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Classification Of Functions-
(1) One - One Function (Injective mapping): A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different f  images in B.  

Diagrammatically an injective mapping can be shown as 

Remark:
(i) Any function which is entirely increasing or decreasing in its domain, is one-one.
(ii) If any line parallel to x-axis cuts the graph of the function atmost at one point, then the function is one-one.

(2) Many-One function: A function f : A → B is said to be a many one function if two or more elements of A have the same  f image in B . Thus f : A → B is many one if for

Diagrammatically a many one mapping can be shown as

Remark:
(i) A continuous function f(x) which has atleast one local maximum or local minimum, is many-one. In other words, if a line parallel to x-axis cuts the graph of the function atleast at two points, then f is many-one.
(ii) If a function is one-one, it cannot be many-one and vice versa.
(iii) If f and g both are one-one, then fog and gof would also be one-one (if they exist).

Ex.16 Show that the function  is not one-one.
Sol. Test for one-one function
A function is one-one if f(x₁) = f(x₂) ⇒ x₁ = x₂

⇒

⇒

⇒

⇒
Since f(x₁) = f(x₂) does not imply x₁ = x₂ alone, f(x) is not a one-one function.

Ex.17 Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) for all x, y ∈ R.
if f(4) = 65 and f(0) ≠ 2, then show that f(x) - 1 = x³ ∀ x∈ R.
Sol.   Given that f(x) f(y) + 2 = f(x) + f(y) + f(xy) ......(i)
Putting x = y = 0 in equation (i), we get f(0) f(0) + 2 = f(0) + f(0) + f(0)
or (f(0))² + 2 = 3f(0) or (f(0) – 2) (f(0) – 1) = 0  or  f(0) = 1  (∴ f(0) ≠2) ....(ii)
Again putting x = y = 1 in equation (i) and repeating the above steps, we get
(f(1) - 2) (f(1) - 1) = 0
But f(1) ≠ 1 as f(x) is injective. ∴ f(1) = 2
Now putting y = 1/x in equation (i), we get

or  

or  

or  

or  

or  

Let f(x) – 1 = g(x)

 ⇒  

from equation (iv), we get g(x) g(1/x) = 1,which is only possible when

G. General Definition-
(1) Identity function: A function  is called the identity of A & denoted byIdentity function

Every Identity function is a bijection.
(2) Constant function: A function f : A → B is said to be constant function. If every element of set A has the same functional image in set B i.e.  is called constant function.
(3) Homogeneous function: A function is said to be homogeneous w.r.t. any set of variables when each of its term is of the same degree w.r.t. those variables.
(4) Bounded Function: A function y = f(x) is said to be bounded if it can be express is the form of
 where a and b are finite quantities.
Ex: -1≤ sin x ≤ 1; 1≤{x} < 1; -1≤sgn(x) ≤ 1 but e^x is not bounded.

Ex: Any function having singleton range like constant function.
(5) Implicit function & Explicit function: If y has been expressed entirely in terms of `x' then it is called an explicit function.

If x & y are written together in the form of an equation then it is known as implicit equation corresponding to each implicit equation there can be one, two or more explicit function satisfying it
Ex: y = x³ + 4x² + 5x → Explicit function,
Ex: x + y = 1 → Implicit equation,
Ex: y = 1 - x → Explicit function.

H. Even & Odd Functions
Function must be defined in symmetric interval [-x, x]
If f (-x) = f (x) for all x in the domain of 'f ' then f is said to be an even function.
e.g. f (x) = cos x ; g (x) = x² + 3.
If f (-x) = -f (x) for all x in the domain of `f' then f is said to be an odd function.
e.g. f (x) = sin x ; g (x) = x³ + x.
Remark:
(a) f (x) - f (-x) = 0 ⇒ f (x) is even & f (x) + f (-x) = 0 ⇒ f (x) is odd.
(b) A function may be neither even nor odd.
(c) Inverse of an even function is not defined.
(d) Every even function is symmetric about the y-axis & every odd function is symmetric about the origin.
(e) A function (whose domain is symmetric about origin) can be expressed as a sum of an even & an odd function. e.g.
(f) The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.
(g) If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd and other even then f.g will be odd.

