FUNCTIONS. Note: Example of a function may be represented diagrammatically. The above example can be written diagrammatically as follows.

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1 FUNCTIONS Def : A relation f from a set A into a set is said to be a function or mapping from A into if for each A there eists a unique such that (, ) f. It is denoted b f : A. Note: Eample of a function ma be represented diagrammaticall. The above eample can be written diagrammaticall as follows. A Def : A relation f from a set A into a set is a said to be a function or mapping from a into if i) A f ( ) ii), A, f ( ) = f ( ) Def : If f : A is a function, then A is called domain, is called codomain and f( A) = { f( ): A} is called range of f. Def 4: A function f : A if said to be one one function or injection from A into if different element in A have different f-images in. Note: A function f : A is one one if f (, ) f,(, ) f =. Note: A function f : A is one one iff, A, f ( ) f ( ) Note: A function f : A is one one iff, A, f ( ) = f ( ) = Note: A function f : A which is not one one is called man one function Note: If f : A is one one and A, are finite then na ( ) n ( ). Def 5: A function f : A is said to be onto function or surjection from A onto if f(a) =. Note: A function f : A is onto if A f( ) =. Note: A function f : A which is not onto is called an into function. Note: If A, are two finite sets and f : A is onto then n ( ) na ( ). p q r Note: If A, are two finite sets and n () =, then the number of onto functions that can be n( ) defined from A onto is A. Def 6: A function f : A is said to be one one onto function or bijection from A onto if f : A is both one one function and onto function.

2 Theorem: If f : A, g: C are two functions then the composite relation gof is a function a into C. Theorem: If f : A, g: C are two one one onto functions then gof : A C is also one one be onto. Sol: i) Let, A and f( ) = f( )., A, f : A f ( ), f ( ) f ( ), f ( ), C, f ( ) g[ f ( )] = g[ f ( )] ( gof )( ) = ( gof )( ), A,( gof )( ) = ( gof ): A C is one one =, A, f( ) = f ( ) =. f : A Is one one. is onto : ( ) ii) Proof: let z C, g: C A f ( ) = G {f()} = t (g o f) = t z C A ( gof)( ) = z. g is onto. Def 7: Two functions f : A, g: C g = z f : A is onto D are said to be equal if i) A = C, = D ii) f( ) = g( ) A. It is denoted b f = g Theorem: If f : A, g: C, h: C D are three functions, then ho( gof ) = ( hof ) of Theorem: if A is set, then the identif relation I on A is one one onto. Def 8: If A is a set, then the function I on A defined b I( ) = A, is called identif function on A. it is denoted b I A. Theorem: If f : A and IA, I are identif functions on A, respectivel then foi = I of = f. A Proof: IA : A A, f : A foia : A f : A, I : Iof : A ( foi A)( ) = f { I A( )} = f ( ), A. fi 0 A = f ( Iof )( ) = I{ f( )} = f( ), A Iof = f fo I A= Iof = f Def 9: If f : A is a function then {(, ) A:(, ) f} is called inverse of f. It is denoted b f. Def 0: If f : A is a bijection, then the function f( ) = is called inverse function of f. f : A defined b f ( ) = iff

3 Theorem: If f : A is a bijection, then f of = I, fof = I A Proof: Since f : A is a bijection f ( ) = f( ) = f : A f : A, f : A f of : A A Clearl IA : A A such that IA ( ) =, A. Let A A, f : A f( ) Let = f() = f( ) f ( ) = ( f of)( ) = f [ f( ) = f ( ) = = IA( ) ( f of)( ) = IA ( ) A f : A, f : A fof : is also a bijection and = f of I A Clearl I : such that I ( ) = Let f, : A= f( ) A Let f ( ) = f ( ) = f( ) = ( fof )( ) = f [ f ( )] = f ( ) = = I ( ) ( fof )( ) = I ( ) = fof I Theorem: If f : A, g: C are two bijections then ( gof ) = f og. Proof: f : A, g: C are bijections gof : A C is bijection ( gof ) : C A is a bijection. f : A is a bijection f : A is a bijection g: C Is a bijection g : C is a bijection g : C, g : A are bijections f og : C A is a bijection Let z C z C, g: Cis onto g( ) = z g ( z) =, f : A is onto A f( ) f ( ) = = ( )( ) [ ( )] ( ) ( ) ( ) gof = g f = g = z gof z = = = = = ( gof ) ( z) f ( ) f [ g ( z)] ( f og )( z) ( gof ) = f og

