Mathematical Logic. Target. Chapter 01: Mathematical Logic

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1 Trget Publictions Pvt. Ltd. Mthemticl Logic Chpter : Mthemticl Logic. + is n open sentence. It is not sttement. option (C) is correct.. Since p q is flse, when p is true nd q is flse. p (q r) is flse, p is true nd q r is flse p is true nd both q nd r re flse.. Since, contrpositive of p q is ~q ~p. contrpositive of (~p q) ~r is ~(~r) ~(~p q) r (p ~q). ~p: Rohit is short. The given sttement cn be written smbolicll s p (~p q). 5. Let p: is comple number q: is negtive number Logicl sttement is p q converse of p q is q p option (B) is correct. 6. Consider option (C) p q r ~q p ~q (p ~q) r T T T F F T T T F F F T T F T T T T T F F T T F F T T F F T F T F F F T F F T T F T F F F T F T (p ~q) r is contingenc option (C) is correct. 7. Consider option (A) p q p q p q ~(p q) (p q) (p q)) T T T T F F T F F T F F F T F T F F F F F F T F (p q) (~(p q)) is contrdiction. option (A) is correct p q ~q p q p ~q ~(p ~q) T T F T F T T F T F T F F T F F T F F F T T F T The entries in the columns nd 6 re identicl. ~(p ~q) p q sttement-l is true. Also, ll the entries in the lst column of the bove truth tble re not T. ~(p ~q) is not tutolog. sttement- is flse. option (B) is correct. 9. Consider option (C) (p q) (p r) (T T) (T F) T T T option (C) is correct.. The sttement Sumn is brillint nd dishonest iff sumn is rich cn be epressed s Q (P ~R) The negtion of this sttement is ~(Q (P ~R)). (q) (p) is contrpositive of p q. p q (~q) (~p) option (D) is true.. (~p ~q) (p q) (~p q) ~p (~q q) (p q) (~p T) (p q) ~p (p q) (~p p) (~p q) T (~p q) ~p q option (B) is correct.. Since, inverse of p q is ~p ~q. inverse of (p ~q) r is ~(p ~q) ~r i.e., ~p q ~r

2 Trget Publictions Pvt. Ltd. Mtrices Chpter : Mtrices. A A eists. () (7) + (9) 7 9 dj A 5 6 A sum of the elements of A T ( ). (dj A) A A I n A. Let A A A does not eist. tn tn sec sec tn tn tn tn T 6. tn tn tn sec tn tn tn sec tn tn cos sin sin cos B equlit of mtrices, we get cos, b sin z R R R, R R R z 5z 5 z + + z 6 7. Let A Mtri will not be invertible if A () + (9) () 9 8. Given, A nd B AB A B nd A B A B A B... A A AB nd A B re singulr.

3 Std. XII : Triumph Mths 9. (AB) B A B A. (A 8A)A A.A.A 8A.A A 8I Since, A(dj A) A.I Replcing A b dj A, we get dj A (dj(dj A)) dj A I A. A (dj(dj A)) dj A I. A (dja) A A (dj (dj A)) A.I.[ dj A A n ] A (dj (dj A)) I A (dj (dj A)) I Given, A (dj (dj A)) ki k. det A det (dj (dj A)) ( ) (det A). dj(dja) A (det A) () (n ). A. (dj A).I Since, A(dj A) A.I A From (i), A. dj A 6.(i) dj A 6 6 Also, dj (dj A) A (n ) dj(dja) dja ( ) A ()

4 Trget Publictions Pvt. Ltd. Trigonometric Functions Chpter : Trigonometric Functions. sin sin sin sin (sin sin ) cos cos cos cos (cos cos ) Given, cos cos + sin sin cos (cos + cos ) + sin ( sin sin ) cos cos cos +sin sin sin + cos sin + (cos cos + sin sin ) cos 6 + cos cos cos + cos cos cos (n + ) (n + ) But cos > ( must be in st or th Qudrnt) the possible vlues re,. 8 8 Cse II: If cos <, sin (cos ) sin 5, 7 5 7, 8 8 The vlues of stisfing the given eqution 5 7 between nd re,,, These re in A.P. with common difference.. sin cos 6 6 sin sin sin Let t 6 6 sin 6 sin. sin 8cos sin cos. 8 sin cos Cse I: If cos >, sin cos sin sin 9,,,.[ (, ), (, )] 9,,, t + 6 t t + 6 t t t + 6 (t ) (t 8) t or t 8 sin 6 sin or sin 6 8 sin or sin or sin sin or sin sin sin 6 or sin sin n ± or n ± 6 7 5,,, or,,, There re 8 solutions in [, ].

5 Std. XII : Triumph Mths. The mimum vlue of sin + b cos is b. Mimum vlue of sin + cos is nd the mimum vlue of + sin is. The given eqution will be true onl when sin + cos nd + sin If sin + cos cos + sin cos cos + sin sin cos n, n +.(i) + sin sin sin sin n + () n. n + () n..(ii) The vlue of [, ] which stisfies both (i) nd (ii) is. 5. sin + cos sin cos (sin + cos ) sin cos sin cos ( sin cos ).sin cos sin sin sin + sin (sin + ) (sin ) sin.[sin ] sin sin n + () n n + () n 5 The vlue of in [, ] re nd. There re solutions. 6. tn sec + tn ( + tn ) + tn tn + tn tn + + (tn ) 7. cos + sin 5 tn tn + tn 5 tn Let tn t t + 8t 5 + 5t 8t 8t + t t + (t ) t tn tn n + n + 8. tn + tn + tn tn tn + tn + tn tn tn ( tn ) (tn )( tn ) (tn ) ( tn ) tn 9 tn tn tn tntn tn tn tn tn n + (n + )

6 Chpter : Trigonometric Functions tn A > tn B A B A > B A B tn A B tn tn A B tn tn 5 tn A B C 85 tn C tn 5 cot C 5 tn C 5 Since, 7 > 5 tn B tn C B C B > C A > B > C > b > c. A(ABC) 9 bcsina 9 bc 9 bc 8 b c cos A bc. sin A sin. cos (b c) bc bc ( ) cm B Let B, C 5 A 5 sina sinb sinc b c sin5 sin sin 5 b c b c sin sin5 sin5 sin5 sin5 sin5 A(ABC) bc sin A sin5 sin5 sin5 5 A sin (65 ) 5 C

7 Std. XII : Triumph Mths. Let b c c b ( b c) k 6 b c 8.(B propert of equl rtio) b + c k, c + k, + b k, + b + c 8 k 7k, b 6k, c 5k b c cos A bc 6k 5k 9k (6k)(5k) k 6k 5 cos A 5. A n n (n + ) (n ) n or n But n cnnot be negtive. n The sides of the re, 5, 6.. E A O r 7 r D C In ODC, OD OC r, DOC A(ODC) r.r. sin 7 r sin 7 B n + n A Are of pentgon 5 r sin 7 B n + C Let AC n, AB n +, BC n + Lrgest ngle is A nd smllest ngle is B. A B Since, A + B + C 8 B + C 8 C 8 B sin C sin(8 B) sin B sin A sin B sin C n n n sin B sin B sinb n n n sinbcosb sin B sinbsin B n n n cosb n n sin B n cos B n, sin B n n n ( cos B) n n n + n n n n n + n n n n + n + n + n + n A Are of circle r A A r 5 r sin7 5cos8 sec 8 sec Let k, b 5k, c 6k Now, s b c k 5k 6k 5k s(s )(sb)(s c) 5k 5k 5k 5k k 5k 6k 5k 7k 5k k 5 7 k B sine Rule, sin A R sin A R bcsina bc R bc R R bc k.5k.6k 8 5 7k 7 k Also rs, where r Rdius of incircle of ABC

8 Chpter : Trigonometric Functions 5 7 k r 7 k s 5k R 8 r 7 k 7k 6 7 R 6 r 7 b c 6. cos A bc cos [ ] bcsina sin s b c rs r s ( ) b + c (b + c ) + b + c b c + b + c b + b c c b c b c b + c bc b c cos A bc A B C 5 8 bc bc bc A p Let length of ltitude p Since, A + B + C 5 A A 8 8 Are of p bc sin A p bc sin p bc bc p B sine rule, sin b sin 8 c 5 sin 8.(i) sin b 8 sin 8 5 sin c 8 sin 5 8 From (i), 5 sin. sin p 8 8 P 5 sin sin sin sin cos cos cos cos 5

