Scribd
Upload
29 views10 pages

Continuity PDF

Uploaded by

sheriffnasreen800
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
29 views10 pages

Continuity PDF

Uploaded by

sheriffnasreen800
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
You are on page 1/ 10
ta point : ” jssaid to be continuous at'g' if defined inaneighbourhood of'a'and 4 s()=£(2) - ait gn est F(a) then f(x) is left \i - a. “it Ht f(x)=F(@) then f(x x) isright J sss a. ® function 4 t x1) be ay x#0is continuous © gixe0- Then the value of f(0) is e (ay saciven (0 Dein ee al ail Maca x=) Ge aja } Afunction f is said to be continuous in an open jnterval (a,b) if itis continuous at each and every point in the interval (a,b) > Afimetion / ispaidto be continuouson [4,5] if {fis continuous at each point of (a,b) i fisright continuous at x =a (ii) fisleft continuous at x = b WE-2 Let f be a continuous function on 1,3]. If f takes only rational values for = x and f(2)=10 then f(1.5) is equal is continuous function on [1,3] and takes ational values then f(x) is constant Discontinuity : > If (x) is not continuous at x= that /'(2x) isdiscontinuousat x= a. J (x) willbe discontinuous at x= a inany of the following cases : i) im f(x) and lim f(x) exist but are not equal. Ci) Zim) ana equal butnot equal to f(a) 1 Gi) £(@) isnotdefined (iv) Atleast one of the limit doesn't exist. W.E-3: Let f(x) be defined in the interval I-x, 0sxSl [0,4] such that 7) eee 4-x, Usxs4 f (x) existand are then number of points where f(/(x)) i discontinuous is Sol: f(x) is discontinuous at x= 1 andx=2 = (f(x) is discontinuous when f(x)= 1&2 Now 1-x=1>x=0, where f(x) is continuous x42=1>x=-1e(L2) “g-x=1>x=3e[2,4] Now, 1-x=2>x=-1¢[0,1) xt+2=2=> x=0¢ (1,2) 1] 4-x=2 => x=2€[2,4] Hence, f(f(x)) is discontinuous at two points, x =2, 3. > If f and g are continuous functions of x at x=a, then the following functions are continuousat x =a. )ftgif-g fg iv) of if ceR oF ite(a)eo en it and g are hot continuous at ed PIR AR, J 8 area APS iscontinnonsat y= cand g iscantinnous at f(a).then (gof) iscontinuousal y= a > AFF iscontiunous in [a,)] then itis bounded in [ad]. ie there exist 4 and yn such that KS f(x)Sm,Vre[ab] whore k and m areminimum and maximum values of / (\) respectively in the interval [4] Inthis case f takes every re k and m at least once go? may he continous value between Thus range of Fis[k,m] ( > If Fiscontinuous on [a,b] such that f(a and f(b) areofopposite signs then there exist atleast one solution forthe equation f(x) =0 intheinterval (a,b) Remembering method fx) gs) [Ane [fxg [Anigin) 2) provided g(a) 40 continu- | continu | continu: | continuous | continu- ous fous | ous ous eontinu- | discon- | discon- | maybe | may be ous | tinuous tinuous | continuous | continu ‘ordiscon- | ous or tinvous | discon- tinuous iscon- | diseon- | may be | maybe | may be tinuous | tinuous | continu- | continuous | continu- ousor | ordiscon- | ous or discon- | tinuous | discon- tinuous tinuous ‘Types of discontinuity > Discontinuity of first kind (or) Remo abl discontinuity : If lim f(x) exists butis not equal to f(a) (or) f(a) not difined then the f is said to have aremovable discontinuity at x= a. Itis also called discontinuity of the I*kind, In this case we can redefine the function by making lim f (x)= (4) andmake itcontinuous at x=a. Removable discontinuities are of two types 1) Missing point discontinuity 2) Isolated point discontinuity » y Vv Mivalng polit discantinulty ; fin £(¥) exists finitely and May defined * (2-H gy ample, (0) ) ee inissing point discontinuity at y ~ 9 linn (x) existsfinitely and f(a) ig