4 Theorem: If f : A, g: f : A is a bijection and A are two functions such that gof = I A and fog = I then f g =. Proof: Let, A, f( ) = f( ), A, f : A f ( ), f( ) f( ), f( ), f( ) = f( ), g = A g[ f ( )] = g[ f ( )] ( gof )( ) = ( gof )( ) I A ( ) =, A, f( ) = f( ) =. f : A is one one Let., g: A g( ) A Def : A function f : A is said tobe a constant function if the range of f contain onl one element i.e., f( ) = c A where c is a fied element of Def : A function f : A is said to be a real variable function if A R. Def : A function f : A is said to be a real valued function iff R. Def 4: A function f : A is said to be a real function if A R, R. Def 5: If f : A R, g: R then f + g: A R is defined as ( f + g)( ) = f( ) + g( ) A Def 6: If f : A R and k R then kf : A R is defined as ( kf )( ) = kf ( ), A Def 7: If f : A, g: R then fg : A R is defined as ( fg)( ) = f ( ) g( ) A. Def 8: If f : A R, g: C = { A : g( ) 0}. f R then : C R is defined as g f ( ) = g f( ) g ( ) C where Def 9: If f : A R then f ( ) = f( ), A Def 0: If n Z, n 0, a0, a, a,... a n R, an 0, then the function f : R R defined b n f ( ) = a + a + a a R is called a polnomial function of degree n. 0 n Def : If f : R R, g: R Rare two polnomial functions, then the quotient f/g is called a rational function. Def : A function f : A R is said to be bounded on A if there eists real numbers k, k such that k f( ) k A Def : A function f : A R is said to be an even function if f( ) = f ( ) A Def 4: A function f : A R is said to be an odd function if f( ) = f ( ) A.

5 Def 5: If a R, a > 0 then the function f : R R defined as f( ) = a is called an eponential function. Def 6: If a R, a > 0, a then the function f :(0, ) R defined as f( ) = log a is called a logarithmic function. Def 7: The function f : R R defined as f() = n where n Z such that n < n+ R is called step function or greatest integer function. It is denoted b f () = [] Def 8: The functions f() = sin, cos, tan, cot, sec or cosec are called trigonometric functions. Def 9: The functions f( ) = sin,cos, tan,cot,sec or cos ec are called inverse trigonometric functions. Def 0: The functions f() = sinh, cosh, coth, sech or cosech are called hperbolic functions. Def : The functions f( ) = sinh,cos, tanh, coth,sech or iverse hperbolic functions Function Domain Range. a R (0, ). log a (0, ) R. [X] R Z 4. [X] R [0, ) 5. [0, ) [0, ) 6. sin R [-, ] 7. cos R [-, ] 8. tan π R {(n+ ) : n Z} R 9. cot R { nπ : n Z} R 0. sec π R {(n+ ) : n Z} (, ] [, ). cosec R { nπ : n Z} (, ] [, ) Sin Cos Tan Cot Sec [-, ] [ π /, π /] [ -, ] [0, π ] R ( π /, π /) R (0, π ) (, ] [, ) [0, π / ) ( π /, π] 7. Cosec (, ] [, ) [ π /,0) (0, π /] 8. sinh R R 9. cosh R [, ) 0. tanh R (,). coth (,0) (0, ) (, ) (, ). sech R (0, ]. cosech (,0) (0, ) (,0) (0, ) cos ech are called

