + 1)2
statement -2 is true
>
> , > . . . . . + +......
RESONANCE SOLUTIONS (XI) # 100
ADVANCE LEVEL PROBLEM
1. Statement p q and its contrapositive ~q ~p are logically equivalent and give same meaning.
2.
3.
4.
5. p : Wages will increaseq : there is an inflationr : cost of living will increaseA : p qB : q rC : pS : rA : T, B : T C : T S : T S valid
6. Here = = 8 +12 + 13 + 15 + 22 = 70
& = 64 + 144 + 169 + 225 + 484 = 1086
2 = � = � = � = 21.2
RESONANCE SOLUTIONS (XI) # 101
7. If a xi b a b
xi � b � a
(xi � )2 (b � a)2 n(b � a)2
so var(x) (b � a)2
8. Total money per kg. = = so total kg per rupee = = 1.92
11. Let P(n) ; sin + sin2 + .......+ sin n = sin
P(A) is trueLet P(k) is also true
sin + sin2 + ..........+ sin k = sin
add sin(k + 1) both sidessin + sin2 + ............+ sin k + sin(k + 1)
= sin sin cosec + sin(k + 1)
= sin = sin
= . cosec
P(k + 1) is true
SOLUTION OF TRIANGLEEXERCISE # 1
PART - ISection (A) :
A-1. (i) L.H.S. = a sin (B � C) + b sin (C � A) + c sin (A � B)
= k sin A sin (B � C) + k sin B sin (C � A) + k sin C sin (A � B)
= k (sin2 B � sin2 C) + k (sin2C � sin2 A) + k (sin2 A � sin2 B)= 0 = R.H.S.
(ii) L.H.S. =
first term = =
= k2 sin (B + C) sin (B � C)
= k2 (sin2 B � sin2 C)
Similarly = k2 (sin2 C � sin2 A)
and = k2 (sin2 A � sin2 B)
L.H.S. = k2 (sin2 B � sin2C + sin2C � sin2A + sin2 A � sin2 B)= 0 = R.H.S.
RESONANCE SOLUTIONS (XI) # 102
(iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C= b2 + c2 � a2 + a2 + c2 � b2 + a2 + b2 � c2
= a2 + b2 + c2
= R.H.S
(iv) L.H.S. = a2 � 2ab
= a2 + b2 � 2ab cos C
= a2 + b2 � (a2 + b2 � c2)= c2 = R.H.S.
(v) L.H.S. = b2 sin 2C + c2 sin 2B= 2b2 sin C cos C + 2c2 sin B cos B= 2k2 sin2 B cos C sin C + 2k2 sin2 C sin B cos B (b = ksin B, c = ksin C)= 2k2 sin B sin C [sin B cos C + cos B sin C]= 2(k sin B) (k sin C) sin (B + C)= 2bc sin A
(vi) R.H.S = c = a cos B + b cos A,
b = c cos A + a cos C
= =
= = L.H.S.
A�4. =
sin(B + C) sin(B � C) = sin(A + B) sin(A � B)
sin2 B � sin2 C = sin2 A � sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c2 a2, b2, c2 are in A.P.
A�7. x3 � Px2 + Qx � R = 0
a2 + b2 + c2 = Pa2b2 + b2c2 + c2a2 = Q
a2b2c2 = R abc = + + = [a2 + b2 + c2] =
Section (B) :
B�1. (i) L.H.S. = 2a sin2 + 2 c sin2
= a(1 � cos c) + c(1 � cos A)
= a + c � (a cos C + c cos A)
= a + c � b
= R.H.S.
(ii) L.H.S. = + +
= . + . + .
= = .
RESONANCE SOLUTIONS (XI) # 103
(iii) L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C)= 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C
= 2 + a2 + b2 + c2 = (a + b + c)2
= R.H.S.
