Question 1:Show that the function f: R* → R* defined byis 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 Nwith co-domain being same as R*?

It is given that fR* → R* is defined by
One-one:
f is one-one.
Onto:
It is clear that for y R*, there existssuch 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.

Question 2:Check the injectivity and surjectivity of the following functions:(i) f: N → N given by f(x) = x2(ii) f: Z → Z given by f(x) = x2(iii) f: R → R given by f(x) = x2(iv) f: N → N given by f(x) = x3(v) f: Z → Z given by f(x) = x3

(i) fN → N is given by,
f(x) = x2
It is seen that for xy ∈Nf(x) = f(y) ⇒ x2 = y2 ⇒ x = y.
f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

(ii)  fZ → Z is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ is not injective.
Now,−2 ∈ Z. But, there does not exist any element Z such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iii) fR → R is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ is not injective.
Now,−2 ∈ R. But, there does not exist any element ∈ R such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.

(iv) fN → N given by,
f(x) = x3
It is seen that for xy ∈Nf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.
∴ f is not surjective
Hence, function f is injective but not surjective.

(v) f→ Z is given by,
f(x) = x3
It is seen that for xy ∈ Zf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.

Question 3: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.

fR → 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.

Question 4: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.

fR → 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.

Question 5:Show that the Signum Function f: R → R, given byis neither one-one nor onto.

fR → 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.

Question 6: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.

It is given that A = {1, 2, 3}, = {4, 5, 6, 7}.
fA → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.

Question 7: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) f→ R is defined as f(x) = 3 − 4x.
.
∴ f is one-one.
For any real number (y) in R, there existsin R such that
is onto.
Hence, is bijective.
(ii) fR → R is defined as
.
.
does not imply that
For instance,
∴ f is not one-one.
Consider an element −2 in co-domain R.
It is seen thatis positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f(x) = −2.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Question 8:Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.

fA × B → B × A is defined as f(ab) = (ba).
.
∴ f is one-one.
Now, let (ba) ∈ B × A be any element.
Then, there exists (ab) ∈A × B such that f(ab) = (ba). [By definition of f]
∴ f is onto.
Hence, f is bijective.

Question 9:Let f: N → N be defined byState whether the function f is bijective. Justify your answer.

fN → 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 4+ 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.

Question 10: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 is onto if there exists x ∈A such that f(x) = y.
Now,
Thus, for any ∈ B, there existssuch that
Hence, function f is one-one and onto.

Question 11: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

fR → R is defined as
Let x∈ 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.

Question 12: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

fR → R is defined as f(x) = 3x.
Let x∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
is one-one.
Also, for any real number (y) in co-domain R, there exists  in R such that.
is onto.
Hence, function f is one-one and onto.

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