6 Sinh Cosh Tanh Coth Sech Coseh R [, ) [0, ) (-, ) R (, ) (, ) (,0) (0, ) (0, ] [0, ) (,0) (0, ) (,0) (0, ) R PROLEMS VSAQ S. If : R {0} is defined b Sol. Given that f() = f = f() + f = + = 0 f() =, then show that f() + f = 0.. If f : R [±] R is defined b + Sol. f() = log + f log + = (+ ) = log = log + ( ) + f() = log, then show that f f() = = log = log = f (). If A = {,, 0,, } and f : A is a surjection defined b f() = + +, then find. Sol. Given that f() = + + f( ) = ( ) + = 4 + = f( ) = ( ) + = + = f(0) = (0) 0 + = f() = + + =

7 f() = + + = 7 Thus range of f, f(a) = {,, 7} Since f is onto, f(a) = = {,, 7} 4. If A = {,,, 4} and f : A R is a function defined b Sol. Given that + f() = + + f() = = + + f() = = = f() = = f(4) = = Range of f is,,, f() = + then find the range of f. 5. If f( + ) = f(), R then prove that f is a constant function. Sol. f( + ) = f() Let f(0) = k then f() = f( + 0) = f( 0) = f(0) = k f( + ) = k f is a constant function. 6. Which of the following are injections or surjections or bijections? Justif our answers. + i) f : R R defined b f() = ii) f : R (0, ) defined b f() =. iii) f : (0, ) R defined b f() = log e iv) f : [0, ) [0, ) defined b f() = v) f : R [0, ) defined b f() = vi) f : R R defined b f() = + i) f : R R defined b f() = is a bijection. + Sol. i) f : R R defined b f() = a) To prove f : R R is injection Let, R and f( ) = f( )

8 + + = + = + = = f : R R is injection b) To prove f : R R is surjection Let R and f() = + = + = = = Thus for ever R, an element R such that + f + = = = f : R R is both injection and surjection f : R R is a bijection. ii) f : R (0, ) defined b f() =. a) To prove f : R R + is injection Let, R and f( ) = f( ) = = f : R + R is injection. b) To prove f : R R + is surjection Let R + and f() = = = log R Thus for ever R +, an element log such that log f(log ) = = f : R R + is a surjection Thus f : R R + is both injection and surjection. f : R R + is a bijection.

9 iii) f : (0, ) R defined b f() = log e Eplanation : a) To prove f : R + R is injection Let, R + and f( ) = f( ) log = log = f : R e e + R is injection. b) To prove f : R + R is surjection Let R and f() = log e = = e R + Thus for ever R, an element e R + such that e f(e ) = log e = log e= f : R + R is surjection Thus f : R + R is both injection and surjection. f : R + R is a bijection. iv) f : [0, ) [0, ) defined b f() = Eplanation : a) To prove f : A A is injection Let, A and f( ) = f( ) = = ( 0, 0) f : A A is injection b) To prove f : A A is surjection Let A and f() = = = A Thus for ever A, an element A f = = f : A A is a surjection Thus f : A A is both injection and surjection. f : [0, ) [0, ) is a bijection. Such that ( ) e

10 v) f : R [0, ) defined b f() = Eplanation : a) To prove f : R A is not a injection Since distinct elements have not having distinct f-images For eample : f() = = 4 = ( ) = f( ) ut b) To prove f : R A is surjection Let A and f() = = = ± R Thus for ever A, an element ± Rsuch that f( ± ) = ( ± ) = f : R A is a surjection Thus f : R A is surjection onl. vi) f : R R defined b f() = a) To prove f : R R is not a injection Since distinct element in set R are not having distinct f-images in R. For eample : f() = = 4 = ( ) = f( ) ut f : R R is not a injection. b) To prove f : R R is not surjection R, suppose f() = = = R f : R R is not surjection. 7. If g = {(, ), (, ), (, 5), (4, 7)} is a function from A = {,,, 4} to = {,, 5, 7}? If this is given b the formula g() = a + b, then find a and b. Sol. Given that A = {,,,4} and = {,, 5, 7} and g = {(, ), (, ), (, 5), (4, 7)} () Clearl ever element in set A has unique g-image in set. g : A is a function. Consider, g() = a + b g() = a + b g() = a + b g() = a + b g(4) = 4a + b