(iv) L.H.S. = (b � c) + (c � a) + (a � b)
(b � c) cot = k(sin B � sin C)
= 2k cos sin
= 2k sin sin
= k [cos C � cos B]
similarly (c � a) cot = k[cos A � cos C]
and (a � b) cot = k[cos B � cos A]
L.H.S. = k[cos C � cos B + cos A � cos C + cos B � cos A]
= 0= R.H.S.
(v) L.H.S. = 4 (cot A + cot B + cot C)
= 4
= 2bc cos A + 2 ca cos B + 2ab cos C= a2 + b2 + c2 = R.H.S.
(vi) L.H.S. = cos . cos . cos
=
= = = R.H.S.
B�3.
Let ADB =
we have to prove that tan =
if we aply m � n rule, then
(1 + 1) cot= 1.cot C � 1.cotA.
= � = �
RESONANCE SOLUTIONS (XI) # 104
=
= [2(a2 � c2)]
2cot = tan =
Section (C) :
C�2. (i) r. r1 .r
2 .r
3 = = 2
(ii) r1 + r
2 � r
3 + r = 4R cosC
L.H.S. =
=
=
=
=
= =
= c =
cos C =
L.H.S. =
= = = 4RcosC
(iii) L.H.S. =
= [s2 + (s � a)2 + (s � b)2 + (s � c)2]
= [4s2 � 2s(a + b + c) +a2]
= = R.H.S.
(iv) L.H.S. =
RESONANCE SOLUTIONS (XI) # 105
= (s + s � a + s � b + s � c)2 = 4 =
R.H.S. =
= · (s � a + s � b + s � c) = =
(v) =
=
= =
= =
= = = r
similarly we can show that = = r
C�4. = 24 sq. cm .... (i)2s = 24 s = 12 .... (ii)
r1, r
2, r
3 are in H.P.
are in A.P..
are in A.P..
a, b, c are in A.P. 2b = a + c 2s = 24 a + b + c = 24
3b = 24 b = 8 a + c = 16
But = = 24 × 24 = 12 × (12 � a) × 4 × (12 � c) 2 × 6 = 144 � 12 (a + c) + ac
12 = 144 � 192 + ac
ac = 60 and a + c = 16 a= 10, c = 6 or a = 6, c = 10 and b = 8
Section (D) :
D�1. (i) = ,= , = =
R.H.S. =
RESONANCE SOLUTIONS (XI) # 106
=
= L.H.S. = R.H.S.
(ii) = = =
R.H.S. = =
=
L.H.S.= R.H.S.PART - II
Section (A) :A�4. (a + b + c) (b + c � a) = kbc (b + c)2 � a2 = kbc
b2 + c2 � a2 = (k � 2) bc = = cos AA
In a ABC �1 < cos A < 1 �1 < < 1
0 < k < 4.
Section (B) :
B�2. b cos2 + a cos2 = c. b + a = c.
[ s � a + s � b] = c × c = c
= a + b = 2c
a, c, b are in A.P.
B�5. = (a + b � c) (a � b + c)
= 4(s � c) (s � b) =
tan = tan A = tan A =
B�6*. (A) tan = cot .........(i)
tan2 = = =
tan = a = 5 and b = 4
from equation (i), we get
= cot = cot cot =
RESONANCE SOLUTIONS (XI) # 107
cos C = = = =
cos C = c2 = a2 + b2 � 2ab cos C c = 6
(B), (C) Area = ab sinC cosC = sinC = =
Area = × 5 × 4 ×
Area = sq. unit. From Sine rule
= = sinA = =
sinA =
Section (C) :
C�3. =
=
= 4
= .
C�5*. (A) + +
= + +
=
(B) + +
= + + =
(C) = = cot A = cot B = cot C
A = B = C
true for equilateral triangle only
(D) = =
RESONANCE SOLUTIONS (XI) # 108
= =
cot A = cot B = cot C A = B = C true for equilateral triangle only
Section (D) :
D�1. = 2
D�4*. a = cos
(A) correct(B) incorrect
(C) = = = cos
(D) cosec = . = . = cos
EXERCISE # 2PART - I
3.