efiney but lim f(x) 4 f(a) Example: /(«)=[*]+[ x} has isolateg point discontinuities at all integers (or) (function /()issaid to have discontinuity of the second kind at =a if lim f(x) does not exist. Irremovable discontinuities are of thre types 1) Finite discontinuity (or) jump discontinuity 2) Infinite discontinuity 3) Oscillatory discontinuity Ke lim f(x), lim /(*) are both finite and are not equal Example: /(x)=-—'— at x=0 1+2 If at least one of the limits L¢ f(x) and E(x) be 450, then f(x) has infinite discontinuity at x= a cos.x Example: /(x)=— aty=0. y atx=0 Oscillatory Discontinuity The discontinuity is said to be oscillatory when the limits osciallte between two finite quantities. Consider the function £( x)=sinZ, ici x0,2->0 sin(>) can take any eee =] to 1 or we can say when y_,9, sin oscillates between-L and I as shown in x the figure. “I 2 ase discontinuity ‘of the second kind, the bo ifference between the value of the gut x24 and LHL at x=a iscalled the jp 0D nity. fiction having finite | aan pie ise continuous of Sectionally continuous jnthis interval. , The jumP of discontinuity of the 2x-3] a 2x-3 ee .. Jump ofdiscontinuity =2 Allpolynomials, Trigonometrical functions, feponental and Logarithmic functions are SF qsusintheirrespective domains Jatermedial > suppose f(x) isc anda, bare any Wo ontinuous on a interval I, points of L. If y, isa ) and f(b) then there mumberbetween f (a een a and b such that nists a number ¢ betw f(c)=¥% Single Point Continuity: Functions which are continuow pointare said to exhibit single port contin behaviour. fx, xe is Example! HG) =), ip x€0 s only at one ity continuous only at x =0- 1 1 B= where = x1 P+i-2 the number of points of discontinuous of y=f(x),x€R is ee isi f= is discontinuous at x=1.Also is discountinuous at t=-2 and Continuity at a pol When f=1, Aljarsh oo So, y= f(x) is discontinuoys at three 1 ints, x= 1 5 2 points, 2 1, The function cos3x=0084* for x #0 W0)=|,* iB forx 1) continuous 2) discontinuos 3)leftcontinuows __4)seht continuous The function defined by 1 xsin— = x for x#0 f(x) atx=0 is 0 forx=0 1) continuous 2) Only right continuous 3) Only left continuous 4)can not be determined e* —cosx y=; for x40 x If the function f(x) is continuous at x=0 then f(0)= 172 232 3)2 _4)18 ‘The value of (0) so that the function ae we(1+2)-re('-5] {=< x is continuous at x=0 is fesbl ve ee iD ) 4) (#0) ab a-b a+b » ab 3) a+b 1-2: tevBsine ip gH m-4x Iff(x)= i fx=5 is continuous at * = then a= 14 22, Ais ala, “Directional continuity: (left and right continuity) 6 Let f:R>R be define d by sel if x30 x meee ne | yf x0 [x] denotes the integral partof x. If f where si a continuous at x=0,then B-@= eg 30° «42 1. If [x] denotes a greatest integer not exceeding x andifthefunction f defined a+2Cosx ifx<0 S(2)= 4 pl Sa} if x20 continuous at x=0, then the ordered pair (a8) is 1) (-21) 2) (-2.-1) 3) (-143) 4) (2-N3) a a f)=PHs>ig>tre0, s(0)=1 res* is left continuous at x=0 then 1) p=0 2)p=r 3)p=q 4) p¥q Continuity of piecewise functions: x-4 peut! 9, Let fid=; a+b, x4 x4 a Then f(x) iscontinuows at x= 4 when 1) a=0,b=0 3) a=-1,b=1 wi f() fj+, x1 8(z)= -fe-2), x>1' thenthenumber of discontinuties of f(x)+g(x) is__ 11. ‘The function x+aVdsinx Osrcp, 2xeotxtb *4saey) acos2x-bsinx , rea 1) is continuous for 0 23) py of composite Kc 2)4)(0,) nection: g(r) =sin.x+ cos x, then points of inuity of (fog)(x) in (0,277) are xb) +x The function f(x) =(x- atb takes the value = for some value of ce[a.5] For a=3, b=5 Then the value of fois x-2 aif ste-{q : umber of discontinuity of f(f(a))is xs0 then the Or = 2005 then ve_ BM Tsing? x 