11 g = {(, a + b), (, a + b), (, a + b), (4, 4a + b)} () Comparing () and () a + b = a = b a = + = a + b = [ b] + b = b + b = b = b = 8. If f() =, g() =, h() = for all R, then find [fo(goh)()]. Sol. fo(goh)() = fog [h()] = fog () = f [g()] = f (4 ) = fo(goh)() =. 9. Find the inverse of the following functions. i) If a, b R, f : R R defined b f() = a + b (a 0) ii) f : R (0, ) defined b f() = 5 iii) f : (0, ) R defined b f() = log. Sol. i) Let f() = a + b = b a = b = a b Thus f () = a ii) Let f() = 5 = = log 5 Thus f () = log 5 iii) Let f() = log = = f () = 0. If f() = for < then show that Sol. f() = for < = ( ) b inomial theorem for rational inde = = = = = f () = f () =.

12 . If f : [, ) [, ) defined b f() = ( ) then find f (). Sol. f() : [ ) [ ) f() = ( ) ( ) f() = = ( ) = log log = 0 b± b 4ac = a ± + 4log = ± + 4log f () =. f() =, g() = + for all R, find gof(). Sol. gof() = g[f()] = g( ) + = = = gof() =. Find the domain of the following real valued functions. i) ii) iii) f() = ( )( )( ) f() = log( ) f() = 4 iv) f() = v) Sol. i) f() = 5 vi) f() = [] vii) f() = [] f() = ( )( )( ) ( )( )( ) 0 0, 0, 0,, R {,,} Domain of f is R {,,}

13 ii) f() = log( ) > 0 and > and < and Domain of f is (,) (, ) iii) f() = (4 ) Since the coefficient of is ve Domain of f is [0, 4] iv) f() = > 0 ( )(+ ) > 0 < < Since the coefficient of is ve Domain of f is (, ). v) f() = ( 5)( + 5) 0 5 or 5 Since the coefficient of is +ve Domain of f is (, 5] [5, ) vi) f() = [] [] 0 [] It is true for all R Domain of f is R. vii) f() = [] [] 0 [] It is true onl when is an integer Domain of f is Z.

14 4. Find the ranges of the following real valued functions. i) log 4 ii) [] sin π[] iii) + [] iv) 4 v) 9+ Sol. i) f()= log 4 Domain of f is R {, } Range = R ii) f() = [] Domain of f is Z Range of f is {0} sin π[] iii) + [] Domain of f is R Range of f is {0} Since sin nπ = 0, n Z. 4 iv) f() = Domain of f is R {} Range of f is R {4} v) f() = > 0, R Domain of f is R Range of f is [, ) 5. If the function f : R R defined b f( + ) + f( ) = f() f(). Sol. Given that + + f() = and f() =. We have f( + ) = ( ) + (+ ) + + f( ) = L.H.S. = f( + ) + f( ) SAQ S + f() =, then show that

15 + (+ ) ( ) + + = + + (+ ) ( )...() = R.H.S. : f() f() = = ( ) (+ )...() = From () and () L.H.S. = R.H.S. f( + ) + f( ) = f() f() 4 6. If the function f : R R defined b f() =, then show that 4 + f( ) = f(), and hence deduce the value of f + f + f 4 4. Sol. Given that 4 f() = We obtain, f( ) = = = =...() f() = = = () From () and () : f( ) = f() We have f( ) = f() Now, f( ) + f()= Put = ¼, then f( /4) + f(/4)= f(/4) + f(/4)= () f( ) + f()= put =/ then f( /) + f(/)= f(/) + f(/)= => f(/) = (4) ()+(4) => f(/4) + f(/4)+ f(/) =

