Classification Of Functions JEE Notes | EduRev

Mathematics (Maths) Class 12

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(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. Thus for x1, x2 ∈ A &  f(x1), f(x2) ∈ B,  f(x1) = f(x2)  ⇔  x1 = x2   or  x1 ≠ x2  ⇔  f(x1) ≠ f(x2).

Diagrammatically an injective mapping can be shown as

Classification Of Functions JEE Notes | EduRev

OR  Classification Of Functions JEE Notes | EduRev

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 → 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
x1, x2 ∈ A, f(x1) = f(x2) but  x1≠ x2 

Diagrammatically a many one mapping can be shown as

Classification Of Functions JEE Notes | EduRev

OR

Classification Of Functions JEE Notes | EduRev

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) = (x2 - 8x + 18)/(x2 + 4x + 30) is not one-one.
Solution. Test for one-one function
A function is one-one if f(x1) = f(x2) ⇒ x1 = x2

Classification Of Functions JEE Notes | EduRev
Classification Of Functions JEE Notes | EduRev
Since f(x1) = f(x2) does not imply x1 = x2 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) ∀ x, y ∈ R.
If f(4) = 65 and f(0) ≠ 2, then show that f(x) – 1=  x3 for x ∈ R
Solution. 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 ...(iii)
Now putting y = 1/x in equation (i), we get

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev
Let f(x) – 1 = g(x)

Classification Of Functions JEE Notes | EduRev
from equation (iv), we get g(x) g(1/x) = 1 which is only possible when
g(x) = ± xn
f(x) = ± xn + 1
or  f(x) = ± xn + 1  or 65 = ± 4n + 1
or 4n = 64 = (4)3
n = 3
f(x) = x3 + 1  or  f(x) – 1 = x3 (neglecting negative sign)

(3) Onto-function (Surjective mapping): If the function f: A → 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 → B is surjective if ∀ b ∈ B, ∃ some a ∈ such that f (a) = b.

Diagramatically surjective mapping can be shown as 

Classification Of Functions JEE Notes | EduRev

OR

Classification Of Functions JEE Notes | EduRev

Note that: if range ≡ co-domain, then  f(x) is onto.

(4) Into function: If f: A → 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

Classification Of Functions JEE Notes | EduRev

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)

Classification Of Functions JEE Notes | EduRev

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

Classification Of Functions JEE Notes | EduRev

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

Classification Of Functions JEE Notes | EduRev

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

Classification Of Functions JEE Notes | EduRev

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 → A is nn & 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 → 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.
Solution. f is many one

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Example 19. Find the linear function(s) which map the interval [0 , 2 ] onto [1 , 4 ].
Solution. Let f (x) =  a x + b
f (0) = 1  &    f (2) = 4 ⇒      b = 1   &   a =3/2
or f (0) = 4  &    f (2) = 1 ⇒      b = 4   &   a = -3/2
f(x) = 3x/2 + 1 or f(x) = 4 - 3x/2

Example 20. 
(i) Find whether f(x) = x + cos x is one-one.
(ii) Identify whether the function f(x) = -x3 + 3x2 - 2x + 4 ; R → R is ONTO or INTO
(iii) f(x) = x2 - 2x + 3; [0, 3] → A. Find whether f(x) is injective or not.
Also find the set A, if f(x) is surjective.
Solution. 
(i) The domain of f(x) is R. f'(x) = 1 – sin x.
f'(x) ≥ 0 for all x ∈complete domain and equality holds at discrete points only
∴ f(x) is strictly increasing on R. Hence f(x) is one-one
(ii) As codomain ≡ range, therefore given function is ONTO
(iii) f'(x) = 2(x – 1); 0 ≤ x ≤ 3

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

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]
∴ A ≡ [2, 6]

Example 21. If f and g be two linear functions from [-1, 1] onto [0, 2] and Classification Of Functions JEE Notes | EduRev  be defined by

Classification Of Functions JEE Notes | EduRev
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(–1) = 0 and h(1) = 2 ⇒ –a + b = 0 and a + b = 2 ⇒ a = 1 & b = 1.
Hence h(x) = x + 1.
If a < 0, then h(x) is a decreasing function & h(–1) = 2 and h(1) = 0 ⇒  –a + b = 2 and a + b = 0 ⇒ a = –1 & b = 1.
Hence h(x) = 1 – x
Now according to the question f(x) = 1 + x & g(x) = 1 - x
or f(x) = 1 – x & g(x) = 1 + x
Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev  Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev, Classification Of Functions JEE Notes | EduRev
In both cases, |Ø(Ø(x)) + Ø(Ø(1/x)) |  
Classification Of Functions JEE Notes | EduRev

(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)   ⇒  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) = akx , a > 0
(iv) f(x + y) = f(x) + f(y)  ⇒  f(x) = kx,  where k is a constant.

Example 22. 
(a) If f(x + y + 1) 
Classification Of Functions JEE Notes | EduRev
Determine f(x).
Solution.
Classification Of Functions JEE Notes | EduRev
Classification Of Functions JEE Notes | EduRev
Similarly, f(x) = (x + 1)2

(b) Let f : R - {2} → R function satisfying the following functional equation,
Classification Of Functions JEE Notes | EduRev
 Determine f(x).

Solution. We have,
Classification Of Functions JEE Notes | EduRev
Replacing x by Classification Of Functions JEE Notes | EduRev in the given functional equation we get,

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev...(ii)
putting (ii) in (i), we get,
Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev
Classification Of Functions JEE Notes | EduRev

Example 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 n  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)  ⇒ 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 – 1) nf (n) = n(n + 1) f(n)
⇒ f(1) = 2n f(n)

Classification Of Functions JEE Notes | EduRev

Classification Of Functions JEE Notes | EduRev

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