1) [-1,0] 2) [-2.0] §(0.2] 4) [5] 4%, The function f(x)=x° /4-sinzx+3 takes the value a for x¢[-2,2] then oo io 19 , 50 1 re 3 3 ve Single point continuity and dirichlet ions: 4M fms 22-2 fae c x ‘continuous at y=, then f(0)= ie 22 2-2 43 els. =I for’ #1 and fis continuous | £(0)= Hig £ (2) = tim x. cosar-cosbe fa! HOE al lin, SO a en a (2) iscontinuous at x= 0 Use L-Hospital rule 7 ‘ 3 (0) = lim De + sit “5 2x Use L-Hospital rule Use L-Hospital rule a+0=2=f-1 a=f-1>f-a=| f(0)=5 to define a=-2>b=-1 Put x=—h ash +0 Aty=4 LC=a-land RC=b+1 Here each of the branch functions of f(x) interval. £(0") ; and g(x) are continuous in the given i Butand lim f(x) = lim x=0.So0, f(x) is discontinuous at x =1- Now, f(x) iscontinuous at x=1 and g(x) is continuous at y=0 Hence f (x)+g(x) isdiscontinuousat x =0 and x=1- Ne enya 4 4 4 Qn qd RC=—cot— +b ani roots ax nt < Ate=23LC =2x2cotZ + 2 pin and RC = acos 2 —bsin > . Atx=0 LC=e" and ROsa ie aus ie 0 me (9) dicen reall 1))is abe discontiniyg é Mao, A(1(9)) 8 Aiscontinugy a eee (1) co et? ta Hence, F(x) is discontinuows when ew Is Gisconteacn 10m? 1s i asemeeens +1 MW pa bet p(n) (1 sin? x)? l,) Thus, f is continuous and (0) 1g asl 16, F(x) [2sgn(2a)]42 and f(2)22"" Hence, from , intermediate value theorem, there Cris, Na *)=foxije2=4 #{0)=2 4(0")-Pal+ rumbercin (0,2) such that fc) «hg A0)-PLa2=4 *Y3,_ f(x) iscontinuous for x [2,9] Ao")=1(0)# 0 Now f(-2)=1 and f(2)=5 By intermediate value theorem f tg kes values between | and 5 al 2.x) hasremovable discontinuity 17. f(x) is discontinuous when y’-r=0 [>-ve(*)] =r=0,-1,1 24, (0)=limx] 3-lo ia 18. Function continuous forall x ae | =-2 s = 2sinx+a#0> sinx = 3. payee =e! =[pisec-aine>2 EXERCISE- Il 19. bir tere crer 1. HfiR>R is defined by sf xer-(4,9} Sigix))=) 0, if if -L if as ea. (e'-1) sored eee ete teen in I(*)=) sin*(x/a)(log(1+(x/a))) exists at least one ce[a,b] such that a fe Bs, and f isa continuous at x=0, then the i= ( 2 38 44 e pa rear continuous atx => thenk= 1 i 10. m2 3 7 ) 6 um Dg 4, 1f/(@) bea continuous function for at values of x and satisfies y+x(f(#)-2)+2V3-3-V3f(x)=0 wreR then the value of f(V3) is uu. yy 2)2(3-1)3) 25-1 4) 2(1-v5) (256+ ax)'®-2 eer ey = 5, Let fe) (B2+6x)'5-2 * continuous at x = 0 , then the value of g/} is 3 12. 2) 2 (0) If f is 8 1) 3f Let f(x)= — . secx=008 x x40. Then the value of {(0) sothat / iscontinuous at =0 1S 1 2)0 3)2 1. Ifthe function f(x)= ~ vx vk j Continuous for every xe R Then 1) ke[-2,0) 2) ke (0,2) 3) ke (-2,0) AkeR & Thevalues of p and q for which the sin(p+!)x+sinx if x<0 J(x)= « , if x=0 ; is 7x70 continuous for all x in Ris DPS 14. {(x) is continuous at x=0 then 1) 2loglal=b 2) Dloglb|=e 3) loga=2logib] 4) a=b The function F(x)eafe+t]+ole—H> (a#0,b#0) where [+] is the greatest integer function is continuous at x 1) a=2 2a=b 3) a+b=0 4) a+2b=0 Let f(x) =[2x° 6] when [2] is ere x then the alif atest integer less than or equal to number of points in (1,2) where f is B ry es discontinuous is If f(x 0, for[x]=0 _ I)continuous at x=0 Aediscontinuous at x=0 3) LH.L-0 4)R.H.L=1 ee fie If (x)= oi esn ey sents the greatest integer function, repre is continuous at x=0 Then B 1) A=-3,B=-V3. (9 4=3.B > 9 A= 8 issih | 3cos? x It f(s)=42 a(l-sinx) (4-2) ” so that f(x) is continuous at x=F then 1) a=1/2 1s. 16. 18. 19. 