MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial equatio is y y y y / Squarig ad multiplyig by y y y Hece order is ad degree is y 6 The differetial equatio of the family of curves for which the legth of the ormal is equal to a costat k, is give by ( ) ( ) (d) ( ) The legth of ormal is give by, y y k y k y y k y k y 6 The solutio of the differetial equatio ( ) is y ay y( ay) a y ay ( a) O itegratig both sides, we get logy log( ay) log( a) logc y c( a) ( ay) or c( a)( ay) y (d) Noe of these
6 The solutio of the differetial equatio is (d) Noe of these Here y y log It is homogeeous equatio So ow put y v ad v dv (i) O itegratig, we get log(logv ) log logc y c log c y e, the the equatio (i) reduces to 65 The geeral solutio of is ( ) ( ) ( ) (d) ( ) y y y y Put v / y y vy v y dv dv dv v y v v v y v dv y ta ( / y) logy c ta ( v ) logy C dv v logv 66 The solutio of the equatio is (d) Noe of these y y y It is liear equatio, therefore IF Hece the solutio of the equatio is e y e y ( y ) e c ce y y e y
67 The solutio of the give differetial equatio is (d) ( ) y ; IF e ( ) e Required solutio is ye c or y ce 68 A solutio of the differetial equatio ( ) is (d) Give equatio ca be writte as y If p, the y p p Differetiatig wrt, we get dp dp dp dp p p p ( p) Itegratig wrt, we get p c c ; y c c If c, the y 69 A cotiuously differetiable fuctio satisfyig is ta y (d) y Itegratig both sides, y ta y c (d) Not possible At, y, the c At,, y the ta c c ta y y ta ( ) Therefore, solutio is y ta But ta is ot cotiuous fuctio i (, ) Hece, () is ot possible i (, )
7 The rate of icrease of bacteria i a certai culture is proportioal to the umber preset If it double i 5 hours the i 5 hours, its umber would be 8 times the origial 6 times the origial times the origial (d) 6 times the origial Let P be the iitial populatio ad let the populatio after t years be P The dp dp kp k dt dt P O itegratig, we have log P kt c At, t P P P logp c ; logp kt logp log kt P Whe t 5 hrs, P P P log 5k P Whe t 5 hours, we have log P P log 5 log P log K ; log t 5 P 5 5 5 log log ; P P 7 The solutio of the differetial equatio is equal to y cosec( y) y cosec ( y) (d) ( y ) cosec ( y) d( y) cosec ( y) d( y) cosec ( y) Itegratig both sides, si( y ) d( y) d ( y) cosec ( ) y cos( y) c ; cos( y) c d y 7 The solutio of the equatio l, whe, y ad is
(d) (l ) l (l) l (l) l (d) (l) l d y log At,, y, (log ) c c log log y (log ) 7 If y cos cos y, the y () is (d) y cos cos y Differetiate both sides with respect to, we get y si cos y ( siy) y cos y Agai differetiate with respect to ysi y cos cos y si y siy y [cos y( y) siy y] siy y Puttig, we get y y siy y y y y siy Sice at, y ; (y ) 7 If is the greatest of the defiite itegrals e I I e cos,, I I e cos e /, the I I I I I I (d) I I (d) For, we have, so that e e Hece Also cos e, cos e cos Therefore e / e e Hece I is the greatest itegral cos I
75 Let f () be a fuctio satisfyig f ( ) f( ) with f ( ) ad g () be the fuctio satisfyig f( ) g( ) The value of itegral f ( ) g( ) is equal to ( e 7) ( e ) ( e ) (d) Noe of these f'( ) f( ) (d) We have f '( ) f( ) log f( ) logc f( ) ce Sice f ( ), therefore ce c Thus ( ) e f Hece g( ) e f ) g( ) e ( ( e ) e e 76 Let f be a positive fuctio Let k ) k I k ) k f (, I f ( whe k The I / I is / (d) k k I f{ ( )} k k ( k k ) f[( k k ){ ( k k )}] ( f ( ) f( a b ) ) k k k k ( ) f{ ( )} k f{ ( )} f{ ( )} I I I k I I I 77 If is ay iteger, the e cos ( ) cos (d) Noe of these Sice cos( )( ) cos[( ) ( ) ] = cos( ) ad cos ( ) cos So that f( a ) f( ), ad hece by the property of defiite itegral e cos cos ( ) b a b a
