CONTINUI TY. JEE-Mathematics. Illustration 1 : Solution : Illustration 2 : 1. CONTINUOUS FUNCTIONS :

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1 J-Mathematics. CONTINUOUS FUNCTIONS : CONTINUI TY A fuctio for which a small chage i the idepedet variable causes oly a small chage ad ot a sudde jump i the depedet variable are called cotiuous fuctios. Naively, we may say that a fuctio is cotiuous at a fied poit if we ca draw the graph of the fuctio aroud that poit without liftig the pe from the plae of the paper. a cotiuous a discotiuous a discotiuous a discotiuous a discotiuous Cotiuity of a fuctio at a poit : A fuctio f() is said to be cotiuous at = a, if Lim f( ) f( a). Symbolically f is cotiuous at = a if Lim f a h Lim f a h f a ( ) ( ), h > 0 h0 h0 a i.e. (LHL = a = RHL =a ) equals value of f at = a. It should be oted that cotiuity of a fuctio at = a ca be discussed oly if the fuctio is defied i the immediate eighbourhood of = a, ot ecessarily at = a.. Cotiuity at = 0 for the curve ca ot be discussed. o Illustratio : Solutio : Illustratio : si, If f() the fid whether f() is cotiuous or ot at =, where [ ] deotes [] greatest iteger fuctio. si, f() [], For cotiuity at =, we determie, f(), Now, f() = [] = f() = si so f() = f() = f() f() is cotiuous at = Cosider f() becomes cotiuous at = 0. 8 f() ad f(). = si = ad f() = [] = 8 4, 0 e si k4, 0 Defie the fuctio at = 0 if possible, so that f() h h h h h h ( ) ( ) Solutio : f(0 + ) = h0 = h h0 h h h (4 ) ( ) = = 4. h0 h h f(0 ) = e si k 4 = k4 0 f() is cotiuous at = 0, f(0 + ) = f(0 ) = f(0) 4. k4 k = f(0) = (4)() Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

2 Illustratio 3 : Let f() = Solutio : a( si ) b cos c d J-Mathematics 0 If f is cotiuous at = 0, the fid out the values of a, b, c ad d. Sice f() is cotiuous at = 0, so at = 0, both left ad right its must eist ad both must be equal to 3. Now Lim 0 If sice Now Do yourself - : a( si ) b cos 5 = 0 Lim (a b 5) a... 0 f() eists the a + b + 5 = 0 ad a b 0 0 c d ( d) = 3 0 So e d = 3 d = 3, eists ( d) d d 0 = e d Hece a =, b = 4, c = 0 ad d = 3. b c d = 3 (By the epasios of si ad cos) = 3 a = ad b = 4 3 = 0 c = 0 (i) (ii) If If cos ; 0 ƒ() fid the value of k if ƒ () is cotiuous at = 0. k; 0 ; the discuss the cotiuity of ƒ () at = ; ƒ() ta ( ) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65. CONTINUITY OF TH FUNCTION IN AN INTRVAL : ( a ) A fuctio is said to be cotiuous i (a,b) if f is cotiuous at each & every poit belogig to (a, b). ( b ) A fuctio is said to be cotiuous i a closed iterval [a,b] if : (i) f is cotiuous i the ope iterval (a,b) (ii) f is right cotiuous at a i.e. Lim f f a a (iii) f is left cotiuous at b i.e. Lim f f b b Note : 9 = a fiite quatity a fiite quatity (i) Obseve that ƒ() a ad ƒ() b do ot make sese. As a cosequece of this defiitio, if ƒ() is defied oly at oe poit, it is cotiuous there, i.e., if the domai of ƒ() is a sigleto, ƒ() is a cotiuous fuctio. (ii) ample : Cosider ƒ() = a a. ƒ() is a sigleto fuctio defied oly at = a. Hece ƒ() is a cotiuous fuctio. All polyomials, trigoometrical fuctios, epoetial & logarithmic fuctios are cotiuous i their domais. (iii) If ƒ () & g() are two fuctios that are cotiuous at = c the the fuctio defied by : 3 F f g ; F K f, where K is ay real umber ; F f. g are also cotiuous at = c. f Further, if g(c) is ot zero, the F4 is also cotiuous at = c. g

