CONTENT Preface MATHEMATICS CLASS : XI Page No.. Trigonometric ratio identities & Equations Eercise 0-7. Fundamentals of Mathematics - II Eercise 8-8. Straight Line Eercise 9-70 4. Circle Eercise 70-9 5. Mathematical Reasoning, Induction & Statistics Eercise 9-0 6. Solution of Triangle Eercise 0-5 Copyright reserved 0-. All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.
Section (A) : A-. c 80º TRIGONOMETRIC RATIO, IDENTITIES & EQUATIONS A-4. (a) + + 6 EXERCISE # PART - I (b) + + 4 0 (c) + 0 5 5 A-6. ( cos)cos cot sin ( sin) ( cos )cos cot sin ( sin) A-9. tan 5 sin 5 < < and cot 5 LHS Section (B) : sin cot cos ec cos ec sin cot cos ec 5 5 5 8 8 B-4. LHS cos + cos ( + ) { cos cos sin sin cos cos } cos cos ( + ). cos ( ) cos cos + sin sin RHS B-6. (i) sin A sin B sin A cos A sinb cosb sin(a sina B) sin(a B) sinb cos(a 5º ) (ii) cot (A + 5º) tan (A 5º) sin(a 5º ) RHS sin(a B) sin(a B) cos(a B) sin(a B) sin( A 5º ) cos( A 5º ) tan (A + B) cos(a 5º )cos(a 5º ) sin(a 5º )sin(a 5º ) sin(a 5º )cos(a 5º ) (cos A sin 5º ) (sin (sin A sin0º ) A sin 5º ) cos A 4cosA sina sina B-7 A + B 45 tan(a + B) tan(45º) tan A tanb tan A tanb tana + tanb + tana tanb ( + tana) ( + tanb) put A B ( + tan º ) tan º º SOLUTIONS (XI) #
Section (C) : C-. LHS tan tan 4 cos 4 cot 4 sec 9 cos cos cot 4 sec 9 cos cos4 sin sec sin 4 9 sin 4 cos 4 cos sin 4 sin sec 9 sin 4 cos 9 9. sec cosec 4 RHS C-. (ii) cos A sin A cos A sin A cos A sin A cos A sin A 4 sin A cos A cos A sin A sina cosa tan A C-9 tan tan(60 + ) tan(60 ) tan LHS tan tan tan tan tan tan tan tan tan tan tan tan Put 0 tan 0 tan 0 tan80 tan 40 tan60 Section (D) : D-. Let y cos.cos cos y 4 cos cos cos cos y cos 4 y cos cos cos y [cos + cos cos] 4 y 4 cos cos y min 4 and y ma 4 D-. (i) y 0 cos 6 sin cos + sin 5 ( + cos ) sin + cos (ii) 4 cos sin + 6 a b a cos + b sin a b y ma 5 + 6 y min 5 + 6 y + sin + cos y + sin + sin y ( sin sin ) SOLUTIONS (XI) #
y (sin sin + 9 9 ) y sin 0 9 sin + 6 y ma, ymin 9 + (iii) y cos + 5 cos + y cos. sin + 5 cos + y cos sin + 5 cos + y cos sin + Section (E) : y ma y min 69 7 + 7 + 0 4 4 69 7 + 7 + 4 4 4 E-. (i) cos A cos eca sin A sec A cos A sin A cos A sin A sina cos A(cos A sina) cos A sina cos A sina (ii) sec tan cos cos cos cos sin cos sin cos ( sin)cos sin sin sin ( sin)cos cos cos cos sin sin cos ( sin)cos sin( sin) ( sin)cos sin cos (iii) Section (F) : F-. (i) LHS cos cos A sin A cos sin A cos A + sin A sina cosa + cos A + sin A + sin A cosa cos A sin A cos A sin A 4 6 4 cos cos cos cos cos 7 7 7 7 7 7 4 5 (ii) LHS cos cos cos. cos. cos 4 8 6 cos cos cos cos cos 8 sin 7.sin 7 8 RHS sin 5.sin sin.sin RHS SOLUTIONS (XI) #
F-. LHS sin + sin + sin +... + sin n cos cos 4 cosn... n [(cos + cos4 + cos6 +...+ cos n)] n n() sin sin. cos n n sinn. cos(n ) sin RHS F-7. cos (S A) + cos (S B) + cos (S C) + cos S S A B cos cos B A S C + cos C cos C B A A B C C cos cos + cos cos cos Section (G) : G-4. sin cos cos cos cos A B cos n ± ± n n n, 5 n, 5 n G-8. tan tan sin sin cos cos Section (H) : 0 cos (n + ) (n + ) 6. H-4. cos + cos + cos cos cos 4 cos 6 + + cos + cos4 + cos6 cos4 cos cos cos 0 or cos4 + cos 0 (n +) n Now (n + ) + 6 6 or cos cos 0 (k + ) + 6,(k + ) + 6, (k) + 6 k, (k + ) 5 k, 6, 6 k 6 k 6 may also be written as (n ), (n + ), (n + ) 4 6 ( k is same as (n + ) ) m 6 SOLUTIONS (XI) # 4
H-5. Section (I) : sin n sin (n ) sin sin(n ) sin sin m, sin(n ) sin 0 sin 0 or sin (n ) sin (n ) cosnsin 0 n (p ), ( n ) m,, p n n I-. tan ( + ) tan + 0 tan, n + 4, n +. I-. 4 cos sec tan sin 4 cos cos cos 4 cos sin 4 4 sin sin 4 sin + sin 0 sin 4 6 8 5 sin 4, 5 4 8 5 4 5 cos 6º, sin 8º sin 54º, sin 8º sin, sin 0 0 n + ( ) n 0 or n ( ) n. 0 Section (J) : J-. sin cos sin cos sin 6 sin 6 sin 4 6 n + ( ) n 4. J-. 5 sin + cos 5 5 sin + 9 9 cos 5 9 sin sin + cos cos 5 9 cos ( ) sin cos n ± n ± + n ±, n ± + n +, n + For n +, SOLUTIONS (XI) # 5
We have n + 4 n + 5 tan tan 5 n + tan 5 n + tan 7 n + or n + where tan 7 PART - II Section (A) : A-. {cos 4 + sin 4 } {cos 6 + sin 6 } { sin cos } { (cos 4 + sin 4 sin 4 cos )} 6 sin cos { sin cos } 6 sin cos + 6 sin cos A-6. cos cos cos cos 0 0 0 0 cos 0 cos 0 sin. sin 0 0 5 4 5 4 4 6 6 Section (B) : B-. sin 5 sin sin sin 5 sin sin sin sin 8 tan tan 4 B-7. cot (A + B) cot 5 Now cot A cot B + cot A + cot B cot A.cotB cot A cotb cot A cotb cot A cotb ( cot A cotb) cot A cot A cotb cotb Section (C) : C-. tana 4 A III rd quadrant 5 sin A + sina + 4 cosa 0 sina cosa + sina + 4 cosa 0 sina cosa + sina + 4 cosa 0 SOLUTIONS (XI) # 6
C-6. tan tan +... (i) cos + sin tan tan + sin tan + sin tan sin + sin 0 which is independent of C-7*. sin t + cos t 5 tan ( tan ) + sin t tan tan tan t t 5 0 tan t + 5 5 tan t + tan t 6 tan t 0 tan t 4 0 tan t 6 tan t + tan t 0 Section (D) : D-. D-*. tan t f() sin 4 + cos sin ( cos ) + cos sin + cos sin cos f() 4 sin f() ma t t tan + tan t 0 tan 0 sin f() min /4 Range is, 4 4 + 4 sin+ cos 4 sin+ cos[ 5, 5] Ma. + 5 6 Min. 5 4 Section (E) :, tan t E-. E-5*. square & add a + b 9 + 6 5 radian ~ 57º (appro.) sin > sin cos > cos SOLUTIONS (XI) # 7
Section (F) : F-. A tan 6 tan 4 B cot 66 cot 78 A tan 6 tan 4 tan 66 tan 78 B A B tan6tan(60 6) tan(60 6) tan54. tan 78 tan 4 A B tan 8. tan(60 8)tan(60 8) tan54 tan 54 tan54 A A B B 4 8 6 F-5*. cos cos cos cos cos 0 0 0 0 0 5 sin 0 5 sin 0 sin 0 sin 0 sin 0 sin 0 sin cos 0 0 sin 0 6 cos 0 64 0 5 F-7*. cos + cos y + cos z cos cosy cosz (Given + y z) + cos ( + y) cos ( y) + cos z cos cosy cosz + cosz [cos ( y) + cos ( + y)] cos cosy cosz + cosz. cos cosy cos cosy cosz cos ( + y z) F-9*. tan A + tan B + tan C 6, tan A tan B In any ABC, tan A + tan B + tan C tan A tan B tan C 6 tan C tan C tan A + tan B + 6 tan A + tan B & tan A tan B Now (tan A tan B) (tan A + tan B) 4tan A tan B 9 8 tan A tan B ± tan A tan B or tan A tan B tan A + tan B tan A + tan B on solving on solving tan A tan A tan B tan B Section (G) : G-. tan + tan + tan tan tan n +, n SOLUTIONS (XI) # 8