20. % - x<-l | f= agua sles Is av -hr-2, f 22-1 continuous everywhere. Then the equation whose roots are a and b is i 1) @a3r-2-0 0) xi-3x+2=0 3) fa3re2-0 4) x7 -5x4+6=0 2 log(I+2") ii Let f(0) =F 6425 1) f iscontinuous on [* 10 2) f iscontinuous on [-2,2 3) f iscontinous on [-6,6] 4) f iscontinuous on [1, 7 ; If[.] denotes the greatest integer function then the number of points where 1 2 eo)-te|x4} [+5 | he 8 discontinuous for x <(0,3) isn then — is if f:[22]>R is defined by ite-ViKe gy 9 for continuous on [~2,2] then c = 1)3 z 2 3 + = ), d5 areas (S)n2", ifx<0 2 S(x)= mes, Px20x0 0 2 x75 then in order that f be continuous at Be x=0, the value of — is , Let f(x)=sen(x) and g(x)=3(x?-5r+6). The funetion £(g(-)) isdiscontinuous at 1) infinitely many points 2) exactly onepoi 3) exactly three points 4) no point a . The value of 3 . If f(x)=lim 25a an ma =lin— it Ss) = ims te a OI is continuous at x = is x" =sin x" = lim =——, for x> Let (+) = in sin x? (OX? Oxy and f(1)=0. Then the function S (jy x= is 1) Continuous 2) Left continuous 3)Rightcontinous 4) Discontinuous . uf £(x)= Lt (sinx)”, then f(x) is 1) discontinuous at x = 7/2 2) continous at x= 7/2 3) discontinuous at x =—7/2 4) discontinuous at infinite number of points l+x then f(x) is discontinuous at 2) x=—Lonly 4) no point S(x)= ae aya is discontinuousat 1) x=0 only 2) 2 only 3) x=0 and2 4) x=0,1,2 only . The number of points of discontinuity of 28. 1) Discontinuous due to vertical asymptote 2) Removable discontinuity 3) Missing point of discontiuity 4) Oscillating discontinuity 1 The funetion (x)= sa( 4) 1) Jump discontinuity 2)Removable discontinuity 3) Missing point of discontiuity 3 4) Oscillating discontinuity “ x” a 2 4) continuous only at x= 0 continous VCE? . Thefunction f(x) =[x} -L of the following functions e+] e4g 2) a 3) Ysanvt A point where the function /(x) is not continuous when f(x)=[sin[x}] in (0.24) (L1denotes greatest integer < ,)is (0) 29(20) 3)(10) 4) (4,-1) rhe function defined by /(x)=(-1)") where ul denotes the greatest integer function, is 1 13 Where n isany Discontinuous for integer 2) 2) (5) 3) fi(x)=lfor-1R is a function defined by f= 25) where dee thegreatest integer function, then f is 1) Continuous for every teal x 2) discontinuous only at x =0 3) discontinuous only at non-Zero integral values of x. 36, Let / be a continuous 37. 3 39. MIE (r)=1/(1 0), the poll of Wiscontinulty of the composite functio® 1(1(/(1))) are yt 2 44 function on Rael Ho +7, Then wel oa (Leslee? the value of /(0) Is 1 HI 7 0 4)2 3 3) ) l+x, 05x82 h Let /(*) ee ee function g(x)=S((*)) js discontinuous 1) x=1only only 3) x=1,2 only 2 only Let f(x) be defined in thei erval [0.4] (I-x o[0.1] be # continuous 3)3 44 function, Then for some ¢ € {0.1] 1) f(c)=e 2) f(e)= -¢ 3) Both (1) and (2) 4) Neither (1) nor (2) . Let f be continuous on the interval [0.1] to R such that (0) = f (1). Then there 1 exists a point ¢ in [a3 such that f(cta)=f(c+A)- The value of a +B is "finn, if ational f | rf : x is irrational then the function Is n 1) diseontinousat 84 | ” 2) continuous al VT 4 2) discontinuousatall 4) continuousatall x ycos(I/x), if x40 . It H-{ : 0 » Fx=0 iscontinuous at x =0 then Na<0 2Ya>r03)a=0 4)a20 . The function f(2)= ¥—ac+3,if xis rational 2-x, if xis irrational continuous at exactly two points then the possible values of ‘a’ are 1) (2.0) 2) (-», 3) 3) (-»,-1)U(3, ») YR . Let f:R—> Rbe given by _{ox, if xeQ ao 1) f iscontinuous at y= and x =3 2) f isdiscontinuous at += and y=3 3) f iscontininuousat x = 2 butnotat x=3 4) f iscontinuous at x = 3 butnotat y=7 EXERCISE-II : KEY O1)3 02)4 03)4—04)4—05)3 06)1 O7)1 —08)2—09)1— 103 M13 12)2 13)3 141 15)2 16)1__ 17)21.3318)1 —19)9,60. 20)3 201250 22)3 23)4 243 25)3 26)2 273 28)4 294304 31)1 323 33)4 34) 35)3 36)1 37)3 39)3 40) 050 1 F(x) is not continuous at — 2, lim f(x)= a tl ty( vl x poe ee array (loging since. (9) =lim (x) gy at 16

You might also like