3 J-Mathematics (iii) Some cotiuous fuctios : Fuctio f() Iterval i which f() is cotiuous Costat fuctio, is a iteger 0 ( ) ( ), is a positive iteger ( ) {0} a ( ) p() = a 0 + a + a a ( ) p(), where p() ad q() are polyomial i ( ) { : q() = 0} q() si, cos, e ( ) ta, sec ( ) {( + )/ : I} cot, cosec ( ) { : I} (0) (iv) Some Discotiuous Fuctios : Fuctios Poits of discotiuity [], {} ta, sec very Iteger 3,,... cot, cosec 0,,,... si, cos,, e/ = 0, 3, 0 Illustratio 4 : Discuss the cotiuity of f() = 3, , 3, 3, 0 Solutio : We write f() as f() = 3, , 3 As we ca see, f() is defied as a polyomial fuctio i each of itervals (, ), (,0), (0,3) ad (3,). Therefore, it is cotiuous i each of these four ope itervals. Thus we check the cotiuity at =,0,3. At the poit = f() = ( ) = + = f() = ( + 3) =. ( ) + 3 = 30 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

4 J-Mathematics Therefore, f() does ot eist ad hece f() is discotiuous at =. Do yourself - : At the poit = 0 0 f() = 0 ( + 3) = 3 0 f() = 0 ( + 3) = 3 f(0) = = 3 Therefore f() is cotiuous at = 0. At the poit = 3 3 f() = 3 ( + 3) = = 3 f() = 3 ( 3 5) = = f(3) = = Therefore, f() is cotiuous at = 3. We fid that f() is cotiuous at all poits i R ecept at = (i) If ; 0 a ƒ() ; b 4b ; the fid the value of a & b if ƒ () is cotiuous i [0,) Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 3 ; 0 (ii) Discuss the cotiuity of ƒ () = si ; log ; 3 3. RASONS OF DISCONTINUITY : ( a ) Limit does ot eist i.e. Lim f( ) Lim f( ) a a ( b ) f() is ot defied at = a ( c ) Lim f( ) f( a) a Geometrically, the graph of the fuctio will ehibit a break at = a, if the fuctio is discotiuous at = a. The graph as show is discotiuous at =, ad 3. i [0,3) Lim f() f() Lim f() does ot eist f() is ot defied at = 3 3

5 J-Mathematics 4. TYPS OF DISCONTINUITIS : Type- : (Removable type of discotiuities) : - I case Lim f( ) eists but is ot equal to f(a) the the fuctio is said to have a removable discotiuity or discotiuity of the first kid. I this case we ca redefie the fuctio such that Lim f f a classified as: a ( a ) Missig poit discotiuity : Where Lim a a ( ) & make it cotiuous at = a. Removable type of discotiuity ca be further f( ) eists but f(a) is ot defied. ( b ) Isolated poit discotiuity : Where Lim f( ) eists & f(a) also eists but; Lim f( ) f( a). a a, 0 Illustratio 5 : amie the fuctio, f() = / 4, 0. Discuss the cotiuity, ad if discotiuous remove, 0 the discotiuity by redefiig the fuctio (if possible). Solutio : Graph of f() is show, from graph it is see that y 0 f() = 0 f() =, but f(0) = /4 Thus, f() has removable discotiuity ad f() could be made cotiuous by takig f(0) = f() =, 0, 0, 0 /4 O y = f() before redefiig Do yourself -3 : ; 0 3 ; 4 (i) If ƒ(), the discuss the types of discotiuity for the fuctio. 5 ; 4 / 4 ; 4 Type- : (No-Removable type of discotiuities) :- I case Lim f( ) does ot eist the it is ot possible to make the fuctio cotiuous by redefiig it. Such a a discotiuity is kow as o-removable discotiuity or discotiuity of the d kid. No-removable type of discotiuity ca be further classified as : (i) Fiite type discotiuity : I such type of discotiuity left had it ad right had it at a poit eists but are ot equal. (ii) Ifiite type discotiuity : I such type of discotiuity atleast oe of the it viz. LHL ad RHL is tedig to ifiity. 3 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

6 J-Mathematics ( i i i) Oscillatory type discotiuity : e.g. f si at = 0 si f y y=si( /) f() has o removable oscillatory type discotiuity at = 0 ample : From the adjacet graph ote that (i) f is cotiuous at = (ii) f has isolated discotiuity at = (iii) f has missig poit discotiuity at = (iv) f has o removable (fiite type) discotiuity at the origi. - 0 Note : I case of o-removable (fiite type) discotiuity the o-egative differece betwee the value of the RHL at = a & LHL at = a is called the jump of discotiuity. A fuctio havig a fiite umber of jumps i a give iterval I is called a piece wise cotiuous or sectioally cotiuous fuctio i this iterval. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 Illustratio 6 : Show that the fuctio, f() = = 0. Solutio : We have, f() = e 0, ; whe 0 / e / ; whe 0 0 f() = h0 f(0 + h) = h0 0 f() = h0 0 f() = 0 f() 0 / h e / h e / e / ; whe 0 e 0, ; whe 0 h e h = 0 0 e = h0 e e / h / h has o-removable discotiuity at = [ e /h ] = [ h 0 ; e /h 0] f(). Thus f() has o-removable discotiuity. cos {cot } Illustratio 7 : f() = ; fid jump of discotiuity, where [ ] deotes greatest iteger & [] { } deotes fractioal part fuctio. 33