Section (H) : H-. sin 7 + sin 4 + sin 0 H-4.* Section (I) : sin 4 cos + sin 4 0 sin 4 0 or(;k) cos 4 n or n ± n n, 4 9 sin sin + sin sin sin cos sin 0 or cos n or m options (A), (B), (C), (D) are all a part of n. n, m 4 0,,,,. 4 9 9 I-4. cos + cos 0 cos + cos 0 cos 9 8 4 4 7 7 As cos a cos only n ± where cos 4 I-5. sin + 7 cos 5 t t + 7 t t 5 where t tan 7 4 t + 7 7t 5 + 5t tan is root of t t 0 or 6t t 0. Section (J) : J-. tan cos 7, in [0, ] 4 4 7, in [0, ] common value is 4 4 7 4 7 general solution is n, n I. 4 J-.* Let E sin cos E sin + sin sin + sin 9 sin 4 assumes least value when sin n + ( ) n. 6. sin + sin y a...() cos + cos y b...() EXERCISE # PART - I y y sin cos y y cos cos b a SOLUTIONS (XI) # 9
SOLUTIONS (XI) # 0 tan y b a sin y b a a, cos y b a b sin ( + y) sin y cos y b a ab Now for tan y () + () + + cos ( y) a + b cos ( y) b a tan y y) cos( y) cos( tan y b a b a tan y ± b a b a 4 6. tan q p LHS (p cosec q sec) q p q p q p sec q p q cosec q p p q p p sin, q p q cos q p cos sin sin cos cos sin q p sin4 ) sin( q p sin4 sin4 ( 6) 7. (i) cot 7 o tan 8 o o o sin7 cos7 o o sin5 7 cos ) 0 sin(45 ) 0 cos(45 ) )( ( 6 6 4
( ) ( ) o o (ii) tan 4 cot 5 o tan5 tan 45 7 o tan7 o tan7 o cos7 sin7 o o cos7 sin7 o o o cos7 sin7 cos5 o sin5 cos5 ( )( ) [ ( ) ( ) ] [ ( ) (4 )] [ ( ) ( )] 6 + 6 9. (i) tan9 tan7 tan 6 + tan8 (tan 9º + tan8º) (tan 7º + tan 6º) sin90º cos9º cos8º sin90º cos7º cos6º sin9º cos9º sin7º cos 7º sin8º sin54º 5 4 5 4 8( 5 5 ) 4 4 (ii) cosec 0 sec0 cos0 sin0 4 sin0º cos0º sec 5 cos 40 (iii) sin 0 sin5 sin5 sin5º cos5º sec 5º sin5º cos5ºcos 40º º sin5º sin5º sin0 (sin5º + cos45º + cos 5º cos 5º + cos 45º) (sin5º + cos45º + sin 0º sin ( 5º)) ( ) 4 (iv) cot 70 + 4 cos 70 cos70º sin40º sin70º cos70 cos 70º 4cos70º sin70º 4cos70º sin70 sin70º (cos70º sin40º ) sin40º (sin0º sin40º ) sin40º sin70º sin70º SOLUTIONS (XI) #
sin80º cos 60º sin40º sin0º cos 0º sin70º sin70º (v) tan 0º tan 50º + tan 70º tan 0º tan (60º 0º) + tan (60º + 0º) tan 0º tan0º tan0º tan0º tan0º 9 tan0º tan tan 0º 0º tan0º tan 0º tan 0º tan 0º. A A A A + A A 4 OA OA OA OA 4 r (say) A OA, A OA n sin + sin n n sin n sin n sin sin sin n n 4, A OA n 4. sin n n 6 n sin n cos.sin sin n n sin n n sin cos sin n n n 4 4 sin sin n n n n 4 n n 7. P n P n cos n + sin n cos n sin n cos n (cos ) +sin n (sin ) cos n ( sin ) +sin n ( cos ) ( sin cos ) {cos n 4 +sin n 4 } ( sin cos ) P n 4 put n 4 P 4 P ( sin cos ) P 0 P 4 P sin cos sin cos similarly we can prove the other result also. 5. tan + tan. tan tan + tan. tan (tan tan ) + 4 tan tan tan tan 0 (tan tan ) + 4 tan tan (tan tan ) 0 ( tan )( tan ) SOLUTIONS (XI) #
4 tan tan (tan tan ) 0 ( tan )( tan ) (tan tan ) ( tan. tan ) 0 tan tan or tan. tan L.H.S. tan + tan. tan tan + tan tan. ( tan ) R.H.S. tan + tan. tan tan + tan tan. ( tan ) 9. 8tan 6 tan...() 8 tan 6 tan + 9 6 tan tan, 6 Put in () tan is correct n + tan n +, +, +, + in (, ). tan + sin...() As tan + cos 4 7 7 9 7 sin + cos + sin sin + cos 4 4 4 sin sin n + ( ) n 6 from (), tan sin n +. 4. a cos + b sin c a t bt + c where t tan t t (c + a)t bt + (c a) 0 t + t b, t c a t c a c a cos cos cos + cos + [cos + cos ] + t t t t simplifying and using values for t, t we get ac cos + cos + a b a b ac. a b SOLUTIONS (XI) #
7. RHS + + Minimum value whereas LHS 4()() 4 4() 8 > no solution. PART - II. For dodecagon A 'OB' 0 OA 'B' OB 'A' 75 R o R R sin 75 sin 0 A B R ( )( ) R For heagon AOB 60 6 AOB is equilatecal AB R 6. A + B + C C sin A C sin k C sin A C sin A C sin C sin k k A C A cos sin A C A sin cos k k B A k tan tan k 8. 4 cos 4 + 4 4 sin 4 sin cos 4 cos 4 + sin 4 cos 4 sin cos sin 0. cos A cos ecb cos ecc cos A sinb sinc + cosb sin A sinc + cosc sin A sinb cos A sin A cosb sinb coscsinc sin A sinb sinc SOLUTIONS (XI) # 4
sina sinb sinc sin A sinb sinc 4 sin A sinb sinc sin A sinb sinc (using conditional identity). (cos 6 cos 4) 5(cos 4 cos ) 0(cos ) cos5 5cos 0 cos cos5 cos 5 cos cos 0 cos cos5 5cos 0cos cos5 5cos 0cos cos cos5 5cos 0cos cos 5. cos 4 cos cos ( + cos ) cos cos ( ) (cos + sin ) + (cos + sin ) + cos ( ) a + b cos cos ( ) a + b a cos + b 0. sin 4 sin sin sin 4 sin ( sin ) ( sin sin 4) sin 4 sin sin (cos cos 6) 4 sin (cos cos 6) or sin 0 ( cos ) cos cos 6or sin 0 cos 6 cos 6 or sin 0 sin 0 or cos 6 n or n 6 n 9 0,,,, 9 9 So eight solutions., 9 9. cos sin cos sin sin cos cos (sin cos ) cos sin 4 see from graph or we can put values given in options to verify. 5. tan 5 sec 0 (sec ) 5 sec 0 6 sec 5 sec 0 sec,, SOLUTIONS (XI) # 5
sec sec cos 7 solutions in 5 0, n 5. 8. 4cos 4cos + cos 0 (4 cos + ) (cos ) 0 cos n solutions in the interval [0, 5] are 0,, 4,..., 00 9. h 0 4... 00 arithmatic mean 50 5 {cos cos cos( )} {sin sin sin( )} h [4 cos ( + ) (cos ( ) + ) + 4 sin ( + ) (cos ( ) + ) ] / h [4{cos ( ) + } {cos ( + ) + sin ( + )}] / h ( + cos ( )) h cos h 4 cos. y a cos + b sin cos + c sin & tan z a sin b sin cos + c cos y + z a + c and y z (a c) cos sin + 4b sin cos b a c (a c) cos + b sin ( b (a c) tan ) sin (a c) [cos + tan.sin] (a c) cos sin cos ( a c)cos( ) cos (a c).. A B A B cos.cos A B A B cos.sin n + A B A B sin.cos A B A B sin.sin n A B cot n A B + ( ) n cot n SOLUTIONS (XI) # 6
4. sin 6 + cos 6 a (sin + cos ) (sin 4 + cos 4 sin cos ) a (sin + cos ) sin cos a sin cos a 4 sin a 4( a ) sin 0 4 ( a ) a 0 and 4 4a a 8. cos 5 sin 5 and a 4 a and a or a a,, cos 5 cos 5 or cos 5 5 n ± 5 or 5 n ± 5 n n +, n, +, n 0 40 5 0 n n and, n and, n 5 0 0 40 40. sin + sin cos cos 0 case-i : cos 0 tan + tan 0 tan, n + tan ( ), n + 4 case-ii : cos 0 + 0 0 0 not true. EXERCISE #. (A) sin + cos cos + cos cos cos + 0 cos, cos ( cos ) 0 in [, ] No. of solution sin 4 (B) sin. tan 4 cos sin. cos cos 4 sin4 sin cos4 cos 0 cos5 0 5 (n + ) / (n + ) /0 5 7 9,,,, in (0, ) 0 0 0 0 0 So there are five solutions. (C) ( tan ) sec + tan 0 ( tan 4 tan ) + 0 ( ) + 0 where tan y O y y SOLUTIONS (XI) # 7
tan tan Number of solutions (D) [sin ] + [ cos] in, [sin ] and [ cos] < < and cos cos cos 5 5 for [0, ], [0, ] 4 4 5 5 4 0 < sin < [sin] 0 4. (A) Number of solutions 6 (B) sin 8 ± sin As sin takes at least four values in [0, n] n 4 (C) + sin 4 cos L.H.S. and R.H.S. L.H.S. R.H.S. sin 4 0 and cos n and m n and m n and n m 5 5,, 0,, in, Number of solutions 5. (D) A, B, C are in A.P. B 60º As sin (A + B) A + B 0º or 50º A 0º or 90º A 90º A 45º 5 C 80º A B 75º p. Comprehension # (5, 6, 7) 5. Given cos + cos a cos and sin + sin b cos a... (i) SOLUTIONS (XI) # 8