If we apply Sine-Rule in ABD , we get
= AB = = ...(i)
sin = and cos =
from equation (i), we get
AB = AB =
7.
required distance = inradius of ABC
RESONANCE SOLUTIONS (XI) # 109
2s = a + b + b + c + c + a
= 2 (a + b + c)
s = a + b + c
=
=
required distance
= = =
=
8. (i) L.H.S. = (r3 + r
1) (r
3 + r
2) sin C
= sin C
= sin C
= sin C
=
= = 2 sr3
R.H.S. = 2r3
= 2r3
= 2sr
3
L.H.S. = R.H.S.
(ii) L.H.S. = �
= �
= = R.H.S.
(iii) First term = (r + r1) tan
= cot
= . .
= b � c
similarly second term = c � a & third term = a � b
L.H.S. = b � c + c � a + a � b = 0 = R.H.S.
RESONANCE SOLUTIONS (XI) # 110
(iv) r1 + r
2 + r
3 � r = 4R
(r1 + r
2 + r
3 � r)2 = r
12 + r
22 + r
32 + r2 � 2r (r
1 + r
2 + r
3) + 2(r
1r
2 + r
2r
3 + r
3 r
1) ........(i)
r(r1 + r
2 + r
3) = ab + bc + ca � s2
and r1r
2 + r
2r
3 + r
3r
1 = s2
from equation (i)16R2 = r2 + r
12 + r
22 + r
32 � 2 (ab + bc + ca � s2) + 2s2
r2 + r1
2 + r2
2 + r3
2 = 16 R2 � 4 s2 + 2 (ab + bc + ca)= 16R2
� (a + b + c)2 + 2 (ab + bc + ca)
= 16R2 � a2 � b2 � c2
11. (i) EFA is a cyclic quadrilateral
= A
A = r cosec A/2 EF = r cosec A/2.sin A
= 2 r cos A/2similarly DF = 2 r cos B/2and DE = 2r cos C/2.(ii) ECD is a cyclic quadrilateral
CE = DE =
similarly DF = BF =
FDE = =
= �
(iii) area of DEF = FD . DE sin FDE
=
= 2r2 = 2r2
= = =
= =
= = .
PART - II
3. ED = � c cos B
= � c
= �
RESONANCE SOLUTIONS (XI) # 111
=
=
5. f = RcosA , g = R cos B, h = R cosC.
+ + = + +
= 2
= 8 + + =
= .8 =
9. MNA is a cyclic quadrilatral
= AA MN = r cosec sin A = 2r cos
M = N = r
x = = , =
similarly y = and z =
xyz = = = r2 R
12. r1 + r
2 =
(r1 + r
2) = =
= = = 4Rs2
= 4
14. A, C1 , G and B
1 are cyclic
BC1 . BA = BG . BB
1
. c =
= (2c2 + 2a2 � b2)
c2 + b2 = 2a2
RESONANCE SOLUTIONS (XI) # 112
16. a = 1 2s = 6
2s = 2
R = 1 = 2R sin A =
A =
18. sin C = 1 cos (A � B) 1
cos (A � B) = 1 A � B = 0 A = B
sin C = = 1 C = 90º
20. if we apply m-n Rule in ABE, we get
(1+1) cot = 1.cot B � 1.cot
2 cot = cot B � cot
3 cot = cot B
tan = 3 tan B ..........(1)
Similarly, if we apply m-n Rule in ACD, we get
(1+1) cot (�) = 1.cot � 1.cotC.
cotC = 3 cot tan = 3 tanC .......(2)
form (1) and (2) we can say that
tan B = tan C B=C
A + B + C =
A = � (B + C)
= � 2B B = C
tan A = � tan2B
= � = �
tan A =
RESONANCE SOLUTIONS (XI) # 113
22. r1 � r = � = = a tan
(r1 � r) = abc tan tan tan
= abc tan
= abc
= =
= = = 4Rr2
EXERCISE # 32. Match the column
(A) AA1 and BB
1 are perpendicular
a2 + b2 = 5c2
c2 = = 5 c = (
cos C = = =
sin C = = ab sin C =
2 = 11
(B) G.M. H.M.