7 J-Mathematics cos {cot } Solutio : f() = [] f() = cos {cot } = cos cot h cos tah h0 h0 = f() [] h h0 jump of discotiuity = = Do yourself -4 : (i) Discuss the type of discotiuity for ; ƒ() ; ( ) ; 5. TH INTRMDIAT VALU THORM : Suppose f() is cotiuous o a iterval I, ad a ad b are ay two poits of I. The if y 0 is a umber betwee f(a) ad f(b), there eists a umber c betwee a ad b such that y f(b) y 0 f(a) f(c) = y 0 Note that a fuctio f which is cotiuous i [a,b] possesses the followig properties : 34 0 a c b The fuctio f, beig cotiuous o [a,b] takes o every value betwee f(a) ad f(b) ( i ) If f(a) & f(b) posses opposite sigs, the there eists atleast oe root of the equatio (ii) f() = 0 i the ope iterval (a,b). If K is ay real umber betwee f(a) & f(b), the there eists atleast oe root of the equatio f() = K i the ope iterval (a,b). Note : I above cases the umber of roots is always odd. Illustratio 8 : Show that the fuctio, f() = ( a) ( b) +, takes the value a b Solutio : f() = ( a) ( b) + f(a) = a f(b) = b & a b (f(a), f(b)) Illustratio 9 : Let f : [0, ] for some 0 (a, b) By itermediate value theorem, there is atleast oe 0 (a, b) such that f( 0 ) = a b [0, ] oto [0, ] be a cotiuous fuctio, the prove that f() = for atleast oe. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

8 J-Mathematics Solutio : Cosider g() = f() g(0) = f(0) 0 = f(0) 0 0 f() g() = f() 0 g(0). g() 0 g() = 0 has atleast oe root i [0, ] f() = for atleast oe [0, ] Do yourself -5 : (i) If ƒ () is cotiuous i [a,b] such that ƒ (c) = ƒ(a) 3ƒ(b) 5, the prove that c (a,b) 6. SOM IMPORTANT POINTS : ( a ) If f() is cotiuous & g() is discotiuous at = a the the product fuctio ( ) f( ). g( ) will ot ecessarily be discotiuous at = a, e.g. L M N si f g M 0 & 0 0 f() is cotiuous at = 0 & g() is discotiuous at = 0, but f().g() is cotiuous at = 0. ( b ) If f () ad g () both are discotiuous at = a the the product fuctio ( ) f( ). g( ) is ot ecessarily be discotiuous at = a, e.g. f g L NM 0 0 f() & g() both are discotiuous at = 0 but the product fuctio f.g() is still cotiuous at = 0 ( c ) If f () ad g () both are discotiuous at = a the f() ± g() is ot ecessarily be discotiuous at = a ( d ) A cotiuous fuctio whose domai is closed must have a rage also i closed iterval. (e) If f is cotiuous at = a & g is cotiuous at = f (a) the the composite g[f()] is cotiuous at = a. eg. si f & g cotiuous at = 0 are cotiuous at =0, hece the composite ( gof) si will also be Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 Illustratio 0 : If f() = ad g(), the discuss the cotiuity of f(), g() ad fog () i R. Solutio : f() = f() is a ratioal fuctio it must be cotiuous i its domai ad f is ot defied at =. f is discotiuous at = g() = g() is also a ratioal fuctio. It must be cotiuous i its domai ad g is ot defied at =. g is discotiuous at = Now fog() will be discotiuous at = (poit of discotiuity of g()) Cosider g() = (whe g() = poit of discotiuity of f()) = = 3 fog() is discotiuous at = & = 3. 35

