# CLASS-11RELATION & FUNCTION - TYPES OF FUNCTION - ONE-ONE-ONTO FUNCTION - PROBLEM & SOLUTION

ONE-ONE-ONTO FUNCTION - PROBLEM & SOLUTION -

Example.1) If A = R – {3} and B = R – {1} and f : A → B is a mapping

(x – 2)

defined by f(x) = -----------                                                                                                                                                 (x – 3)

Show that ‘f’ is one-one onto function.

Ans.)  One-one onto function : Let, x, y be any two elements of A. Then,

f(x) = f(y)

(x – 2)         (y – 2)

=> --------- = ----------

(x – 3)         (y – 3)

=>  (x – 2) (y – 3) = (x – 3) (y – 2)

=>  xy – 3x – 2y + 6 = xy – 2x – 3y + 6

=>      3y – 2y  =  3x – 2x

=>         y  =  x

Since, f(x) = f(y)

=>  x = y for all x, y ∈ A, so f is one-one.

Onto : - Let y be any arbitrary element of B. Then

f(x) = y

(x – 2)

=>  ------------  = y

(x – 3)

=>  y(x – 2) = (x – 3)

=> x = (2 – 3y)/(1 – y)

2 – 3y

Clearly,  --------- is a real number for all y ≠ 1.

1 – y

(2 – 3y)                                      (2 – 3y)

Also, ----------- ≠ 3 for any y, for, we take ----------- = 3, then

(1 – y)                                        (1 – y)

we get 2 = 1, which is wrong                                                                                                                         (2 – 3y)

Thus every element y in B has its pre-image x in A given by x = ---------------. So, ‘f ’ is onto.

(1 – y)

Thus, ‘f ’ is both one-one and onto.

Example.2) Let f = N → N be defined by f(x) = x² + x + 1, x ∈ N, then prove that ‘f ’ is one-one onto

Ans.) Let x, y ∈ N such that f(x) = f(y).

Then f(x) = f(y)

=> x² + x + 1 = y² + y + 1

=>  (x² - y²) + (x – y) = 0

=>   (x + y) (x – y) + (x – y) = 0

=>   (x – y) (x + y + 1) = 0

=>  (x – y) = 0  or  (x + y + 1) = 0

=>    x = y  or  x = (- y – 1) ∉ N.

So, f is one-one

Again, since for each y ∈ N, there exists x ∈ N

So, ‘f ’ is onto.

(x – m)

Example.3) Let f : R → R be a function defined by f(x) =  ---------------, m ≠ n.  then show that ‘f ’ is

(x – n)

One-one but not onto.

Ans.) For any x, y ∈ R, we have f(x) = f(y)

(x – m)                 (y – m)

=>  ----------------  =  ----------------

(x – n)                  (y – n)

=>          x = y

So, ‘f ’ is one-one

Let, α ∈ R such that f(x) = α

(x – m)

=>  -------------- = α

(x – n)

=>  (x – m) = (x – n) α

=>   x – m = αx – αn

=>   x – αx = m – αn

=>  x(1 – α) = (m – αn)

(m – αn)

=>  x =  ----------------

(1 – α)

Clearly, x ∉ R for α= 1, so f is not onto

Example.4) The function f : R → R be defined as f(x) = x⁴. Choose the correct answer  -

(a) ‘f ’ is one-one onto                                 (b) ‘f ’ is many-one onto

(c) ‘f ’ is one-one but not onto                  (d) ‘f ’ is neither one-one nor onto

Ans.)  Given f(x) = x⁴

Let a₁, a₂ ∈ A such that f(a₁) = f(a₂)

Then, a₁⁴ = a₂⁴

=>  a₁⁴ - a₂⁴ = 0

=>  (a₁² - a₂²) (a₁² + a₂²) = 0

=>  (a₁ + a₂) (a₁ - a₂) (a₁² + a₂²) = 0

=>  a₁ = - a₂, i.e., a₁ ≠ a₂

=> ‘f ’ is many-one function  ……………….(1)

Now, let y ∈ B so  y = x⁴

=>  x = y1/4

Since, y ∈ R so y1/4  ∈ R

Consider y = - 1 in B. its pre image (- 1)1/4 ∉ R and so does  not exist in A.

Hence, f : R → R is not an onto function. ………….(2)

From  (1) & (2), we conclude that f is neither one-one nor onto.

Hence, the correct answer is (d) ‘f ’ is neither one-one nor onto.