(r1 r
2 r
3)1/3 (r
1 r
2 r
3)1/3 3r 27
(C) tan2 = a = 5, b = 4 2s = 9 + c
= = = c2 = 36 c = 6
(D) 2a2 + 4b2 + c2 = 4ab + 2ac. (a � 2b)2 + (a � c)2 = 0 a = 2b = c
cos B = =
8 cos B = 7
RESONANCE SOLUTIONS (XI) # 114
COMPREHENSION # 2 (Q. No. 7 to 10)7. Clearly
8. Let 3
1
2 =
Then angle of pedal trinagle = � 2 = A
=
9. Side of pedal triangle = I2I
3cos = BC
I2I
3 =
I2I
3 = 4Rcos
10. 1 = 4 R sin
I2I3 = 4 R cos
12 +
2
32 = 16R2
12. 1
2 = 4R cos if we apply Sine-Rule in
1
2
3 , then
2 Rex
= =
=
2Rex
= 4R Rex
= 2R ABC is pedal triangle of I
1 I
2 I
3
statement - 1 and statement - 2 both are correct and statement -2 also explains Statement - 1
14. sin = =
similarly sin =
3 sin � 4 sin3 =
� = r2 = r = a. cm.
19. ax2 + bx + c = 0 ...(1)
x2 + x + 1 = 0 ...(2)
roots of (2) are imaginary and a, b, c are real
= = = k cos C = = = C =
RESONANCE SOLUTIONS (XI) # 115
EXERCISE # 4PART - I
1. We have a2 a2 � (b � c)2 = (a + b � c) (a � b + c)
a2 (2s � 2c) (2s � 2b) = 4(s � b) (s � c)
similarly b2 4 (s � c) (s � a)
and c2 4 (s � a) (s � b).
Multiplying the above inequalities, we geta2b2c2 64 (s � a)2 (s � b)2 (s � c)2
(a + b + c) abc 16 s (s � a) (s � b) (s � c) = 162
Equality occurs if and only if(b � c)2 = 0(c � a)2 = 0
and (a � b)2 = 0i.e if and only if a = b = c.
2. (A) a, sin A, sin B are given one can determine
b = c = So the three sides are unique. So option (a) is incorrect option
(B) The three sides can uniquely determine a triangle.So option (b) is incorrect option.
(C) a , sin B, R are given one can determine b = 2R sin B,
sin A = . So sin C can be determined. Hence side c can also be uniquely determined
(D) for a, sin A, R
= 2R
But this could not determine the exact values of b and c
3. n = 2n × area of OA
1
1
n = 2n × × AA
1
1 × O
1
n = n × sin × cos
n = sin . .........(1)
On = 2n × area of OB
1O
1
On = 2n × × B
1O
1 × O
1O = n × tan × 1 = n tan
On = n tan ......(2)
Now R.H.S. = =
= × 2 cos2 = On. cos2
= n tan .cos2 = sin = n = L.H.S
RESONANCE SOLUTIONS (XI) # 116
4. Let angle of the triangle be 4x, x and x .Then 4x + x + x = 180° x = 30°
Longest side is opposite to the largest angle.Using the law of sines
= 2R
a = R, b = R, c = 2S = =
5. Clearly the triangle is right angled. Hence angles are 30º, 60º and 90º are in ratio 1 : 2 : 3
6. Consider =
= = =
7. Clearly P is the incentre of triangle ABC.
r = =
Here 2s = 7 + 8 + 9 s = 12
Here r = =
8. = . b . b . sin 120º = b2 .........(1)
Also a = .........(2)
and = and s = (a + 2b)
= (a + 2b) ..........(3)
From (1), (2) and (3), we get =
9.* We have ABC = ABD + ACD
bc sin A = c AD sin + b × AD sin
AD =
Again AE = AD sec
RESONANCE SOLUTIONS (XI) # 117
= AE is HM of b and c.