9 J-Mathematics Do yourself -6 : (i) Let ƒ () = [] & g() = sg() (where [.] deotes greatest iteger fuctio), the discuss the cotiuity of ƒ () ± g(), ƒ ().g() & ƒ() at = 0. g() (ii) If ƒ () = si & g() = ta the discuss the cotiuity of ƒ () ± g() ; ƒ() & ƒ () g() g() 7. SINGL POINT CONTINUITY : Fuctios which are cotiuous oly at oe poit are said to ehibit sigle poit cotiuity Illustratio : If ƒ Solutio : if Q if Q, fid the poits where ƒ() is cotiuous Let = a be the poit at which ƒ() is cotiuous. ƒ() ƒ() a a through ratioal through irratioal a = a a = 0 fuctio is cotiuous at = 0. Do yourself -7 : (i) If g if Q, the fid the poits where fuctio is cotiuous. 0 if Q (ii) If ƒ() ; Q ; Q, the fid the poits where fuctio is cotiuous. ANSWRS FOR DO YOURSLF. (i) (ii) discotiuous at =. (i) a= & b= (ii) Discotiuous at = & cotiuous at = 3. (i) Missig poit removable discotiuity at =, isolated poit removable discotiuity at = ( i ) Fiite type o-removable discotiuity at =, 6. (i) All are discotiuous at = 0. (ii) ƒ () g() & ƒ () ± g() are discotiuous at ( ) ; I ƒ() g() is discotiuous at = ; I 7. (i) = 0 (ii) 36 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

10 J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). If, whe f 4, whe 3 5, whe 3, the correct statemet is - (A) f f 3 (B) f() is cotiuous at = 3 (C) f() is cotiuous at = (D) f() is cotiuous at = ad 3, 0 f e, the - 0, 0. If / (A) 0 (B) f f 0 0 (C) f() is discotiuous at = 0 (D) f() is cotiuous 3. If fuctio f() = 3, is cotiuous fuctio, the f(0) is equal to - (A) (B) /4 (C) /6 (D) /3 4. If a a, f, is cotiuous at =, the a is equal to - (A) 0 (B) (C) (D) 5. If f() = log( a) log( b), 0 k, 0, is cotiuous at = 0, the k is equal to - Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 6. If (A) a + b (B) a b (C) b a (D) a + b [] [ ], f(), f is cotiuous at = the is (where [.] deotes greatest iteger) -, (A) (B) 0 (C) (D) 7. If f() = cos 4, 0 a, 0, the correct statemet is - 6 4, 0 (A) f() is discotiuous at = 0 for ay value of a (B) f() is cotiuous at = 0 whe a = 8 (C) f() is cotiuous at = 0 whe a = 0 (D) oe of these 37

11 J-Mathematics 8. Fuctio f() = log is discotiuous at - (A) oe poit (B) two poits (C) three poits (D) ifiite umber of poits 9. Which of the followig fuctios has fiite umber of poits of discotiuity i R (where [.] deotes greatest iteger) (A) ta (B) / (C) + [] (D) si [] 0. If f() = ta,, 0, 4 4 is a cotiuous fuctios, the f(/4) is equal to - (A) / (B) / (C) (D). The value of f(0), so that fuctio, f() = by - a a a a a a becomes cotiuous for all, is give (A) a a (B) a (C) a (D) a a. If e cos f(), 0 is cotiuous at = 0, the - (A) f(0) = 5 (B) [f(0)] = (C) {f(0)} = 0.5 (D) [f(0)].{f(0)}=.5 where [] ad {} deotes greatest iteger ad fractioal part fuctio. 3. Let f() = ( a cos ) b si, 0 ad f(0) =. The value of a ad b so that f is a cotiuous fuctio are - 3 (A) 5/, 3/ (B) 5/, 3/ (C) 5/, 3/ (D) oe of these 4. f is a cotiuous fuctio o the real lie. Give that ( f ) 3. f( ) The the value of fe 3j is - (A) ( 3 ) 3 38 (B) ( 3) (C) zero (D) caot be determied SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS) 5. The value(s) of for which ƒ () = e si 4 9 is cotiuous, is (are) - (A) 3 (B) 3 (C) 5 (D) all (, 3] [3, ) 6. Which of the followig fuctio(s) ot defied at = 0 has/have removable discotiuity at the origi? (A) f() (B) cot (C) f() si (D) f() = si f() cos Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

12 J-Mathematics 7. Fuctio whose jump (o-egative differece of LHL & RHL) of discotiuity is greater tha or equal to oe, is/are - (A) / (e ) / ƒ () (e ) ( cos ) ; 0 ; 0 (B) g() = / 3 / ; ; ( ) (C) u() = si ; 0, ta 3 si ; 0 (D) v() = log 3 ( ) ; log / ( 5) ; 8. If ƒ(), the 7 66 ƒ is discotiuous at = (A) (B) 7 3 (C) 4 (D) 6, 0; Z 9. Let ƒ () = [] & g(), the (where [.] deotes greatest iteger fuctio) - ; R Z (A) Lim g() eists, but g() is ot cotiuous at =. (B) Lim f() does ot eist ad ƒ () is ot cotiuous at =. (C) gof is cotiuous for all. (D) fog is cotiuous for all. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 CHCK YOUR GRASP ANSWR KY X R CI S - Que A s. C C C A A A B C B A Que A s. B D C B A, B B,C,D A,C,D A,B,C A,B,C 39