9 Std. XII : Triumph Mths 9. tn A nd tn B re the roots of the qudrtic eqution tn A + tn B 5 6, tn A. tn B 6 A B 5 tn A tn tn 6 A B tn tn 6 tn A B 6 A B A + B C ABC is right ngled tringle. csin B. r s ( b c) r c bc [ sin B sin 9 ] c cb c b cb c( c b) c( c b) ( c) b c cb c b Dimeter + c b. A 55, B 5, C sin 55 b sin5.[ + c b ] c sin k k sin 55, b k sin 5, c k sin c k sin k sin 55 k (sin + sin 55) (sin sin 55) k sin cos sin cos k sin 65 sin 55 k sin 5 sin 55 (k sin 55) (k sin 5) b. A, B, C re in A.P. A + C B Also, A + B + C 8 B 6 sin A sin B sin C k b c sin A k, sin B bk, sin C ck c sin C + c sin A c ( sin C cos C) + c (sin AcosA) c ( ck cos C) + c (k cos A) k cos C + kc cos A k( cos C + c cos A) kb.[ b cos C + c cos A] sin B.[ B 6]. cot tn tn + tn tn tn 9 tn 6 8 tn cot cot tn tn tn tn tn tn tn tn 7

10 Chpter : Trigonometric Functions. Let cos b cos b cos b tn cos b + tn cos b tn + tn tn tn tn tn tn tn tn tn tn tn cos b tn b 5. cos cos cos Given, cos cos cos cos cos cos Squring on both sides, we get ( ) cos cos + + cos cos + + cos cos + cos sin 6. sin + sin sin sin sin sin sin cos (sin ) cos sin (sin ) cos (sin ). (i) Let sin sin cos cos (sin ).(ii) From (i) nd (ii), we get 5 5 (squring both sides) (From the given reltion it cn be seen tht is positive) 7. L.H.S. sin sin 7 + cos cos 7 + tn tn cot cot 8 sin sin cos cos 7 + tn tn 8 + cot cot 8 sin sin cos cos tn tn 8 + cot cot [ cos () cos ]

11 Std. XII : Triumph Mths 7 b, b 7 + b sin sin + 6 sin 65 sin sin 6 65 sin sin 65 8 sin sin 65 sin sin 65 cos 6 + sin cos sin >.[.7] tn ( ) > tn.[ tn is n incresing function] tn ( ) > A >.(i) sin sin sin sin ( sin sin ) Put sin sin sin sin sin 7 sin 7 sin (.85).7.866,.85 <.866 sin (.85) < sin (.866).[ sin is lso n incresing function] sin < sin sin <...(ii) sin 5 sin (.6) < sin sin 5 <.(iii) From (ii) nd (iii), we get B sin + sin 5 < + B <.(iv) From (i) nd (iv), A > B. cot + cot + cot z tn + tn + tn z tn + tn + tn z tn (tn + tn + tn z) tn Let A tn, B tn, C tn z tn A Btn C tn (A + B + C) tn(ab)tnc tn A + tn B tn C tn A tn B tn A tn B tnc tna tnb tn A tn Btn C tn A tn Btn C tn A tn B tn B tn C tn C tn A

12 Chpter : Trigonometric Functions tn (A + B + C) tn A + tn B + tn C tn A tn B tn C tn (tn ) + tn(tn ) + tn(tn z) tn(tn ) tn(tn ) tn(tn z) + + z z. cos 9 9 cos sin cos 9 9 cos cos sin sin cos 9 cos cos 58 cos cos cos cos cos cos 7 cos nd 7 7 Principl vlue is 7. (6 + ) (8 + 6) ( 7 6) ( ) ( + ),, But >,. cot + sin 5 tn + 5 tn 5. sin tn tn tn tn + tn. tn + tn + tn.[ > ] + tn () tn tn + tn + tn. tn + tn tn tn tn 6 8 9

13 Trget Publictions Pvt. Ltd. Pir of Stright Lines Chpter : Pir of Stright Lines. L : + h + b Eqution of n line pssing through origin nd perpendiculr to L is given b b h +.(interchnging coefficients of nd nd chnge of sign for term) The required eqution of pir of lines is i.e h Here, mm b...(i) nd mm b (m m ) (m + m ) m m h b b h h.[ b h (given)] b h b h mm b...(ii) On solving (i) nd (ii), we get h h m nd m b b m : m :. The lines re prllel, if f bg f 9g f g Let g nd f bc + fgh f bg ch (9) (c) + () () (6) () 9() c (6) c is n number.. Given eqution is., b, c, f, g, h This eqution represents pir of stright lines, if bc + fgh f bg ch The given eqution of pir of lines is +, b, h Now, + b + () The lines re perpendiculr 6. The joint eqution of the lines through the point (, ) nd t right ngles to the lines + h + b is b( ) h( )( ) + ( ) joint eqution of pir of lines drwn through (, ) nd perpendiculr to the pir of lines 7 + is ( ) + 7( )( ) + ( ) 7. The given equtions re nd + 6 The joint eqution is given b ( ) ( + 6) Let the eqution of one of the ngle bisector of the co-ordinte es be + m Given eqution of pir of lines is + h + A, H h, B Now, m m b m Also m + m h b h h 5 9. The given eqution of pir of lines is + + 5, b, c, f 5, g, h Now bc + fgh f bg ch 5 75

14 Std. XII : Triumph Mths 5 5 ( 5)( + 5) 5 5 or. Let m be the common line nd let m nd m be the other lines given b + + nd + b respectivel. Then, m + m, mm, nd m + m b, mm (mm ) (mm ) m (m m ) 9 m 9.[ m m (given)] m When m, mm nd mm m nd m m + m nd m + m b 5 nd b When m, mm nd mm m nd m m + m nd m + m b 5 nd b. Given eqution of pir of lines is 8 +, h, b tn tn 57 6 tn ( ) 6

15 5 Vectors Chpter 5: Vectors. Since, b nd b c re colliner with c nd respectivel b tc (i) b c s (ii) From (i) nd (ii), we get c tcs (s) c( t) But nd c re non-colliner + s, + t s, t Substituting vlue of t in (i) nd vlue of s in (ii), we get b c nd b c Hence, bc.. Given, r r r r b c ( ) ( )b ( )c,, 7,, +. Since, the given vectors re coplnr c c c b Appling C C C, c c b ( b) + c (c) c b Hence, c is the geometric men of nd b.. ˆ ˆ, bˆb ˆ, cˆc ˆ, ˆ ˆ ˆ bbcˆcˆ ˆ ˆ ˆ ˆbˆ ˆcˆ bc bˆ ˆ bˆbˆ bˆcˆ cˆ ˆ cˆbˆ cˆcˆ 5. Since, bc cubic units b c Appling R R R nd R R R, we get b c (b)(c) ()(c) ()(b) Dividing b ( )( b)( c), we get b c Consider, b c.(i).[from (i)] 6. Volume of the prllelopiped formed b vectors is i.e., V +

16 Std. XII : Triumph Mths dv d + dv, 6 d For m. or min. of V, dv d dv 6 > for d V is minimum for 7. Given,.b b.c c. The sclr triple product of three vectors is [bc] ( b).c.b b ngle between nd b is 9 Similrl, [ bc] b n.c ˆ where ˆn is norml vector. ˆn nd c re prllel to ech other [bc] b n. c ˆ 8. Given, r b c b r c b r c is prllel to b r c b for some sclr r c b r. c. + b c. + b b c..(i). r(given).c.b Substituting the vlue of in (i), we get.c r c b.b.c r.b c.b (b.b).b ( ) r.b 9 9. Let ci ˆj ˆ kˆ cc b cb c bc b c Let b c b c 9. 9 b (i ˆj ˆ k) ˆ b. 7i ˆj ˆk ˆ ( 6 ) Now,. Given, mb nc b mc n c m nb l l l. mb nc n b mc m nb c l l l l m n n l m bc m n l l m n n l m... bc m n l l + m + n lmn (l + m + n) (l + m + n lm mn nl) l + m + n B P M D O A H Let point O be the circumcentre of ABC. Let, b, c, p, d, h, m be the position vectors of the respective points. Since, h + b + c.(stndrd formul) C

