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Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 . 5 ) b ( ) a ( IMPORT (iii) (ii) (i) : 1 - yourself Do oe ht l plnma fntos r Agbac u te ovre n o tu. functions A true. not in converse the but Algebraic are functions polynomial all form the that of equation Note x. in polynomial a polynomial satisfies are it then x, of function P algebraic an is y If = as f(x) : (such Examples operations algebraic using constructed be can it if function algebraic an called : is ‘f’ function A function Algebraic (iv) (iii) (ii) (i) : Note a = : f(x) by called is ‘f’ function a If function Polynomial () ( – a – (x = f(x) a numbers real 'n' Given (d) (a) : function following the of range the Find (a) : functions following of domain the Find that is not algebraic is called called is algebraic not is that polynomials. with starting radicals) taking and division, multiplication, subtraction, addition, a and 0 (x)y N T ANT 0 n , a , (ii) ag o od ere oyoil s wees ag o a ee dge plnma i nvr R. never is polynomial degree even an of range whereas R is polynomial R degree is odd of i.e. are function Range They f(1/x). polynomial function. + a f(x) of = linear Domain x f(1/x) odd = f(x). an f(x) relation; the called satisfying is (1) functions, term polynomial two are constant There no with one a ax, = f(x) degree of polynomial A = f(x) sin3x = f(x) log – 1 = y (iii) P + 1 , a , PS F FUNCTIO OF YPES 1 (x)y 1 log – 2 < 0 consider Now 1 = f(x) log = y Let ) 2 o 2x cos 2 ,....a ( – a – (x + n–1 6 sin 16 .... + P + ...... + 10 n o o 1 i 1) x sin (16 log 2 log are real numbers and a and numbers real are x 1 x h rne s (– is range the – log – 2 0 2 2 0 n n 2 y = sin = 1 & 1 + 2 2 4 2 2 < < ) 2 2 2 log 1 x + 1 1 + x ... ( – a – (x .....+ + a , y i o 2 3 x cos x sin (b)
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Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 ouin : Solution ouin : Solution lutain : 4 Illustration lutain : 3 Illustration (i) (i) : 2 - yourself Do ) h ( Note : : Note : n o x ti number real nc a of u f modulus) (or value e absolute The alu v te lu o s b A e g n a R n i a m o D follows as defined is (x) sgn = y function Signum function Signum (vi) (v) (iv) (iii) (ii) are (i) function value absolute of properties The that number real conditions. the negative satisfies non a is |x|) (written eemn te aus f stsyn te qaiy |x equality the satisfying x of values the Determine o 0 x for 0 y ,x0 x 0, x x n u cs h eult wl hl re o te au o x t hc x and which at sign x of same value the the x for have Hence true b hold and will a if equality the only case and our In if true holds |a|–|b| = |a| b| – |a equality The |x equality the satisfying x of values the Determine sign, same the have summands both if only and if valid is |b| + |a| = b| + |a equality The equality. the |(x satisfying x of values the Determine |x = |xy| y| ± |x y| ± |x |x| inequality the |x| inequality the = |x| x| | x y| | y 0 x , f(x)= 2 4 + ) (x 3| |x = 3)| – (2x + 9) + 4x + : : : x at which 2x – 3 – 2x which at x x |b|. R 2 –, , 1} 0, {–1, 2) + (x = 9 + 4x + x| | 2 1 fx0 x if x – 2 2 – , (y , fx0 x if x x + |y| + |x| o 0 x for 1
