excersie 1.2

Question 1:

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_*?

solutions =

 

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.

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

^{SOLUTIONS =} 

(i) f: N → N is given by,

f(x) = x²

It is seen that for x, y ∈N, f(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) f: Z → Z is given by,

f(x) = x²

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x² = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) f: R → R is given by,

f(x) = x²

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x² = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) f: N → N given by,

f(x) = x³

It is seen that for x, y ∈N, f(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 → Z is given by,

f(x) = x³

It is seen that for x, y ∈ Z, f(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-once nor onto, where [x] denotes the greatest integer less than or equal to x.

SOLUTIONS =

 

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.

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 and is − x, if x is negative.

SOLUTIONS =

 

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.

 

Question 5:

Show that the Signum Function f: R → R, given by

is neither one-one nor onto.

SOLUTIONS =

 

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.

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.

SOLUTIONS =

 

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f: A → 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

^{SOLUTIONS =}

 

(i) f: R → R is defined as f(x) = 3 − 4x.

.

∴ f is one-one.

For any real number (y) in R, there exists i n R such that

∴f is onto.

Hence, f is bijective.

(ii) f: R → 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 that is  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.

SOLUTIONS =

 

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.

 

Question 9:

Let f: N → N be defined by

State whether the function f is bijective. Justify your answer.

SOLUTIONS=

 

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.

 

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.

SOLUTIONS=

 

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,

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

SOLUTIONS=

 

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.

 

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

SOLUTIONS=

 

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.