17 m p h p b c DM m d p b c b c DM PA p p p p.[ O is circumcentre, OA OP i.e., p] DM is perpendiculr to PA. 7. Chpter 5: Vectors m n m n mz nz D,, mn mn mn ( 9) (5) (6) () ( ) (),, ,, ,, ,, A(,, z ) (l,, ) (,, n) 5. Let position vector of Q be r Since, p divides PQ in the rtio : 6. r (p q) p 7 p r+ p+ q 5 p q r r 5p q B distnce formul, AB (5 ) ( ) ( ) 9 A(,, ) B(5,, ) D C ( 9, 6, ) B(,, z ) (, m, ) C(,, z ) + l, +, + On solving we get l, l, l +, + m, + On solving we get m, m, m z + z, z + z, z + z n On solving we get z n, z n, z n A(l, m, n), B(l, m, n), C(l, m, n) B distnce formul, AB (l l) + (m m) +(n + n) m + n BC (l + l) + (m m) +(n n) l + n CA (l + l) + (m m) +(n n) l + m AB BC CA l m n m n l n l m l m n l m n 8 8 l m n 8. A(,, ) AC ( 9) ( 6) ( ) Point D divides seg BC in the rtio of : B section formul, B(, 7, ) D C(, 5, ) Let D be the foot of perpendiculr nd let it divide BC in the rtio : internll

18 Std. XII : Triumph Mths 57 D,, AD d 5 7 ˆ i ˆ j kˆ ˆi kˆ 5 7 ˆ i ˆ j kˆ i ˆ 5j ˆ k ˆ i ˆ 7 ˆ j k ˆ BC ˆij ˆ kˆ Since, AD BC. AD. BC ( ) 5 7 ( ) () D,, ,, ,,

19 6 Chpter 6: Three Dimensionl Geometr Three Dimensionl Geometr., Since, cos + cos + cos Since, AD is equll inclined to the es p 5 r p 7, r cos + cos + cos cos + cos cos + cos 6 7 cos () 8 If is cute, then cos is positive. cos 7 8. The d.r.s of AB re,, 6 5 i.e.,, Let, b, c,, d.r.s. of BC re, 5, 5 i.e.,,, Let, b, c,, + b b + c c () + ()() + () + + AB nd BC re perpendiculr. mabc 9. cos l l + m m + n n sin cos. cos l m n l m n (l l + m m + n n ) ll lmlnl mmmmn +ln mn nn ll mm nn ll mmmmnnnn ll l m ll mm l m mn mmnn mn ln ll nn l n (l m l m ) + (m n m n ) + (n l n l ). Given, A(,, 7), B(,, ), C(p, 5, r) Let D be the midpoint of BC. p 5 r D,, p r,, d.r.s. of AD re p r,, 7 i.e., p 5 r,, 5. The given equtions re 6mn nl + 5lm, nd l + m + 5n.(i) m l 5n.(ii) Substituting vlue of m in eqution (i), we get 6(l 5n)n nl + 5l( l 5n) 8ln n nl 5l 5nl 5l + 5ln + n l + ln + n (l + n)(l + n) l n or l n If l n, then m n l n nd m n l m n d.r.s. of the st line re,,. If l n, then m n l n nd m n

20 Std. XII : Triumph Mths l m n d.r.s. of the nd line re,,. cos cos 6 ( ) ( ) ( ) ( ) cos l m n cos 6. Since, (l m) l lm + m l + m lm.(i) Similrl, m + n mn.(ii) nd n + l nl.(iii) Adding (i), (ii) nd (iii), we get (l + m + n ) (lm + mn + nl) lm + mn + nl The mimum vlue of lm + mn + nl is. 7. Let A (,, ), B (, b, 7) nd C (,, 5) d.r.s of AB re, b, d.r.s of BC re 6, b, Since the points re colliner b 6 b, b 8. Let the d.r.s of the line perpendiculr to both the lines be, b, c. d.r.s of lines is,, nd,, b.(i) b + c.(ii) On solving (i) nd (ii), we get b c d.r.s of the line re,, the required d.c.s re,, 9. Since cos + cos + cos cos + cos + cos.[ ] cos

21 Trget Publictions Pvt. Ltd. 7 Line Chpter 7: Line z 7. Let r 5 r, 5r +, z r + 7 Co-ordintes of n point on the line re (r, 5r +, r + 7). This point lies on the curve 5 + z ( r ) 5(5r + ) + (r + 7) r + r + 5r 6r 7 + r + 8r + 9 r 55r 66 r + 5r + 6 (r + )(r + ) r or r If r, then the point is (,, ) nd if r, then the point is (,, ) option (A) is correct.. The given eqution of line is + 5, z 6. It cn be written s 5 z 6 r, s co-ordintes of the n point on the line re (r + 5, r, r 6). This point is t distnce of 6 from the point (5,, 6) (r + 5 5) + (r ) + (r 6 + 6) 6 6r + r + 9r 6r r 9 r If r, then the point is ( + 5,, 6) (7,, ). Let the components of the line vector be, b, c. + b + c (6).(i) Also, b c k, s 6 k, b k, c 6k Substituting vlue of, b nd c in eqution (i), we get 9k + k + 6k 6 9k 6 6 k k 9 Since, the line mkes obtuse ngle with X-is component long X-is is negtive. k 9 The components of the line vector re k, k, 6k i.e., 7, 8, 5. Let M be the foot of the perpendiculr drwn from the point P(,, ) to the given line. z Let, +, z + M (, +, + ) d.r.s. of PM re, +, 8 Since, PM is perpendiculr to the given line ( )() + ( + )() + ( 8)() M (, 5, 7) length of perpendiculr (PM) ( ) ( 5) ( 7) When squre is folded co-ordintes will be D(,, ), C(,, ), A(,, ), B(,, ). X A Y D B Y z Eqution AB is, z nd eqution of DC is C X

22 Std. XII : Triumph Mths shortest distnce ( ) ( ) ( ) ( ) ( ) 6. Given eqution of motion of rocket is t, t, z t z i.e., the eqution of the pth is z i.e., Thus, the pth of the rocket represents stright line pssing through the origin. For t sec. we hve,,, z Let M(,, ) OM z km Rocket will be t 6 km from the strting point O(,, ) in seconds. 7. d.r.s. of L re,, nd d.r.s. of L re,, ˆi ˆj kˆ vector perpendiculr to L nd L ˆ i() ˆ j(9 ) k(6 ˆ ) ˆi7j ˆ 5kˆ ˆi7j ˆ5kˆ ˆi7j ˆ5kˆ unit vector Let S be the foot of perpendiculr drwn from P(,, ) to the join of points A(, 7, ) nd B(, 5, ) P (,, ) Let S divide AB in the rtio : 57 S,,.(i) Now, d.r.s. of PS re, 5 7, i.e., 7, i.e., +, 5 + 7, Also, d.r.s. of AB re,, Since, PS AB ( + )() + (5 + 7)() + ()() 7 Substituting the vlue of in (i), we get S,, 9. Eqution of the line pssing through the points (5,, ) nd (, b, ) is b z 5 b.(i) 7 The line psses through the point,, 7 b b.[from (i)] nd + b 7 b 5b b 6, b A(,7,) S B(,5,)

23 Trget Publictions Pvt. Ltd. 8 Plne Chpter 8: Plne. Given plnes re c bz.(i) c + z.(ii) b + z.(iii) Eqution of plne pssing through the line of intersection of plnes (i) nd (ii) is c bz + k(c + z) ( + ck) (c + k) (b k)z.(iv) Now, plnes (iii) nd (iv) re sme for some vlue of k, ck c k (b k) b ck c k b + ck bc bk k(b + c) ( + bc) bc k b c c k Also, b k bc c bc b c b b c bcc bc b + bc + + bc c + b + bc + b + c + bc. Let, b, c be the intercepts form b the plne on co-ordinte es. Since, b c b c The point (,, ) stisfies the eqution of the z plne. b c the required point is (,, ).. Given eution of line nd plne re ˆiˆj i ˆˆj kˆ, nd r r. ˆi j ˆkˆ b i ˆˆj kˆ nd n ˆij ˆ kˆ Consider b n i ˆ ˆ jk ˆ ˆ i j ˆ k ˆ + the line lies in the plne.. The eqution of the given line is + t, + t, z t z The given line psses through the point,, nd it s d. r.s re,, The eqution of the given plne is + + 6z d.r.s of the norml to the plne re,, 6 b cz d p b c () () , 5 5(9) 5

