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**(1) One-One Function (Injective mapping):** A function f: A â†’ B is said to be a one-one function or injective mapping if different elements of A have different f images in B. Thus for x_{1}, x_{2} âˆˆ A & f(x_{1}), f(x_{2}) âˆˆ B, f(x_{1}) = f(x_{2}) â‡” x_{1} = x_{2} or x_{1} â‰ x_{2} â‡” f(x_{1}) â‰ f(x_{2}).

**Diagrammatically an injective mapping can be shown as**

OR

**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 at most at one point, then the function is one-one.**(2) Many-One function: **A function f: A â†’ 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 â†’ B is many one if for

x_{1}, x_{2} âˆˆ A, f(x_{1}) = f(x_{2}) but x_{1}â‰ x_{2}

**Diagrammatically a many one mapping can be shown as**

OR

**Remark: ****(i)** A continuous function f(x) which has at least 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 at least 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).**Example 16. Show that the function f(x) = (x ^{2} - 8x + 18)/(x^{2} + 4x + 30) **

A function is one-one if f(x

Since f(x_{1}) = f(x_{2}) does not imply x_{1} = x_{2} alone, f(x) is not a one-one function.**Example 17. Let f be an injective function such that ****f(x) f(y) + 2 = f(x) + f(y) + f(xy) âˆ€ x, y âˆˆ R.****If f(4) = 65 and f(0) â‰ 2, then show that f(x) â€“ 1= x ^{3} for x âˆˆ R**

Putting x = y = 0 in equation (i), we get f(0) f(0) + 2 = f(0) + f(0) + f(0)

or (f(0))

Again putting x = y = 1 in equation (i) and repeating the above steps, we get (f(1) â€“ 2) (f(1) â€“ 1) = 0

But f(1) â‰ 1 as f(x) is injective.

âˆ´ f(1) = 2 ...(iii)

Now putting y = 1/x in equation (i), we get

Let f(x) â€“ 1 = g(x)

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

g(x) = Â± x^{n}

f(x) = Â± x^{n} + 1

or f(x) = Â± x^{n} + 1 or 65 = Â± 4^{n} + 1

or 4n = 64 = (4)^{3}

n = 3

f(x) = x^{3} + 1 or f(x) â€“ 1 = x^{3} (neglecting negative sign)**(3) Onto-function (Surjective mapping):** If the function f: A â†’ B ^{ }is such that each element in B (co-domain) is the f image of atleast one element in A, then we say that f is a function of A 'onto' B. Thus ^{ }f: A â†’ B is surjective if âˆ€ b âˆˆ B, âˆƒ some a âˆˆ^{ }such that f (a) = b.

**Diagramatically surjective mapping can be shown as **

OR

**Note that:** if range â‰¡ co-domain, then f(x) is onto.**(4) Into function:** If f: A â†’ B ^{ }is such that there exists at least one element in co-domain which is not the image of any element in domain, then f(x) is into.

**Diagramatically into function can be shown as**

**Remark: ****(i)** If a function is onto, it cannot be into and vice versa.**(ii) **If f and g are both onto, then gof or fog would be onto (if exists).

Thus a function can be one of these four types:**(a)** One-one onto (injective & surjective)

**(b)** One-one into (injective but not surjective)

**(c)** Many-one onto (surjective but not injective)

**(d)** many-one into (neither surjective nor injective)

**Remark: ****(i)** If f is both injective & surjective, then it is called a **Bijective** function. Bijective functions are also named as invertible, non singular or biuniform functions.**(ii)** If a set A contains n distinct elements then the number of different functions defined from A â†’ A is n^{n} & out of it n ! are one-one.**(iii)** The composite of two bijections is a bijection i.e. if f & g are two bijections such that gof is defined, then gof is also a bijection.**Example 18. ****A function is defined as , f: D â†’ R f(x) = cot ^{-1} (sgn x) + sin^{-1}(x - {x}) (where {x} denotes the fractional part function) Find the largest domain and range of the function. State with reasons whether the function is injective or not. Also draw the graph of the function.**

**Example 19. Find the linear function(s) which map the interval [0 _{ }, 2_{ }] onto [1 , 4_{ }].**

f (0) = 1 & f (2) = 4 â‡’ b = 1 & a =3/2

or f (0) = 4 & f (2) = 1 â‡’ b = 4 & a = -3/2

f(x) = 3x/2 + 1 or f(x) = 4 - 3x/2

f'(x) â‰¥ 0 for all x âˆˆcomplete domain and equality holds at discrete points only

âˆ´ f(x) is strictly increasing on R. Hence f(x) is one-one

f(x) is a non monotonic continuous function. Hence it is not injective.

For f(x) to be surjective, A should be equal to its range. From graph, range is [2, 6]

âˆ´ A â‰¡ [2, 6]**Example 21. If f and g be two linear functions from [-1, 1] onto [0, 2] and **** be defined by**

**Solution.** Let h be a linear function from [-1, 1] onto [0, 2].

Let h(x) = ax + b, then h'(x) = a

If a > 0, then h(x) is an increasing function & h(â€“1) = 0 and h(1) = 2 â‡’ â€“a + b = 0 and a + b = 2 â‡’ a = 1 & b = 1.

Hence h(x) = x + 1.

If a < 0, then h(x) is a decreasing function & h(â€“1) = 2 and h(1) = 0 â‡’ â€“a + b = 2 and a + b = 0 â‡’ a = â€“1 & b = 1.

Hence h(x) = 1 â€“ x

Now according to the question f(x) = 1 + x & g(x) = 1 - x

or f(x) = 1 â€“ x & g(x) = 1 + x

,

In both cases, |Ã˜(Ã˜(x)) + Ã˜(Ã˜(1/x)) | **(5) 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) â‡’ f(x) = k ln x or f(x) = 0 .

(ii) f(xy) = f(x). f(y) â‡’ f(x) = x^{n}, n âˆˆ R

(iii) f(x + y) = f(x). f(y) â‡’ f(x) = akx , a > 0

(iv) f(x + y) = f(x) + f(y) â‡’ f(x) = kx, where k is a constant.**Example 22. ****(a) If f(x + y + 1) ****Determine f(x).****Solution.**

Similarly, f(x) = (x + 1)^{2}

**(b)** **Let f : R - {2} â†’ R function satisfying the following functional equation, Determine f(x).**

Replacing x by in the given functional equation we get,

...(ii)

putting (ii) in (i), we get,

**Example 23. Let f be a function from the set of positive integers to the set of real numbers i.e., f : N â†’ R such that****(i) f(1) = 1;****(ii) f(1) + 2f(2) + 3f(3) + ... + nf(n) = n (n + 1) f(n) for n **â‰¥** 2 then find the value of f (1994).****Solution.** 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) then we get

(n + 1) f(n + 1) = (n + 1) (n + 2) f (n + 1) - n (n + 1) f(n) â‡’ nf(n) = (n + 1) f(n + 1)

From which we conclude that 2f(2) = 3f (3) = 4f (4) = ... = nf(n)

Substituting the value of 2f(2), 3f(3), .... in terms of nf(n) in (1), we have

f(1) + (n â€“ 1) nf (n) = n(n + 1) f(n)

â‡’ f(1) = 2n f(n)

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