13 J-Mathematics XRCIS - 0 BRAIN TASRS SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS) if 0. Cosider the piecewise defied fuctio f() 0 if if 4 the cotiuity of this fuctio - (A) the fuctio is ubouded ad therefore caot be cotiuous (B) the fuctio is right cotiuous at = 0 choose the aswer which best describes (C) the fuctio has a removable discotiuity at 0 ad 4, but is cotiuous o the rest of the real lie (D) the fuctio is cotiuous o the etire real lie. f() is cotiuous at =0, the which of the followig are always true? (A) Lim f() 0 (B) f() is o cotiuous at = 0 (C) g() = f() is cotiuous at = 0 3. Idicate all correct alteratives if, f (A) ta (f ()) & f (C) ta (f ())& f () are both cotiuous (D) Lim (f() f(0)) , the o the iterval [0,] are both cotiuous (B) ta (f()) & f (D) ta (f()) is cotiuous but are both discotiuous 4. If f() = sg(cos si + 3), where sg ( ) is the sigum fuctio, the f() - (A) is cotiuous over its domai (C) has isolated poit discotiuity cos cos si e 5. f() ; g() ( ) 8 4 h() = f() for </ = g() for >/ the which of the followigs does ot holds? (A) h is cotiuous at = / f is ot (B) has a missig poit discotiuity (D) has irremovable discotiuity. (C) h has a removable discotiuity at = / (D) f g (B) h has a irremovable discotiuity at =/ 6. The umber of poits where f() = [si + cos] (where [ ] deotes the greatest iteger fuctio), (0, ) is ot cotiuous is - (A) 3 (B) 4 (C) 5 (D) 6 7. O the iterval I = [, ], the fuctio f() = the which oe of the followig hold good? ( )e ( 0) 0 ( 0) (A) is cotiuous for all values of I (B) is cotiuous for I (0) (C) assumes all itermediate values from f( ) & f() (D) has a maimum value equal to 3/e 8. If f() = cos cos ; where [] is the greatest iteger fuctio of, the f() is cotiuous at - (A) = 0 (B) = (C) = (D) oe of these Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

14 9. Give f() / cos e cot for 0 for 0 J-Mathematics where { } & [ ] deotes the fractioal part ad the itegral part fuctios respectively, the which of the followig statemet does ot hold good - (A) f (0 ) = 0 (B) f(0 + )=3 (C) f(0)=0 cotiuity of f at = 0 (D) irremovable discotiuity of f at = 0 0. Let f be a cotiuous fuctio o R. If f ( / 4 ) (si e )e (A) ot uique (B) (C) data sufficiet to fid f(0) (D) data isufficiet to fid f(0). Give f() = b ([] + []) + for = si ( ( a)) for < the f(0) is - where [] deotes the itegral part of, the for what values of a, b the fuctio is cotiuous at =? (A) a (3 / ); b R ; I (B) a 4 ; b R ; I (C) a 4 (3 / ) ; b R ; I (D) a 4 ; b R ; I [] log (+) for < < 0. Cosider f() = le e { } j where [*] & {*} are the greatest iteger fuctio & for 0 < < ta fractioal part fuctio respectively, the - (A) f(0) = l f is cotiuous at = 0 (B) f(0) = f is cotiuous at = 0 (C) f(0) = e f is cotiuous at = 0 (D) f has a irremovable discotiuity at = 0 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 3. Let a si for 0 ad f() m b cos for 0 ad m the - (A) f(0 ) f(0 + ) (B) f(0 + ) f(0) (C) f(0 ) f(0) (D) f is cotiuous at = 0 4. Cosider f Lim si si (A) f is cotiuous at = (B) f has a fiite discotiuity at = for 0, f()=0 the - (C) f has a ifiite or oscillatory discotiuity at = (D) f has a removable type of discotiuity at = BRAIN TASRS ANSWR KY X R CI S - Que A s. D C, D C, D C A,C,D C B,C,D B, C B, D B, C Que. 3 4 A s. A,C D A B 4