24 Std. XII : Triumph Mths 5. Let be the vector long the line of intersection of the plnes 7 5z nd 8 + z. the d.r.s of the normls to the plnes re, 7, 5 nd 8,,. ˆi ˆj kˆ ˆ i( 55) ˆ j(6 ) k( ˆ 56) 69i ˆ 6ˆj kˆ Similrl, let b the vector long the line of intersection of the plnes 5 + z + nd 8 + z the d.r.s of the normls to the plnes re 5,, nd 8,, ˆi ˆj kˆ b 5 8 ˆ i( 6 ) ˆ j( ) k( ˆ 55 ) 7i ˆj ˆ 9kˆ Consider, 69i ˆ 6j ˆ k ˆ. 7i ˆ j ˆ 9k ˆ.b (6) ndbre perpendiculr 9 sin sin 9 6. The eqution of the given plne is ( + ) + z + z ( ) ( z ) 7. ( ) ( z) The plne psses through the point of intersection of the plnes nd z A(,, ) M B z 9 Let A (,, ), AM be to the given plne nd let B (,, z) be the imge of A in the Plne. the d.r.s. of the norml to the plne re,, The eqution of the line AM is z k, s k +, k, z k + Let M (k +, k, k + ) eqution of plne becomes (k + ) (k ) ( k + ) 9 k 7 6 M,, ,, Since, M is the mid point of AB. 7, z, , Imge of A is 5, z B,, Since, nd b re coplnr, b is vector perpendiculr to the plne contining nd b. Similrl, c d is vector perpendiculr to the plne contining c nd d. The two plnes will be prllel, if their normls b nd c d re prllel. bcd 9. Eqution of the plne contining the given lines is z 5 ( ) (5 6) ( ) ( ) + (z ) (8 9) ( ) () ( ) () + (z ) () + + z + + z + z.(i)

25 Chpter 8: Plne Given eqution of plne is A + z d.(ii) The plnes given b eqution (i) nd (ii) re prllel. A distnce between the plnes (D) is. D d d 6 6 d 6 Q d 6 Since, direction cosines of PQ re equl nd positive the d.r.s. of PQ re,, The eqution of the line PQ is z + z k, s Co-ordinte of the point Q re (k +, k, k + ) The point Q lies on the plne + + z 9 (k + ) + k + k + 9 k k Q (,, ) PQ. Let A (,, ), B (, b, ), C (,, c) b c G (,, z),,, b, c z, b, c z P(,, ) + + z 9.(i) The eqution of the plne is z b c Since, this plne is t distnce of unit from the origin, b c b c.[from (i)] 9 9 9z z k 9 9. Let the eqution of the plne OAB be + b + cz d This plne psses through the points A(,, ) nd B(,, ) + b + c, (i) nd + b + c (ii) on solving (i) nd (ii), we get b c 5 Similrl, let the eqution of the plne ABC be ( + ) + b( ) + c(z ) Substituting the co-ordintes of A nd B, we get + b c, nd + c b c 5 If is the ngle between two plnes, then it is the ngle between their normls. 5 ( ) ( 5) ( ) ( ) cos cos 9 5

26 Std. XII : Triumph Mths. The eqution of the given plne cn be written z s 5 Let the plne intersects the, nd z es in the points A(,, ), B(, 5, ), C(,, ) i,b ˆ 5j ˆ, nd c kˆ Volume of tetrhedron b c Given lines re coplnr. 5 k k ( + k) ( + k ) + ( k) k k + k k k k(k + ) k or k

27 9 Liner Progrmming Chpter 9: Liner Progrmming. Let no. of model M nd no. of model M, Constrints re + 8 +, Mimize z + The corners of fesible region re O(, ), A(, ), B(.5, 5), C(, 6) At A (, ), z () + 6 At B (.5,5), z (.5) + (5) 7.5 At C (, 6), z + (6) 8 z is mimum t B(.5, 5).. Objective function P + The corner points of fesible region re B(, ), C(,), D(, ), E(, ), F(8, ) At B P B () + () 6 At C P C () + () 5 At D P D () + () 9 At E P E () + () 7 At F P F (8) + () 7 P is mimum t F(8, ).. For (, ), >, for (5, ), 5 + >, nd for (, ), + > Similrl, other inequlities stisfies the given points. Option (D) is the correct nswer. 5. Y X C(, 6) O X Y Y O (,) Y (, ) Y B(,) C(,) B(.5, 5) A(, ) F(8,) E(,) D(,) (9,) + (, ) + X X (,5) (,) A(,6) B(8,6) C(,5) 6 X O Y (,) X D(5,) + + 5

28 Std. XII : Triumph Mths OABCD is the fesible region O(, ), A(, 6), B(8, 6), C(, 5), D(5, ) z + At point C nd D, z is mimum. M z 5 Infinite optiml solutions eist long CD. 6. Consider option (C) + () () + () () + 5() All the bove three in-equlities hold for point (, ). Option (C) is the correct nswer. 7. Let the mnufcturer produce nd bottles of medicines A nd B. He must hve + 66, + 5,,,,,. the number of constrints is Let the compn produce telephones of A tpe nd telephones of B tpe. Constrints re + 8 +, + Mimize z + Y (, ) (, ) + (, ) X O (,) Y + the fesible region of the LPP is bounded. X 9. Given tht + 8, + 5 the fesible region lies on origin side of + 8 nd + 5. Also,, the fesible region lies in first qudrnt. option (C) is correct. X. Objective function z + The corner points of fesible region re O(, ), A(, ), B(, ), C, 7 nd D(, ) At B(, ) nd C, 7, z is mimum. M z Infinite number of solutions eists long BC. X + D(,) O C, 7 A(,) X B(, ) X +

29 Chpter 9: Liner Progrmming. Objective function z + The corner points of fesible region re 5 A,, 5 B, 6 6, C(, ), D(, ), E(, ), 5 7 F, 5 At A z A.5 5 At B z B At C z C () + () At D z D () + () 9 At E z E () + () At F z F.5 Mimum vlue of z t (,) is 5. X O Y Y 5 (,6) F(5/, 7/) E(,) A(/, 5/) B(/6,5/6) D(,) C(,) + (6,) X + 6

30 Trget Publictions Pvt. Ltd. Continuit Chpter : Continuit cos(cos ). f() lim sin sin lim sin sin sin lim sin sin lim 8. f is continuous t. log( ) log( ) f () lim sec cos lim cos cos log( ) log( ) cos log log lim sin log log cos lim sin (cos ). For f() to be continuous t, we must hve f() limf( ) lim lim lim lim f() lim sin 5 ( ) 7 ( ) lim sin 5 7 lim sin / / 5 (log ) log 7 It is discontinuous t nd it is removble. lim lim sin log tn e 5 sin log( ) tn 5 e 5 5

31 Std. XII : Triumph Mths (). () () () For f to be continuous t, f() lim ( - lim e 7. Given function is continuous t (, 6). t nd, function is continuous. If the function f() is continuous t, then limf( ) limf( ) + sin + b. Given, f() [] [ ] < <, f() (), f() < <, f(), f(), f( ), f( ), f( ), f( ), f( ), f ( ) 5, f( ) 5, f( ) 5 Hence, the given function is discontinuous t ll integers ecept. + b...(i) If the function is continuous t, then lim f( ) lim f( ) + b 6 tn + b 6...(ii) From (i) nd (ii),, b 8. Since, nd re continuous for ll. + is continuous for (, ). 9. For f() to be continuous t, we must hve lim f() f() lim f() lim f() f() lim tn /tn lim e tn tn lim e e f() b e lim f() lim ( sin ) f() lim b e / sin lim sin sin e f() e e e

32 Differentition Chpter : Differentition Now, log( + ) log( ). 5 log log( + ) log( ) log log 8 log d 8 8. (cos + i sin ) (cos + i sin ).(cos(n ) + i sin(n )) i Since, cos + i sin e e i e i e i5 i(n ). e i[ (n )] e in e d in in e d d in ine n. f 5 d f d 5 d d f f tn 5 6 () cos + sin Since, d d d cos cos d sin (cos ) + d sin d (sin ) d cos sin ( sin ) + cos sin cos When, cos cos, cos nd sin sin, sin d +

33 Std. XII : Triumph Mths n 5.. d. n +. n n When, + + Ecept st term ll terms re. ( ) ( ) ( ). ( n) d ( ) ( ) n (n )!, 6. f (), Lf () lim f( ) f() lim Rf () lim f() is differentible t nd f (). 7. f() sin(log ) f () cos(log ) f d f d d coslog. 66 coslog 9 cos log 8. d tn blog d tn + b log d log tn, b b 9. f() cos cos cos cos 8 cos 6 6 ( sin cos cos cos sin cos 8 cos 6) 6 (sin cos cos cos 8 sin cos 6) 8 (sin cos cos 8 sin cos 6) (sin 8 cos 8 cos 6) sin sin 6 cos 6 sin sin sin f () sin cossincos sin f

