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Show that the function f: R* → R* defined by
is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
It is given that f: R* → R* is defined by
One-one:
∴f is one-one.
Onto:
It is clear that for y∈ R*, there exists
such that
∴f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function g: N → R*defined by
We have,
∴g is one-one.
Further, it is clear that g is not onto as for 1.2 ∈R* there does not exit any x in N such that g(x) =
.
Hence, function g is one-one but not onto.
Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
Show that the Modulus Function f: R → R given by
, is neither one-one nor onto, where
is x, if x is positive or 0 andis − x, if x is negative.
f: R → R is given by,
It is seen that
.
∴f(−1) = f(1), but −1 ≠ 1.
∴ f is not one-one.
Now, consider −1 ∈ R.
It is known that f(x) = is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = = −1.
∴ f is not onto.
Hence, the modulus function is neither one-one nor onto.
Show that the Signum Function f: R → R, given by
is neither one-one nor onto.
f: R → R is given by,
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴f is not one-one.
Now, as f(x) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.
f : A B defined as f (1) =4,f (2) = 5, f (3) = 6 It is seen that each image of distinct elements of A under f have distinct elements, so according to the definition, f is a function.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x2
(i) it is given that f : R → R defined by f (x) = 3 – 4x
.
⇒ f is one- one
we know on thing if a function f(x) is inversible then f(x) is definitely a bijective function. means, f(x) will be one - one and onto.
Let's try the inverse of f(x) = 3 - 4x
y = 3 - 4x
y - 3 = 4x => x = (y - 3)/4
f?¹(x) = (x - 3)/4
hence, f(x) is inversible .
so, f(x) is one - one and onto function.
hence, f(x) is bijective function [ if any function is one -one and onto then it is also known as bijective function.]
(ii) it is given that f :R→R defined by f(x) = 1 +x²
.
now, f(1) = f(-1) = 2
so, f is not one - one function.
also for all real value of x , f(x) is always greater than 1 . so, range of f(x) ∈ [1,∞)
but co-domain ∈ R
e.g., Co - domain ≠ range
so, f is not onto function.
also f is not bijective function.
Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
f: A × B → B × A is defined as f(a, b) = (b, a).
.
∴ f is one-one.
Now, let (b, a) ∈ B × A be any element.
Then, there exists (a, b) ∈A × B such that f(a, b) = (b, a). [By definition of f]
∴ f is onto.
Hence, f is bijective.
Let f: N → N be defined by
State whether the function f is bijective. Justify your answer.
f: N → N is defined as
It can be observed that:
∴ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
∴n = 2r + 1 for some r ∈ N. Then, there exists 4r + 1∈N such that
.
Case II: n is even
∴n = 2r for some r ∈ N. Then,there exists 4r ∈N such that
.
∴ f is onto.
Hence, f is not a bijective function.
Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by
. Is f one-one and onto? Justify your answer.
A = R − {3}, B = R − {1}
f: A → B is defined as.
.
∴ f is one-one.
Let y ∈B = R − {1}. Then, y ≠ 1.
The function f is onto if there exists x ∈A such that f(x) = y.
Now,
Thus, for any y ∈ B, there exists
such that
Hence, function f is one-one and onto.
Let 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
f: R → R is defined as
Let x, y ∈ R such that f(x) = f(y).
∴
does not imply that
.
For instance,
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
The correct answer is D.
Let f: R → R be defined as f(x) = 3x. 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.
f: R → R is defined as f(x) = 3x.
Let x, y ∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
∴f is one-one.
Also, for any real number (y) in co-domain R, there exists 
in R such that
.
∴f is onto.
Hence, function f is one-one and onto.
The correct answer is A.
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