15 J-Mathematics XRCIS - 03 TRU / FALS. [] is discotiuous at ifiite poits. ([ ] deotes greatest iteger fuctio). si + si is ot cotiuous for all. 3. If f is cotiuous ad g is discotiuous at = a, the f().g() is discotiuous at = a. 4. There eists a cotiuous oto fuctio f : [0, ] [0, 0], but there eists o cotiuous oto fuctio g : [0, ] (0, 0) ta( / 4 ) 5. If f() = for, the the value which ca be give to f() at = so that the fuctio cos 4 4 becomes cotiuous every where i (0, /) is /4. 6. The fuctio f, defied by f() = is cotiuous for real. ta 7. f() = si is cotiuous at =. MISCLLANOUS TYP QUSTIONS 8. If f() is cotiuous i [0, ] ad f() = for all ratioal umbers i [0, ] the f =. MATCH TH COLUMN Followig questios cotais statemets give i two colums, which have to be matched. The statemets i Colum-I are labelled as A, B, C ad D while the statemets i Colum-II are labelled as p, q, r ad s. Ay give statemet i Colum-I ca have correct matchig with ON OR MOR statemet(s) i Colum-II.. Colum-I Colum-II (A) If f() = si{}; cos a ; where {.} deotes (p) the fractioal part fuctio, such that f() is cotiuous at =. If k = a (4 ) si 4 the k is (B) ( cos(si )) If the fuctio f() = is (q) 0 cotiuous at = 0, the f(0) is (C), Q f() =, the the values, Q (r) of at which f() is cotiuous (D) If f() = + { } + [], where [] ad {} (s) represets itegral ad fractioal part of, the the values of at which f() is discotiuous 4 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

16 . Colum-I Colum-II J-Mathematics (A) If f() = /( ), the the poits at which (p) the fuctio fofof() is discotiuous (B) f(u) = u u, where u=. (q) 0 The values of at which 'f' is discotiuous (C) f() = u, where u =, 0, 0 (r) The umber of values of at which 'f' is discotiuous (D) The umber of value of at which the (s) fuctio f() = discotiuous ASSRTION & RASON is Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 These questios cotais, Statemet I (assertio) ad Statemet II (reaso). (A) Statemet-I is true, Statemet-II is true ; Statemet-II is correct eplaatio for Statemet-I. (B) Statemet-I is true, Statemet-II is true ; Statemet-II is NOT a correct eplaatio for statemet-i. (C) Statemet-I is true, Statemet-II is false. (D) Statemet-I is false, Statemet-II is true.. Statemet-I : f() = si + [] is discotiuous at = 0 B e c a u s e Statemet-II : If g() is cotiuous & h() is discotiuous at = a, the g() + h() will ecessarily be discotiuous at = a (A) A (B) B (C) C (D) D si(a cos ) if (0,). Cosider ƒ () = 3 if 0 a b if 0 3. Let Statemet-I : If b = 3 ad a = 3 B e c a u s e the ƒ () is cotiuous i (, ) Statemet-II : If a fuctio is defied o a iterval I ad it eist at every poit of iterval I the fuctio is cotiuous i I. (A) A (B) B (C) C (D) D / cos e, 0 ƒ() 3 0, 0 the Statemet-I : ƒ () is cotiuous at = 0. B e c a u s e Statemet-II : / cos e. 0 4 (A) A (B) B (C) C (D) D 43

17 J-Mathematics 4. Statemet-I : The equatio B e c a u s e 3 si 3 has atleast oe solutio i [, ] 4 3 Statemet-II : If f:[a, b] R be a fuctio & let 'c' be a umber such that f(a) < c < f(b), the there is atleast oe umber (a, b) such that f() = c. (A) A (B) B (C) C (D) D e 5. Statemet-I : Rage of ƒ () = e e e 4 is ot R. B e c a u s e Statemet-II : Rage of a cotiuous eve fuctio ca ot be R. (A) A (B) B (C) C (D) D A B 6. Let ƒ () = 3A B (, ] 4 Statemet-I : ƒ () is cotiuous at all if A = 3 4, B = 4. B e c a u s e Statemet-II : Polyomial fuctio is always cotiuous. (A) A (B) B (C) C (D) D COMPRHNSION BASD QUSTIONS Comprehesio # If S ()... ( )( ) ( )( )...( ) ad > S () a b, 0 g(), 0 h : R R h() = O the basis of above iformatio, aswer the followig questios :. If g() is cotiuous at = 0 the a + b is equal to - (A) 0 (B) (C) (D) 3. If g() is cotiuous at = 0 the g'(0) is equal to - (A) 3. Idetify the icorrect optio - (B) h(6) 44 (C) a b (A) h() is surjective (B) domai of g() is [ /, ) (C) h() is bouded (D) = (D) does ot eist Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