34 Chpter : Differentition ( + ) ( + + ) ( + ) d d d d ( + ) + b, b. Let , d 5 d 5 5 ( ) 5 d ( d ) + d d d Dividing both sides b, we get d ( d ) d k 5. + d d 5 tn e Diff. w.r.t., we get d e tn..(i) d () 5 d d d. e + d d d d Diff. w.r.t., we get d d + + d d d d + d + d d d d ( ) d d d.(ii) tn + d d From (i), when, e From (ii), when, d d d e d d e d f g h. f() f g h f g h f g h f () f g h + f g h. f g h f g h f g h + d d f g h f g h f g h + +.[ f, g, h re polnomils of nd degree, f g h ] cos sin cos sin cos 5 sin 6 cos 7 sin 8 cos.[ C C ]

35 Std. XII : Triumph Mths 5. sin cos {sin(cos )} sin cos sin sin sin[cos (cos(sin )] sin(sin ) d d 6. 8 f() + 6f + 5.(i) Replcing b, we get 8f 5 6f() + 8f 5.(ii) (i) 8 (ii) 6 gives 6 f() 6 f() f() 8 6 Given, 6 f() (8 6 + ) d 8 ( 6 + ) ( 6 ) d f( ) 5 Diff. w.r.t., we get f ( ). 5 f ( ) 5 f (7) f ( ) 5 () 5 8. Since, g() is the inverse of f(). f[g()] f g( ) g( ) f g() g () g() f (g()).(i) f() + e / f() f () g().[ g() f ()(given)] From (i), we get g() f() Now, f() + e / f () + e/ f () g() / 9. f( ) d f ( ). tn( ) z g( 5 ) dz d g (5 ).5 5 sec( 5 ) d tn 5 dz dz 5 sec tn 5 sec d. Put sin nd sin 6 6 ( ) 5 sin sin (sin sin ) cos + cos (sin sin ) cos cos.sin cos cot cot sin sin constnt Diff. w.r.t., we get 6 6 d d 6 6

36 Chpter : Differentition. Let f() p + q + r f() f() p + q + r p q + r q. f() p + r f () p f () p, f (b) bp nd f (c) cp Since,, b, c re in A.P. p, bp, cp re in A.P. f (), f (b), f (c) re in A.P. d sec tn + sin d n n nd n sec.sectn n cos.( sin ) d n n n sec tn n cos sin n n d n sec tn n cos sin d d sectn sin d Dividing N r nd D r b tn, we get d seccos n n n(sec cos ) n n n (sec cos ) d (seccos ) (seccos ) sec.cos n n [(sec n cos ) n sec n cos ] n( ) ( + ) d. f() f () + n ( + ) sin cos tn sin 5 sin cos tn sin 5 cos cos sec cos 5 + sin sin tn sin 5 f( ) sin cos tn sin 5 cos cos + sec cos 5 + sin sin tn sin 5 f( ) lim Since, g is the inverse of f. f[g()] Diff. w.r.t., we get f (g()) g() g() + [g()] 5 f(g( )) sin sin sin sin sin sin sin sin nsin(n ) sin( ) sin( ) sin((n ) n )... sin sin sin sin sin nsin(n ) sin cos cos sin sin cos sin sin sin sin sin sin cos sin sin(n ) cos n cos(n ) sin n... sin sin sin nsin (n ) sin nsin (n ) cot cot + cot cot +. + cot n cot(n+ ) cot cot(n + ) d cosec [ cosec (n + )] (n + ) (n + ) cosec (n + ) cosec 6. If r <, + r + r +. + r sin + sin + sin 6 sin +. sin sin cos tn e tn d tn e. tn sec tn e tn sec 5

37 Std. XII : Triumph Mths 7. tn + tn + tn +. to n terms 57 tn ( ) + tn ( + )( + ) + tn +. to n terms ( + )( + ) tn ( ) ( ) + ( ) ( + ) tn ( )( + ) + tn ( ) ( ) +. to n terms ( )( + ) tn ( + ) tn + tn ( + ) tn ( + ) + tn ( + ) tn ( + ) +. + tn ( + n) tn ( + (n )) tn ( + n) tn d ( n) d n n n n n 8. sin(b + c) cos(b + c).b b sin b c b sin(b + c).b b sin( + b + c) b cos(b + c).b b sin b c b ( sin(b + c).b) b sin( + b + c) b sin b c In generl, n b n sin n b c f()! f ()! f ()! n C n() f() f()! n n n C n n(n )()! n(n )(n )()! + f ()! n(n )! n f ()! n C n(n )(n )! +.+() n n f () n! n C n C + n C n C +. + () n n C n. p cos + b sin dp d. cos ( sin ) + b. sin cos (b ) sin dp d (b ) cos (b ) (cos sin ) dp p + d cos + b sin + (b ) (cos sin ) cos ( + b ) + sin (b b + ) cos ( + b ) + sin ( + b ) ( + b ) (cos + sin ) + b ( + b ) c.[ + b c (given)] 9. f() n f () n n f () n(n ) n f ( ) n(n ) (n ) n f() n n C 6

38 Applictions of Derivtives Chpter : Applictions of Derivtives. Let f() + b + c + d f() nd f(). + b. + c. + d b + 9c + d (7 + 9b + c + d) f() f() f() is polnomil function, it is continuous nd differentible. Now, f () + b + c + d B Rolle s theorem, there eist t lest root of the eqution f () in between nd.. The eqution of the curve is + b + c. d.(i) Since, the curve touches the line t (,). [ + b] (, ) () + b b Substituting the vlue of b in eqution (i), we get d Since, grdient is negtive. d < < <. The eqution of the prbol is 8. d d 8 d 8 m Slope of given line, m m m Since, tn m m tn or 8 Putting in the eqution of the curve, we get the point of contct is,.. f() tn log f () ( ) ( ) Now, f () f() tn log..785 Since, we re finding mim on n intervl,. We hve to find the vlue of f() t nd f tn + log log 6. log f ( ) tn log log the gretest vlue of f() is + log. 6

39 Std. XII : Triumph Mths 5. + cos cos sin Let cos cos cos sin sin cos. cos d Now, d cos d d Also, sin < d is mimum when it is mimum t 6. Let P(, ) be the point on the curve t which tngent is drwn. The eqution of the curve is c. d d d (, ) The eqution of the tngent is ( ) + + The tngent meets the X-is in the point A(, ) nd the Y-is in the point B(, ) P is the mid point of AB The rtio is : b 7. Let f() c d f () + b + c Now, f() + b b 6c 6d + c + d 6 f() 6d d.[ + b + 6c ] 6 Also, f() d f() f() f() is polnomil function, it is continuous nd differentible. There eists t lest one vlue of in (, ) t which f () + b + c one root of the eqution + b + c hs vlue between nd. 8. f() sin ( + cos ) sin + sin cos f() sin + sin f () cos + cos cos cos f () cos or cos or or f () sin sin <, onl when The mimum vlue of function is t f 9. f() sin + cos (sin + cos ) sin cos f() sin f () ( sin cos ) f () sin cos Now, f () sin or cos or Since, f () sin cos f () sin f() cos For, f() < For f() > At, f() is minimum Minimum vlue of f() ()

40 Chpter : Applictions of Derivtives. ( ) 7 is minimum when 7 is minimum. Since, ( ) ( 9 + 7) 9 7, for ll Minimum vlue of ( ) + 7 is. Minimum vlue of ( ) 7. f() cos 6 + b cos 6 + b.[ cos ( ) cos ] f () sin 6 The function f() is incresing for ll R. f () > sin 6 > 6 < sin < sin <. Let f() sec + b cosec f (). sec sec tn + b. cosec ( cosec cot ) sec tn b cosec cot Now, f () sec tn b cosec cot. cos sin cos b sin cos sin sin cos b tn b tn b nd cot b Also, f () sec.sec tn.sec sec tn cosec ( cosec ) b cot.cosec ( cosec cot ) sec + sec tn b cosec cosec cot f () > for ll. f() is minimum when tn b Minimum vlue of f() ( + tn ) + b ( + cot ) b + b b b b b b ( + b) + b( + b) ( + b) b b. ( )( ) 5 d ( 5) (b)(5) ( 5) For etreme (i.e., mimum or minimum) d ( 5 + ) ( + b) ( 5) Since, hs n etreme t P(, ) stisfies bove eqution ( + ) ( + b) () + + b b, stisfies the eqution of the curve () b, b. Let f() tn f () sec + tn f () > for, f() is incresing in the intervl, Since, < < < f() < f() tn < tn < tn tn