Ex.29 Which of the following functions is odd ?
(A) sgn x + x²⁰⁰⁰ 
(B) | x | - tan x 
(C) x³ cot x 
(D) cosec x⁵⁵
Sol. Let's name the function of the parts (A), (B), (C) and (D) as f(x), g(x), h(x)  respectively. Now
(A) f(–x) = sin (–x) + (–x)²⁰⁰⁰ = –sgn x + x²⁰⁰⁰ ≠f(x) & ≠ - f(x) ∴ f is neither even nor odd
(B) g(–x) = |–x| – tan (–x) = |x| + tan x  ∴ g is neither even nor odd.
(C) h(–x) = (–x)³ cot (–x) = –x³ (–cot x) = x³ cot x = h(x) ∴ h is an even function  
(D) Ø(–x) = cosec (–x)⁵⁵ = cosec (–x⁵⁵) = –cosec x⁵⁵ = – Ø(x)  ∴ Ø is an odd function.
Alternatively
(A) f(x) = sin (x) + x²⁰⁰⁰ = O + E = neither E nor O
(B) g(x) = E – O = Neither E nor O
(C) h(x) = O × O = E
(D) f(–x) = O o O = O
∴ (D) is the correct option

Ex.30

(A) an even function
(B) an odd function
(C) neither even nor odd function
(D) none of these
Sol.

Ex.31 Let f: [–2, 2]⇒R be a function if 

(i) f is an odd function (ii) f is an even function  (where [ ] denotes the greatest integer function)
Sol.

(i) If f is an odd function then f(x)  

(ii) If f is an even function  

Ex.32 Let f(x) = e^x + sin x be defined on the interval [-4, 0]. Find the odd and even extension of f(x) in the interval [-4, 4].
Sol. Odd Extension : Let g₀ be the odd extension of f(x), then

E. Functional Equation
Functional Equation is an equation where the unknown is a function. On solving such an equation we obtain one or more functions as solutions. If x, y are independent variables, then :

(i) f(xy) = f(x) + f(y)  ⇒  f(x) = k ln x  or   f(x) = 0.
(ii) f(xy) = f(x) . f(y)  ⇒  f(x) = xⁿ ,   n∈ R
(iii) f(x + y) = f(x) . f(y)    ⇒  f(x) = a^kx , a > 0
(iv) f(x + y) = f(x) + f(y)  ⇒  f(x) = kx,  where k is a constant.

Ex.22 Determine f(x).
Sol.
Given f(x+y+1) =
Putting x = y = 0; then f(1) =   = (1+1)²  = 2²
Again putting x = 0, y=1 then f(2) = =(1+2)² = 3²
and for x=1, y=1; f(3) = = (2+2)² = 4²
similarly, f(x) = (x+1)²
(b) Let f : R – {2} → R function satisfying the following functional equation, 

Sol.

Ex.23 Let f be a function from the set of positive integers to the set of real numbers i.e. f : N → R such that
(i) f(1) = 1; 
(ii) f(1) + 2f(2) + 3f(3) + ... + nf(n) = n (n + 1) f(n) for  then find the value of f (1994).
Sol.
Given f(1) + 2f(2) + 3f(3) + .....+ nf(n) = n(n+1)f(n)        ..... (1)

Replacing n by (n + 1) then
f(1) + 2f(2) + 3f(3)+ ...... + nf(n) = n(n+1)f(n+1) = (n+1)(n+2)f(n+1)   ....(2)

Subtracting (1) from (2).

F. Composite Functions
Let f: X→Y₁ and g:Y₂→Z be two functions and the set D = {x∈ X:f(x)∈ Y₂ }. If D ≠φ then the function h defined on D by h(x) = g{f(x)} is called composite function of g and f and is denoted by gof. It is also called function of a function.
Remark : Domain of gof is D which is a subset of X (the domain of f). Range of gof is a subset of the range of g.
Properties of composite functions :
(i) The composite of functions is not commutative i.e.
(ii) The composite of functions is associative  i.e.  if f, g, h are three functions such that fo (goh) &  (fog) oh  are defined, then fo (goh) = (fog) oh.

Ex.24 Let f(x) = e^x ; R⁺ → R and g(x) = sin^-1 x; [-1, 1] → . Find domain and range of fog (x)
Sol. Domain of  Range of g(x) :

The values in range of g(x) which are accepted by f(x) are

Ex.25    f³ (x) = f{f²(x)},.....f^{k + 1} (x) = f{f^k(x)}. for k = 1, 2, 3,...., Find f¹⁹⁹⁸ (x).
Sol. 

Thus, we can see that f^k(x) repeats itself at intervals of k = 4.