sin cos by (i) & (ii) tan b a sin + cos b... (ii) tan a b b a b a + b a b a a ab b a + a b b a b a b b ab b(a b) + a b n 6. sin n A sin A sin A sin A sin A sin 4A sin A ( sin A cos A) ( sin A 4 sin A) (4 sin A cos A ( sin A)) 8 sin 4 A ( sin A) ( sin A) ( 4 sin A) If we put sin A, then given epression is a polynomial of degree 5 in. 7. If p 5 sin + (p 5), cos, tan sin, cos, tan are in G.P. cos sin tan cos sin cos cos cos + cos taking cube both sides cos 9 + cos 6 + cos 5 cos 9 + cos 6 + cos 5 0 Comprehension # 4 4. sin 6 + cos 6 < 6 7 sin cos > 6 sin cos < 6 7 sin > 4 cos 4 > 4 cos4 > cos4 < General value is 4 Principal is value 4, 4 4 n, n n, 6 n, n SOLUTIONS (XI) # 9
5. cos + 5 cos + 0 cos + 5cos + 0 (cos + )( cos + ) 0 cos + 0 ( cos + > 0) cos, 6. sin + 4 cos 0 cos + cos 0 cos sin cos sin sin + n, n 6 7 6 5 5 n, n 6 n, n 4 7 5,, 4, 4 in, 9. Statement- : cos A cosec B cosec C cos A sinb sinc sin A cos A sin A sinb sinc sina sin A sinbsinc 4 sin A sinb sinc sin A sinb sinc Statement- : tan A tan B iff the triangle ABC is right angled Statment is false 0. y tan tan tan tan tan tan y tan tan y tan y 0 y, (, ) statement- and statement- both are true and statement- eplains statement-. cos. sin y Either cos and siny or cos and siny 5 5 (,y) 0,, 0,,,,, or (, y),,, Number of pairs 6 4. log [cos ( ) cos ( ) coscos] log [cos ( ) sin ( ) coscos] log [ + cos.cos cos cos ] log 0 SOLUTIONS (XI) # 0
0. sin ( sin 4 sin ) ( sin cos ) + 8 sin 6 sin 4 sin + 0 ( sin )(4 sin ) 0 sin, ± For sin ±, cos + 0 so given equation becomes undefined sin only n +( ) n 6, n.. sin. 8cos sin. cos. cos cos tan y ( + sin z) 4 8 sin cos. cos +, + tan y, + sin z cos cos, tan y 0, sin z n, n I. 6. cos 0 + sin 55 sin 65 cos 0 + cos 0 sin 65 + sin 65 sin 45 sin 65 + sin 65 sin 65 4. sin sin cos + cos cos sin 4 cos 4 4sin cos + 4cos sin 4(cos sin ) sin + sin cos cos cos sin sin sin sincos cos sin sin cos cos cos cossin cos sin cos sin sincos sin sin sin cos sincos( sin ) sin cos either sin 0 or cos sin cos n or sin + cos + cos cos cos n n or (n + ) But at (n + ), + cos 0 (n + ). cos 4. 6 cos 7 sin + cos 0 6 cos 7 7cos + cos 0 7cos cos + cos 0 7cos cos cos (so cos 0) 7 cos cos cos 6 cos cos 0 ( cos ) ( cos + ) 0 cos, SOLUTIONS (XI) #
But cos 0 cos cos cos where cos cos cos( ) n ± ( ). 4. + + 4 + sin 0 + + 4 sin...() when 0, 0 0 0 is the solution when [0, ), + + 4 > 0 where as sin < 0 no solution for (0, ) when [, ], + + 4 + + 4 > whereas 0 sin no solution for [, ] so given equation has only one solution in [0, ] and that solution is 0.. Clearly 0º and (60º, 90º) Hence + lies in (90º, 0º). EXERCISE # 4 PART - I. Let y sin t 5 y (y 5) (y ) (y + ) 0 R, D 0 y y 0 y 5 or y 5 sin t 5 4 or sin t 5 4 range of t is, 0, 0. O BD, BD OD cot 0º BD similarly EC BC AB AC + area of ABC 4 ( ) 4 ( ) 4 6 + 4 sq. unit 4. 0, or 0 cos e SOLUTIONS (XI) #
This is true for '4' value of '', '' If and and cos ( + ) (No solution) similarly if and again no solution results 5. 0, 4 tan in 0, and 0 < tan < 4 cot in 0, and cot > 4 Let tan and cot + where and are very small and positive, then ( t ( ), t ( ), t ) ( ), t 4 ( ) t 4 > t > t > t 6. sin 5sin + > 0 (sin )(sin ) > 0 sin < From graph, we get [ sin ] 5 0,, 6 6 O y / 7. sin cos 0...(i) sin cos sin 0 sin sin + 0 sin,...(ii) So sin is the only solution at, 6 5 6 8.* sin 4 cos 4 + 5 sin 4 ( sin + ) 5 sin 4 4 sin + 5 6 5 sin 4 sin + 4 sin 5 sin 4 0 sin + 4 0 5 (5 sin ) 0 sin 5, cos 5 9. f() tan sin sincos 5cos and cos 5( cos) sin sin 8 cos 8 + 8 7 5 6 sin 4cos f() ma 6 5 SOLUTIONS (XI) #
0. sin n sin sin n n cos sin n n sin sin n n sin n sin 4 sin n n 4 ( ) k n + k, k n If k m m n m, not possible n If k m + 7 (m + ) n 7, m 0 n. tan cot 5 sin cos cos5 sin5 6 (n + ) cos 6 0 (n + ) ; n 5,, 5,,, 4 4...() sin cos4 sin sin sin + sin 0 sin, (4m ), p + ( ) p 6 p (4m ), + ( ) p ; m, p I 4 5,,...() 4 From () & () 5,, 4 Number of solution is.. P {: sin cos cos } sin ( + ) cos tan + n+ ; n I 8 Q {: sin + cos sin } cos ( ) sin tan P Q + n+ ; n I 8 SOLUTIONS (XI) # 4
. As tan( ) > 0, < sin <, [0, ] 5 < < Now cos( sin) sin ( tan / + cot /)cos cos( sin) sin cos cos + sin( + ) 5 As, < sin( + ) < cos + (, ) < sin( + ) < 5 As + [0, 4] +, 6 6 7 or +, 6 6 5 7 < < or < < 6 6 6 6 7 5,, 6, correct option is (A, C, D) PART - II. u a cos b sin + a sin b cos u a cos + b sin + a sin + b cos + a cos b sin a sin b cos u (a + b ) + a b a sin a b a cos u a b + 4 a a b a b a sin cos u b a + b a a b sin. min(u ) a + b + ab a b and ma(u ) a + b + a b a b Now, ma(u ) min(u ) (a b) 7. sin + sin and cos + cos 65 65 squaring and adding, we get sin + sin + sin sin + cos + cos + cos. cos 65 + 7 65 70 + cos ( ) 45 cos 70 4 45 9 0 cos 0 ( < < < < ) SOLUTIONS (XI) # 5
. tan P and tan Q are the roots of equation a + b + c 0 tan P + tan Q a b and tan P tan Q a c P Q R + + P Q P Q 4 R ( P + Q + R ) ( R ) P tan Q P Q tan tan P Q tan.tan b / a c / a c a + b 4. cos + sin tan tan / / + tan / tan /, Let tan t t t t + t t 4t 0 t 7 as 0 < < 0 < < tan is positive t tan 7 Now tan tan / tan / t t 5. Given equation is sin + 5 sin 0 ( sin ) (sin + ) 0 sin ( sin ) 7 tan 7 4 7 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 + sin tan tan + tan tan SOLUTIONS (XI) # 6
t Let tan t t t + t t 4t 0 t 7 As 0 < < 0 < < tan is positive. t tan 7 Now, tan tan tan tan 7 7 t t 7 7 7 tan 7 tan 4 7. 7. {cos ( ) + cos ( ) + cos ( )} + 0 (cos + cos + cos ) + (sin + sin + sin ) 0 cos 0 sin 8. tan tan (( + ) + ( )) Hence correct option is () tan( ) tan( ) tan( )tan( ) 5 4 5. 4 (9 5)4 48 5 4 4 56 9. A sin + cos 4 sin + ( sin ) sin 4 sin + sin + 4 4 A 0. sin P + 4 cos Q 6...(i) 4 sin Q + cos P...(ii) Squaring and adding (i) & (ii) we get sin (P + Q) 5 P + Q or 6 6 5 R or 6 6 5 If R then 0 < P, Q < 6 6 cos Q < and sin P < sinp + 4 cosq < So R 6 SOLUTIONS (XI) # 7
Fundamentals of mathematics - II Section (A) : A-5. log p log p (p) p n OBJECTIVE QUESTIONS n logp p log p p n n independent of p. log 5 log A-7*. N log log 5 405 5 log 7 log 5 5 log 5 log 5 log log 5.log 405 log 5 log (8 5) ( + log 5) ( + log 5) log 5(4 + log 5). A-8*. log >, log 0 < log > log 0 log 6 5 <, log 7 8 > log 6 5 < log 7 8 log 6 <, log 9 > log 6 < log 9 log 6 5 <, log 0 > log 6 5 < log 0 Section (B) : B-4*. (log 5 ) + log 5 5 (log 5 ) + log 5 5 log 5 (log 5 ) + log 5 log 5 log 5 5 5 log log 5 log 5 5 5 (log 5 ) + log 5 log5 log 5 Let log 5 t t t ( t) t t + t t t t + t + t + t t + t t 0 t(t + t ) 0 t(t ) (t + ) 0 t 0,, log 5 0,,, 5, 5 B-6*. log log 5 9 log 9 log + 5 log log 9 log + 5 log Let log t 9 t t + 5 t t 9t + 0t 0 t satisfies it t 9t + 0t t (t ) 7t(t ) + (t ) (t ) t 7t (t ) (t ) (t ) SOLUTIONS (XI) # 8