41 Std. XII : Triumph Mths 5. The point of intersection of the given curves is (, ). Now, log d d (,) log m (s) Also, 5 5 log5 d log 5 m (s) d (,) m m log log5 tn m m log log5 6. Let f() + b + c f () + b since, nd re roots of the eqution + b + c f() f() f() being polnomil function in, it is continuous nd differentible. There eists k in (, ) such tht f (k) k + b, k b But k [, ] < k < < b < 7. f() tn (sin + cos ) f () (cos sin ) (sin cos ) cos (sin cos ) For f() to be incresing, f () > cos > cos > < + < < < f() is n incresing function in,. 8. f() + 6 Diff. w.r.t., we get f () () + 6 () + 6 Now, f () ( ) ( 6) or 6 Also, f () 6 [f ()] < [f ()] 6 6 > Mim t p nd minim t q 6 p q.(given) (6) The functions e, sin, cos re continuous nd differentible in their respective domins. f() is continuous nd differentible Also f f 5 Now, f () e (sin cos ) + e (cos + sin ) e ( sin + cos + cos + sin ) e cos Also, f () cos 5,..(i) Diff. w.r.t., we get d d d d slope of the norml Since, the norml to the given curve mkes equl intercepts with the is. Substituting in (i) nd solving, we get 8 the point, 9 7.

42 Integrtion Chpter : Integrtion. Let I 5 6 d 5 5 Put t. d 5. d 5() d dt d I t dt 5 dt 5 / t. + c 5 / 5 75 / + c. Multipling N r nd D r b sin, we get cos5 cos d cos sincos5sincos d sin sin cos sin cos5 cos d sinsin 6 9 sin cos cos cos d 9 cos sin cos cos d cos cos d sin sin + c. Let I sin d Put tn t d tn t sec t dt tn t I sin tn t sec t dt tn t sin (sin t) tn t sec t dt ttntsec tdt d ttntsec tdt (t) tntsec tdtdt dt tn t tn t t.. dt ttn t (sec t)dt ttn t tntt c, where t tn tn tn + c. Let I cosec d sin d sin d sin sin sin d sin sin cos d sin sin Put sin t cos d dt I dt t t dt t t

43 Std. XII : Triumph Mths t log t t t c, where t sin dt log sin + sin sin + c 5. Let I tn d Put tn t sec d tdt t d dt t t I t. dt t t dt t t t dt t t t dt + dt t t I + I (s).(i) t I dt t t dt t t dt t t t, where t t tn tn t t t tn t t I dt t t dt t t dt t t t dm, where t + m t m log m t m log t t t log t t t t From (i), I t tn + t log t t t t + c tn tn tn tn tn + log + c tn tn / 5/ 6. Let I / d Put t 5 d / 5.( 5/ ) /. / d 5/ dt, / d 5 dt I (t ).t. dt 5 t 5/ t / t / 5 dt 7/ 5/ / t t t c, where t 5/ / / 5/ 7/ 5/ 5/ / + c

44 tn 7. Let I tn tn d tn sec tn d sin cos d sin cos cos sin cos sincos d sin d sin sin sin d sin sin d sin d I (s).(i) I sin d Put tn t sec d dt d t tn sin tn t t I dt t t t dt t t dt t t dt t t tn + c dt From (i), tn tn + c I tn tn + c A A 8. Let I log d Put log t I d dt A 9. Let I d dt t. dt t c Chpter : Integrtion log + c d d ( ) Put + tn d sec d sec I d (tn ) sec d sec + cos d ( cos ) d sin c ( + sin cos ) + c tn ( ) c tn ( ) + c

45 Std. XII : Triumph Mths. Let I d 6 6 cos sin Since, + b ( + b) b( + b) cos 6 + sin 6 sin cos.[ + b cos + sin ] I sin cos d d sin sin d cosec d cosec cosec d (cot ) ( cosec ) d cot Put cot t cosec d dt I dt t tn (t) + c tn ( cot ) + c tn cos sin + c sincos tn (cot tn ) + c tn (tn cot ) + c. f( )sincosd log[f ( )] + c (b ) d log f ( ) c d (b ) f() sin cos f () f() sin cos (b ) f ( ) f( ) (b ) sin cos [f ( )] Integrting on both sides, we get f( ) d (b ) sincosd f( ) f( ) (b ) sincosd b sincosd sincosd b ( cos ) (sin ) f( ) b cos sin f() sin b cos sin. Let I e log(sin ) cosec cosd Put sin t cos d dt t I e logt dt t t e logt dt t t t t e logt c t. d. log t dt t t t sin e log(sin ) c sin sin e log(sin ) cosec c cos sin tn log log cos sin tn Since, log tn secd log tn d log tn sec.(i) d Integrting the given epression b prts, we get sin I logtn sin secd.[from (i)] sin logtn tn d sin logtn log(sec ) c. Let t ( ) () t ( ) () t

46 Chpter : Integrtion 6 t 8 t t t + t t + t6t6 t t t Given, f + t f(t) t t 5 t t t t 8 (t ) f() 8 ( ) f( )d 5. Let I log log ()d 8 d ( ) 8 log + c d log( ) d log( ). d log dd d d log( ) + d log( ) + I.(i) Now, I Put t, d t dt d I t t t.t dt dt t t dt t t dt t tdt t t log( t) t + c log( ) + c From (i), I log( ). log ( ) + c c 6. P() d, Q() d P() + Q() d d d + I.(i) I ( ) d A Put ( ) C + A B ( ) C( ).(ii) Putting in (ii), C Putting in (ii), A Putting in (ii), B I ( ) log log log + 5

47 Std. XII : Triumph Mths From (i), P() + Q() + log log + + c (P + Q) () P() + Q() + log log + + c 5 5 log + c. 5 (PQ)() c log P() + Q() + log log + + log P() + Q() + log log + + log + log 8 sin bsin 7. Let I d (bcos ) ( bcos ) sin d (bcos ) Put b + cos t sin d dt sin d dt t b b I t dt btb dt t ( b )t bt dt b b + c t t b b + + c t t 8. Let I ( ) d Put t, d t I dt t t t dt dt t t t dt t5t dt 5 t t 5 5 dt 5 t t dt 5 t 5 5 t log t t + c log c 5 5 ( ) 5( ) 5 9. Let I 5 5 5( ) log cos sin e d sin e ( cos ) sin sin e e d Put e sin t [e sin cos + e sin ()] d dt e sin ( + cos ) d dt I dt dt t( t ) t( t)( t) + c Put A B + t( t)( t) t t + C t A( t) ( + t) + Bt( + t) + Ct( t).(i) Putting t in (i), we get A Putting t in (i), we get B 6

48 Chpter : Integrtion I Putting t in (i), we get C / / dt t t t log t log t log + t + c t e log c log t e sin sin tn(sin )d + c. Let I tn tn d d Put t, d dt, d dt + t t + t t d (t ).t (t ). dt (t ) t t d dt (t ) t t I dt t (t) t t dt t(t ) t t Put A B + t(t ) t t + C (t ) t t + A(t ) + Bt(t ) + Ct Putting t in (i), we get A Putting t in (i), we get.(i) I dt t dt t + c + c t 5/ 7/. Let I sec cosec d 5/ 7/ cos sin d 5 Now 7 5 Multipling nd dividing b cos, we get 5/ 7/ I cos cos sin sec d 7/ tn ( + tn )sec d Put tn t, sec d dt 7/ I t (+ t )dt 7/ / t +t dt / / t t + c / / (tn ) (tn ) + c. Let I Put + d t t C Putting t in (i), we get A + B + C B B I. dt t (t) (t) log t log(t ). t + c log t log(t ) t + c, where t + nd t + P, Q, R. Let I ( ) Put t, d dt t I t dt t t d 7

49 Std. XII : Triumph Mths tdt t t Put t m t dt m dm, t dt m dm mdm m m I m dm m tn c t tn c tn c tn + c cot + c tn + c tn. Let I d Dividing N r nd D r b 5, we get 5 I d + c Put t d dt 5 I dt t c t c log 5. d ( ) log.( ) d log.. d log + d ( ) log + n 6. I n sin d n sin.sin d sin n sin d d log + log log + + c d n sin sin d d d sin n ( cos ) n (n )sin cos ( cos )d sin n n cos + (n ) sin cos d sin n n cos + (n ) sin sin d sin n n n cos + (n ) sin sin d sin n n cos + (n ) sin d n (n ) sin d I n sin n cos + (n ) I n (n ) I n I n + (n )I n (n )I n sin n cos ni n (n ) I n sin n cos 7. u f () sin + f () cos du f () sin f () cos d + f () cos f () sin f () sin f () sin v f () cos + f () sin dv f () sin + f () cos d + f () cos + f () sin f () cos + f () cos du d + dv [ f () sin f () sin ] d + [f () cos + f () cos ] 8