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Then the function gof : A A : gof function the Then functions. two be C O O ORML x=1 f y=e . = –2 y=e udr h fnto gf s h giae f -mg o x. of f-image of g-image the is gof function the under A f(x) x & NO & Y |x| x x – x x 1 2 g |x| , () s n lmn o te oan f g s ta w cn ae its take can we that so ‘g’ of domain the of element an is f(x) A, 2 o(x2 1 2) cos(2x x 1) + (x N-UNIF 2| – g (f(x)) 17 (0,–4) ORML y O DFND FUNCTIO DEFINED Y x=1 (0,1) 2 x 1). + (x y= y |x–1| O –4 =e–2| y=|e C defined by (gof ) (x) = g(f(x)) g(f(x)) = (x) ) (gof by defined C y=e x fog. |x| |x| x y=2 N: JEE-Mathematics (–1,0) (0,–2) y 0 ½ c o s 2 – ½ y=2– (3,0) x x |x–1| 4 A is A x JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com . 1 1 JEE-Mathematics lutain 1 : 21 Illustration ouin : Solution ouin : Solution lutain 0 : 20 Illustration ouin : Solution lutain 9 : 19 Illustration OOEEU FNTOS : FUNCTIONS HOMOGENEOUS (ii) (i) : 6 - yourself Do ooeeu fnto o dge n. degree of function the homogeneous of is terms 5x its of examples each For when variables. variables those of to set any respect to with respect with degree homogeneous same be to said is function A f () = f(x) If (d) (a) x = f(x) A 1 (A) (gof) then function, modulus the be g and function integer greatest the be f If f(f(1)) and g(x) that clear is It graph. the means draw that should we variable the becomes g(x) Here, = f(g(x)) = f(x) Let 1} {–2, is h(x) of Range and = h(x) get we f(x), of graph From = g(f(x)) = h(x) = f(x) where g(f(x)), = h(x) of range and domain the Find (gof) Given 3 g f x gx sn x find 2x, sin = g(x) & x – |;2x3 x 2 ; x| | ;0x2 x 0 1; x 2 3y + f(g(x)) = = f(g(x)) g(x) < 1 2] [–2, is h(x) of Domain 4 ,1x2 x 1 1, x i(x1,1x2 x 1 1), x sin( x,2x1 x 2 [x], gx ,1gx 2 g(x) 1 1, 2g(x) [],2x1 x 2 [[x]], 2 ()1 ()1 g(x) 1, g(x) x i hmgnu i x y Smoial i, (x t) t = ty) f(tx, if, Symbolically y. & x in homogenous is xy – x1 2 x 1 1, 2x 3 ,x1 x 1, x 5 x1 2 x 1 1, 2x ; 2 ,1x1 x 1 1, x – (fog) (fog) – i((),0f(x) 0 sin(f(x)), 2 2 te fn fof(x). find then , fx] ()0 f(x) [f(x)], (e) (b) ; 1 B –1 (B) x x g(f(1)) f(f(–1)) and g(x) = g(x) and 3 5 (1, (1, and g(x) = = g(x) and = [sin3, 1] [sin3, x x g f f g 2 18 ] [–1, 1] [–1, i ,0x 0 x, sin x,x0 x [x], 3 3 5 5 ,2x3 x 2 2, x 2 x 1 , x (f) (c) 2 C 2 (C) g g g f [.] , = g(–2) – f f – g(–2) = 2 2 denotes the greatest integer function. integer greatest the denotes fn (fog) find , –2 –2 –1 –1 5 3 2 3 n 1 5 D 4 (D) 4 +3 (, ) hn (, ) is y) f(x, then y) f(x, y = 2 – 1 = 1 1 = 1 – 2 = 0 –2 –1 +1 3 5 x 1 2 2 f(x)=– f(x)= (fog) – 2 x + 2 3 Ans.(A) 3 5 x = E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 . 4 1 . 3 1 . 