18 J-Mathematics Comprehesio # A ma leaves his home early i the morig to have a walk. N He arrives at a juctio of road A & road B as show i figure. He takes the followig steps i later jourey : (a) km i orth directio Road B W S (b) chages directio & moves i orth-east directio for kms. Road A (c) chages directio & moves southwards for distace of km. (d) fially he chages the directio & moves i Home south-east directio to reach road A agai. Visible/Ivisible path :- The path traced by the ma i the directio parallel to road A & road B is called ivisible path, the remaiig path traced is visible. Visible poits :- The poits about which the ma chages directio are called visible poits ecept the poit from where he chages directio last time Now if road A & road B are take as -ais & y-ais the visible path & visible poit represets the graph of y = f(). O t he basis of above i format io, a swer t he fol low i g que st ios :. The value of at which the fuctio is discotiuous - (A) (B) 0 (C) (D) 3. The value of at which fof() is discotiuous - (A) 0 (B) (C) (D) 3 3. If f() is periodic with period 3, the f(9) is - (A) (B) 3 (C) 9 (D) oe of these Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 MISCLLANOUS TYP QUSTION ANSWR KY XRCIS -3 Tr ue / False. T. F 3. F 4. T 5. F 6. F 7. F 8. T Match the Colum. (A) (p, r); (B) (s); (C) (s); (D) (p, q, r). (A) (q, s); (B) (p, r, s); (C) (q); (D) (q) Assertio & Reaso. A. C 3. A 4. C 5. A 6. B Comprehesio Based Questios Comprehesio # :. D. B 3. C Comprehesio # :. A. B,C 3. A 45

19 J-Mathematics XRCIS - 4 [A] CONCPTUAL SUBJCTIV XRCIS. If. Let, whe 0 5 4, whe 0 f() 3 4, whe 4 3, whe si for f a si b for cos for, discuss the cotiuity of f() i R. 3. Determie the values of a,b & c for which the fuctio is cotiuous at = 0. If f is cotiuous o, the fid the values of a & b. 46 si a si for 0 f c for 0 b / / 3 / for 0 b 4. Determie the kid of discotiuity of the fuctio y / at the poit = 0 f, 3 5. Suppose that 3 fbg 3 4 ad h 3 the K 3 (a) fid all zeros of f (b) fid the value of K that makes h cotiuous at =3 (c) usig the value of K foud i (b) determie whether h is a eve fuctio. 6. Draw the graph of the fuctio f, & discuss the cotiuity or discotiuity of f i the iterval. 7. si 3 A si B si If fbg b 0g is cotiuous at = 0, the fid A & B. Also fid f(0) (a) Let f( + y) = f() + f(y) for all, y & if the fuctio f() is cotiuous at = 0, the show that f() is cotiuous at all. (b) If f(. y) = f(). f(y) for all, y ad f() is cotiuous at =. Prove that f() is cotiuous for all ecept at = 0. Give f() amie the cotiuity at = 0 of the sum fuctio of the ifiite series : Show that : (a) a polyomial of a odd degree has at least oe real root (b) / a polyomial of a eve degree has at least two real roots if it attais at least oe value opposite i sig to the coefficiet of its highest-degree term. CONCPTUAL SUBJCTIV XRCIS ANSWR KY X R C IS - 4 ( A ). cotiuous every where ecept at = 0. a = b = 3. a = 3/, b 0, c = / 4. o-removable - fiite type 5. (a),, 3 (b) K= 5 (c) eve 6. f is cotiuous i 7. A = 4, B = 5, f( 0) = 9. discotiuous at = 0 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

20 J-Mathematics XRCIS - 4 [B] BRAIN STORMING SUBJCTIV XRCIS. Give fbg F H G I r N r K J F H G I ta sec r K J ;, g r l f ta f ta. si ta Lim ; / 4 f ta K ; / 4 F HG I K J. Fid the value of k, if possible, where [ ] deotes the greatest iteger fuctio ad the domai of g() is 0, so that g() is cotiuous at / 4. Also state the poits of discotiuity of g() i b0, / 4g, if ay. 3 / 3, 0 ( ), 0. Let f()= ; g() /, 0 ( ), 0 3. Discuss the cotiuity of f i [0,] where f greater tha. Also draw the graph L N l 4. Discuss the cotiuity of the fuctio f Lim si Discuss the cotiuity of g(f()). for M 4 5 [ ] ; where [] is the greatest iteger ot [cos ] for at = a a a for 0 5. Cosider the fuctio g()= a where a > 0. a a for 0 Fid the value of a & g(0) so that the fuctio g() is cotiuous at = 0. Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 si {}.si {} 3 6. Let f() = {} {} 7. Cosider aother fuctio g(); such that g() = f() for 0 f() for < 0 for 0 for 0 Discuss the cotiuity of the fuctios f() & g() at = 0. si ta a a f() ta si = ( ) ( ) sec cos for 0 where {} is the fractioal part of. for < 0, if f is cotiuous at = 0, fid a ow if g()= a.cot( a) for a, a 0, a > 0. If g is cotiuous at = a the show that g(e ) = e 47