t t log log t log / 7. B-9. Number of digits in integral part number of digit in 60 before decimal P 60 logp log 60 log 60 [log 6 + ] [log + log + ] [.00 +.477 + ] [.780].6 number of digits in integral part B-0. log 6 4 Section (C) : 6 /4 8. C-. log ( ) >...() (i) When 0 < < 0 < < So no common range comes out. (ii) When > < 0 but > here, also no common range comes out., hence no solution. Finally, no solution C-6. log 0. ( 8)( ) 0 (log 7 5 ) 0 For ( 8)( ) to be defined (i) ( 8) ( ) 0 ( ) ( 8) 0 8 0 0 Now Let say y log (log 0. 5 log 7 ) log (log 0. 5/) 7 Let y < 0 (assume) 0 then log (log 0. 5/) < 0 7 0 7 log 5/ > log 7 5/ > 0 So y < 0 so denominator is ve and numerator is +ve, so inequality is not satisfied, thus ( 8)( ) 0, 8...(i) Now > ( ) > log > + log 4.9 (appro) > 7.9 8 C-8. Domain + 4 5 0 (, 5] [, ) Case I : (, 5] [, ) ve < + ve alsways true (, 5] [, )... () Case II : [, ).. (i) 5 > (7/0) which is true SOLUTIONS (XI) # 9
< 4 5 6 + 9 < + 4 5 > 5 7... (ii) (i) (ii) [, )... () () () (, 5] [, ) Ans. (A) Section (D) : D-. z ( i) 4 4 ( i)( i) ( i) 4 4 i ( i) 4 4e i i / i e e i / z amp (z) z amp(z) 4 (D) D-6*. z + z z + z z z zz 0 z z z z z z z z z so amp z 0 is may be or z is purely imaginary z D-8. D-9. z / a ib z (a ib) + iy (a ab ) + i(b a b) a a y a b b 4(a b ) k 4 y b a b EXERCISE #. (i) / log 7 5 5 log 0 0. (7 + ) / (ii) log /4 log / / 8 log /4 log () /4 SOLUTIONS (XI) # 0
(iii) 49 log 7 (7 log7 7 log7 ) 4 ) log 7 ( 7 7 log7 (4) 96 & log 5 7 / 5 7 log 5 5 7 7 + 96 (iv) log 5 7 + log 5 7 log 7 5 7 log 5 log 5 7 + log 5 7 log 5 7 7 log 5 {using property log c b a log c a b } 0 5. (i) log 0 ( + 6) (i) + 6 > 0 ( 6) > 0 R {6} (ii) + 6 00 64 0 ( 6) ( + 4) 0 6, 4. (ii) log 4 log log 0 log log log 8. (iii) log log9 9 log 9 + + 9 log 9 + + 9 9 log 9 9 / (iv) log 4 (4 ) 4 log ( ) (i) 4 > 0 < 4 (ii) > 0 < (iii) log (4 ) 4 log ( ) log (4 ) ( ) 4 (4 ) ( ) 6 8 + 6 4 0 ( 6) ( + 4) 0 6 (not possible), 4. (v) log 0 + log 0 log 0 log 0 + log 0 + log 0 (log 0 + ) log 0 log 0 + ± log 0 and 0 5 (vi) log 4 (log ) + log (log 4 ) log (log ) + log (log 4 ) log ( log 4 ) + log (log 4 ) (vii) log + log (log 4 ) + log (log 4 ) log (log 4 ) log (log 4 ) log 4 4 6. log5 log 5 log 0 5 log 5 + log SOLUTIONS (XI) #
log + log 5 0 (log + 5)(log ) 0 log 5, log 0 5, 0. (viii) Domain > 0 and + > 0 and y > 0 > > < 7 (, 7)...(i) log ( ) log ( + ) + log (7 ) log ( ) + log (7 ) (7 ) log + 4 5 0 ( + 7) ( ) 0 or 7 (rejected) (7 ) 6. (a) log 0 0.00. log 0 0.477 let 6 5 log 0 5 log 0 6 5(log 0 + log 0 ) 5(0.00 + 0.477).675 characteristic of 6 5 is number of digits in 6 5 is. (b) let 00 log 0 00 log 0 47.7 number of zeroes immediately after the decimal in 00 is 47. 0. (i) 4 6 log 0 5 4 6 > 0, (0, )...(i) 4 6 & 0 (, ] (0, )...(ii) (i) (ii), (ii) log (4. + 7) > 5 4. + 7 > 0 ( ). + 7 > 0 R and 4. + 7 > ( ). 5 > 0 ( + ) ( 5) > 0 < or > 5 or > log 5 (log 5, ) (iii) (log) log 0 > 0...(i) (log ) (log +) 0 log or log 0 or 00...(ii) 0 (i) (ii) 0, 00, SOLUTIONS (XI) #
(iv) log 0.5 ( + 5) > log / ( ) ( + 5) > 0 R { 5}...(i) ( ) > 0 R...(ii) ( + 5) < ( ) 8 6 4 > 0 > 0 ( ) ( + ) > 0 (, ) (, )...(iii) (i) (ii) (iii) gives (, 5) ( 5, ) (, ) (v) log + > > 0 R {0} < ( + ) / + > 4 ( ) ( + ) > 0 (, ) (, ) (vi) log ( + ) < + > 0 > Case-I : when 0 < < (, 0) (0, ) then + > < 0 (, ) {0} Case-II : > > + < > 0 (, ) (, ) Hence, (, ) (, 0) (0, ) (, ). (i) < Case-I : < 0 <...(i) ( ) (, ) (5, )...(ii) (i) (ii) (, )...(A) Case-II : > 0 >...(iii) ( ) < 4 + 4 6 + 5 > 0 (, ) (5, )...(iv) (iii) (iv) (5, )...(B) (A) (B) (, ) (5, ) (ii) < Case-I : < 0...(i) 0 + 0...(ii) (i) (ii) [, 0)...(A) Case-II : 0...(i) 0...(ii) < + < + + 4 < 4 5 5 < 4 5 < < 5...(iii) SOLUTIONS (XI) #
5 (i) (ii) (iii) 0,...(B) (A) (B), 5 (iii) 6 8 Domain + 0 6 + 8 0 ( ) ( 4) 0 or 4 Domain [, ] [4, ) squaring 6 + 8 + 7 + 7 0 7 7 7 0 4, (iv) 8 > 6 (a) 8 + 0 [, 4]... (i) case - I when (i) 6 0... (ii) so 8 + > 6 + 9 6 0 8 + 8 < 0 5 9 + 4 < 0 (5 4) ( )< 0 4,... (iii) 5 by () and () and () (, ] Case - 6 < 0 > + ve > ve so >... (iv) by () and (4) (, 4] so by case () and () (, 4] (v) 7 + 0 0 and 4 0 0 ( ) ( 5) 0 and ( ) ( 5) 0...(i) so or 5 now check for 9 log 4 9 4 9 9 which is true hence is a solution now check 5 9 5 log 8 5 log 8 (.6) 4 4.096 4 which is false so only solution is SOLUTIONS (XI) # 4
(vi) Domain > 0 log + log 0 log (log + ) 0 log or log 0 0 < 4 or 0, [, )...(i) 4 Case-I 4 log < 0 positive < negative (false) Case-II 4 log 0 log 4 log log < (4 log ) Let log t t + t < (4 t) t 8t + > 0 (t 6) (t ) > 0 t < t > 6 log < log > 6 (Rejected) log < < 4...(ii) by (i) and (ii) 0, [, 4) 4. Square root of 7 + i ± where 7 + 4 i 5 5 7 i 5 7 5. (i) z R ( + i)+m + i 0 + m 0 & + 0 8 6 + m 0 m (ii) If one root is i then other is i Let forth root is. 4 ±(4 + i) a a + i + ( i) + 4 4 0. (i) z + e 5 z 8 9 i i e 5 9 i 5 9 cos 5 e i e 9 5 e 9 i 5 z (ii) z 9 9 cos Arg z 5 5 e i e i/ 6 e i5/ 6 z Arg z 5. 6 SOLUTIONS (XI) # 5
(iii) z tan sec Arg z Arg(tan i) (iv) z z (i ) sin sin i cos 5 5 5 cos ec 5 sin 5 Arg(z) 4 5 0. 4...(i) squaring both sides ( + ) + ( ) 4 EXERCISE # PART - I ( ) squaring both sides + 4 4 4 4...(ii) 4 5 4 5 does not satisfy equation (i) No solution. log log + log / log log (log ) log log log log log log log Let log y y y 0 (y ) (y + ) 0 y, log,, but log > 0 log is not possible 8. (a) z z z z z z z z z Given z z z z z z z z z z z z (b) arg (z) < 0 arg ( z) Hence (A) arg ( z) arg (z) ( ) SOLUTIONS (XI) # 6
4. log /4 log 8 ( + 7) + log / log /4 ( + 7) log log /4 log ( + 7) log let log ( + 7) t ( 7) t t t log /4 log + 0 log /4 t + log 0 log /4 4 t log 4 t log ( + 7) 4 this gives ±. t 4 t 4 5. log ( ) log ( ) ( ) 6 + 9 7 + 0 0 ( 5) ( ) 0 but 5 zz 6. Let z z z z < z z ( z z ) ( z z z) + z z z z < 0 ( z ) + ( z ) z < 0 ( z ) ( z ) < 0 which is true because of z < < z. 7. () n (y) n n n() n n(y) n (n + ny)... () also n ny n n ny n... () by () n n() n (n + ny) n. n () n n n n (n + n) n n (n) 0 nn n n n 0 8. Let 4 4... t 4 t t 4 t t t + t 4 0 t + t 0 t 4 7 t 8, 66 6 SOLUTIONS (XI) # 7