2 1 ouin : Solution Illustration ouin : Solution Illustration rpris f nes fnto : function ) a ( inverse of Properties f = g Thus y = f(x) A : f : Let FUNCTION A OF INVERSE (ii) (i) : 8 - yourself Do an an called called is is it variable then alone dependent x the of for solved x not equations the equation e.g. an by : defined FUNCTION function A EXPLICIT & IMPLICIT (i) : 7 - yourself Do |f(x)| if bounded be to said is function : A FUNCTION BOUNDED ) d ( ) c ( ) b ( 23 22 If f & g are two bijections f : A A : f bijections two bijection. are g & a f If also is bijection a of inverse The I & A : f If unique. is bijection a of inverse The Find the boundness of the the of boundness the Find ovr te mlct om no h epii fnto : (a) function explicit the into form implicit the Convert log(x = y (C) 0 = y) + cos(x – xy (A) function implicit is function following the of Which : B : –1 –1 g(y) = x, x, = g(y) r iett fntos n h st A B epciey I ff I te f s nes o itself. of inverse is f then I, = fof If respectively. 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But we get the point (b, a) from (a, b) by reflecting about the line y = x. = y line the about reflecting by b) (a, from a) (b, point the get we But . 1 1 1 x x x < , 2 1 x x –1 e e e e (a, b) 1 2 x /2 – e e e e x x x x 2 –1 2 2 1 1 x fo te nw gah f = f(x) = y of graph known the from (x) x x x x x 1 2 2 1 x w hv t frt f l fn te nevl n hc te ucin s bijective is function the which in interval the find all of first to have we (x) 2 –1 ad x and R –1 (b) = a, the point (a, b) is on the graph of ‘f’ if and only if the point (b, a) (b, point the if only and if ‘f’ of graph the on is b) (a, point the a, = (b) 1 : : –1 –1 –x –1 x. We know that y = f(x) = sinx is invertible if f : : f if invertible is sinx = f(x) = y that know We x. (x). ,1 , 1] 1, [ /2) 2 –x < Bcue ae > 1) > e base (Because 1) > e base (Because y=x 1 < /2 y 1 x 1 2 (0,–1) O 20 f –1 x –e 2 2 ) ( ,/2 1, –x y 0 /. s () netbe I s, id t inverse. its find so, If ? invertible f(x) Is )/2. 1 . f(x f 1 ffaotteln x. y line the about f of h < f(x < ) y=x /2 ,1) ( /2 ...... y=sinx 2 ie f s one-one. is f i.e. ) x x (ii) (i) y=f(x) (–1,0) y 0 2 2 (0,–1) ,1] 1, [ , = (x) y=f –1 y=x x E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 lutain 5 : 25 Illustration ouin : Solution (i) : 9 - yourself Do Let f : [–1, 1] 1] [–1, : f Let (b) Hence Hence y = f(x) Let Given = f(x) function the of inverse the Find (c) aig oaih, e ae f have we logarithm, Taking Hence Since f Let invertible. is f f(x) find onto. To onto, is and f(x) one-one Therefore both R. is set f(x) the Since as same is f of range the Hence x as Similarly and f(x) does so values x when larger and larger to tends x As : Onto y ,1y16 y 1 y, (y) f ,1y4 y 1 y, x e e f(x) = = f(x) x)f( ) (x f ) (x f f 1 –1 1 1 (x) = = (x) –1 –, ] eie b fx = || fn f find x|x|, = f(x) by defined 1] [–1, ,y1 y y, 464 64 y y e h ivre ucin f , hn y ue f dniy fof identity of rule by then f, of function inverse the be 2 e 2 2 e x) (x f 4 , x 1 x e e –1 x) (x f 1 1x4 x ;1 x ;x4 x x; 8 ;1 x; ,y1 y y, 64 x)f( 2 ) (x f ) (x f y 1 : 2 1 1 2 0 hne eaie in s ue ot and out ruled is sign negative hence 0, > 64 ;1 x; x 16 y , ; 16 x x;1 f(x) f(x) = = 2 16 x ; x4 4 4x 2x = f = x x 1 x x 0 1 2xe e x , f(x) f(x) , –1 f( x) (x f ) (x 2f 2 (y) 1 1 2 2 –1 (x) = = (x) 21 i.e. i.e. 1x4 x ;1 x ;x4 x x; 8 ;1 x; 2 ( ) x 1 n(x < f(x) < f(x) < –1 (x). 2 so long as x x as long so . –1 x = x = (x) ( JEE-Mathematics ) s. ns A JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com . 