EF = ED + DF = 2DE = 2 × AD tan = × cos × tan an
= sin
As and DE = DF and AD is bisector AEF is isosceles.
Hence A, B, C and D are correct answers.
10. In ABC , by sine rule
= = C = 45º, C = 135º
When C = 45º A = 180º � (45º + 30º) = 105º
When C = 135º A = 180º � (135º + 30º) = 15º
Area of ABC = AB . AC.sin BAC = × 4 × sin (15º) = × = 2
Area of ABC = AB . AC .sinA = × 4 × sin (105º) = 2
Absolute difference of areas of triangles = | 2 � 2 | = 4
Aliter
AD = 2 , DC = 2 Difference of Areas of triangle ABC and ABC = Area of triangle ACC
= AD × CC = × 2 × 4 = 4
12. cos B + cos C = 4 sin2 2 cos cos = 4 sin2
2 sin = 0
cos � 2 cos = 0 as sin 0
� cos cos + 3 sin sin = 0
tan tan =
=
= 2s = 3a b + c = 2a
Locus of A is an ellipse
RESONANCE SOLUTIONS (XI) # 118
11. sin 2C + sin 2A = (a cos C + c cos A) = = 2 sin B = 2 sin 60º =
12. cos =
=
=
= (x2 + x + 1) = 2x2 + 2x � 1
( � 2) x2 + ( � 2) x + ( + 1) = 0
on solving
x2 + x � = 0 we get
x = + 1, �
At x = � , Side c becomes negative. x =
13. Area of triangle = ab sin C = 15
. 6 . 10 sin C = 15 sin C =
C = (C is obtuse angle )
Now cos C =
� = c = 14
r = = = r2 = 3
14. a = 2 = QR
b = = PR
c = = PQ
s = = = 4
= = = = tan2
= = = =
RESONANCE SOLUTIONS (XI) # 119
PART - II1. Let a = 3x + 4y, b = 4x + 3y and c = 5x + 5y
as x, y > 0, c = 5x + 5y is the largest side C is the largest angle. Now
cos C =
= < 0
C is obtuse angle ABC is obtuse angled.
2. r1 > r
2 > r
3 > >
s � a < s � b < s � c �a < �b < �c; a > b > c
3. tan = ; sin =
r + R = r + R = .cot
4. a =
= a + b + c = 3b.
a + c = 2b a, b, c are in A.P.
5. AD = 4
AG = × 4 =
Area of ABG = × AB × AG sin 30º
= × × × = Sin 60º = AB = =
Area of ABC = 3(Area of ABG) =
6. cos = = � = 120º
7. C = /2
r = (s � c) tan C = 90º
r = s � 2R
2r + 2R = 2 (s � 2R) + 2R.
= 2s � 2R
= (a + b + c) � C = 90º
= a + b + c � c
= a + b
RESONANCE SOLUTIONS (XI) # 120
8. are in H.P.
are in A.P. a,b,c are in A.P.
9. = cos
Let cos = for some n 3, n N
As cos cos cos
3 n < 4, which is not possibleso option (2) is the false statementso it will be the right choiceHence correct option is (2)
ADVANCE LEVEL PROBLEMSPART - I
1. From figure, AD = c sin B
Hence number of triangle is 0 if b < c sin B
one triangle for b = c sin B
two triangles for b > c sin B
2. C = 60°
Hence c2 = a2 + b2 � ab
= = 2 cos
3. Using properties of pedal triangle,
we have MLN = 180° � 2A
LMN = 180° � 2B
MNL = 180° � 2C
Hence the required sum = sin2A + sin2B + sin2C
= 4sinA sinB sinC
4.
From figure, we can observe that OGD is directly similar to PGA
5. BD = s � b, CE = s � c and AF = s � a
Hence BD + CE + AF = s
6.