16 Therefore, f + f + f 4 4. =. 7. If the function f : {, } {0, }, defined b f() = a + b is a surjection, then find a and b. Sol. Domain of f is {, } and f() = a + b f( ) = a + b f() = a + b Case I : Suppose f = {(, 0), (, )} () and f = {(, ( a + b)), (, (a + b))} () Comparing () and () a + b = 0 a = b a + b = b + b = ( a = b) b = b = ; a =. Case II : Suppose f = {(, ), (, 0)} () and f = {(, ( a + b)), (, (a + b))} (4) Comparing () and (4) we get a + b = a = b a + b = 0 b = a Thus a a = a = a = b = ( ) = Thus a =, b =. 8. If f() = cos (log ), then show that Sol. Given that f() = cos (log ) Consider, f f = cos log cos log = cos(log )cos(log ) = [ cos(log )][ cos(log )] = cos(log )cos(log ) f f = cos(log )cos(log ) () Again f f f + f() = 0.

17 f + f () = cos log + cos log() = [ cos(log log ) + cos(log + log ) ] = cos(log )cos(log ) = cos (log ) cos (log ) [ cos(a ) + cos(a+) = cos A cos ] f + f () = cos(log )cos(log ) () () () : f f f + f() = If f() = and g() = + Sol. Given that f() = and g() = + fog() = f[g()] = f + + = + = = + + fog() = then show that (fog)() =. 0. If f : R R and g : R R are defined b f() = + and g() = then find (i) (fog)() (ii) gof() (iii) fof (0) (iv) go(fof)() Sol. i) fog() = f[g()] = f( ) = ( ) + = [9 + 4 ] + = = (fog)() = ii) gof() = g[f()] = g( + ) = ( +) = 6 + 9

18 = (gof)() = iii) fof(0) = f[f(0)] = f[(0) + ] = f() = () + = 9 + = 8 + = fof(0) = iv) go(fof)() = gof[f()] = gof() = g[f()] = g[() + ] = g[ (44) + ] = g[88 + ] = g(885) = (885) = 655 = 65 go(fof)() = 65.. If f : R R, g : R R are defined b f() =, g() = +, then find (i) fof( + ) (ii) fog(), (iii) gof(a ). Sol. i) fof( + ) = f [f( + )] = f [( + ) ] = f [ + ] = f [ + ] = ( + ) = = fof( + ) = ii) log() = f[g()] = f( + ) = f(5) = (5) = 5 = 4 fog() = 4 iii) (gof)(a ) = g [f(a )] = g[(a ) ] = g[6a 9 ] = g[6a 0] = (6a 0) + = 6a a + = 6a 0a + 0 (gof)(a ) = 6a 0a + 0

19 . If f() =, ±,show that fof ()=. + Sol. Given that f() = + Let = f() + = = + f + () = + f () = fof () = f[f ()] + + = f = = = = + + fof () =. If f : R R, g : R R defined b f() =, g() = + then find (i) gof (), (ii) gof( ). Sol. i) Given that f() = Let = f() = + = + f () = gof () = g[f ()] + 4 = g = g = + = + = 9 9 ii) gof( ) = g[f( )] = g[( ) ] = g[ ] = g[ 5] = ( 5) + = = 9 + 6

20 4. Let f = {(, a), (, c), (4, d), (, b)} and g = {(, a), (4, b), (, c), (, d)}, then show that (gof) = f og. Sol. Given that, f = {(, a), (, c), (4, d), (, b)} f = {(a, ), (c, ), (d, 4), (b, )} g = {(, a), (4, b), (, c), (, d)} g = {(a, ), (b, 4), (c, ), (d, )} L.H.S. : gof = {(, ), (, ), (4, ), (, 4)} (gof) = {(, ), (, ), (, 4), (4, )} R.H.S. : f og = {(, ), (4, ), (, ), (, 4)} L.H.S. = R.H.S. 5. Let f : R R, g : R R are defined b f() =, g() = + 5 then find (fog) (). Sol. Given that, f() = and g() = + 5 fog() = f[g()] = f( + 5) = ( + 5) = + 0 = + 7 fog() = + 7 Let = fog() = = = 7 (fog) () = 7 7 (fog) () = / + 6. If f() = ( ± ) then find (fofof)() and (fofofof)(). + Sol. Given that, f() = (fofof)() = (fof)[f()]