5 1 JEE-Mathematics lutain 7 : 27 Illustration ouin : Solution lutain 6 : 26 Illustration (v) (iv) (iii) (ii) (i) : Note satisfies function it & odd an domain is it's it in also –f(x) function is = '–x' even f(–x) an domain is it's it in is f(x) = 'x' : whenever f(–x) that NS IO such T is NC U function F a If EN EV & DD O h ol fnto wih s eie o te nie ubr ie ee ad d a te ae ie s () = f(x) is time same the at odd and even & line number entire the on defined is which function only The = f(x) e.g. sum the as expressed be can domain, it's in origin. is the 'x' about whenever domain symmetric it's is in '–x' function has odd which function. every function & one Every – y-axis the many about is it symmetric as is function defined, even not Every is function even. even an nor of odd be Inverse neither may function A f n vn a od ucin . function odd an & even an of even even odd odd ƒ(x) (i) : neither or even odd, as functions given the Identify (v) (iv) (iii) (ii) (i) (iv) : neither (i) or odd even, (are) is functions following the of Which even even odd odd g(x) f(x) + f(-x) f(x) - f(-x) f(x) = = f(x) = = = f(–x) (–x) = f(–x) cosx – sinx = f(x) x = f(x) f(–x) = = f(–x) odd. nor even neither is f(x) cosx. Hence – –sinx = cos(–x) – sin(–x) = f(–x) log = f(–x) EVEN neither odd nor even nor odd neither even nor odd neither 2 1 e x 2 x x e e e e ƒ (x) + g(x) + ƒ (x) sinx x x 1 x x 1 )(x )(x) ( x) ( 1 x) ( x) ( 1 x x ) x ( x + 2 even odd i(x = –x = sin(–x) x) ( 1 x) ( 1 2 2 2 ODD 2 2 2 1 (v) (ii) log = 2 2 2 ix –f(x). = sinx neither odd nor even nor odd neither even nor odd neither f(x) = = f(x) = f(x) x 1 x 1 f(x). = 22 ƒ(x) – g(x) – ƒ(x) (ii) –f(x). = e e even odd x x 1 x x 1 x x –f(x). = 2 f(x + y) = f(x) + f(y) for all x, y y x, all for f(y) + f(x) = y) + f(x ec fx i odd. is f(x) Hence 2 2 ec fx i even is f(x) Hence odd is f(x) Hence odd. is f(x) Hence ƒ (x) . g(x) . ƒ (x) even even odd odd ƒ(x)/g(x) (iii) even even odd odd () log = f(x) (goƒ even even even odd R ) (x) ) x 1 x 1 (ƒog)(x) even even even odd 0 E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 . 6 1 ouin : Solution lutain 8 : 28 Illustration ouin : Solution (vi) (v) (iv) (iii) (ii) 2 (i) f(x). over of periodic domain are : the both Note within cosx x & of sinx function values the of function all period The for the called e.g. f(x), >0) = T(T T) number + positive least f(x a that exists there such if periodic : called is N f(x) IO function T A NC U F DIC O I ER P : (i) 10 - yourself Do f () a a eid & () lo a a eid te i de nt en ht () g(x) + (a f(x) T/|a| period that a has mean b) + not f(ax does then T it period then has T f(x) If period a then p, has period has also f(x) If g(x) & T defined. not period is a period its has but f(x) periodic, is exist. If not function does constant period. function Every the is periodic a 'T' of where Inverse f(–T), = f(0) = f(T) (c) (a) : neither or odd even, (are) is functions following the of Which o proi fnto : function periodic For ut ae pro T eg fx = sn | |o x|. |cos + x| |sin = f(x) e.g. T. period a have must (i) (i) (vii) (v) (iii) function integer greatest (i) the denotes [.] where functions, following the of periodic) (if periods the Find = f(x) x = f(x) (ii) ec fx i a od function. odd an is f(x) Hence get we –x, 2f(0) = by y f(0) get Replacing we zero, by y x, Replacing f(y) + f(x) = y) + f(x function. even an is f(x) Hence = = f(–x) have, We {0}. ~ R is f(x) of domain Clearly = f(x) = 5) – (3x cosec e of Period e = f(x) = f(x) cos(cosx) + cos(sinx) = f(x) = f(x) e = f(x) e e e – e 3 i 3x sin x – x – x x e 1 e i o x cos x sin 1 1 x i o x cos x sin x o o2x.... o n cos ...... x cos2 x cos [x] x n(sinx) n(sinx) x e1) (e e 1 1 e eid 2 = Period x x x x x 1 e 1 e xx( ) x 1)x 1 (e x .x e x x tan + tan + nsinx x x x x x x 2 2 = f(x) 1 2 2 2 3 3 2 1 – oe(x 5) – cosec(3x – x 5) – cosec(3x – x 3 2 2 and and 1 1 1 tan , for all x, y y x, all for 3 f(x) x = = x (b) (d) x x lo a a eid p. period a has also 23 () f–) f0 = 0 = f(0) = f(–x) + f(x) () = f(x) () x = f(x) R f(x) = e e tn x tan & 2 (vi) (iv) (ii) x – x 2 + 2 2 2x x f(x) = = f(x) tan = f(x) b b], – [x – x = f(x) 0). is periodic over over periodic is 1csx( o cx) ec cos x)(1 cos (1 1snx( e x) sec x)(1 sin (1 2 [x] () –f(–x) = f(x) 0 = f(0) JEE-Mathematics . R JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com JEE-Mathematics ouin : Solution lutain 9 : 29 Illustration (iii) (ii) (ii) (i) (i) function integer greatest the denotes [.] where functions, following the of periodic) (if periods the Find (vii) (vi) (v) (iv) o eid f () il e ... f l pro = 1 = period all of L.C.M. be will f(x) of period So |cosn of Period ...... |cos2 of Period |cos of Period 1 = [x] – x of Period = f(x) period has f(x) Hence = f(x) = T comparing On cos = cos(1) + cos(0) = cos(cosT) + cos(sinT) then 0 = x If f(x) = T) + f(x then periodic is f(x) Let tan tan = f(x) = cosx| + |sinx of period Since = f(x) b} – {x + b = b] – [x – x = f(x) with with of Period of Period 2 = sinx of period e of Period e = f(x) 2 x T= tan T]= + [x e 1csx( o cx) ec cos x)(1 cos (1 i o x cos x sin i o x cos x sin s t period its as ... f w dfeet id o irtoa nme de nt exist. not does number irrational periodic. of not kinds different two of L.C.M. exist. not does number periodic. irrational not an and rational of L.C.M. x o o2x.... o n cos ...... x cos2 x cos [x] x () tanx = f(x) cos(cosx) + cos(sinx) = T)) + cos(cos(x T))+ + cos(sin(x 2 = Period 2 = T 1 = Period 1snx( e x) sec x)(1 sin (1 x–[x] 2 sinx + sin cos [x] x–[x] 1 = | 1 = x| 2 x 3 x x| = = x| x| = = x| 2 = = = 2 [x] [x] 3 / 2 2 / 2 1 n . 2 1 cos (ii) 2 2 = = x T = n = T] + [x 3 2 2 2 24 cos + and period of |sinx| + |cosx| is is |cosx| + |sinx| of period and o ( i )1csx) cos x)(1 sin x(1 cos 1snx( o )snx sin x) cos x)(1 sin (1 f(x) = = f(x) sin i cos sin + + 2 3 2 x x 2 [x] (iii) 2 f(x) = = f(x) . Hence f(x) is periodic is f(x) Hence . i cos sin 3 2 3 x x E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65 . 