t, 8 and is rejected so 6 + log / 8 4 6 + log / 9 6 + log / 6 4 PART - II. Let z r e i and w r e i z r e i Given, z r r and arg (z) arg () Then, z r i i e.re r r i( ) e From Eqs. (i) and (ii), we get z. i / e cos i sin z i....(i) i i e.re r. i i ( i) ( i) ( i) ( i) ( i) i i i i (i) (given) (i) (i) 4n, where n is any positive integer. 4n.. Since, z + i w 0 z i w z iw w iz Also, arg(zw) arg( iz ) arg( i) + arg(z) + arg (z) arg( i) arg (z) arg (z). 4 4. Let z i z i i i. 5. Let roots be p + iq and p iq p, q R root lie on line Re(z) p product of roots p + q + q (, (q 0, roots are distinct) Ans. SOLUTIONS (XI) # 8
STRAIGHT LINE Section (A) : EXERCISE # PART - I A(0,0) 0 5 6 0 A-._ (i) centroid, (7, 8) 0 B(5,) C(6,) (ii) Let coordinates of circumcentre is O (, y). Therefore OA OB OC + y ( 5) + (y ) ( 6) + (y ) + y ( 5) + (y ) 0 + 4y 6g ( 5) + (y ) ( 6) + (y ), y 8 0 5 0 6 0 0 (iii), 0 0 (7, 9) 5 0 6 0 0 (iv), 0 0 (7, ) A-4. Let coordinates of P(,y) given PA PB ( ) + (y 4) ( 5) + (y + ) 4 y 4 y...(i) 5 y 4 0 6 + y 6 ± 0 + y ± 0 + y...(ii) + y...(iii) Solving (i) and (ii) we get (7, ) Solving (i) and (iii) we get (, 0) Section (B) : B-. Let equation of line is + my + n 0...(i) given a a, a a, b b, b b and c c, c c are collinear t t, t t is general point which satisfies line (i) t t + m t t + n 0 t + m t + nt (m + n) 0 SOLUTIONS (XI) # 9
m a + b + c n ab + bc + ac m n abc Now LHS abc (ab + bc + ac) + (a + b + c) ( m n) n m + 0 B-5. Let point is P(, y) and A(ae, 0) and B( ae,0) Given PA PB a ( ae) y ( ae) y a Let ( ae) y A, ( ae) y B Hence A B a A B (A + B) (A B) A + B e Hence A a e A a e ( ae) + y (a e) a a y (e ) Section (C) : C-. C-4. Obvious By parametric form Q 4 cos, sin it lies on y 0 + cos sin 0 p(4, ) y 0 + cos squaring both sides 9cos + sin 6 sin cos 8(sin + cos ) cos 6sin cos 7 sin 0 7tan + 6tan 0 tan, 7. Section (D) : Hence required line are + y 5, 7y + 0 sin 0 cos sin D-. foot of perpendicular image y ( 4) ( ) y ( 4) ( ) 9 (, y), 0 0 4 (, y), 5 5 SOLUTIONS (XI) # 40
slope of line perpendicular to the line y 4 is hence its equation y ( ) + y 0 D-5. L : 4 + y 7 0 L : 4 + 7y 0 a a + b b 4 4 + 7 > 0 Hence + sign gives obtuse angle bisector and sign gives acute angle bisector Now, put origin in both 4 0 + 0 7 < 0 4 0 + 7(0) < 0 Hence sign gives that bisector in which origin lies. Hence origin lies in obtuse angle bisector 4 Now, equation of bisector Section (E) : + sign y + 0 sign + y 0 y 7 5 4 7y ± 5 E-4. 0y + y + 5y + 0 This represents pair of straight lines if abc + fgh af bg ch 0 we get Now 0y + y + 5y + (6 y + p) ( y + q) compair both sides p + 6q p q 5 solving both we get p 4, q Hence required lines are 6 y + 4 0 y + 0 y + 0 4 y + 0 solving both equations we get point of intersection Now angle between both lines, 5 tan m m m m tan 7 7 Now equation of pair of angle bisector 5 y 5 y 5 + 4y y + 6 4y + 7 0 E-5. Homogenize + y a by y m + c y m we get + y a c This equation represents pair of lines passing through origin. That will be right angle if coeff. + coeff. y 0 c a ( + m ) Section (F) % F-_. (i) (, 5, 8) (ii) ( 5, 4, ) (iii) (, 0, 7) (iv) (8,, 5) SOLUTIONS (XI) # 4
F-_. (i) (ii) Section (A) : PART - II A-*. AB 4 9 BC 6 6 CD 4 9 AD 6 6 AC 64 65 BD 6 49 65 its rectangle A-4. If H is orthocentre of triangle ABC, then orthocentre of triangle BCH is point A Section (B) : B-*. Since A, B, C are coffe. near Slope of AB Slope of BC A(k, k) B( k, k) C( k 4, 6 k) k k k k k 6 k k k 4 4k k 4k 6 5 0 0k (4k 6)(k ) (4k 6)(k ) + 0(k ) 0 k, SOLUTIONS (XI) # 4
B-. AP (y 4) BP (y 4) AP BP 6 AP BP ± 6 (y 4) (y 4) ± 6 On squaring we get the locus of P 9 7y + 6 0 Section (C) : C-. + y 5... (i) 4... (ii) co - ordinates of G are (4, ) 4...(iii) y y and ¼rFkk½... (iv) solving above equations, we get B & C. C-4. Let coordinates of point P by parametric P( + r cos 45º, + r sin 45º) It satisfies the line y + 9 0 r r + 9 0 r 4 Section (D) : D-. a + a by + 0 origin and (, ) lies on same side. a + ab + > 0 a R D < 0 b 4 < 0 b (, ) but b > 0 b (0, ) 64 5 D-4. p ( 6) 9 60 64 8 4 5 q 64 8 p < q Hence 6y 5 - is acute angle besectory 64+8y+5 0 q(, 4) p B : 6y 5 0 SOLUTIONS (XI) # 4
Section (E) : E-. m + m 0 m m a given m 4m m, m 8, a 6 E-5. Homogenize given curve with given line + 4y 4( + y) + ( + y) 0 + 4y 8 4y + 4 + y + 4y 0 + 4y + y coeff. + coeff. y 0 Hence angle is 90º + y + 4y 4 + 0 Section (F) : F-_. + y + y + z + z + 6 ( + y + z ) 6 y z F-4_. The two numbers are and + (a) > 0 (b) + > 0 > 8 (c) + + < 4 < X < 6 Now must be between 0 < < 6 (, ), (, 5) F-6_. F-8_. Let the third PH reading is 7.48 8.4 7.4 < < 8.. < 5.90 + < 4.6 6. < < 8.7 PH range should be in between 6. to 8.7 Standard result. EXERCISE # PART - I. A, S, B are collinear 0 0 0 SOLUTIONS (XI) # 44
SOLUTIONS (XI) # 45... () B, R, C are collinear 0 0 + 0... () Solving (i) and (ii) we get 4 9 Hence 0, 4 9 Q, 0, P, 4, 4 9 S, 4, R 6. (i) D is mid point of BC Hence co-ordinates of D are y y, Therefore, equation of the median AD is y y y y 0 Applying R R y y y y 0 y y y + y y y 0 (using the addition property of determinats) (ii) Let P(, y) be any point on the line parallel to BC Area of ABP Area of ACP y y y y y y y y y y y y 0 This gives the equation of line AP. (iii) Let AD be the internal bisector of angle A, DC BD CA BA b c D b c by cy, b c b c
Let P(,y) be any point on AD then P,A,D are collinear y y c b cy by 0 b c b c R (b + c) R c b cy y y by b c 0 y c y cy + c y y 0 (Addtion property) b by b c y y y + b y This is the equation of AD. 9. equation of line L is 5 y. ( ) y 0 y or y 0 or 4 y 0 equation of line L is y 5 ( ) or y + 0 Point C is mirror image of point A w.r.t line L ( ) 4 y () ( 8 6 ) 0 C(4, 0) similarly B is mirror image of A in line L 0 ( ) (y ) ( ) B(, ) D(, ) ; E (0, ) median through B is 5 / (y + ) ( ) 5 + y 8 median through C is (y ) ( 0) 4 + 4y 4. a + b c... (i) Let L is (, y ) L is foot of perpendicular from point P(a, b) on line AB equation of AB is b + ay ab 0 a b y b a (ab ab ab) a b SOLUTIONS (XI) # 46
a b y b a ab c ab a(c b ) a a /c c c a c... (ii) similarly b c y... (iii) using these relations (ii) & (iii) in equation (i), we get required locus. 4. Given pair of lines are a +h(a + b)y + b y 0 a +hy + by 0 Equation of pair of bisectors of first pair is a y b y ha b y y a b h Which is also bisector of second pair. Hence both pair are equally inclined. 5. Let equation of chords h + ky By homogenisation y (h + ky) + 4y (h + ky) 0 it makes 90º. Hence coeff. + coeff. y 0 h + 4k 0 h k Hence all chords are concurrent at (, ) Similarly homogenize y + 4y 0 + y (h + ky) + 4y (h + ky) 0 again coeff. + coeff. y 0 h k + h + 4k 0 h k Hence, all chords passes though,.. here tan 5 5 required line y PART - II 5 tan 5 5 4. p 0 0 a 5 a 5 tan 45º p p Hence area ()(p) p p a/5 SOLUTIONS (XI) # 47