21 J-Mathematics 8. Let [] deote the greatest iteger fuctio & f() be defied i a eighbourhood of by [] ep 4 4 6, f() 4 6 cos( ) A, ( ) ta( ) Fid the value of A & f() i order that f() may be cotiuous at =. 9. If g : [a, b] oto [a, b] is cotiuous show that there is some c [a, b] such that g(c) = c. e j e j 0. Let y... ad y Lim y. Discuss the cotiuity of b g y,, 3... ad y () at = 0 BRAIN STORMING SUBJCTIV XRCIS ANSWR KY X R C I S - 4 ( B ). k 0 ; g L NM b g l ta if 0 4. Hece g() is cotiuous everywhere. 0 if 4. gof is discotiuous at = 0, ad 3. the fuctio f is cotiuous everywhere i [0,] ecept for = 0,, & 4. discotiuous at = 5. a =, g(0)= 8 6. f(0 + ) = ; f(0 ) = 4 7. a = e 8. A = ; f() = / f is dicotiuous at = 0 ; g(0 + ) = g(0 ) = g(0) = g is cotiuous at = 0 0. y () is cotiuous at = 0 for all ad y () is discotiuous at = 0 48 Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65

22 J-Mathematics XRCIS - 05 [A] J-[MAIN] : PRVIOUS YAR QUSTIONS. If f() = Q, the f is cotiuous at- [AI 00] Q () Oly at zero () oly at 0, (3) all real umbers (4) all ratioal umbers. If f() = e, 0 0, 0 the f() is- [AI 003] () discotiuous everywhere () cotiuous as well as differetiable for all (3) cotiuous for all but ot differetiable at =0 (4) either differetiable or cotiuous at = 0 3. Let f() = ta 4,, 0, 4, If f() is cotiuous i 0,, the f 4 is- [AI 004] () () / (3) / (4) 4. The fuctio f : R/{0} R give by f() = e ca be made cotiuous at = 0 by defiig f(0) as- [AI 007] () () (3) 0 (4) si(p ) si, 0 q, 0 5. The values of p ad q for which the fuctio f() =, 0 3 is cotiuous for all i R, are:- [AI 0] () p = 3, q = () p =, q = 3 (3) p =, q = 3 (4) p = 5, q = Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65 6. Defie F() as the product of two real fuctios f () =, IR, ad si, if 0 f () 0, if 0 as follows: f ().f () if 0 F() [AI 0] 0, if 0 Statemet- : F() is cotiuous o IR. Statemet- : f () ad f () are cotiuous o IR. () Stateme- is false, statemet- is true. () Stateme- i s true, statemet - i s true; Statemet- is correct eplaatio for statemet-. (3) Statemet- is true, statemet- is true, statemet- is ot a correct eplaatio for statemet- (4) Statemet- is true, statemet- is false 7. Cosider the fuctio, f() = + 5, R. Statemet : f'(4) = 0. Statemet : f is cotiuous i [, 5], differetiable i (, 5) ad f() = f(5). [AI 0] () Statemet is true, Statemet is false. () Statemet is false, Statemet is true. (3) Statemet is true, Statemet is true ; Statemet is a correct eplaatio for Statemet. (4) Statemet is true, Statemet is true ; Statemet is ot a correct eplaatio for Statemet. PRVIOUS YARS QUSTIONS ANSWR KY Q u e A s XRCIS-5 [A] 49

23 J-Mathematics XRCIS - 05 [B] J-[ADVANCD] : PRVIOUS YAR QUSTIONS. Discuss the cotiuity of the fuctio /( ) e /( ), ƒ() e, at =. [R 00 (Mais), 3]. For every iteger, let a ad b be real umbers. Let fuctio ƒ : IR IR be give by a si, for, ƒ(), for all itegers. b cos, for, If ƒ is cotiuous, the which of the followig holds(s) for all? [J 0, 4] (A) a b = 0 (B) a b = (C) a b + = (D) a b = PRVIOUS YARS QUSTIONS ANSWR KY. Discotiuous at = ; f( + ) = ad f( ) =. B,D 50 XRCIS-5 [B] Node-6\:\Data\04\Kota\J-Advaced\SMP\Maths\Uit#04\g\0.CONTINUITY\CONTINUITY.P65