21 + + = fof = f f = f = f = f = f() = + (fofof )() = (fofofof )() = f[(fofof )()] + = f = = = = (fofofof )() = 7. If f and g are real valued functions defined b f() = and g() = then find (i) (f g)() (ii) (fg)() (iii) Sol. Given that f() =, g() = i) f = ( ) = 6 g() = g = f g (f g)() = f() g() = 6 = + 6 = [ 6 + ] ii) (fg)() = f()g() = ( ) = f f() iii) () = = g g() iv) (f + g + )() = f() + g() + = + + = + + = = ( + ) + ( + ) = ( + )( + ) = ( + ) () (iv) (f + g + )()

22 8. If f = {(, ), (, ), (, )} then find (i) f, (ii) + f, (iii) f, (iv) f. Sol. Given that f = {(, ), (, ), (, )} i) f = {(, ), (, ), (, )} = {(, 4), (, 6), (, )} ii) + f = {(, +), (, +), (, +)} = {(, 4), (, ), (, )} iii) f = {(, ), (, ( ) ), (, ( ) )} = {(, 4), (, 9), (, )} iv) f = {(, ) } 9. Find the domains at the following real valued functions. i) ii) iii) iv) f() = + f () = log( 4 + ) f() = f() = + + log 0 (4 ) v) f() = 4 [] + vi) vii) f() = log ( ) 0. f() = + Sol. i) ii) f() = ( )( ) 0 or Since the coefficient of is +ve Domain of f is (, ] [, ) f () = log( 4 + ) 4+ > 0 ( )( ) > 0 < or > Since the coefficient of is +ve Domain of f is R [, ]

23 iii) f() = Domain of f is [, ] {0} iv) f() = log 0 v) 0 (4 ) 4 > 0 &4 4 4> < 4 Domain of f is (, ) (, ) (, 4) or Domain of f is (, 4) {, } f() = 4 [] + Case I : 4 0 ( + )( ) 0 [, ] () Since the coefficient of is ve Also [] + > 0 [] > [, ) () From () and () [, ] Case II : ( + )( ) 0 (, ] [, ) () Since the coefficient of is +ve Also [] + < 0 [] < (, ) (4) From () and (4) (, ) From case-i and case-ii Domain of f is (, ) [, ]

24 vi) f() = log ( ) log ( ) 0 0 ( ) (0.) > 0, R () > 0 < 0 ( ) < 0 0< < Since the coefficient of is +ve (0, ) () From () and () Domain of f is R (0, ) = (0, ) (or) Domain of f is (0, ) vii) f() = It is not holds good when (, 0] Domain of f is (0, ) = R Prove that the real valued function f() = + e Sol. f() = + () e Let R {0} Consider f() = + + e = + + e e e = + + = + + e (e ) = e e + + Consider f() f( ) () is an even function on R {0}.

25 e = + e e e = e (e ) = (e ) = = 0 f() f( ) = 0 f( ) = f() f is an even function.. Find the domain and range of the following functions. tan π[] i) f() = + sin π [] + [ ] ii) f() = iii) f() = + + tan π[] Sol. i) f() = + sin π [] + [ ] Domain of f is R ( tan nπ = 0, n Z) Range of f is {0} ii) f() = 0 Domain of f is R = = ( ) = + = (+ ) = = + + 0

26 Range of f is R. iii) f() = + + Domain of f is R Range of f is [, ). Determine whether the function f : R R defined b surjection or a bijection. Sol. Since >, we have f() = Since <, we have f() = 5() = and have same f image. Hence f is not an injection. Let R then > (or) If > take = R so that f() = = If take + + = R and = < f() = 5 = 5 = 5 f is a surjection Since f is not an injection, it is not a bijection. if > f() = 5 if is an injection or a. If f : R R, g : R R are defined b f() = 4 and g() = + then find a+ (i) (gof)() (ii) (gof ) (iii) fof() (iv) go(fof)(0) 4 Sol. i) (gof)() = g(f()) = g(4 ) = (4 ) + = = a+ a ii) (gof ) g + = f 4 4 a+ = g 4 4 = g(a) = a + iii) fof() = f[f()] = f(4 ) = 4[4 ] = 6 4 = 6 5