8. Image of A(, 0) in + y 6 0 y 0 y 0 A' ( 5, 6) 6 0 6 4 6 Equation of A'B is y ( 4) 5 4 0. By geometry a + b (a + b) By section formula h a b y ( 4) y 9 + 4 + y 0...(i) n(a b) a k a b Put value of and in (i) k(a b) b a h (a b) h k + a b + b k (a b) (a + b) Locus of (opchim) is a. py y 0 y + b pair of angle bisector of this pair y ( ) y p y + p y 0 compare this bisector pair with qy y 0 4. y 6 0 y + 0 + 4y 0 Hence [, ] [, ] q pq. p SOLUTIONS (XI) # 48
6. Both A & B are same side of line y 9 0 Now, perimeter of APB will be least when points A, P, B were collinear. Let B' is image of B Then 0 y 4 0 9 ( ) 84 74 B', 74 Now equation of AB' is y ( + ) 0 point of intersection of given line & Q is P 7,. 7 7 EXERCISE #. (A) Slope of such line is ± y (B) (C) area of OAB 4 6 sq. units To represent pair of straight lines c B ( 4,0) O A (0, ) 0 c (D) Lines represented by given equation are + y + a 0 and + y 9a 0 distance between these parallel lines is 0a 5 a Comprehension # (5, 6, 7) Slopes of the lines + 4y 5 is m 4 and 4 y 5 is m 4 m m given lines are perpendicular and A Now required equation of BC is m tan( / 4) (y ) ( )...() mtan( / 4) where m slope of AB 4 equation of BC is (on solving ()) 7y + 0 and 7 + y 9 0 L 7y + 0 L 7 + y 9 0 5. c + f 4 6. Equation of a straight line through (, ) and inclined at an angle of (/) with y-ais ((/6) with -ais) is cos( / 6) y sin( / 6) y Points at a distance c + f 4 units from point P are ( + 4 cos (/6), + 4 sin (/6)) ( +, 5) SOLUTIONS (XI) # 49
and ( 4 cos (/6), 4 sin (/6)) (, ) only (A) is true out of given options 7. Let required line be + y a which is at b a 5 4 4 4 units from origin required line is + y 4 6 0 (since intercepts are on positive aes only) 8. a + b y + cy + dy 0 since this is homogeneous pair represent there straight lines passing through origin a + b y + cy + dy (y m )(y m )(y m ) or put y m in given equation we get m d + cm + bm + a 0 m + m + m c d m m + m m + m m a m m m d b d given two lines + hence m m m a/d eliminate m from remaining equation 0. S is standard result. equation of angle bisectors of lines given in S are 4y 4 y ± y 0 and 7 + 7y 4 0 5 5 4._ Let R(5, ) divides line segment joining P(,0) and Q(6, ) in : 6 5 Hence Harmonic conjugate divides in : eternaly 8 6 0 Hence required part is, (8, -8) 9. Required point is foot of perpendicular from (0, 0) on the given line which is EXERCISE # 4 PART - I. A(,y ), B(, y ), C(, y ),, and y, y, y are in G.P. of common ratio r. r, r, y y r, y y r 0 0 4 () 5 Area of triangle ABC y r y r r y r 0 A, B & C are collinear.. Let m be the slope of PQ then tan 45 0 m ( ) m( ) m m SOLUTIONS (XI) # 50
m ± m m + m or + m m + m or m PR makes 45 0 with PQ equation of PQ y ( ) + y 5 0 equation of PR is y ( ) y 5 0 combined equation of PQ and PR ( + y 5) ( y 5) 0 y + 8y 0 0y + 5 0. S is the mid point of Q and R 7 6 S,, slope of PS m / 9 equation of line passing through (, ) and parallel to PS is y + 9 ( ) + 9y + 7 0 4. BC AB AC Hence ABC is an equilateral triangle. In equilateral triangle incentre coincides with centroid. Thus 0, 0 0, 5. p (h, k) be a general point in the first quadrent such that d(p, A) d(p, O) h + k h + k h + k [h and k are positive point (h, k) being in quadrent] If h <, k <, then (h, k) lie in region If h >, k <, then (h, k) lie in region If h >, k >, then (h, k) lie in If h <, k >, then (h, k) lie in IV 5 In region h + k h + k h + k In region h + k h + k k possible not y O IV +y5/ / SOLUTIONS (XI) # 5 I III A(,) II (0,0) (5/,0)
In region h + k h + k 5 0 Not possible In region IV h + k h + k h Set consist line segment + y 5 of finite length In Ist region and in the IV region. 6. c ac a a aby ac ab a ac a y b ay a by c cy b c a cy b a by c c c + bc + cc a (a b c ) (a b c )y (a b c ) ay b by c a b cy c a cy b a by c a y ay b by c a b cy c a b cy c a by as a + b + c c c bc, c c cc a y ay c a cy a b a by R R a y ay c a cy a b a by R R + yr + R a y y 0 c a cy 0 b a by ( + y + ) (a + by + c) Given 0 a + by + c 0 which represent a straight line 7. The -coordinate of intersection of lines + 4y 9 and y m + is For being an integer + 4m should be divisor of 5 i.e.,, 5 or 5 + 4m m (Not integer) 4m + m (Interger) + 4m 5 m (Not an integer) + 4m 5 m (integer) there are two integral value of m 5 4m SOLUTIONS (XI) # 5
8. In parallelogram OABC B(0,) and point A in the point of intersection of y m and y n + m and y m n m n Now area of parallelogram (OAB) m n m n 9. y y + Region is clearly square with vertices at the point (,0), (0,), (,0), (0, ). So, its area. 0. Let XOS and XOT let p(cos, sin ), then TOP let Q be the image of P in OT. Then QOT QOX XOQ Q is image of P in the line whose slope is tan. The line segment QR makes an angle 60º with the positive direction of -ais. hence bisector of angle PQR will make 0º with +ve direction of -ais. Its equation y 0 tan0º ( 0) y y 0 SOLUTIONS (XI) # 5
. as OPA ~ OQC OP OA 9 / OQ OC 4 4 m. The line y m meets the given lines in P, m m and Q m,. Hence equation of L m m is y m m y m m...(i) and that of L is y m m y + m 6 m...(ii) Form (i) and (ii) y y y + 5 0; which is a straight line 4. equation of line y m( 8) where m < 0 P 8, 0 and Q (0, 8m) m Now OP + OQ 8 + 8m m 0 + ( m) + 8( m) 0 + 8 ( m) m 8 5. The number of integral points that lie in the interior of square OABC is 0 0. These points are (, y) where, y,,..., 0. Out of these 400 points 0 lie on the line AC. Out of the remaining eactly half lie in ABC. number of integeral point in the triangle OAC [0 0 0] 90 Alternative Solution There are 9 points that lie in the interior of ABC and on the line, 8 point that lie on the line and so on. Thus, the number of desired points is 9 + 8 + 7 +... + + 0 9 90. SOLUTIONS (XI) # 54