50 Chpter : Integrtion [f ()] sin + f () f () sin + [f ()] sin + [f ()] cos + f () f () cos + [f ()] cos [f ()] + f () f () + [f ()] [f () + f ()] /.[ sin + cos ] du dv d d d f ( ) f ( ) d f () + f() + c 8. Let I ( ) d ( ) d ( ) d d sin cos sin cos sin cos d sin sin cos + cos sin cos (sin + cos ) d Put sin cos t (cos + sin )d dt I t dt sin (t) c sin (sin cos ) + c Put + t d t dt t I dt (t ) t dt t tn t + c tn + c d sin cos d cos sin sin + cos sin cos d 9. Let I tn cot (sincos ) d sin cos sin cos sincos d 9

51 Trget Publictions Pvt. Ltd. Definite Integrls Chpter : Definite Integrls log( ). Let I d Put tn t d sec t dt When, t nd when, t I I log( tn t log( tn t) dt tn t) sec t dt log tn t dt tnt log dt tnt log dt tnt [log log( tn t)]dt (log )dt I I logt / log I 8 log. 5 f( )d + f( )d + f( )d + + f( )d f( )d + 5 d d d d 5 f( )d ( ) + ( ) + 9( ) + 6(5 ) Let I I d. (i) f( ) d f( ) d f( ) f( ) f( ). f( )d f( )d.[ f() + f( ) (given)] d. (ii) Adding (i) nd (ii), we get I I f( ) d + f( ) f( ) f( ) d f( ) log. Let I d ( ) d d log log d d ( ) ( ) I + I (S).(i) log I d ( ) Put d When, nd when,

52 Std. XII : Triumph Mths I log d log ( ) log ( ) log d ( ) I I From (i), I I + I 5. n [ ]d. log log n [ ]d + [ ]d + [ ]d + + [ ]d 6. I n n d + d + d + + n (n ) n n (n )d ( ) + ( ) +. + (n )(n n + ) + () +. + (n )() (n ) (n )n n(n ) I I / sin d sin / sin d sin / / / sin d sin (sin sin ) sin (sin cos ) sin d / cos d d / (9 sin 6sin )d I +I I I, I, I re in A.P. 7. Let I d ( )( ) Put sin t + cos t d (. sin t cos t +. cos t ( sin t))dt ( ) sin t cos t dt When, sin t + ( sin t) + ( ) sin t sint, t When, ( cos t) + cos t + ( )cos t cos t, t ( ) ( ) ( sin t + cos t ) ( sin t cos t) [ cos t ( sin )] [( cos t) sin t] ( ) cos t ( ) sin t Since, > ( )( ) ( ) sin t cost I ( )sintcost dt ( )sintcost ( )dt dt t / 8. Let h() f() e e h() () e e e e e e h() h() h() is n even function.

53 Chpter : Definite Integrls tf(t)dt h(t)dt h(t)dt 9. f(m, n) t f(t)dt f( )d. f( )d m n (log ) d m n d m n d (log ) d (log ) dd n m m n (log ) m(log ) d n n m m n (log ) d n.[ log ] m f(m, n) n. () 7 6 (sin t cos t)dt () sin + cos If 7, 6, then is in the third qudrnt. sin nd cos re both negtive. () sin + cos < 7 () is decresing on the intervl, 6 7 Minimum (lest) vlue of () on, 6 / is (sin t cos t)dt 7 /6 cost sint / 7 /6 7 7 cos cos sin sin Let I b f( )d Put (b )t + d (b )dt When, t nd when b, t I f (b)t (b)dt (b ) f (b )t dt (b ) f (b ) d b. If <, then <, [ ] If <, then <, [ ] If.5, then.5, [ ].5.5 d d + d + d d + d d + (.5 ) +. f f().(given) f() f sec sec Let I f( )d f d I cos cos Put t, d dt When cos, t sec nd when sec, t cos cos sec I + I I I sec f(t)dt f(t)dt cos sec f( )d I cos

54 Std. XII : Triumph Mths. 5. n n lim n n n n n n n... n n(n) lim... n (n ) n n n n (n) lim n n r d r n ( ) () lim n n r r r n n n lim n r n r r n n n n lim n r n r r n n d Put + t. d dt d dt When, t nd when, t I dt t t 5 6. f( ) f( ) f( ) f( ) Integrting on both sides, we get logf ( ) logc f ( ) ce f() c c f() e Now, f() + g() g() e f( )g( )d 7. Let I I e( e)d e e d e e e e e e ( sin cos )d ( sin cos )d ( sin cos )d sin cos is periodic... function with period (sin cos ) d sin d+ cos d [I + I ] (S).(i) Where I sin d ( cos )( sin ) d Put cos t, sin d dt When, t cos nd when, t cos

55 Chpter : Definite Integrls I ( t )dt ( t )dt t t I 8. I cos d cos d sin d I From (i), I 8 sin cos d sincos cos sin d sincos... f( )d f d sin cos d I sincos I I I cos cos 6 6 d d.[ cos 6 is periodic f n with period ] 6 6 (cos + cos ( ))d.. f( )d [f( ) f( )]d cos 6 d I I I sin d log d log d log d.[ sin is n odd function]. log d f( )df( )d I I I I I I I, but I 9. Let I n sin n sin cos n I sin n d ( )sin ( ) n n sin ( ) cos ( ) n sin d I n n sin cos n sin n n sin cos n sin cos n n sin d n n sin cos n d d n is periodicf with period 5 f( )d f( )d., if f ( ) f ( )

56 Std. XII : Triumph Mths. 6 bsin sin cd cos i.e., I + I + I I sin d sin d sin d ( sin )dsin d cos cos / / I bsin d cos blog cos / /. cos cos c I c d c I + I + I becomes + c.[ I ] The given eqution is reltion between nd c. log. Let I cose I e d Put e t e d dt When, t e When log, t log / e cos tdt sin t / sin sin sin. f() f( + ) f() + f() + f() + 6 sin tdt + 6 sin tdt 6 sin (u 6 sin u du 6 sin tdt f( + ) f() + f(). Let f() + b + c f () + b f () f() c f () b 7 f () 8 f() sin tdt sin tdt )du, where t u + f() + f() ( 7 )d f( )d A f( )d A Asin B d A A cos B A B A B

57 Chpter : Definite Integrls Now, f() A sin B f () A cos. f A. 5. f() 6. A A sin sin sin sin sin sin sin sin sin sin sin sin sin (C C C C ) sin sin ( sin ) sin sin sin f( )d sin d cos / cos cos cost cos t sin t cos t dt / sint sintcostsint ( ) () ( ) ( ) + + from the given condition, + + ( + 5) ( ) 5 The positive integer vlues of stisfing the bove inequlit re,,,. There re such vlues. 7. Since, sin sin 5 6 [sin ]d [sin ]d [sin ]d [sin ]d [sin ]d 7 6 ()d ()d ( )d ( )d Appling R Rsec R, we get sec cot cosec cos f ( ) cos cos cosec cos cos (sec + cot cosec cos ) (cos cos ) (sec + cot cosec cos ) ( cos sin ) sin cos + cos sin sin cos ( sin ) 5 sin cos 5 f( )d (sin cos )d

58 Std. XII : Triumph Mths 9. d < /. + + < + 6 < < < < But cnnot be negtive nd ccording to the problem, < < < <. Let I Put I sin e d t d dt 6 sin t e t 6 sin e 6 [f ( )] dt d f(6) f() k 6 sin e d sin d e. [f ( )] d 8

59 5 Chpter 5: Applictions of Definite Integrl Applictions of Definite Integrl. Y. Y + 5 X O X (, ) (, ). X Required re sin d Y O sin d sin d cos cos sin d sin d cos cos() (cos cos) () ( ) + () sq. units Y Y sin X X Required re (, ) O (, ) (, ) X Y 5 d ( ) d 5 5 sin 5 ( )d + 5 sin 5 5 sin 5 sin sin Y Required re d sin / cos ( ) sq.units + 5 sin sin 5 sin sin sin cos 5 5. sin cos 5 5 sq.unit