Ex.26 Let g: R → R be given by g(x) = 3 + 4x. If gⁿ(x) = gogo....og(x), show that fⁿ(x) = (4ⁿ - 1) + 4ⁿx if g^-n (x) denotes the inverse of gⁿ (x).
Sol. Since g(x) = 3 + 4x

g²(x) = (gog) (x) = g {g (x)} = g (3 + 4x) = 3 + 4 (3 + 4x) or g²(x) = 15 + 4²x = (4² – 1) + 4²x

Now g³(x) = (gogog) x = g {g² (x) } = g (15 + 4² x) = 3 + 4 (15 + 4² x) = 63 + 4³ x = (4³ –1) + 4³x
Similarly we get gⁿ(x) = (4ⁿ – 1) + 4ⁿx
Now leg gⁿ (x) = y ⇒ x = g–n(y)   ....(i)
y = (4ⁿ – 1) + 4ⁿx or x = (y + 1 – 4ⁿ)4^–n  ....(ii)
From (1) and (2) we get g^–n (y) = (y + 1 – 4ⁿ) 4^–n. Hence g^–n (x) = (x + 1 – 4ⁿ) 4^–n

Ex.27  If f(x)  = | |x – 3| – 2 | ; 0 £ x £ 4 and g(x) = 4 – |2 – x| ; –1 ≤ x ≤ 3 then find fog(x).
Sol. 

I. Periodic Function
A function f(x) is called periodic if there exists a positive number T (T > 0) called the period of the function such that f (x + T) = f(x), for all values of x and x + T within the domain of f(x). The least positive period is called the principal or fundamental period of f.

e.g. The function sin x & cos x both are periodic over
Remark:
(a) A constant function is always periodic, with no fundamental period.
(b) If  f(x) has a period p, then 1/f(x) and √f(x) also has a period p.
(c) if  f(x) has a period T then f(ax + b) has a period  T/a  (a > 0).
(d) If f(x) has a period T₁ & g(x) also has a period T₂ then period of f(x) +_- g(x) or f(x)/g(x) s L.C.M of T₁ & T₂ provided their L.C.M. exists. However that L.C.M. (if exists) need not to be fundamental period. If L.C.M. does not exists then f(x) ± g(x) or f(x) . g(x) or f(x)/g(x) is non periodic e.g. |sin x| has the period π, |cos x| also has the period π
|sin x| + |cos x| also has a period p. But the fundamental period of |sin x| + |cos x| is π/2.

(e) If g is a function such that gof is defined on the domain of f and f is periodic with T, then gof is also periodic with T as one of its periods. Further if
# g is one-one, then T is the period of gof
# g is also periodic with T' as the period and the range of f is a subset of [0, T'], then T is the period of gof
(f) Inverse of a periodic function does not exist.

Ex.33 Find period of the following functions

Sol.
(i) Period of sin x/2 is 4π while period of cos x/3 is 6π. Hence period of sin x/2 + cos x/3 is 12π {L.C.M. or 4 & 6 is 12}
(ii) Period of sin x = 2π; Period of {x} = 1; but L.C.M of 2π & 1 is not possible. ∴it is aperiodic
(iii) f(x) = cos x.cos 3x; Period of f(x) is L.C.M of  = 2π
but 2π may or may not be the fundamental period. The fundamental period can be 2π/n, where n ∈ N. Hence cross-checking for n = 1,2,3, .... we find π to be fundamental period f(π+x) = (-cos x (-cos 3x) = f(x)
(iv)  Period of f(x) id L.C.M of 

Ex.34 If f(x) = sin x + cos ax is a periodic function, show that a is a rational number.
Sol. Given f(x) = sin x + cos ax

J. Inverse Of A Function

Let f : A → B be a one-one & onto function, then their exists a unique function

g :  B → A  such that    Then g is said to be inverse of f.  Thus g = f^-1 :  B → A =  {(f(x), x) | (x,  f(x))∈ f }

Properties of inverse function :
(i) The inverse of a bijection is unique, and it is also a bijection.

(ii) If  f :  A → B  is a bijection & g :  B →A is the inverse of f, then  fog = IB and
gof = I_A ,  where  I_A &  I_B are identity functions on the sets A & B respectively.

(iii) The  graphs  of  f & g  are  the  mirror  images  of  each  other  in  the line  y = x.

(iv) Normally points of intersection of f and f^–1 lie on the straight line y =x. However it must be noted that f(x) and f^–1(x) may intersect otherwise also.

(v) In general fog(x) and gof(x) are not equal. But if either f and g are inverse of each other or atleast one of f, g is an identity function, then gof = fog.

(vi) If f & g are two bijections  f :  A →B ,  g :  B → C  then the inverse of gof exists and  (gof)^-1 = f^-1 o g^-1

Ex.38 Find the inverse of the function   and assuming it to be an onto function.
Sol. 

Given f(x) = ln (x² + 3x + 1)

∴  

which is a strictly increasing function. Thus f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Now let y = f(x) = ln (x² + 3x + 1) then x = f^–1 (y) ...(1)

and y = ln (x² + 3x + 1) ⇒ e^y = x² + 3x + 1 ⇒ x² + 3x + 1 – e^y = 0

From (1) and (2), we get