27 iv) go(fof)(0) = go(fof) = g[6 0 5] = g[ 5] = ( 5) + = 5 + = 7 4. If f : Q Q is defined b f() = for all Q, show that f is a bijection and find f. Sol. Let, Q, f( ) = f( ) = = 5 = f is an injection. 4 Let Q, then = Q and f() = f = 5 + 4= 5 5 f is a surjection, f is a bijection. f : Q Q is a bijection. We have fof () = () f[f ()] = 5f () + 4 = 4 f () = for all Q 5 5. Find the domains of the following real valued functions. i) f() = ii) f() = Sol. i) f() = > 0 > (,0) R Domain of f is (, 0) ii) f() = 0, which is true R Domain of f is R.

28 6. If f = {(4, 5), (5, 6), (6, 4)} and g = {(4, 4), (6, 5), (8, 5)} then find (i) f + g (ii) f g (iii) f + 4g (iv) f + 4 (v) fg (vi) f g (vii) f (viii) f (i) f ()f. Sol. Domain of f = A = {4, 5, 6} Domain of g = = {4, 6, 8} Domain of f ± g = A = {4, 6} i) f + g = {(4, 5 4), (6, 4 + 5)] = {(4, ), (6, )} (ii) f g = {(4, 5 + 4), (6, 4 5)] = {(4, 9), (6, 9)} (iii) f = {(4, 5), (5, 6 ), (6, 4 )} = {(4, 0), (5, ), (6, 8)} 4g = {(4, 4 4), (6, 5 4), (8, 5 4)} = {(4, 6), (6, 0), (8, 0)} Domain of f + 4g = {4, 6} f+4g = {(4, 0, 6), (6, 8+0)} = {(4, 6), (6, )} (iv) f + 4 = {(4, 5+4), (5, 6+4), (6, 4+4)} = {(4, 9), (5, 0), (6, 0)} (v) fg = {(4, (5 4)), (6, 4 5)} = {(4, 0), (6, 0)} (vi) f g = 5 4 4,, 6, 4 5 (vii) f = {(4, 5 ), (5, 6 ), (6, 4 )} = {(4, 5), (5, 6), (6, 4)} (viii) f = {(4, 5),(5, 6) } (i) f = {(4, 5), (5, 6), (6, 6)} () f = {(4, 5), (5, 6), (6, 64)} 7. Find the domain of the following real valued functions. i) f( ) = [] [] ii) f( ) = log( ) = iii) f( ) log0 v) f( ) Sol. i) f( ) = = + + R [] [] iv) ( ) f = + + log( )

29 [] [] > 0 ([] + )([] ) > 0 [] < (or) [] > 0 ut [] < [] =,, 4,... < [] > [] =,4,... Domain of f = (, ) [, ) = R [, ) f = log( ) R ii) ( ) [] > 0 > [] is a non-integer Domain of f is R Z. f = log R log0 0 and > 0 iii) ( ) = and > 0, > 0 and 0< < and 0< <, (0,) = 0, Domain of f is 0,. f = + + R log( ) + 0 and > 0 and and > and 0 [, ) (, ) {0} [, ) {0} Domain of f is [, ) {0}. iv) ( ) + + = R + 0, 0, 0 v) f( ), 0 [,] {0} Domain of f is [, ] {0}.

FREE Download Study Package from website: &

FREE Download Study Package from website: & SHORT REVISION (FUNCTIONS) THINGS TO REMEMBER :. GENERAL DEFINITION : If to every value (Considered as real unless otherwise stated) of a variable which belongs to some collection (Set) E there corresponds