60 Std. XII : Triumph Mths. Y B(,) X (, ) O A(,) X + (, ) + Y Required re (re of OAB) d d sq. units 5/ + 5/ ( ) sin sin sin ( ) sq.unit 6. The point of intersection of the curve nd the line re (, ) nd (, ). 5/ 5. The given eqution cn be written s Y ( ) + ( ) This is circle with centre t (, ) nd rdius. X O (, ) (, ) X Y X Required re O Y (, ) 5/ (, ) 5 5 d ( ) d () + () X Y Required re [( ) ( )]d ( )d 9 sq. unit (, )

61 Chpter 5: Applictions of Definite Integrl 7. Y ( + ) X O, (, ) Y or cos, X 9. + () sq.unit + Y Required Are cos d cos d (, ).[ cos is n even f n ] sin (sin ) sin X O Y X 8. X O Y cos sin 5 X Required re ( )d Required re cossin d Y. Y sq. unit (cos sin )d+ (sin cos )d 5 X O X + (cos sin )d 5 / 5 / sin cos cos sin / 5 / + / sin cos Y Putting in the given eqution, we get or

62 Std. XII : Triumph Mths Required Are d ( ) 8 Required re d + d / sq. units /. Y (, ) sq.units ( ). + + X B A C Y O D(, ) X. Required re + X O Y (, ) (, ) (6 ) 6 sq. units O Y Y (, ) (, ) (5, ) + (, ) X X + + d d + () + + d (,) The eqution of the tngent t the point (, ) is ( ) i.e.,. It psses through origin. Required re re of the region OABCO + re of the region OCDO d ( )d ( )d ( )d ( )d ( )d Y sq. unit 6

63 Chpter 5: Applictions of Definite Integrl. The eqution + is of ellipse with centre t origin nd the eqution is of prbol with verte t origin. Solving the equtions, we get + ( + )( ) But is not possible, since both points of intersection lie on the right hnd side of Y-is. nd ± The points of intersection re A, nd X B,. Required re + d d 8 O Y Y B, sin ( ).[ the function is even] A, X 8 5. X sin sq. unit Y O Y Slope of tngent d d + ( )d + + c The curve psses through the point (, ). + + c c The eqution of the curve is +, which is prbol s shown in the figure. Required re ( )d 5 sq. unit 6 6. Drw AP to X-is. A A(OAP) A Are bounded the curve OA nd the lines OP nd AP d d Required re A A sq. unit 6 + X 5

64 Std. XII : Triumph Mths X 6, C X Y Required re re of the rectngle OABC re of the region OBCO O Y sin sq. units + Q (, ) Y The eqution of the prbol is ( ) Diff. w.r.t., we get ( ) d d d ( ) d ( ) (,) B, A, O Y sin (, ) / cos Eqution of tngent is ( ) 6 + It cuts the X-is t the point Q (, ) nd the prbol cuts the X-is t the point R(5, ). X R (5, ) ( ) X required re d ( ) () 6 9 d sq. units 9. Required re sin sin d cos cos sin sin. cos cos cos sin cos sin cos sin cos sin cos sin cos sin d cos sin cos sin d tn tn d tn tn tn tn tn d d tn tn Put tn t sec d dt required re tn 8 t t t t ( t ) t dt dt. tn 8

65 6 Differentil Equtions Chpter 6: Differentil Equtions. The given eqution is ( ) Put sin, sin The eqution becomes cos + cos (sin sin ) cos cos sin cos cot cot sin sin cot Differentiting w.r.t., we get d d d d Degree nd order re both.. Since, the given differentil eqution cnnot be epressed s polnomil in differentil coefficients, so its degree is not defined.. The eqution of tngent t n point P(, ) is Y d (X ) d d This meets the X-is t A,. d Similrl, it meets the Y-is t B,. d According to the given condition, P is the mid-point of AB. d nd d d + d nd + d d Both of these equtions reduce to d Integrting both sides, we get log + log log c log () log c c, which is the eqution of rectngulr hperbol.. A Put tn, tn The eqution becomes sec + sec A (tn sec tn sec ) sin sin A cos cos cos cos cos cos cos cos sin sin A cos cos cos cos cos + cos A (sin sin ) cos cos cot A cot A tn tn cot A Differentiting w.r.t., we get d d A. sin cos Degree nd order of the differentil eqution re both.

66 Std. XII : Triumph Mths 5. f d f Put v d v + dv d The given eqution becomes, v + dv d v + f(v) f(v) d f(v) dv f(v) Integrting on both sides, we get log log f(v) + log K f(v)k Kf f K c, where c K 6. The given eqution is d f( ) f( )f( ) d f( )d f( ) I.F. e e the required solution is.e f () f( ) e.f( )f ( )d t e.tdt, where f() t t t t.e e.dt te t e t + c.e f () f() e f () e f () + c f () f() + ce 7. The given eqution is ( + ) f () ( e + ) f() If f(), the eqution is e, which is liner eqution d ( ) d I.F. e e the required solution is.e d + c c ( ) When, 5 c 6.e f() e 8. d ( + ) d + d d ( + ) d Dividing b, we get d d d d Integrting both sides, we get c + c When,, c the required solution is i.e.,, which is the eqution of hperbol. d 9. d + + (d + ) + d d( ) d tn Integrting both sides, we get c ( ) tn + + tn c tn c tn c tn c c tn

67 Chpter 6: Differentil Equtions. d d tn ( ). d d +. tn ( ) d d ( ) tn ( ) dz d tn z, where z d cot zdz Integrting both sides, we get log (sin z) + c log(sin ) + c When,, z log + c, c the required solution is log (sin ) + log (sin ) sin ( ) e. d + tn e sec Dividing b sec, we get tn e sec d sec cos d d + ( sin ). e Put sin t, cos d d dt d the eqution becomes, dt e t, which is liner eqution d d log log I.F. e e e the required solution is t e. d t e e + c sin. e e + c When, c sin. e e sin ( ) e.. We hve, d d tn sec d d tn sec d d tn sec Put t, dt d d The eqution becomes, dt (tn ) t d sec, which is liner eqution. tn d I.F. e log (sec ) e sec the required solution is t sec sec d c sec tn + c sec (c + tn ). ( ) d d d Dividing b, we get d d Put t d dt The eqution becomes dt t, which is liner eqution d I.F. e e log the required solution is t d + c t log + c log + c (log + c) The curve psses through the point (, ). ( + c), c the required solution is (log ).

68 Std. XII : Triumph Mths. d e ( e ) e d e ( e ) e d e e.e e d + ee e Put e dt t, e d d The given eqution becomes dt e.t e, d which is liner eqution. e d e I.F. e e the required solution is e t.e e e.e d e t.e z e.dz e z + c e e c e e e.e e c, where e z 5. The eqution of the tngent to the curve f() t P (, ) is Y d (X ) d d This meets the X-is t,. d According to the given condition, d d d d d, which is homogeneous d.e. Put v d v + dv d the eqution becomes, v + dv d v v v v dv d v v v v v v v v dv d v Integrting both sides, we get v dv d v + c log v log + c v log log + c 6. log + log log + c log c This curve psses through (, ). c log + log log e e e e.e d d d + c ( + c) ( + c) + Rdius is fied, which is nd the centre is (c, ) which is vrible centre on the X-is.

69 Chpter 6: Differentil Equtions 7. (A) t t f(t, t) t t t t f(, ) Homogeneous of degree. 8. t (B) f(t, t) (t ) (t ) tn (C) t (t) tn t f(, ) Homogeneous of degree. f(t, t) t log t t log t t t tlog t e t t log log e t f(, ) Homogeneous of degree. (D) + t e f(t, t) t t t log log(tt ) t t t + t tn t t + tlog t tn ( ) Non-Homogeneous. sin d tn d tn sin d sin tn sin, which is liner eqution cosec d log(tn ) log (tn ) I.F. e e e log cot e cot t t the required solution is tn cot cot c sin sin d c tn cos sin cos sec tn + c cot tn c The curve psses through,. + c, c the eqution of the curve is tn 9. The eqution of hperbol is m d (slope of tngent to the hperbol) m d slope of tngent to the required d fmil of curves. The curves re intersecting orthogonll, m m d d Integrting both sides, we get + c, 6 which is the eqution of required fmil of curves.. The given eqution is d ( ) ( ) d ( ) d ( ) d ( ) (+ ) d, which is liner d ( ) eqution d I.F. e e log 5