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.
Your second block of text...