7 1 lutain 1 : 31 Illustration ouin : Solution lutain 0 : 30 Illustration iclaeu Ilsrto : Illustration Miscellaneous ) d ( ) c ( ) b ( : then variables, ) a ( independent are y x, If : GENERAL : (i) 11 - yourself Do f(x + y) = f(x) + f(y) f(y) + f(x) = y) + f(y) . f(x f(x) = y) + f(x f(y) . f(x) = f(xy) f(y) + f(x) = f(xy) (d) (c) (b) functions. (a) following the of periodic) (if periods the Find (iii) If the function f(x) satisfies the functional rule, f(x + y) = f(x) + f(y) f(y) + f(x) = y) + f(x rule, functional the satisfies f(x) function the If : Case-II : Case-I situations different two as are = vertex, x There a at at area A with this square Find the x. of of segment function the a of area the Express sides. adjacent two cuts A side of square a is ABCD = f(x) e = f(x) (cosx) sin + (sinx) sin = f(x) 2 = f(x) 1 n m f(n) of Period of Period i cos sin n as rv ta () s d function. odd is f(x) that prove also and x–[x] n(cosx) , 2 2 x x () = s(x) ar( – ar(ABCD) = ar(ABEFDA) i.e., OA, > AP = x when AP = x when r E)x2 x x.2x AEF) ar( . dnts raet nee function integer greatest denotes [.] tan + Periodic with period = = period with exist. number Periodic irrational similar two of L.C.M. the required function s(x) is as follows : follows as is s(x) function required the () x = f(x) () k = f(x) 2 2 2 2 2 2 o 3 4 3 2 cos sin 2 x , – x – x 2 2 () a = f(x) 3 () k, hr k s constant. a is k where kx, = f(x) x 0 , x x. 2 1 2 3 2 / 3 2 x x 2 2 x 2 2 x 2 3 / 3 n 2 x 1 2 , n n , 2 x x n 22x 2 .2 x 2 x kx OA, i.e., i.e., OA, A ieprle o h daoa B a adsac '' rm h vertex the from 'x' distance a at BD diagonal the to parallel line A . 2 2 r () 0 = f(x) or o fx = 0 = f(x) or R 2 2 25 x 3 4 x ( 2 / 1 E P = P x) = AP = PF = PE 2 CFE) 2 2 but but and at x = 2, when when 2, = x at and [ ae o sx = s(x) of area ; 2 x P2x 2 CP ] x,y x,y (2–1 tx2 x at 1) – 2 8( 2. = 1 1 2 tx at R & f(1) = 5, then find then 5, = f(1) & R JEE-Mathematics A D A D F x 2 P E O O F P –x 2 s. ns A . ns A E B B C C JEEMAIN.GURU Downloaded fromphysicsbyfiziks.com JEE-Mathematics ouin : Solution 1 : 11 : 10 : 9 : 8 : 7 : 6 : 5 : 4 : 3 : 2 : 1 (i) (i) (i) (i) (i) (i) (i) (i) (i) (i) (iii) (ii) (i) ) a ( ) a ( A ) a ( onto not ) a ( (A) = x ) a ( ) a ( (x) f ,) 8, ( ) 8 , ( x 0, 1 2 1 . a ... a a no 1] [–1, x 2 even 0 (q), (B) (q), n 2 1 f(y); + f(x) = y) + f(x Here, Thus, Thus, function. given the in (y) for (–x) putting Also have we function, given the in y=0 x=0, putting Now Hence, (0, (ii) ) b ( (ii) ) b ( ,0x1 x 0 x, n (x = –f(x) = f(–x) f(–x) + f(x) = 0 f(–x) + f(x) = x) f(– f(0) + f(x) = x) – f(x 0 = f(0) f(0) + f(0) = 0) + f(0 5t = f(t) 1)5 – (t + 5 = f(t) 1)5 – (t f(1)+ = f(t) 1)5 – (t + –1)} (t – f{t = f(t) ...... 3(5) ...... + 3) – f(t = f(t) 2(5) + 2) – f(t = f(t) 5 + 5} + 2) – {f(t = f(t) 5 + 1) – f(t = f(t) f(1) + 1) – f(t = f(t) 1 n 1 n m m ,1x0 x 1 x, 1 n ) m 1 n ()(n [ .. m] ..... 3 2 5[1 (5n) f(n) (p), (C) (p), m ) b ( ) b ( ) a ( 0 yes no ) b ( f(n) f(n) ) c ( ) b ( 2 odd = y (c) NWR FR O YOURSELF DO FOR ANSWERS 1 1 2 4 mm1) 5m(m mm1) 5m(m , 1 x (u), (D) (u), yes x 0 (iii) 2 2 (– ) c ( ) c ( ) c ( ) d ( A and f(x) is odd function. odd is f(x) and , 5/2] , (r), (E) (r), 1 odd ) b ( 0 [0, u x t 1 y 1 = y 1, – t = x put ) = y (e) 26 (s), (F) (s), x 1 2 ) d ( ) d ( 0 (d) ) c ( –, 1] [–1, neither ) f ( x mm1) 5m(m (t) (– 2 0 , 0) , .....(i) ....(1) ....(ii) (e) (4, (ii) 2] [–2, ) ,2x3 x 2 2 x 1 1 x, x 1, 0 x 2, x (f) ) d ( 0, x 2 1 (– , 0) , E Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#03\ENG\01.FUNCTION\01.THEORY.p65