78 The value of the defiite itegral 6 such iterval is lies i the iterval [ a, b] The smallest, [, ] 7, (d) Noe of these 7 f ( ) f() The fuctio 7 f( ) is a icreasig fuctio, so Mi f( ) f() ad Ma 6 b Therefore by the property ( b a) f( ) M( b m a) (where m ad M are the smallest ad greatest values of fuctio) 6 7 a m 79 If l( m, ) t ( t) dt, the the epressio for l ( m, ) i terms of l ( m, ) is l( m, ) m m l( m, ) m m l( m, ) ( t) t m m t m ( t) dt ( t) l( m, ) m m m t dt m l( m, ) m m (d) l( m, ) 8 For which of the followig values of m, the area of the regio bouded by the curve y ad the lie y m equals (d) The equatio of curve is y y y This is a parabola whose verte is Y, y = m O m X
Hece poit of itersectio of the curve ad the lie m ( m) ie, or m 9 m ( ( m) ( m) m) ( m) ( m) 6 m m 6 9 / ( m ) 7 m 7 m Also, ( m ) ( m )[( m) 9 ( m)] ( m ) ( m) 9 m m m 9 m 5m 5 5 5 m ie, m is imagiary Hece, m 8 If for a real umber y, [ y] is the greatest iteger less tha or equal to y, the the / value of the itegral [ si ] is / (d) We kow si si 6 I [ si] [ si] 5 [ si] 6 [ si ] 5 6 7 7 [ si] 6 / 5/6 7/6 / 5 6 6 () 5 () ( ) ( 6 7 7 ) 6 5 7 7 6 6 6 6 6 6
8 The area bouded by the curves y l, y l, y l ad y l is sq uit 6 sq uit sq uit (d) Noe of these We kow that log is defied for ad log is defied for all R {} Also log ad log Y X X Required area is symmetrical i all the four quadrats ad is equal to log log,(i (,),log ) = [ log ] ( ) squit, si 8 si Y, ( N) equals lim log (d) si cos cos cos si si cos si cos si cos 5 si si si si si si si si cos cos si cos si si cos cos si si si
8 If I, I, I, I, the I I I I I I (d) I I (d) For, we have ad for, we have for ad for ad I I ad I I 85 ( si cos ) si cos si cos si cos si cos Differetiatio of si cos is cos, the I ( si cos ) Itegrate by parts I ( si cos ) cos ( cos ( si cos ) dt t t cos cos ( si) si cos ) cos si cos cos sec si si cos cos cos si si cos ( si cos )cos si cos ( si ) si cos ( si cos )cos si cos si cos si cos (d) Noe of these
86 e / si e / cos c e / cos c e / si c (d) e / si c (d) Let I e si / si e cos e / / c si e / e / cos si e / Therefore, I e si cos / I e si cos / e / si Trick : By ispectio, d e e si c e / e si c cos e si / / / 87 (si si) cos si e / / si log( cos ) log( cos ) log( cos ) 6 6 log( cos ) log( cos ) log( cos ) 6 log( cos ) log( cos ) log( cos ) (d) Noe of these I si si( cos ) si ( cos ) si ( cos )( cos )( cos ) Now differetial coefficiet of cos is si which is give i umerator ad hece we make the substitutio cos t si dt I dt ( t)( t)( t) We split the itegrad ito partial fractios I dt t t t etc 6( ) ( ) ( )
e 6e 88 If A B log( 9e ) C, the A, B ad C are 9e e (d) 6 A, B, C log costat 5 5 A, B, C log costat 6 5 A, B, C log costat 6 (d) Noe of these e 6e I 9e e 9 9e 9e 8 log( 9e 6 9 e 9e ) log cost 5 log( 9 I e ) log cost 6 Comparig with the give itegral, we get 5 A, B, C log cost 6 89 The fuctio f( ) l( ) l( e ) is Icreasig o [ Decreasig o [ Decreasig o, e ad icreasig o, e (d) Icreasig o Let l( ) f( ) l( e ) l( e ) l( ) f' ( ) e l ( e ), e ad decreasig o, e ( e )l( e ) ( )l( ) l ( e ) ( e )( ) f' ( ) for all, { e} Hece f() is decreasig i [, )
9 If is taget to the curve at (, ), the (d) Give curve y p q (i) Differetiate with respect to, y p p y, p p For give lie, slope of taget p p From equatio (i), 9 8 q q 7