RESONANCE SOLUTIONS (XI) # 121
, as cos = cos
A = B, in either case
7. ,
Using cosine rule in ABO, we get
h =
8. In ABD,
Comprehension # 1
9. + + = b sin B + c sin C + a sin A =
k = 2R
10. cot A + cot B + cot C = (b2 + c2 � a2 + c2 + a2 � b2 + a2 + b2 � c2)
= (b2 + c2 + a2) =
=
= . = k =
11. = = 6
Comprehension # 2 (12 to 14)
12. PG = AD
=
= .ab sin C or
= b sin C ( = ac sin B)
RESONANCE SOLUTIONS (XI) # 122
PG = ac sin B
= c sin B
13. Area of GPL = (PL) (PG)
and Area of ALD = (DL) (AD) PL = DL and PG = ADAD
= = =
14. Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)
Area of PGQ = PG.GQ.sin(PGQ)
= × AD × BE sin ( � C)
= × × sin C
= × bc sin A × ac sin B × sin C
= sin A.sin B.sin C
Similarly Area of QGR = sin A.sin B .sin C and Area of RGP = sin A.sin B.sin C
From equation (1), we get
Area of PQR = (a2 + b2 + c2) sin A.sin B.sin C
15. In CDB , =
Also from same triangle = BD =
16. cosAcosB + sinAsinBsinC = 1
(cosA � cosB)2 + (sinA � sinB)2 + 2sinAsinB(1 � sinC) = 0
A = B & C = 90°
a : b : c = 1 : 1 :
17. We have
a : b : c = 5 : 4 : 3
RESONANCE SOLUTIONS (XI) # 123
18. from figure, OO = ON � ON = R �
ZO = ZM +
= RcosA +
from OZO, using Pythagorous theorem,
we get (R � )2 = (RcosA + )2 +
=
PART - II
1. from ABC , =
AB = 2Rsin(A + )
from ACB, =
AC� = 2Rsin( � A)
BC = 2R(sin (A + ) � sin( � A))
= 4RcossinA = 2acos
similarly CA = 2bcos area ABC =
=
= 4cos2.
2. c2 � 2bc cosA + (b2 � a2) = 0
c1 & c2 are roots of this quadratic equation
Hence (c1 � c2)2 + (c1 + c2)
2tan2A = 4a2
3. Area =
=
=
= 2Rs
=
RESONANCE SOLUTIONS (XI) # 124
4. We know that OA = R, HA = 2RcosA and
applying Appoloneous theorem to AOH, we get
2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2
2.(AQ)2 = R2 + 4R2cos2A �
5. = +
using sine rule, diameter of required circle
= = = 20
radius = 10
6. L.H.S. = (a2 (b + c � a) + b2 (c + a � b) + c2 (a + b � c))
=
=
= abc
= 4R
7. from the parellelogram ABAC, AA = 21 ,
from AAC, AA < b + c
21 < b + c ...(1)
similarly 22 < c + a ...(2)
and 23 < a + b ...(3)
(1) + (2) + (3) gives 1 + 2 + 3 < 2s
8. ZXY =
and
Area of
RESONANCE SOLUTIONS (XI) # 125
= 2
Ccos
2
Bcos
2
AcosR2 2
Area of CsinBsinAsinR2Csinab2
1ABC 2
Area of XYZ = 2R2 cos 2A
cos2B
cos 2C
= r2R
9. Drop a perpendicular from the apex P to the base ABC.
The foot of perpendicular is at circum centre O of ABC
Using given data, we get 52
21RBO
from, right angle POB, we get
POh = 22 OBPB
= 8.83 m
10. from cyclic quadrileteral CQFP, we get
BCFPCQP
from cyclic quadriletral AQMF, we get
FQM = FAM = 90º � B
AQM = 90º + 90º � B = 180º � B
180CQPAQM
P, Q, M are collinear
A
CBP D
N
MQ
EF
similarly P, Q, N are collinear
hence, P, Q, M, N are collinear