6. Refer Figure Equation of altitude BD is. 4 0 slope of AB is 4 4 slope of OE is /4 Equation of OE is y 4. Lines BD and OE meets at (, /4) 7. The lines given by 8 + 0 are and 6. The lines given by y 4y + 45 0 are y 5 and y 9 Centre of the required circle is the centre of the square. Required centre is 6 5 9, (4, 7). 8. y + y ± (y ) Bisector of above lines are 0, y so Area between 0, y and + y squ. units 9. A line passing through P(h, k) and parallel to -ais is y k the other lines given are y and y + Let ABC be the formed by the points of intersection of the lines (i), (ii) and (iii) be A(k, k), B(, ), C ( k, k) k k Area of ABC 4h k k C C C 0 0 k k k 4h ( k) (k ) 4h (k ) 4h k h, k h k h + k h + locus of (h, k) is y + y + 0. R is centroid hence R 4, SOLUTIONS (XI) # 55
. PR OP RQ OQ PR OP RQ OQ 5 but statement is false Ans. (C). P ( sin ( ), cos ) Q (cos ( ), sin ) R (cos ( + ), sin ( )) 0 <,, < 4 R cos ( ) cos sin ( ) sin R Q. cos + P. sin y R sin cos cos sin y R y Q. cos + y P. sin For P, Q, R to be collinear sin + cos sin 4 non collinear not possible for the given interval 0, 4. ( + p) py + p ( + p) 0...() ( + q) qy + q( + q) 0...() on solving () and (), we get C(pq, ( + p) ( + q)) equation of altitude CM passing through C and perpendicular to AB is pq...() q slope of line () is q slope of altitude BN (as shown in figure) is q q equation of BN is y 0 q q ( + p) q y ( + p)... (4) ( q) Let orthocentre of triangle be H(h, k) which is the point of intersection of () and (4) on solving () and (4), we get pq and y pq h pq and k pq h + k 0 locus of H(h, k) is + y 0 4. Let slope of line L m m ( m( ) ) tan 60º m m taking positive sign, m + m m 0 taking negative sign m + + m 0 m SOLUTIONS (XI) # 56
As L cuts -ais m so L is y + ( ). (h a ) + (k b ) (h a ) + (k b ) h(a a ) + k(b b ) + a b a b 0 compare with (a a ) + (b b ) y + c 0 a b a b c.. h a cos t + b sin t k a sin t b cos t squaring and add. (Locus) ( ) + 9y a + b. py y 0 PART - II pair of angle bisector of this pair y ( ) y p y + p y 0 compare this bisector pair with qy y 0 4. Equation of AC q pq. p sin cos y a sin ( acos ) cos sin y(cos + sin ) + (cos sin ) a(sin cos + sin sin cos + cos ) y(cos + sin ) + (cos sin ) a. h k 5. G, h + (k ) h + k 9 Locus + y 9. 6. Let equation of line is y + a b it passes through (4, ) 4 + a b SOLUTIONS (XI) # 57
sum of intercepts is b, a b, a a + b a b 4 + b b 4b b b b b + b 0 b, + y y +. 7. cy 7y 0 sum of the slopes m + m Product of slopes m m 7 c 7 given m + m 4m m c 7 4 7 c. 8. Pair 6 y + 4cy 0 has its one line + 4y 0 y 4 6 + 4 + + 9c 0 c. 9. a + by + b 0 b ay a 0 6ab 6ab y 9 + 4c 0 4 6 b a a Hence point of intersection (0, /) Line parallel to -ais y /. b 0. a, b, c are in H.P. b a + c a b + c 0 y given line + + 0 a b c Clearly line passes through (, ). 7. Centroid is, SOLUTIONS (XI) # 58
. Pair of lines a + (a + b)y + by 0 Area of sector A r A r + 80º given A A 45º, 5º Angle between lines is (a b) a b 4 a b ab a + b + ab a + b + ab 0. ab. Let equation of line is y. a b By section formula a a 6 b 4 b 8 y + 4 + y 4. 6 8 4. Since (, ) and (a, a ) Both lies same side with respect to both lines a a < 0 a a > 0 a(a ) > 0 a (, 0), a a > 0 a a < 0 a (0, ) Hence after taking intersection a,. 5. AB ( h ) (k ) BC AC ( h ) (k ) AB + BC AC (h ) + (k ) + (h ) + (k ) h h Area of ABC (h ) (k ) (K ) 4 k ± k,. 6. SOLUTIONS (XI) # 59
The line segment QR makes an angle 60º with the positive direction of -ais. hence bisector of angle PQR will make 0º with +ve direction of -ais. Its equation y 0 tan0º ( 0) y y 0 7. Bisector of 0 and y 0 is either y or y If y is Bisector, then m + ( m ) m 0 m + m m 0 m m ±. 8. Slope of PQ k Hence equation of to line PQ 7 ( k) y (k ) Put 0 7 ( k) ( k) y + 4 7 + ( k ) 8 k 6 k ±4. Hence possible answer 4. 9. p(p + ) y + q 0 (p + ) + (p + ) y + q 0 are perpendicular for a common line lines are parallel slopes are equal p(p ) (p (p ) ) p PA 0. PB ( + ) + y 9(( ) + y ) + + + y 9 + 9y 8 + 9 8 + 8y 0 + 8 0 + y 0 + 0 4 5 circumcentre, 0 4.. y + 5 b 8 + 5 b b 5 b 0 y 4 y 0 5 0 Line K has same slope c 4 c 4 4 y distance 7 Hence correct option is () SOLUTIONS (XI) # 60
. AD : DB : 5 OD is angle bisector of angle AOB St : true St. false (obvious) Ans.. + y a a y if a > 0 + y a a y ------------------------------------ ( + a) + a as y a It is in the first quadrant so a 0 a a [, ) If a < 0 + y a a y + ------------------------------- ( + a) a a > 0 a a a < 0...() y a a a a a a > 0 a a a > 0 from () and () a {} 4. h k k + third verte on the line + y 9 + 9 (h) + (k + ) 9 h + k + y 0 a < 0...() a SOLUTIONS (XI) # 6
5. 8 4 C, 5 5 8 4 Line + y k passes C, 5 5 8 4 5 5 k 6 k 6. (y ) m( ) OP m OQ m Area of POQ (OP)(OQ) ( m) m 4 m m m 4 4 m m. Condition for concurrency ADVANCE LEVEL PROBLEMS PART - I a a b b 0 4c c So a, b, c are in H.P. + b a c. (sec sin ) y tan + y sin 0 m m ( m m ) 4mm tan sin sec sin 4 sin. Equation of family of curves passing through intersection of C & C is + 4y y 9 + + ( + y 4y + ) 0...(i) It will give the joint equation of pair of lines passing through origin, if coefficient of 0 & Constant 0 put in equation (i), we get + 4y y + 6 + 9y y 0 It will subtend 90º at origin if coeff. of + coeff. of y 0 9 SOLUTIONS (XI) # 6
4. For B and C apply Parametric form y ± 5 cos sin Points are (7, 5) & (, ) 5 O A 5 (, ) B C 5. From figure it is clear that A is orthocentre of ABC 6. p qy y 0 m tan m + m q, tan (+ ) m tan m m p q p 7. ( + ) + ( ) y + (4 ) 0 distance from point A is ( ) ( ) (4 ) ( ) Hence, the required line is y + 0 ( ) 0 8. To find equations of AB and CD AB and CD are parallel to 4y 0 and at a distance of units from (, ) 4y + k 0 and 4 k 5 k, 9 k 0 equations of two sides of the square which are parallel to 4y 0 are 4y + 0 and 4y 9 0 Now the remaining two sides will be perpendicular to 4y 0 and at a distance of unit from (, ) 4 + y + k 0 and 4 k 5 k + 7 0 k, 7 remaining two sides are 4 + y + 0 and 4 + y 7 0 9. Given cos + y sin a...() sin y cos b...() square () and () then add them. + y a + b SOLUTIONS (XI) # 6
0. Let point of concurrency of given family of lines is Q and it can be obtained by solving + 4y + 6 0 and + y + 0 Q (, 0) Now required line will pass through Q(, 0) and perpendicular to PQ. Equation of required line is y 0 4 ( + ) 4 + y + 8 0. (i) After reflection about line y position of point will be (, 4) (ii) After this step (, 4) (iii) (h + ki) ( + 4i) e i/4 ( + 4i) i h 7, k (h, k) Hence the final position will be 7, (, 4). Let the point P be (, y) + y...(i) Case - > 0, y > 0 Equation (i) will become : + y P(, y) B Area (OAB) 9 O A Similarly for each quadrant, a triangle will be formed. Hence area enclosed will be 8.. P ( 4, ) and Q (, 6) Let slopes of PM and QM be m and m respectively. m and m. Let be the acute angle between PM and QM tan m m m m tan 4 SOLUTIONS (XI) # 64
4. For collinearity of points cos 0 sin 0 sin cos 5. For ABC a + b > c, b + c > a, c + a > b + 4 + > + + 8 > 5 + 5 + > + > + 5 + 8 > + + + 5 > 0 R Common to all is > 5. 6. point of intersection of the two ray is P(0, ) Point A is, 0 or, 0 and PO is bisector of the angle between two rays required point is (0, 0) 7. Slope BD, Equation of BD, + y a + b equation of AC y a A (, y ) D (a, b) On solving, we get O B (a + b, 0) m AC tan a b, b B (, y) O a + b b,, C (, y ) OA OD b Apply parametric form for finding A & C b a b y ± A & C are (a, 0) and (a + b, b) b 8. p q r q r p 0 r p q p + q + r pqr 0 (p + q + r) (p + q + r pq qr rp) 0 SOLUTIONS (XI) # 65
9. k u k v 0... (i) k u + k v 0... (ii) equations of bisectors of the angles formed by lines (i) and (ii) are (ak bk k u k ) v (k b ak ) (k a bk (k u k k u k v (k u + k v)... (iii) (i) by taking positive sign in (iii), we get k u k v k u + k v. k v 0 v 0 (ii) by taking negative sing in (iii), we get u 0 ) v) (k b ak PART - II ). cos y ± sin 6 ( (, ) lie below the line) 6 + cos, y + (, y ) lies on + y 4 6 sin + 6 (sin + cos) 4 sin + cos 6 (sin + cos) sin( + ) sin 60º or sin 0º 4 5, 6 is rejected. tan 4 4 4 4 4 7 5 4 7 cosec 5 7 SOLUTIONS (XI) # 66
P is from AB to CD, P is from AD to BC for finding P choose arbitrary point (a, a) on AB 4a a a a P 5 5 for P choose arbitrary point (a/, 0) on AD 0 a a P 5. Slope of BC is ( ) ( ) a 5 Equation of a line parallel to BC is y + c i.e. + y c 0 its distnace from the origin is Area P P cosec a 7 c c Equations of the lines are + y ± 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 + y + 0. 4. AB 64 6 45 BC 576 6 6 AC 56 69 45 AB AC BC, triangle is isosceles and in isosceles triangle O, H, I, G are collinear 5. D is mid point of AB and lies on the line + y 6 + 6 7 + 0...(), multiplication of slope of AB & line ( ) 0...(), satisfies both () & () A 7-y+ 0 +y- 0 6. AB AC B m m BC C (, -0) SOLUTIONS (XI) # 67
The bisectors are 7 y 5 ± ( y ) Their slopes are, Required lines are y + 0 ( ) and y + 0 ( ) i.e. y 0 and + y + 7 0 7. AB d OAPB is a cyclic quadrilateral and OP will be diameter of the circumcircle of this quadrilateral Let Q be the centre of the circle in AQT sin d y... (i) h ab tan a b from (i) and (ii), we get... (ii) h ab a b d y 4d Locus of P(, y ) is ( + y ) (h ab) d {(a b) + 4h } 8. The slopes of the lines AB, BC and CA are, and 7 respectively 7 Let m 7, m, m 7 m > m > m tangent of internal angles of the triangle are 4 tan A, tan B and tan C 4 4 7 interior angles A and B are acute and interior angle C is obtuse internal bisector of B acute bisector of B + 6y 6 0 Eternal bisector of C acute bisector of C 8 + 8y + 7 0 Internal bisector of A acute bisector of A + 6y 0 B + y 5 0 A 7 + y + 4 0 C + 7y 7 0 9. Equation of line passing through P(, ) making angle with + ve direction of -ais is given by y r cos sin, r, r (parametric form) where r, r, r are distances of points A, Q, B from point P respectively. Hence coordinates of A(r cos, r sin + ) SOLUTIONS (XI) # 68
But A lies on -ais Hence r sin + 0 sin r coordinates of point B (r cos, r sin + ) Point B lies on y-ais hence r cos 0 cos r Coordinates of point Q (r cos, r sin + ) Hence h r cos cos and k r sin + sin Now given that r, r, r are in H.P. r r + r h r k r sin r + cos r k r + ( h ) r (k ) + (h + ) 4 k + + h + h k locus y Alt : Use P and Q are harmonic conjugates with respect to A and B. 0. Let P(h, k) be a variable point on the lines passing through the origin. k hy h k (k hy ) (h + k ) locus of P(h, k) is ( y y ) ( + y ) solving it, we get (y ) y y + ( ) y 0.. Let the line (L) through the origin is r cos y r sin as L intersects L at Q and OQ r r sin m r cos + c...() similarly, L intersects L at R and OR r r sin m r cos + c...() Let P (h, k) & OP r r r r...() & h r cos...(4) k r sin...(5) putting the values of r and r from () and () in () r c c (sin m cos ). (sin m cos )...(6) putting the value of cos and sinfrom (4) and (5) in (6), we get r k r m c c h k m r r h r replacing (h, k) by (, y) we get the desired locus as (y m ) ( y m ) c c (k m h) (k m h) c c SOLUTIONS (XI) # 69
. take any point on line + y + 4 0 put 0, we get y Now image of (0, ) in line + y + 0 0 y 0 6 4 9 0 (,) 0, 0 0 4 Hence and y Point of intersection of + y + 0 and + y + 4 0 is (, ) 4 4 / Hence equation of other line y 0 / Section (A) : A-. After simplification, we get 9 + 46y 8 Since BD is diameter of circle Hence ( a) ( 0) + (y 0) (y a) 0 + y a ( + y) CIRCLE 0 EXERCISE # PART - I 4 + y + 0 (0, ) + y + 4 0 A-6. + sin + sin y 4 + cos y 4 cos Squarring and add ( + ) + (y 4) 4 Section (B) : B-4. S (9) + (0) 6 65 > 0 Since (9, 0) lies outside the circle. Hence two real tangents can be drawn. Now S + y 6 S 9 6 Hence pair of tangents SS T ( + y 6) (65) (9 6) 65 + 65y 040 8 + 56 88 6 65y 88 + 96 0 h ab Angle between these tangents (a b) 0 6 65 6 65 8 65 49 B-5. given f g 6 f g g f g + f + g + f + 6 0 g + f + 4g + 4f + 0 Section (C) : RL C-4. Area of triangle formed by pair of tangents & chord of contact is R L Here R a Hence Area L h k a h a h k a k / SOLUTIONS (XI) # 70
C-7. T S y +( ) +4 (y ) + 9 4 + 9 4 + 9 + y + 5 0 Section (D) : D-. S : + y 6y + 9 0 C (, ), r S : + y + 6 y + 0 C (, ), r C C 6 4 0 n + r 4 Hence C C > r + r Both circles are non-intersecting. Hence there are four common tangents. Transverse common tangents : 9 5 coordinate of P, 0, Let slope of these tangents is m y 5 m( 0) m y + 5 0 5 m Now m m m m + m + m m, other tangents is vertical 4 4 Equation of tangents 0 5 y + 0 4y + 0 0 + y 0 4 Direct common tangents 9 coordinate of Q, Q(, 4) Hence equations y 4 m( ) m y + (4 m) 0 m 4 m m Hence equation y 4 0( ) y 4 m m y 4 4 ( ) 4 y 0 + 4m 4m + m m 4m 0 m 0, 4 D-. Equation of circle passing through origin is + y + g + fy 0 This circle cuts the circle + y 4 + 6y + 0 0 orthogonally g( ) + f() 0 + 0 g + f 5 0...() & + y + y + 6 0 also g(0) + f(6) 6 + 0 f SOLUTIONS (XI) # 7